These no bake paleo vegan chocolate hazelnut bars taste like a healthy Ferrero Rocher! Made with just 4 ingredients, this chocolate no bake bar recipe is completely ketogenic friendly and sugar free- The perfect healthy dessert! Low Carb, Gluten Free, Dairy Free.
When it comes to healthy desserts, no bake bars are one of my favorites. Not only are they quick and easy to make, they usually require only a handful of ingredients.
Not to toot my own horn, but I reckon I make some of the best healthy no bake bars around. Some of my favorites include these no bake peanut butter bars, coconut crack bars and my most recent favorite, these no bake cherry chocolate coconut bars.
Not only are these bars secretly healthy, but they are also dietary friendly- All are completely paleo and vegan, with most being suitable for a ketogenic diet too. A ketogenic dessert which is made without dairy and eggs? Count me in!
I wanted to do something a little different today and gave one of my favorite circular candy bars a bar AND healthy makeover.
Ferrero Rochers are one of my all-time favorite boxed chocolates. There is just something special about biting into one and taking your taste buds to the next level.
Funnily enough, they are easy to recreate!
I based these healthy no bake bars off my popular no bake paleo vegan chocolate hazelnut cookies.
Is Nutella good or bad for you?
Everyone's definition of health is different. Personally, I find Nutella to be something to be enjoyed in moderation. It also isn't dietary friendly (it isn't paleo, vegan, gluten-free or keto).
With that said, my healthy homemade Nutella (along with this store-bought chocolate hazelnut butter and this one too) are a much healthier alternative, and diet friendly!
Traditional Nutella is not keto-friendly. This store-bought chocolate hazelnut spread IS keto-friendly, and I have used it before. It is a little more sweet and stiff than traditional Nutella but works well for this recipe.
How do you make homemade healthy Nutella?
In a high-speed blender or good quality food processor, add your roasted hazelnuts and blend until thick and smooth. Ensure you regularly scrap down the sides, especially as the hazelnuts break down.
Add your cocoa powder and syrup and continue blending until smooth and creamy (this could take up to 15 minutes!).
Remove and store in an air-tight container or jar!
You can also use my other healthy "Nutella" recipe here, but it is considerably thicker and firmer.
To make these bars, you'll need to have a batch of homemade healthy Nutella on hand or a packaged option. You'll then simply add coconut flour, chopped hazelnuts and a sticky sweetener of choice and voila- You'll be done.
When it comes to healthy refined sugar free sweeteners, I prefer using maple syrup or agave nectar. They both are mild tasting and have ample sweetness. Honey also works, but it isn't strictly vegan.
To keep it ketogenic and low carb, you will need to use a monk fruit sweetened maple syrup, like this one.
If you have a nut allergy (or to keep it nut free!) simply swap out the healthy Nutella for the nut-free chocolate spread!
To take these Ferrero Rocher no bake bars up a notch, cover it with a layer of chocolate. I love using these paleo vegan chocolate chips, which melt beautifully and taste great. For a sugar free and keto chocolate chip, I enjoy using these ones, which are stevia sweetened.
More delicious no bake keto desserts? Gotcha covered!
2 cups chocolate spread of choice I used homemade "Nutella"
Line a 10 x 10-inch pan with parchment paper and set aside.
In a large mixing bowl, add your chocolate spread of choice. Add your other ingredients, and mix until a thick dough remains. If the mixture is too crumbly, add some liquid (milk or water) until a smooth thick batter remains.
Pour your batter into the lined pan and press firmly in place. Refrigerate until firm. Once firm, cut into bars.
Melt your chocolate chips of choice. Moving quickly, dip each no bake chocolate hazelnut bar in the melted chocolate until evenly coated. Continue until all the bars are covered then refrigerate until firm.
* To keep it 100% refined sugar free, I prefer using maple syrup or agave nectar. To keep it keto and sugar free, I used a monk fruit sweetened maple syrup.
** Only use if the batter is too thick.
No Bake Paleo Vegan Chocolate Hazelnut Bars (Keto, Sugar Free, Low Carb) should be kept refrigerated, in a sealed container. They are also freezer friendly.
YUM!! These bars look amazing! I love how easy it is to make them! Ferrero Rocher is such a wonderful treat – love the idea of making these!
I'm a bit confused, what are those speckles in the chocolate dough? It is the coconut flour? They kind of look like crumbled cookies.
Maple syrup is pure sugar… these are NOT sugar-free.
This recipe doesn't use maple syrup so not sure what you are on about.


Click on the button below to see the anticipated flow from Fall 2017 to Spring 2018. The spreadsheet opens to the BOYS tab. To see GIRLS, click on the "GIRLS" tab at the bottom left hand side of the spreadsheet. Note that for GU16/2006 and older teams, the flow is based on the transition from Spring 2017 to Spring 2018.
This spreadsheet is a "guideline" and the starting point for Spring 2018 pre-alignment.
You can expect Pre-Alignment to be posted to the UYSA website around Feb 8, 2018.
The Final Alignment Public Meeting is set for February 12, 2018. Please consult the www.utahyouthsoccer.net website for updates on time & location.
Your Member Organization is required to submit ONE master alignment spreadsheet to their Region Competition committee representatives no later than February 1, 2018. Please consult with your club as to what your individual team deadline is.
The May UYSA board of directors meeting was held on Thursday the 18th at the UYSA state office.
Below are the policy changes/updates and other important information coming out of that meeting.
1. No promotion/relegation between the fall and spring season. Promotion and relegation in the premier divisions (U11 and older) will only happen once a year, in July, in anticipation for the next seasonal year.
2. No transfer window for Premier Players. The open transfer player will no longer apply to players playing on premier division teams. Premier Players will still be eligible to apply for "exceptional circumstance" transfers at any time during the year. Players playing in non Premier Divisions will still be allowed to transfer into a premier division during the open transfer window by following the standard transfer process and rules.
3. D1 Promotion to P2 will only happen once a year. 3 teams will be moved up each year from D1 to P2. The three teams will be the winner of the fall league, and the 1st and 2nd place teams from the spring league. The winner of D1 in the fall will remain in D1 for the spring season as well.
4. Club pass not allowed for multi-rostered players. In the past, if a player was rostered to more than 1 team the secondary team/club was not allowed to use that player for club passing. The policy was updated so that any player who is multi-rostered is not eligible to club pass in our state league for the primary team/club or the secondary team/club.
5. Alignment Changes. Please read through the attached document in section 6325 Team Placement to see how the alignment process will work for this next seasonal year.
Any questions/clarifications concerning this new policy, please contact our Competition Comte Reps listed in the button at the bottom of this page OR the UYSA State Office 801.307.5150. This is the first I have heard of this and no nothing more than you. We will learn together.
See the below links to check the alignment structure and flow for the Fall 2017 State Competition League (including IRL and X-League). If you have any questions about the document please contact your local competition committee representatives, email the league commissioner at cbaron@utahyouthsoccer.net, or call us at the office at 801.307.5150.
What happens to my team's placement if I switch clubs?
1) The original member organization grants the guaranteed spot to the team that is leaving. Mind you this is NOT the norm and shouldn't be expected. You have to fill out a special form that is submitted with your alignment paperwork no later than July 1st for the Fall or February 1st for Spring alignment. See button below for the form.
2) The member organization you are moving to has an empty alignment spot that they are not using, then your team can move into that spot.


The Markleen MWS skimmer is a high performance weir skimmer for harbour, coastal or offshore oil spills. This skimmer handles all types of hydrocarbons, from light diesel to heavy crude oil, and is suitable for advancing and stationary skimming operations.
Markleen MWS skimmer features a well proven self-adjusting flow-controlled inlet weir, with automatic parallel weir lip movement to water surface. The weir working depth is controlled by the pump flow rate, and determines the quantity of water in the recovered product.
The skimmer incorporates two heavy duty submersible Archimedes Twin Disc Screw pumps which provide 70% higher capacity than traditional screw pumps. The pumps are hydraulically driven and need as such a hydraulic power pack for their operation. The stainless steel pump casing as standard provides high resistance against corrosion and abrasive wear.
The skimmer is mounted on a robust stainless steel frame which sustains the four floats and protects the pumps.
MWS skimmer handles all types of oil spills, from light diesel to heavy crude mixed with debris.
Extremely high recovery capacities at low pump revolutions.
Large 360º inlet opening with strong cutting knives for efficient debris handling.
The pump casing is easily dismantled for maintenance and cleaning.
By removing the floating structure, the skimmer can easily be converted to an efficient submersible transfer or discharge pump.
Twin slow revolving screw pumps with massive torque. All viscosities, if it flows you can pump it!


text:
The Truth About Index Funds
You may think you're diversified by buying an S&P 500 Index fund, but you're making a substantial wager on a handful of stocks.
by: James K. Glassman
Index funds, which are designed to mimic the ups and downs of a specific index, from the S&P 500 Index to the Barclays Capital California Municipal Bond Index, have become a runaway success. Index investing was introduced to the public with mutual funds in the 1970s. The strategy got a big boost in the 1990s with the rise of exchange-traded funds (ETFs), which can be bought and sold like shares of stock.
18 Dirt-Cheap Index Funds to Buy
It wasn't until the turn of the millennium, however, that index funds really caught on. Between 2010 and 2020, they grew from 19% of the total fund market to 40%, and two years ago, the total assets invested in U.S. stock index funds surpassed the assets of funds actively managed by human beings. The 13 largest stock funds all track indexes.
No wonder: Compared with managed funds, index funds offer better average returns, in large part because their expenses are lower. According to fund-tracker Morningstar, the 10-year return of Vanguard S&P 500 (VOO), an ETF linked to the most popular benchmark and carrying an expense ratio of just 0.03%, exceeded the return of 87% of its 809 peers in the large-cap blend category. The index has beaten a majority of those peers in every single one of the past 10 calendar years.
Also, because only a few of their constituent stocks change each year (the annual turnover rate for the Vanguard fund is just 4%), index funds incur minimal capital gains tax liabilities. (Returns and other data are as of Aug. 6; index funds I recommend are in bold.)
Specialized index funds – ETFs such as the iShares MSCI Brazil (EWZ) or the TIAA-CREF Small-Cap Blend Index (TRHBX) – are straightforward. They let you own a country, region, investing style or industry without having to choose individual stocks or bonds. But what if you want to own the market as a whole, or a big chunk of it? The choices can be overwhelming – and not necessarily what they seem.
Begin with the S&P 500, consisting of roughly the 500 largest U.S. companies by market capitalization (number of shares outstanding times price). Like most indexes, the S&P 500 is weighted by capitalization: The bigger the market cap a company has, the more its influence on the performance of the index. Apple (AAPL), for instance, has about 60 times the impact of General Mills (GIS).
15 Best Vanguard Mutual Funds for Investors of All Stripes
Any cap-weighted index fund is a heavy bet on larger companies. Lately, that bet has become extremely heavy because a few stocks have become gigantic. In 2011, for example, the total market cap of the 10 biggest S&P 500 stocks was $2.4 trillion. Currently, it's $13.7 trillion. Apple itself has a cap as large today as all 10 of the largest S&P stocks combined a decade ago.
Or consider simply the five trillionaire stocks I highlighted recently. All by themselves, Alphabet (GOOGL), Amazon.com (AMZN), Apple, Facebook (FB) and Microsoft (MSFT) represent 22% of the value of the S&P 500. In recent years, those stocks have been on a tear, and the index has benefited.
Targeted Bet
You may think you are getting broad diversification by buying an S&P 500 Index fund, but you are actually making a substantial wager on a handful of stocks in the same sector. As of July 31, information technology, Apple's category, and communications services, the sector of Facebook and Alphabet, Google's parent, represent a whopping 39% of the S&P 500. By contrast, energy represents just 2.6%.
Most of the other popular broad indexes are similarly top-heavy and focused on technology. Consider the MSCI U.S. Broad Market Index and other gauges that measure all, or nearly all, of the approximately 4,000 stocks listed on U.S. exchanges.
The five trillionaire stocks represent about 18% of the asset value of Vanguard Total Stock Market (VTI), the most popular of the ETFs based on such indexes; that's only a few percentage points less than the trillionaires' weight in the S&P 500. The iShares Russell 1000 (IWB) is an ETF whose portfolio is based on an index of the 1,000 largest stocks. It has about 20% of its assets in the trillionaires; 36% in tech and communications.
15 Best Fidelity Funds to Buy Now
I still like the trillionaires, and I like technology, but I have decided no longer to deceive myself by thinking that most index funds tracking the S&P 500 are the best way to own the U.S. market.
Most advisers – me included – urge a balanced approach. For example, I have a personal program of putting the same amount of money every month into each of the dozen or so diversified stocks I own. Then, I rebalance at the end of each year by buying and selling so that each stock is valued roughly the same. Such a strategy makes sense for investing in the broad market as well, but most of the popular index funds don't provide it.
Also, although technology is hot now, sector weightings shift back and forth over time. Don't you want your portfolio tilted toward sectors and stocks that are out of favor?
Between 2014 and 2020, technology ranked in the top four of 11 sectors in all but one year, and it ranked number one in three years. By contrast, energy has ranked last in five of the past seven years, and consumer staples stocks such as Procter & Gamble (PG) have finished in the bottom half of the sector rankings for five years in a row. When you buy the S&P 500, you get a lot of tech but little energy and consumer staples – which is the opposite of what bargain-hunting investors want.
The Equal-Weight Solution
There are, however, ways to avoid loading up on a few stocks, or any one sector.
One is the S&P 500 Equal Weight Index. Each stock represents roughly 0.2% of total assets (there are actually 505 stocks in the S&P 500 Index), with rebalancing at the end of each quarter. As a result, every time the index is rebalanced, the trillionaires account for about 1% of assets; technology and communications, 20%. Over the past 10 years, the S&P 500 has beaten its equally weighted cousin by about one percentage point, annualized, but that's hardly unexpected in a great decade for big growth stocks.
Invesco S&P 500 Equal Weight (RSP), an ETF with an expense ratio of 0.2%, offers an easy way to buy the index. Be warned that its turnover, at 24%, is much higher than a standard broad market index fund's turnover, so it's best to own it in a tax-deferred account such as an IRA.
All 30 Dow Jones Stocks Ranked: The Pros Weigh In
A second index alternative is my old favorite, the Dow Jones Industrial Average, composed of just 30 large-cap stocks. The Dow is price weighted. In other words, the higher the price of a share of a stock, the more the company's influence on the value of the index.
As weird as price weighting sounds, it enhances the diversification of the portfolio because stocks whose price runs up quickly tend to split, and new companies take their place leading the index. (Many large tech companies, especially, rarely split their shares. But as a result, the Dow won't let them in.)
You can buy the Dow through the SPDR Dow Jones Industrial Average ETF (DIA), nicknamed Diamonds, with an expense ratio of 0.16%. The fund's 10-year annual average return is nearly two points below the S&P 500's, but that's not bad considering that tech and communications represent just 22% of the portfolio.
I'm not telling you to avoid conventional broad-market funds. Notice that I am still recommending them. I'm just saying there are other ways to get better diversification and come close to really owning the U.S. stock market.
Warren Buffett Stocks Ranked: The Berkshire Hathaway Portfolio
Becoming an Investor
9 Great Growth ETFs for 2022 and Beyond
These growth ETFs offer exposure to higher-risk, higher-reward stocks while lessening the risk of a single stock torpedoing your returns.
5 Beginner Investing Tips for a Healthy Headspace
Lousy with numbers? Bad with money? That's OK. By adopting the right mentality, you can learn not only how to invest, but how to do so effectively.
Alternative Investments for the Rest of Us
These portfolio diversifiers aren't just for the wealthy.
New Leadership at One of Our Funds
Kip 25
Vanguard recently announced that longtime manager Michael Reckmeyer will retire in June.

redpajama_set_name: RedPajamaCommonCrawl


We carry a lifetime of stories in our bodies. Some stories move and rest easily within us, becoming rich grist for the workings of our lives. Some stories get buried because we are not ready to hear them or the world around us rejects them. These stories remain heavy and thick and can cut us off from feeling the full range and richness of life's experiences. When we take time to listen to our bodies and attend lovingly to what we hear we increase our capacity for love, passion and purpose.
During our time together guidance will be offered in different ways of listening to our bodies—guided meditation, movement, and gentle yoga. A safe and sacred space will be created at both ends of the day for the setting of our intentions and for hearing the stories that rise up and ask to be told. Options will be provided throughout the day for working alone or in small groups, to work deeply and silently, or to give voice to experience.
This retreat will be a refreshing opportunity to think less, relax deeply, and perceive core truths stored deep in our own experiences. Expect play, laughter, introspection, and discovery. No experience with yoga or circles is needed. The space will be warm, comfortable, and filled with light.
No Listening to the Stories Our Bodies Hold retreats are currently scheduled. Contact me if you are interested in a custom retreat for you and a few friends.
Please sign up for our newsletter on the right for future retreat announcements.


text:
Shigeru Onishi: A Metamathematical Proposition
Black and White Photography, Classic Photography, Portrait Photography, Street Photography
This book presents an overview of the forgotten avant-garde photographic oeuvre of Japanese mathematician and artist Shigeru Onishi from the 1950s.
Publisher: Steidl 08/2021
Editor: Manfred Heiting
© Shigeru Onishi by courtesy of Steidl
Whether depicting nudes, cityscapes, trees or interiors (or combinations of these realized through multiple exposures or photomontages), most striking about Onishi's photos are his unorthodox printing methods: using a brush to coat the photographic paper with emulsion, fogging, discoloration with acetic acid, creating the effect that the fixing process was incomplete, and color correction by varying the temperature during development.
The painterly results show Onishi's interest to be not conventional representation but, in his words, the visual "formation of ideas," and bringing out "the flavors of the image as they change" by embracing all aspects of chance involved in the photographic process.
"In truth," he argues, "if your photograph consists only of planned elements, it is essentially identical to a drawing of a single equilateral triangle."
Akiko Yano – Music is a gift from the album Music is a gift (Victor Entertainment, 2021).
Book Specs: Shigeru Onishi – A Metamathematical Proposition edited by Manfred Heiting published by Steidl 08/2021 | English/Japanese | Book design: Manfred Heiting |192 pages 195 images | Hardback | 23 x 28 cm | € 58 | ISBN 978-3-95829-706-7. Available here.
Shigeru Onishi (1928–94) was born into a long-established family in Takahashi, Okayama Prefecture. Already by middle school he was engrossed in the study of advanced mathematics while also experimenting with poetry. In 1953 Onishi graduated from Hokkaido University's Department of Mathematics and remained at the university to research topology.
In Hokkaido he began to take photographs, from the beginning seeking to express internal images rather than represent the external world.
His first solo exhibition was at Nabis Gallery, Tokyo, in 1955; in the late 1950s he gradually shifted his interest from photography to abstract ink painting, which French critic Michel Tapié introduced to the European public in the 1960s and '70s with the Gutai Group.
Onishi's work is held in collections including the Museum of Modern Art, New York, Bombas Gens Centre d'Art, València, the National Museum of Modern Art, Kyoto, and the National Museum of Art, Osaka.
Artist: https://mem-inc.jp/artists_e/shigeru-onishi/
Exhibition: Shigeru Onishi The Possibility of Existence at Foam Amsterdam, The Netherlands (until 9 Jan 2022)
In collaboration with MEM Gallery and Bombas Gens, Foam is showing a wide selection of vintage prints in combination with some ink paintings.
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text:
\section{The variational equations and non-integrability}

While proving a dynamical system is integrable is very
challenging, and amounts to essentially finding the first
integrals, or showing the Hamiltonian is separable in some sense
(see for example \'{O} Math\'{u}na \cite{mathuna}), there are a
number of ways of proving a dynamical system is {\it not}
integrable. This is because integrable systems are very
structured, and must admit a number of properties; showing one of
these properties does not hold suffices to prove the system is not
integrable. While Poincar\'{e} famously showed the
non-integrability of the 3-body problem over 100 years ago
\cite{poincare}, a particularly fruitful modern approach is that
due to Morales and Ramis, as expounded in
\cite{morruiz},\cite{mrthm1} and \cite{mrthm2}, and applied to the
3-body problem in \cite{bouch} and \cite{tsy} among others.

This method focuses on the variational equations which arise from
linearizing around a non-constant solution to the dynamical
system. In loose terms (which we will define more precisely
below), if a Hamiltonian system is integrable then the variational
equations will be solvable, in the sense of differential Galois
theory. Consequently, if the variational equations are not
solvable then the Hamiltonian system is not integrable. This means
we need only focus on the variational equations, which is a great
advantage as the variational equations are of course linear and
therefore much easier to analyse. We will give a brief summary of
the required results first.

\subsection{Solvability and the Morales-Ramis theorem}

Consider the system of linear ordinary  differential equations \be
\xi'=A\xi, \quad '=d/dz \label{sysord} \ee where the elements of
$A$ are in some field $K$ (typically the field of meromorphic
functions, $\mathbb{C}(z)$). We may construct the Picard-Vessiot
extension to this field $L/K$ by adjoining to $K$ the solutions to
\eqref{sysord}. Associated with \eqref{sysord} is the differential
Galois group $\mathcal{G}$ which is the group of automorphisms of
$L/K$ which leave fixed the elements of $K$. The linear system is
{\it solvable} if we may go from the base field $K$ to the field
extension $L/K$ by adjoining to $K$
\begin{enumerate} \item integrals, \item exponentiation of
integrals, and \item algebraic functions
\end{enumerate} of the elements of $K$ (\cite{morruiz} gives a
complete introduction to this topic). Thus we can `build' the
solutions to \eqref{sysord} from the elements of the field that
the coefficients of \eqref{sysord} belong to, by performing a
finite number of `allowed' operations. Solutions of this type may
be referred to as closed form or Louivillian solutions, and the
differential equation may be called solvable or Louivillian.

We define the variational equations as follows: for the dynamical
system  $dx/dt=f(x)$ with non-constant solution $\gamma$, we let
$x=\gamma+\xi$ and the first variational equations are given by
\be \der{\xi}{t}=\left(\pder{f}{x}\right)_\gamma \xi. \ee The
variational equations can be separated into a tangential and a
normal component; as autonomous Hamiltonian systems admit the
Hamiltonian itself as a first integral this is `inherited' by the
tangential variational equation which is therefore always
solvable. And now importantly, the normal variational equation
will inherit any other first integrals, {\it if they exist}, and
will therefore also be solvable. This is the Morales-Ramis
theorem, which we will quote from \cite{morruiz}:

\begin{mrthm} For a $2n$ dimensional Hamiltonian system assume there are $n$ first integrals which are meromorphic,
in involution and independent in the neighborhood of some
non-constant solution. Then the identity component of the
 differential Galois group of the normal variational equation is an abelian subgroup of the symplectic group.
\end{mrthm}

In other words, the normal variational equation (NVE) is solvable
in the sense described above. One advantage of the MR theorem is
that it considers the differential Galois group $\mathcal{G}$
associated with the differential equation \eqref{sysord}, which is
a Lie group, rather than the monodromy group used in Ziglin's
theorem, which is discontinuous.

This theorem suggests a procedure for testing the integrability of
a Hamiltonian system: find a non-constant solution; linearize
around it and separate out the NVE; test the NVE for solvability.
If the NVE is not solvable, then the Hamiltonian system is not
integrable. To test the solvability of the NVE, we will use the
Kovacic algorithm.

\subsection{Kovacic algorithm}

The Kovacic algorithm can be used to find the closed form
solutions of a second order linear ODE with rational coefficients,
and is complete in the sense that if the algorithm does not find
such solutions then none exist. The original source for the
algorithm is Kovacic \cite{kovacicmain}, but see also
\cite{duval}, \cite{mrthm2}, \cite{bardin}, \cite{humanez}, to
name a few, for presentation, discussion, and application of the
algorithm. As we will see, the NVE associated with the system
studied in this work will be Fuchsian, which allows for a more
concise version of the algorithm. As such, we will follow most
closely the presentation of the Kovacic algorithm as given in
Churchill and Rod \cite{church}, and note that the algorithm is
also valid in the non-Fuchsian case.

Consider the linear differential equation \begin{align}
\xi''=r(z)\xi, \quad r\in\mathbb{C}(z). \label{DE} \end{align} If
this equation is Fuchsian we can write \be
r(z)=\sum_{j=1}^k\frac{\beta_j}{(z-a_j)^2}+\sum_{j=1}^k
\frac{\delta_j}{z-a_j}, \label{pfe} \ee where $k$ is the number of
finite regular singular points at locations $z=a_j$. When $\sum
\delta_j=0$ then $z=\infty$ is also a regular singular point, with
$\beta_{\infty}=\sum(\beta_j+\delta_j a_j)$. The indicial
exponents are \[
\tau_j^\pm=\tfrac{1}{2}\left(1\pm\sqrt{1+4\beta_j}\right),\quad
\tau_\infty^\pm=\tfrac{1}{2}\left(1\pm\sqrt{1+4\beta_\infty}\right)
\] at $z=a_j$ and $z=\infty$ respectively. Kovacic
\cite{kovacicmain} proved the following theorem:

\begin{kthm}
Let $\mathcal{G}$ be the differential Galois group associated with
\eqref{DE}, and note $\mathcal{G}\in SL(2,\mathbb{C})$. Then only
one of four cases can hold:

\begin{enumerate}
\item $\mathcal{G}$ is triangulisable (or reducible), in which case \eqref{DE} has
a solution of the form $\xi=e^{\int \omega}$ with
$\omega\in\mathbb{C}(z)$ (we could say $\omega$ is algebraic over
$\mathbb{C}(z)$ of degree 1);
\item $\mathcal{G}$ is conjugate to a subgroup of \[ \left\{
\begin{pmatrix} \lambda & 0 \\ 0 & \lambda^{-1}
\end{pmatrix},\lambda\in\mathbb{C}/\{0\}
\right\}\cup\left\{
\begin{pmatrix} 0 & -\beta^{-1} \\ \beta & 0
\end{pmatrix},\beta\in\mathbb{C}/\{0\}
\right\}, \] in which case \eqref{DE} has a solution of the form
$\xi=e^{\int \omega}$ where $\omega$ is algebraic over
$\mathbb{C}(z)$ of degree 2;
\item $\mathcal{G}$ is finite, in which case \eqref{DE} has a solution of the form
$\xi=e^{\int \omega}$ where $\omega$ is algebraic over
$\mathbb{C}(z)$ of degree 4, 6 or 12;
\item $\mathcal{G}=SL(2,\mathbb{C})$, in which case \eqref{DE} is
not solvable.
\end{enumerate}
\end{kthm}

To show the NVE is not solvable, we must check that each of the
cases 1-3 cannot hold. If this is the case, then
$\mathcal{G}=SL(2,\mathbb{C})$ whose identity component (also
$SL(2,\mathbb{C})$) is not abelian, and therefore the original
Hamiltonian system is not meromorphically integrable according to
the Morales-Ramis theorem.

Kovacic provides an algorithm whereby each case is checked in two
stages: first, we consider certain combinations of the indicial
exponents and retain all those leading to a non-negative integer
$d$, i.e.\ $d\in\mathbb{N}_0$. Second, for each of these values of
$d$ we attempt to construct a monic polynomial of degree $d$, or
possibly a set of polynomials, which satisfy certain equations. If
successful, the solution to \eqref{DE} can be found. We will state
the details for each case in turn:

\medskip

\noindent{\bf Case 1} Define the modified indicial exponents
\begin{align*}
&\alpha_j^\pm=\tau_j^\pm\ \textrm{if}\ \beta_j\neq 0;\quad
\alpha_j^\pm=1\ \textrm{if}\ \beta_j=0\ \textrm{and}\ \delta_j
\neq 0;\quad \alpha_j^\pm=0\ \textrm{if}\ \beta_j=\delta_j=0, \\
&\alpha_\infty^\pm=\tau_\infty^\pm\ \textrm{if}\ \beta_\infty\neq
0; \quad \alpha_\infty^+=1,\alpha_\infty^-=0\ \textrm{if}\
\beta_\infty=0.
\end{align*} Find all combinations of these modified indicial
exponents such that \[
d=\alpha_\infty^\pm-\sum_{j=1}^k\alpha_j^\pm \] is a non-negative
integer. For each $d$ found, look for a monic degree $d$
polynomial $P(z)$ satisfying \[ P''+2\theta
P'+(\theta'+\theta^2-r)P=0, \quad \theta=\sum_{j=1}^k
\frac{\alpha_j^\pm}{z-a_j}. \] If found, a solution to \eqref{DE}
will be $\xi=e^{\int \omega}$ where $\omega=\theta+P'/P$.

\medskip

\noindent{\bf Case 2} Define the following sets: \begin{align*}
&E_j=\{2+e\sqrt{1+4\beta_j},e=0,\pm 2\}\cap\mathbb{Z}\
\textrm{if}\ \beta_j\neq 0; \\
&E_j=\{4\}\ \textrm{if}\ \beta_j=0,\delta_j\neq 0;\quad E_j=\{0\}\
\textrm{if}\ \beta_j=\delta_j=0;\\
&E_\infty=\{2+e\sqrt{1+4\beta_\infty},e=0,\pm 2\}\cap\mathbb{Z}\
\textrm{if}\ \beta_\infty\neq 0; \\
&E_\infty=\{0,2,4\}\ \textrm{if}\ \beta_\infty=0.
\end{align*} Find all combinations of $e_j\in E_j$ and
$e_\infty\in E_\infty$, not all even integers, so that \[
d=\frac{1}{2}\Big(e_\infty-\sum_{j=1}^k e_j \Big) \] is a
non-negative integer. For each $d$ found, look for a monic degree
$d$ polynomial $P(z)$ satisfying \[ P'''+3\theta
P''+(3\theta^2+3\theta'-4r)P'+(\theta''+3\theta\theta'+\theta^3-4r\theta-2r')P=0,\quad
\theta=\frac{1}{2}\sum_{j=1}^k \frac{e_j}{z-a_j}.
\] If found, a solution to \eqref{DE} will be $\xi=e^{\int \omega}$ where
$\omega$ is a root of
$\omega^2+\phi\omega+(\tfrac{1}{2}\phi'+\tfrac{1}{2}\phi^2-r)=0$
and $\phi=\theta+P'/P$.

\medskip

\noindent{\bf Case 3} In this case we must follow the procedure
below for each of the subcases $N=4,6,12$. Define the following
sets:
\begin{align} &F_j=\{6+\tfrac{12e}{N}\sqrt{1+4\beta_j},e=0,\pm
1,\ldots,\pm\tfrac{N}{2}\}\cap\mathbb{Z}\
\textrm{if}\ \beta_j\neq 0; \nonumber \\
&F_j=\{12\}\ \textrm{if}\ \beta_j=0,\delta_j\neq 0;\quad
F_j=\{0\}\
\textrm{if}\ \beta_j=\delta_j=0; \label{fsets}\\
&F_\infty=\{6+\tfrac{12e}{N}\sqrt{1+4\beta_\infty},e=0,\pm
1,\ldots,\pm\tfrac{N}{2}\}\cap\mathbb{Z}. \nonumber
\end{align} Find all combinations of $f_j\in F_j$ and
$f_\infty\in F_\infty$ so that \[
d=\frac{N}{12}\Big(f_\infty-\sum_{j=1}^k f_j \Big) \] is a
non-negative integer. Now define \[
\theta=\frac{N}{12}\sum_{j=1}^k \frac{f_j}{z-a_j}, \quad
S=\prod_{i=1}^k (z-a_j). \] Let $P_N=-P$ where $P(z)$ is a monic
polynomial of degree $d$, and recursively define a set of
polynomials $P_{N-1},P_{N-2},\ldots,P_{-1}$ in the following way:
\[ P_{i-1}=-SP_i'+[(N-i)S'-S\theta]P_i-(N-i)(i+1)P_{i+1},\quad
i=N,N-1,\ldots,0. \] If we find $P_{-1}\equiv 0$, then use the set
of polynomials $P_i$ as coefficients in the following polynomial
and find a solution $\omega$: \[ \sum_{i=0}^N
\frac{S^iP_i}{(n-i)!}\omega^i=0.
\] A solution to \eqref{DE} will be $\xi=e^{\int \omega}$.

\medskip

If these three cases fail to produce a closed form solution, it is
because such solutions do not exist. We are now in a position to
derive the NVE and apply the Kovacic algorithm to it.

\subsection{The normal variational equation}

First we must derive a non-constant  solution to linearize around
(there are no constant solutions to geodesic equations). In
general, it is a difficult task to find an exact solution to
geodesic equations, even for simple symmetrical manifolds. In the
case of the sectoral harmonic surfaces however, we are greatly
aided by the existence of certain planes of symmetry, leading to
planar geodesics. These planes of symmetry play the same role in
this context as invariant planes play in more general contexts. In
what follows we will take the equatorial planar geodesic as our
nominal solution (see Appendix A for some comments on the use of
meridianal planar geodesics).

In the equatorial plane $\theta=\tfrac{\pi}{2}, \dot{\theta}=0$,
the geodesic equation is (from \eqref{geoth},\eqref{geoph} and
noting that $\wt{\Gamma}^\theta_{\phi\phi}=0$, where we use a
tilde to denote the value on $\theta=\tfrac{\pi}{2}$) \be
\ddot{\phi}+\wt{\Gamma}^\phi_{\phi\phi}\ \dot{\phi}^2=0 \ee which
admits Hamiltonian \be 2\mathcal{H}=
\wt{g}_{\phi\phi}\dot{\phi}^2=1. \label{redham} \ee To solve for
this equatorial planar geodesic we would need to rewrite
\eqref{redham} as \be \int ds=\int d\phi \sqrt{\wt{g}_{\phi\phi}}
=\int d\phi \sqrt{(1+\vep\cos(n\phi))^2+\vep^2 n^2 \sin^2(n\phi)}
\label{elliptic}\ee and solve to find $s=s(\phi)$, then invert
this relationship to find $\phi=\phi(s)$. The integrals in
\eqref{elliptic} can be written in terms of elliptic integrals of
various kinds, at least for some values of $n$, however the
analysis would benefit from a unified treatment with $n$ as a
parameter. That the integrals in \eqref{elliptic} are daunting
should come as no surprise: we are essentially trying to find the
arc-length parameterization of the curve $r=1+\vep\cos(n\phi)$,
which we could describe as a lima\c{c}on with $n$ `lobes', and
there are few curves for which the arc-length is known exactly.

Fortunately we do not need to solve \eqref{elliptic} exactly;  for
the sake of the NVE we can make a change of independent variable
which requires then only \eqref{redham}, as we will show. Let us
denote the planar geodesic by \be
(\theta,\phi,\dot{\theta},\dot{\phi})=(\tfrac{\pi}{2},\phi_0,0,\dot{\phi}_0),
\ee and let $\theta=\tfrac{\pi}{2}+\xi,\ \phi=\phi_0+\eta$.
Linearizing around this solution we find the equation in
$\ddot{\xi}$ naturally decouples from that of $\ddot{\eta}$, and
since the equatorial/invariant plane is given by
$\theta=\tfrac{\pi}{2}$ the variation normal to this plane is
given by $\xi$, and thus the NVE is \be
\ddot{\xi}=&\left(-\wt{\Gamma}^{\theta}_{\phi\phi,\theta}\dot{\phi}^2\right)\,\xi+
\left(-2\wt{\Gamma}^{\theta}_{\theta\phi}\dot{\phi}\right)\,\dot{\xi},
\label{nve1} \ee where the coefficients are evaluated along the
nominal equatorial geodesic. Now we make the change of independent
variable $z=\vep\cos(n \phi(s))$ so that \be \der{}{s}=-\vep
n\sin(n\phi)\der{\phi}{s}\der{}{z}=-\frac{\vep n
\sin(n\phi)}{\sqrt{\wt{g}_{\phi\phi}}}\der{}{z} \ee and so on. We
may replace instances of $\cos(n\phi)$ with $z/\vep$, to give an
NVE with rational polynomial coefficients (this process is
sometimes referred to as algebrization, and preserves the identity
component of the differential Galois group \cite{simon}), which we
may write as (letting $\xi'=d\xi/dz$ etc.) \be
\xi''+p(z)\xi'+q(z)\xi=0. \label{nve2} \ee Finally, we transform
this equation into the standard form required for the Kovacic
algorithm \[ \xi''=r(z)\xi
\] where \[ r=-q+\tfrac{1}{4}p^2+\tfrac{1}{2}p' \] and we extend $z$ to the complex domain. We are now in
a position to state the main result of his paper:

\begin{thm}
The geodesic equations for sectoral harmonic surfaces with $n>1$
are not integrable.
\end{thm}
\begin{proof}
When $n>1$, the NVE has 6 regular singular points: five finite
regular singular points and one at infinity. The finite regular
singular points are \be
\{a_1,\ldots,a_5\}=\{-1,\vep,-\vep,\rho_+,\rho_-\} \label{app1}
\ee with \be \rho_{+/-}=\frac{1\pm
n\sqrt{1+\vep^2(n^2-1)}}{n^2-1}. \label{app2}\ee The case $n=1$ is
special, and shall be dealt with below. When $n>1$ and $0<\vep<1$
all singular points are real and distinct. The coefficients in the
partial fraction expansion \eqref{pfe} are \be
\beta_i=\left\{0,\frac{-3}{16},\frac{-3}{16},\frac{5}{16},\frac{5}{16}
\right\}, \quad \beta_\infty=\frac{n+1}{n^2}, \ee and the
$\delta_i$'s are given in Appendix B; here we need only mention
that $\sum \delta_i=0$.

We must now check all three cases of the Kovacic algorithm to see
is the NVE solvable. Often in the literature some of the cases can
be ruled out straight away, based on the ``necessary conditions''
given in Section 2.1 of Kovacic \cite{kovacicmain}. However all of
these necessary conditions are satisfied in the present case since
the order of each pole of $r(z)$ is 2 and moreover \be
\sqrt{1+4\beta_i}=\Big\{1,\frac{1}{2},\frac{1}{2},\frac{3}{2},\frac{3}{2}
\Big\}\in\mathbb{Q},\quad\sqrt{1+4\beta_\infty}=\frac{n+2}{n}\in\mathbb{Q};
\ee therefore we must examine each case.

The first stage in each case consists of finding all combinations
of the $\beta$ coefficients such that $d$ is a non-negative
integer. Unfortunately, there are a {\it lot} of such
combinations. We show the combinations in Table 1, where $N=1$ and
$N=2$ represent cases 1 and 2 respectively, and $N=4,6,12$
represent case 3. The values of $n$, the specific sectoral
harmonic under consideration, are shown in the column on the left.
Each entry in the table gives the possible values of $d$ in bold,
and in brackets the number of combinations of indicial exponents
which lead to this value of $d$. As can be seen immediately from
the table, when $n=7,8,9,11$ and $n>12$ there are no combinations
leading to $d\in\mathbb{N}_0$; as such the NVE in these cases is
not solvable.

For the other values of $n$ we must try to construct a monic
polynomial of degree $d$ which solves the required equations.
Clearly there are too many to go through individually; for the
sake of demonstration we will describe the most involved case,
$n=2, N=12$. The relevant sets defined in \eqref{fsets} are \be
&F_1=\{12\},\quad F_2=F_3=\{6,7,5,8,4,9,3\},\quad
F_4=F_5=\{6,9,3,12,0,15,-3\}, \nonumber \\
&F_\infty =\{6,8,4,10,2,12,0,14,-2,16,-4,18,-6\} \ee  It is easy
to see how there can be many combinations leading to
$d\in\mathbb{N}_0$. The largest value comes from the combination
\[ d=18-(12+3+3-3-3)=6. \]
In this case, we let $P=z^6+p_5z^5+\ldots+p_0$, and define
\begin{align*}
P_{12}&=-P, \\
P_{11}&=-S P_{12}'-S\theta P_{12},\\
P_{10}&=-S P_{11}'+[S'-S\theta]P_{11}-12 S^2 r P_{12},\\
P_{9}&=\ldots \end{align*} all the way down to $P_{-1}$. These
computations quickly become very large and cumbersome. A useful
note is that each $P_i$ is simply a polynomial in $z$, albeit one
with very complicated coefficients. The order increases in a
regular fashion: since $S=O(z^k)$, $S\theta=O(z^{k-1})$ and $S^2
r=O(z^{2(k-1)})$ where $k$ is the number of finite regular
singular points, we see that $P_{N}=O(z^d), P_{N-1}=O(z^{k-1+d})$
and so on until $P_{-1}=O(z^{(N+1)(k-1)+d})$. In the present case
this means $P_{-1}=O(z^{58})$, although the leading coefficient
always vanishes. Another useful point is that we only need the
first few coefficients in $P_{-1}$; setting the first equal to
zero gives a value for $p_5$, the second gives a value for $p_4$
and so on until we reach a term which cannot be set equal to zero.
Then we know that $P_{-1}\not\equiv 0$, which means that this
combination for $d$ does not produce a solution.

This procedure has been followed for all combinations listed in
Table 1 (using Mathematica). For each and every case, it is not
possible to construct the required polynomial $P$. As such, the
NVE arising from linearising about an equatorial geodesic on the
sectoral harmonic surfaces with $n>1$ does not have a differential
Galois group belonging to cases 1,2 or 3 in Kovacic's theorem;
instead, $\mathcal{G}=SL(2,\mathbb{C})$ whose identity component
is not abelian. Therefore according to the Morales-Ramis theorem
the geodesic equations are not integrable. This ends the proof.
\end{proof}

\begin{center}
\begin{table}
\begin{tabular}{|c||l|l|l|l|l|}
\hline $n$ & $N=1$ & $N=2$ & $N=4$ & $N=6$ & $N=12$ \\
\hline \multirow{2}{*}{2} & \multirow{2}{*}{{\bf 0}(4)} &
\multirow{2}{*}{{\bf 0}(3), {\bf 1}(1)} & {\bf 0}(4), {\bf 1}(2),
& {\bf 0}(21), {\bf 1}(10), & {\bf 0}(31),
 {\bf 1}(20), {\bf 2}(13), \\
 & & & {\bf 2}(1) & {\bf 2}(3), {\bf 3}(1) & {\bf 3}(8), {\bf 4}(4), {\bf 5}(2), {\bf 6}(1) \\ \hline
\multirow{2}{*}{3} & \multirow{2}{*}{-} & \multirow{2}{*}{-} &
\multirow{2}{*}{-} & \multirow{2}{*}{{\bf 0}(9), {\bf 1}(3), {\bf
2}(1)}
& {\bf 0}(9), {\bf 1}(6), {\bf 2}(3), \\
 & & & & & {\bf 3}(2), {\bf 4}(1) \\ \hline
\multirow{2}{*}{4} & \multirow{2}{*}{-} & \multirow{2}{*}{{\bf
0}(2)} & \multirow{2}{*}{{\bf 0}(2), {\bf 1}(1)} &
\multirow{2}{*}{{\bf 0}(7),
 {\bf 1}(2)} & {\bf 0}(7), {\bf 1}(3), {\bf 2}(2), \\
& & & & & {\bf 3}(1) \\ \hline 5 & - & - & - & - & {\bf 0}(2), {\bf 1}(1) \\
\hline 6 & - & - & - & {\bf 0}(3), {\bf 1}(1) & {\bf 0}(3), {\bf 1}(2), {\bf 2}(1) \\
\hline
7 & - & - & - & - & - \\
8 & - & - & - & - & - \\
9 & - & - & - & - & - \\ \hline 10 & - & - & - & - & {\bf 0}(1) \\
\hline 11 & - & - & - & - & - \\ \hline
12 & - & - & - & {\bf 0}(2) & {\bf 0}(2), {\bf 1}(1) \\
\hline
\end{tabular}
\caption{The possible values of the non-negative integer $d$; see
text for an explanation.}
\end{table}
\end{center}
\vspace{-1cm}

We can now deal with the special case $n=1$:

\begin{thm}
The equatorial NVE on the $n=1$ sectoral harmonic surface is
solvable.
\end{thm}
\begin{proof}
When $n=1$ the equatorial NVE has five regular singular points,
four finite at $a_j=\{-1,\vep,-\vep,\rho\}$ with
$\rho=-\tfrac{1}{2}(1+\vep^2)$, and one at infinity. We calculate
\be \beta_j=\Big\{0,\frac{-3}{16},\frac{-3}{16},\frac{5}{16}
\Big\},\quad \beta_\infty=\frac{45}{16} \ee and hence \be
\alpha_j^\pm=\Big\{
(1,1),\big(\tfrac{3}{4},\tfrac{1}{4}\big),\big(\tfrac{3}{4},\tfrac{1}{4}\big),\big(\tfrac{5}{4},-\tfrac{1}{4}\big),
\Big\},\quad
\alpha_\infty^\pm=\big(\tfrac{9}{4},-\tfrac{5}{4}\big). \ee There
is a combination giving \be
d=\frac{9}{4}-\Big(1+\frac{3}{4}+\frac{3}{4}-\frac{1}{4}\Big)=0
\ee and since $\theta$ thus defined solves the Riccatti equation
$\theta'+\theta^2-r=0$ we have the pair of Louivillian solutions
\be \xi_1=C_1(z+1)(z^2-\vep^2)^{3/4}(z-\rho)^{-1/4}, \quad
\xi_2=C_2 \xi_1 \int \xi_1^{-2}dz. \ee This is of course entirely
as expected since the $n=1$ sectoral surface is a surface of
revolution and thus has integrable geodesic flow.

\end{proof}

\section{The geodesic equations}

One geometrical description of a geodesic curve is as follows: of
all the curves connecting two nearby points on a manifold, the one
that extremizes the length along that curve is the geodesic
(minimizes in the case of a positive definite manifold). An
alternative dynamical description is to imagine a test particle
moving in the manifold free from external forces; if the
particle's speed is constant, then the path it follows is a
geodesic. Consequently, we may define a (dynamic) Lagrangian
$\mathcal{L}$ in the following way: let the coordinates $x^a$
parameterize the $n$-dimensional manifold $(\mathcal{M},\bmt{g})$
and let $s$ parameterize a curve in the manifold, such that
$x^a=x^a(s)$. The length of the tangent vector to this curve is
\bes \left(\sum_{a,b=1}^{n} g_{ab}\ \dot{x}^a\,
\dot{x}^b\right)^{1/2}, \ees but this is simply the speed. Here
$g_{ab}$ are the components of the metric tensor in the $x^a$
coordinate system, and from now on a dot denotes differentiation
w.r.t.\ $s$ and summation over repeated indices is implied. We may
define the kinetic energy as one half the square of this quantity
(assuming unit mass), and therefore the Lagrangian function is \be
\mathcal{L}=\tfrac{1}{2} \ g_{ab}\ \dot{x}^a\, \dot{x}^b. \ee
Putting the Lagrangian function into the Euler-Lagrange equations
we arrive at the geodesic equations, \be
\der{^2x^a}{s^2}+\Gamma^a_{bc}\ \der{x^b}{s}\der{x^c}{s}=0,
\label{geoeqn} \ee where the Christoffel symbols (metric
connections) are given by \be
\Gamma^a_{bc}=\tfrac{1}{2}g^{ad}(g_{bd,c}+g_{cd,b}-g_{bc,d}).
\label{chris}\ee The equivalent Hamiltonian formulation is clear,
by simply letting $\mathcal{H}=\mathcal{L}$. For a more detailed
derivation of the geodesic equations see Paternain
\cite{paternain}, Do Carmo \cite{docarmo} or alternatively Wald
\cite{wald}. We note that some authors prefer to use the
coordinates of the space the manifold is embedded in, and place a
constraint on these coordinates (see for example Davison et al
\cite{davison}). This approach does not suit our purposes here.

\subsection{Surfaces defined in polar form}

Let us consider two-dimensional surfaces given in polar form, that
is \be r=r(\theta,\phi), \ee where $(\theta,\phi)$ are the
standard spherical coordinates with $0<\theta<\pi,\ 0<\phi< 2\pi$.
We may calculate the geodesic equations in the following way: the
line element for $\mathbb{R}^3$ in spherical coordinates is given
by \bes ds^2=dr^2+r^2d\theta^2+r^2\sin^2\theta d\phi^2. \ees On
the surface we have $r=r(\theta,\phi)$ and hence
$dr=r_{,\theta}d\theta+r_{,\phi}d\phi$ and so \be
ds^2=(r_{,\theta}^2+r^2)d\theta^2+2r_{,\theta}r_{,\phi}d\theta
d\phi+(r_{,\phi}^2+r^2\sin^2\theta)d\phi^2, \label{legenr} \ee
from which the metric coefficients may be read off, inserted into
the Christoffel symbols in \eqref{chris} which in turn are put
into the geodesic equations in \eqref{geoeqn}, namely
\begin{align}
\ddot{\theta}+\Gamma^{\theta}_{\theta\theta}\dot{\theta}^2+2\Gamma^{\theta}_{\theta\phi}\dot{\theta}\dot{\phi}
      +\Gamma^{\theta}_{\phi\phi}\dot{\phi}^2=0, \label{geoth} \\ \ddot{\phi}+\Gamma^{\phi}_{\theta\theta}\dot{\theta}^2+2\Gamma^{\phi}_{\theta\phi}\dot{\theta}\dot{\phi}
      +\Gamma^{\phi}_{\phi\phi}\dot{\phi}^2=0. \label{geoph} \end{align}

It will be most informative for later analysis to describe closed
geodesics on these surfaces. For this, we note the following: if a
smooth surface has a plane of symmetry, then the curve formed by
the intersection of the plane of symmetry with the surface will be
a geodesic. Geometrically the reason for this is clear: if the
surface is smooth the plane of symmetry will be normal to the
surface, and any normal curve is geodesic. For the sake of the
analysis to follow however we will make the following argument:
let us arrange the coordinate system so that the plane of symmetry
is given by $\theta=\pi/2$; on this plane $r_{,\theta}=0$. From
\eqref{geoth}, we see $\theta=\pi/2,\dot{\theta}=0$ will be a
solution if $\Gamma^\theta_{\phi\phi}(\theta=\pi/2)=0$. Direct
calculation from \eqref{chris} and \eqref{legenr} gives \[
\Gamma^\theta_{\phi\phi}=\frac{r_{,\theta}(r r_{,\phi\phi}
\sin^2\theta -2r_{,\phi}^2\sin^2\theta-r^2\sin^4\theta
)-\cos\theta(r^3\sin^3\theta+r r_{,\phi}^2\sin\theta
)}{r(r^2\sin^2\theta+r_{,\theta}^2\sin^2\theta+r_{,\phi}^2)},
\] which vanishes on $\theta=\pi/2$. This is equivalent to the `invariant plane' construction typical in dynamics which
we will describe more of below;
  we will subsequently refer to such geodesics as {\it planar} geodesics.\\
      Of course, to describe the geodesic completely we must solve the geodesic equations reduced to this plane, namely
      $\ddot{\phi}+\Gamma^{\phi}_{\phi\phi}(\theta=\pi/2)\dot{\phi}^2=0$;
      we will address this issue in Section 4.

\subsection{Surfaces defined by the spherical harmonics}

The surfaces we will examine in this work are those defined by the
spherical harmonics in the following polar form \be
r(\theta,\phi)=1+\vep\, Y_{lm}(\theta,\phi). \ee Here $\vep$ is a
small parameter and $Y_{lm}$ is the spherical harmonic of degree
$l$ and order $m$ (see Wang and Guo \cite{wang}) given by \[
Y_{lm}(\theta,\phi)=\sqrt{\frac{(2l+1)(l-m)!}{4\pi(l+m)!}}\
P^m_l(\cos\theta) e^{im\phi},\quad m=0,\pm1,\ldots,\pm l, \] where
$P^m_l$ are the associated Legendre polynomials. We may absorb the
orthonormality coefficient into $\vep$ and take the real part of
the exponential to write \be r(\theta,\phi)=1+\vep\,
P^m_l(\cos\theta) \cos(m\phi), \quad 0\leq \vep<1. \label{genpol}
\ee With a slight abuse of language, we shall refer to surfaces
defined in terms of the spherical harmonic functions as `spherical
harmonic surfaces' and so on. We chose to study these surfaces for
the following reasons:
\begin{enumerate}
  \item The spherical harmonics are complete, that is any function
  continuous over the sphere may be decomposed into a (possibly
  infinite) set of spherical harmonics. As such, general
  surfaces of a certain class can be decomposed into a number of
  spherical harmonic surfaces.
  \item The spherical harmonics naturally have the smoothness and
  regularity required for this analysis, and are non-singular as long as
  $\vep<1$.
  \item The small parameter allows for a perturbative approach: with
  $\vep=0$ the surface is a sphere, and as we let $\vep$ grow we
  are deforming the surface away from the sphere.
  \item The spherical harmonic surfaces are compact, enabling the
  Poincar\'{e} section analysis of the next section.
\end{enumerate}

There are three classes of spherical harmonic: the zonal, the
sectoral and the tesseral (see Figure \ref{harmex} for
representative examples).
\begin{description}
  \item[Zonal] These are the spherical harmonics with $m=0$, and thus $r=1+\vep\, P_l^0(\cos\theta)$. Since $r_{,\phi}=0$,
  this means $\partial \mathcal{L}/\partial \phi=0$, which from the Euler- Lagrange equations leads to
  $\partial\mathcal{L}/\partial\dot{\phi}=$constant. This is of course Clairault's integral, as the zonal harmonic surfaces are surfaces
  of revolution; the motion of the geodesics on zonal harmonic surfaces will therefore be regular.
  \item[Sectoral] These are the spherical harmonics for which $l=m$. We shall relabel the degree/order $n$ in this case, and $Y^n_n$ takes
   on a simple form: the sectoral harmonic surfaces are defined by \be r=1+\vep\, \sin^n(\theta)\cos(n\phi). \label{secpol} \ee This describes a two-parameter family of surfaces which admit planes of symmetry; as such they provide an interesting case for analysis and will form the main part of this work.
  \item[Tesseral] These are the remaining members of the family, with $m=\pm 1,\ldots$,$\pm l-1$. We will not examine these surfaces in any
  detail in this work, but discuss them briefly in the
  conclusions.
\end{description}

\begin{center}
\begin{figure}
\includegraphics[width=0.95\textwidth]{sphharm}
\caption{Examples of
spherical harmonic surfaces, for various values of $\vep$: on the
left the $l=2,m=0$ zonal harmonic, in the centre the $l=m=4$
sectoral harmonic, and on the right the $l=7,m=4$ tesseral
harmonic.} \label{harmex}
\end{figure}
\end{center}

\vspace{-0.7cm} The Hamiltonian function for the geodesic
equations on the sectoral harmonic surfaces is given by \be
2\ham=g_{\theta\theta}
\dot{\theta}^2+2g_{\theta\phi}\dot{\theta}\dot{\phi}+g_{\phi\phi}\dot{\phi}^2
\ee where \begin{align*}
  &g_{\theta\theta}=(1+\vep\cos(n\phi)\sin^n(\theta))^2+\vep^2n^2\cos^2(n\phi)\sin^{2n-2}(\theta), \\
  &g_{\theta\phi}=g_{\phi\theta}=-\vep^2n^2\cos(n\phi)\sin(n\phi)\cos(\theta)\sin^{2n-1}(\theta), \\
  &g_{\phi\phi}=\sin^2(\theta)(1+\vep\cos(n\phi)\sin^n(\theta))^2+\vep^2n^2\sin^2(n\phi)\sin^{2n}(\theta).
\end{align*}

It is clear from \eqref{secpol}, or indeed Figure \ref{harmex},
that the $n$th sectoral harmonic surface has $n+1$ planes of
symmetry: $n$ vertical/meridianal planes given by $\phi= \pi i/n,\
i=0,\ldots,n-1$ and 1 horizontal plane given by $\theta=\pi/2$,
the equator. Each of these planes defines a planar geodesic as
described above.

\section{Poincar\'{e} Sections}

This is a much used technique to attempt to visualize the whole of
phase space in the case of a 2-degree of freedom Hamiltonian
system. By fixing a value of $\mathcal{H}=\mathcal{H}_0$, a
3-dimensional submanifold of phasespace, and intersecting this
manifold with a hyperplane (typically given by a coordinate taking
on a fixed value), we consider successive intersections of
solution trajectories with this hyperplane. If the Hamiltonian
system is integrable, then the second integral restricts the
solutions further to a 2-dimensional submanifold of phasespace,
whose intersections with the hyperplane will trace out a closed
curve. On the other hand, if the Hamiltonian system is not
integrable then successive intersections of the solution
trajectory with the hyperplane may instead fill a 2-dimensional
region.

Typically, for a fixed set of parameters a number of Poincar\'{e}
sections for various values of the `energy' $\mathcal{H}_0$ must
be drawn, for example in the H\'{e}non-Heiles system \cite{hh}.
This is not necessary in the current situation, since the
numerical value of the Hamiltonian function may be changed from
$\ham_0$ to $\ham_1$ by the change of independent variable $s\to
\sqrt{\ham_0/\ham_1}\, s$. Therefore the numerical value of $\ham$
signifies the {\it parameterization} of the geodesic curve, rather
than any `energy' concept. Indeed, any affine coordinate
transformation $s\to \alpha s+\beta$ will change the value of
$\ham$ but leaves \eqref{geoeqn} invariant. Therefore, any
qualitative change in the Poincar\'{e} sections will be due to
variations in the surface's parameters ($n$ and $\vep$ for the
sectoral harmonic surfaces), rather than the value of $\ham$. In
what follows, we will assume the geodesic is parameterized by
arc-length (sometimes referred to as `unit-speed curves' in the
literature), and therefore \[ \ham\equiv \tfrac{1}{2}. \]

In choosing the hyperplane to act as the Poincar\'{e} section we
must ensure all solutions intersect the hyperplane transversally;
no tangential grazings are allowed, and a guarantee that the
trajectory will intersect the surface again is preferable. Any of
the $n+1$ planes of symmetry would be suitable candidates,
although for simplicity we will use the equatorial plane (some
comments regarding the use of meridianal planes are given in
Appendix A). First we prove the following lemma:

\begin{lemma}
Along a geodesic, $\theta$ cannot obtain a maximum/minimum in the
northern/southern hemisphere respectively, at least for moderate
values of $\vep$.
\end{lemma}
\begin{proof}
We remind the reader that $\theta$ is measured {\it down} from the
positive $z$-axis, and the northern and southern hemispheres are
given by $\theta\in(0,\pi/2)$ or $(\pi/2,\pi)$ respectively. Due
to symmetry, we need only show that $\theta$ cannot obtain a
maximum for $\theta\in(0,\pi/2)$.\\
From \eqref{geoth}, a maximum w.r.t.\ $\theta$ in the northern
hemisphere would require $\Gamma^\theta_{\phi\phi}>0$ for some
$\theta\in(0,\pi/2)$ (on the sphere,
$\Gamma^\theta_{\phi\phi}=-\cos\theta\sin\theta< 0$ for
$0<\theta<\pi/2$, ruling out this possibility). We can write \[
\Gamma^\theta_{\phi\phi}=-\frac{r\cos\theta\sin^3\theta}{det(g)}\
f(\sin\theta,\cos(n\phi);n,\vep). \] As the surface is positive
definite to prove the lemma we must show $f\geq 0$ in
$\Omega=\{0<\theta<\pi/2$, $0<\phi<2\pi\}$. Letting
$\rho=\sin\theta$ with $0<\rho<1$ in $\Omega$, we can write \[
f(\rho)=1+a_1\rho^n+a_2\rho^{2n-2}+a_3\rho^{2n}+a_4\rho^{3n-2}+a_5\rho^{3n}
\] where the $a_i$ depend on $\cos(n\phi),\ n$ and $\vep$ ($\phi$ here
being the coordinate at which the proposed maximum takes place).
When $\cos(n\phi)>0$, all the $a_i$ are positive and so
$f(\rho)>0$ in $\Omega$.\\ If, on the other hand, $\phi$ is such
that $\cos(n\phi)<0$, Descarte's rule of signs tells us that
$f(\rho)$ can only have 1 or 3 roots in $0<\rho<1$, and since
$f(0)>0$ this is impossible unless $f(1)<0$. Letting $f(1)=0$
gives a cubic in $\cos(n\phi)$, and we can then find explicitly
the value of $\vep$ as a function of $n$ at which the value of
$f(1)$ becomes negative for some values of $\phi$. If we call this
critical value $\vep^*$, then for example when $n=2,3,4$ we find
$\vep^*=0.57,0.497,0.445$.
\end{proof}

Note this lemma rules out the possibility of closed geodesics
which do not intersect the equator at some stage (the only
$\theta=$ constant geodesic is the equator). What it does not rule
out, however, is the possibility of a geodesic asymptotically
approaching the equator. In fact, this cannot occur as the equator
is of elliptic type. A rigorous proof of the non-hyperbolic nature
of the equator is too much of a digression; certainly numerical
evidence can be easily provided using Floquet theory (see
\cite{watquasi} for a detailed description) by calculating the
eigenvalues of the monodromy matrix and showing they remain on the
unit circle in the complex plane for moderate values of $\vep>0$
and various values of $n$. Having said that the most immediate
demonstration of the elliptic nature of the equator is to be found
in the Poincar\'{e} section shown in Fig.\ A.3.

\begin{prop} With the equatorial geodesic of elliptic type, any geodesic which intersects the equatorial
plane transversally must re-intersect the equatorial plane an
infinite number of times, at least for moderate values of $\vep$.
\end{prop}
\begin{proof} First we can see that any trajectory which intersects the
equatorial plane transversally cannot subsequently graze this
plane tangentially as it is an invariant plane. If we consider a
geodesic which crosses the equator with $\dot{\theta}<0$ (i.e.\
entering the northern hemisphere) then there are two
possibilities: 1) the geodesic does not pass through the north
pole, and 2) the geodesic passes through the north pole. In the
event of case 1), the geodesic must attain a minimum w.r.t.\
$\theta$ ($\theta$ is an initially positive decreasing continuous
function of arc-length and so must either: attain a minimum; pass
through zero (case 2); or asymptote to some constant value which
is impossible as the surface is compact and there are no
equilibrium points or closed geodesics contained in a hemisphere).
As the geodesic cannot attain a maximum w.r.t.\ $\theta$ in the
northern hemisphere or asymptote to the equator, it must
subsequently re-intersect the equator. In the event of case 2),
consider a geodesic which starts at the north pole. While there is
a degeneracy in defining $\phi$ at the poles there is no problem
in defining $\dot{\phi}$. Following the geodesic in this direction
it must intersect the equator for the reasons given in case 1),
but the same can be said if we follow the geodesic ``backwards'',
and so it must intersect the equator in both directions.
\end{proof}

A final consideration is the ranges of velocities the geodesics
may explore. On any of the planes of symmetry, either
$r_{,\theta}$ or $r_{,\phi}$ vanishes, thus (from \eqref{legenr})
$g_{\theta\phi}=0$. Therefore on the section we have \be
g_{\theta\theta} \dot{\theta}^2+g_{\phi\phi} \dot{\phi}^2=1, \ee
which is an ellipse of semi-axes
$1/\sqrt{g_{\theta\theta\phantom{\phi}}}$ and
$1/\sqrt{g_{\phi\phi}}$. Since $g_{\phi\phi}(\theta=\pi/2)=(1+\vep
\cos(n\phi))^2+\vep^2 n^2 \sin^2(n\phi)$, when choosing our
initial conditions to generate the Poincar\'{e} section plots we
should therefore choose random values for $\dot{\phi}$ in the
range $\dot{\phi}\in(-\dot{\phi}_m,\dot{\phi}_m)$ with
\begin{equation}
\dot{\phi}_m=\left\{ \begin{array}{cc} \frac{1}{1+\vep} & \quad
\textrm{if}\ \vep<\frac{1}{n^2-1} \\ \\
\frac{1}{\sqrt{\frac{n^2(1+\vep^2(n^2-1))}{(n^2-1)}}} & \quad
\textrm{if}\ \vep>\frac{1}{n^2-1}.
\end{array} \right.
\end{equation}

The initial conditions can now be chosen in the following way:
random values for $\phi$ are chosen in the range $(0,2\pi)$,
random values for $\dot{\phi}$ in the range
$(-\dot{\phi}_m,\dot{\phi}_m)$, and the value of $\dot{\theta}$ is
found from $\mathcal{H}=\tfrac{1}{2}$ (we will always take the
positive value to give a `downward' pointing geodesic). The
geodesic with these initial conditions is followed for a long time
and intersections with the equatorial plane with $\dot{\theta}>0$
are monitored. We show in Figure \ref{psecplots} some of the
resulting Poincar\'{e} sections for the case of the $n=3$ sectoral
harmonic surface. Some interesting points to note are:

\begin{enumerate}
\item There are three main families of {\bf closed geodesics};
these appear as fixed points of the Poincar\'{e} section, and are
as follows: Firstly there are $n$ {\it planar} geodesics as
described in Section 2 leading to fixed points at $\phi=\pi
i/n,i=1,\ldots,n$ and $\dot{\phi}=0$. Secondly on the sectoral
surfaces with $n$ odd there are $n$ {\it perpendicular} geodesics,
which are not planar but intersect the equator at right angles,
leading to fixed points roughly midway between those due to the
planar geodesics. Thirdly there are {\it oblique} geodesics which
are simple (period-1) and give
rise to fixed points off the horizontal axis of the Poincar\'{e} section.\\
Of course, there are an infinite number of closed geodesics on
topological spheres (see Bangert \cite{bangert} and Franks
\cite{franks}) and we make no attempt to classify them all.
\item The {\bf stability} of these closed geodesics is perhaps not what would be expected (Note: we are using only the Poincar\'{e}
  sections to draw conclusions about stability; a more rigorous stability analysis would make an interesting topic for
  future research). Consider first the sectoral surfaces with $n$ even, as shown in Figure 1 (b). The planar geodesics alternate
  between those for which $\partial^2 r/\partial \phi^2\leq 0$ and those for which $\partial^2 r/\partial \phi^2\geq 0$ everywhere
  along the curve. In analogy with the surfaces of revolution \cite{docarmo}, we might therefore expect that these geodesics would alternate
  between stability and instability, respectively. However this is not the case: all planar geodesics are in fact stable, leading to
  elliptic fixed points on the Poincar\'{e} section.\\
      If we now consider the sectoral surfaces for $n$ odd, then $\partial^2 r/\partial \phi^2$
      is both positive and negative along each planar geodesic. In fact, these planar geodesics are all unstable; what's more each is
      homoclinic to itself (at least for small $\vep$, see below) and thus admits two elliptic points within its homoclinic loops, see Figure \ref{psecplots}.
      These elliptic points form a fourth family of closed geodesics, and some numerical experimentation reveals these are also period-1.\\
      The perpendicular geodesics mentioned in 1.\ are all stable, leading to elliptic fixed points in the Poincar\'{e} section,
      and most importantly the oblique geodesics are unstable, leading to hyperbolic fixed points. These hyperbolic fixed points
      are heteroclinic to one another as Figure \ref{psecplots} shows (in fact there are complex heteroclinic networks, where up to $2n$ saddles
      are connected to one another), and this is the crucial ingredient leading to chaotic motion, which we shall describe next.
  \item As is clear from Figure \ref{psecplots}, {\bf chaotic} motion takes place for large $\vep$, that is as we deform the surface away from the sphere.
  The onset of chaos
  as depicted in Figure \ref{psecplots} is straight out of a textbook: the invariant manifolds of unstable fixed points intersect leading to a
  heteroclinic
  tangle, and as $\vep$ increases further rational invariant tori are destroyed. This phenomenon is well described in the literature, see for
  example Jos\'{e} and Saletan \cite{jose}, Calkin \cite{calkin} or Berry \cite{berry}. The value of $\vep$ at which this onset of chaotic motion takes
  place varies for different values of $n$, but note that we see chaotic motion for values of $\vep$ less than those mentioned in Lemma 1.
\end{enumerate}

\begin{center}
\begin{figure}
{(a) \includegraphics[width=0.8\textwidth]{loweeps} \\
(b) \includegraphics[width=0.8\textwidth]{mideeps} \\
(c)
\includegraphics[width=0.8\textwidth]{higheeps}}
\caption{Sample
equatorial Poincar\'{e} sections for the $n=3$ sectoral harmonic
surface, with the following values of $\vep$: (a) $\vep=0.1$; (b)
$\vep=0.2$; (c) $\vep=0.3$. The axes are $\phi$ and $\dot{\phi}$.
See text for discussion.}\label{psecplots}
\end{figure}
\end{center}

The onset of chaotic motion is visible in the Poincar\'{e} section
of all sectoral harmonic surfaces, with the exception of the $n=1$
surface. On this surface the geodesic flow is perfectly regular
for all values of $\vep\in[0,1)$. This is because the $n=1$
sectoral surface is in fact a surface of revolution; it is the
surface formed by revolving the lima\c{c}on around the $x$-axis,
as we may show:

The $n=1$ sectoral surface develops a `dimple' along  the
$x$-axis. Let us rotate the axes (or alternatively the surface)
about the $y$-axis so that the dimple is along the $z$-axis. As
this rotation amounts to $(x,y,z)\to(z,y,-x)$ we can define a new
set of spherical coordinates $(\tb,\pb)$ related to the old
spherical coordinates by \bes \sin\theta=\sqrt{1-\sin^2\tb
\cos^2\pb},\quad \cos\phi=\frac{\cos\tb}{\sqrt{1-\sin^2\tb
\cos^2\pb}}. \ees The $n=1$ sectoral harmonic surface is therefore
given by \be r=1+\vep \sin\theta \cos\phi=1+\vep\cos\tb, \ee the
parametric form of a lima\c{c}on, and clearly $\partial
r/\partial\pb=0$. This surface will therefore have integrable
geodesic flow, a fact that will prove useful in the next section
as a check of the procedure carried out.

The Poincar\'{e} sections provide strong evidence for the
following conclusion: the geodesic equations for the sectoral
harmonic surfaces with $n>1$ are not integrable. However, the
Poincar\'{e} sections cannot be used as proof of this statement,
for the following reasons:\\ 1) as the break up of regular motion
apparently happens only for larger values of $\vep$, there are
possibly small values of $\vep$ for which the equations are
integrable, as in the general Henon-Heiles system for example \cite{mccauley};\\
2) the integrations which generate the plots are run over a finite
time; if the integrations were run over a longer time interval
perhaps a sequence of points which seems to form a closed curve
would begin to wander away. This `stickiness' phenomenon is in
fact quite common in chaotic systems (see for example Murray and
Dermott \cite{murray}), and\\ 3) while great pain may be taken to
ensure accuracy in numerical investigations, there is no escaping
the fact that these are only approximate solutions to a system of
nonlinear ODE's.

Taking these points on board we seek a rigorous and analytical
test of the integrability or otherwise of the geodesic equations.
This will be the goal of the next section.

\section{Introduction}

The study of the qualitative behavior of geodesics goes back to
the beginnings of differential geometry; in fact Clairault's
relation dates from the 1740's. One of the many interesting
features of the geodesic equations is their complexity: how can we
solve such difficult nonlinear differential equations, and indeed
what does it mean to `solve' a differential equation at all? In
the general development of Lagrangian/Hamiltonian dynamics the
issue of integrability came to the fore, and the use of first
integrals to solve a dynamical system via quadratures was
developed. By integrability here we mean intebrability in the
sense of Louiville: a Hamiltonian system with $n$ degrees of
freedom is Louiville integrable if there exist $n$ independent
first integrals of the motion in involution (vanishing pairwise
Poisson brackets). This formalism in turn lead to the consequences
of non-integrability, the onset of chaos and the celebrated KAM
theory. For a particularly interesting introduction to this topic
see Berry \cite{berry}, McCauley \cite{mccauley} or Jos\'{e} and
Saletan \cite{jose}.

The integrability of the geodesic equations on some surfaces, and
therefore the regularity of the geodesic trajectories, has been
well understood for many years. Clairault's relation
\cite{docarmo} provides a second integral for surfaces of
revolution, and Jacobi \cite{kling} integrated the triaxial
ellipsoid. These results have been extended to $n$-dimensional
ellipsoids and to quadrics in general (see Audin \cite{audin}),
where interestingly degenerate cases can prove more difficult to
analyse (see Davison et al \cite{davison}). Recently, the
integrability of the geodesic equations has been demonstrated for
more abstract manifolds, defined in terms of Lie groups (see for
example Thimm \cite{thimm}). Aside from being interesting in their
own right, the geodesics of a manifold also have some useful
applications. For example, Maupertius's principle \cite{bols}
relates the dynamics on an isoenergetic level of any mechanical
system to the geodesics of a manifold, and the relativistic
formulation of gravity dictates that massive particles move along
the timelike geodesics of spacetime \cite{wald}.

There are many powerful results regarding the non-integrability of
the geodesic flow\footnote{Technically, the geodesic flow is
defined on the cotangent bundle whose projection onto the manifold
gives the geodesic curves, however we will use these two tems
interchangeably.} on manifolds of a certain class, for example a
theorem of Kolokol'tsov \cite{kolok} says that two-dimensional
compact surfaces of genus $g>1$ do not admit geodesic flows that
are integrable with a second integral polynomial with respect to
the momenta (see Bolsinov \cite{bols2} for a review and
references). What's more the integrable geodesic flows on surfaces
with $g=0,1$ with second integrals either linear or quadratic in
the momenta are completely described. Nonetheless, new examples of
integrable geodesic flows with second integrals cubic in the
momenta have recently been found (see for example Dullin and
Matveev \cite{dullin}) and the possibility of non-polynomial, and
in particular meromorphic second integrals, is open.

A very significant recent development with regards integrability
has been the theorem of Morales Ruiz and Ramis
\cite{mrthm1,mrthm2}. This theorem provides a link between the
question of whether a nonlinear Hamiltonian system is {\it
integrable} (with meromorphic first integrals) with whether the
variational equations of a non-trivial solution to the Hamiltonian
system are {\it solvable}. This is a great advance as the
variational equations are linear with variable coefficients, and
the solvability of equations of this type may be addressed using
differential Galois theory. This approach has proved very
successful (we give here only some examples): in Morales Ruiz and
Ramis \cite{mrthm2} the authors use this formalism to prove the
non-integrability of certain problems from mechanics and celestial
mechanics; in Morales Ruiz et al \cite{simon} the non
integrability of Hill's problem is shown; in Pujol et al
\cite{sam} the non-integrability of the swinging Atwood's machine
is demonstrated; in Acosta-Hum\'{a}nez et al \cite{humanez} some
problems in celestial mechanics are studied and in Bardin et al
\cite{bardin} the generalized Jacobi problem is studied. In the
majority of these papers, the solvability of the variational
equations is determined using Kovacic's algorithm, and this is the
approach we will follow in this work, with a description of the
algorithm and relevant references given in Section 4.

To the best of our knowledge this powerful Morales-Ramis formalism
has not been applied to the study of geodesics on manifolds and
their integrability properties, and the approach taken in this
work is novel in this respect. In the following we will examine
the properties of geodesics on surfaces defined in terms of the
spherical harmonics, and we shall rigorously prove using the
Morales-Ramis theorem the non-integrability of the geodesic
equations on surfaces defined in terms of sectoral spherical
harmonics. Thus we present a two-parameter family of surfaces
which appear very simple and symmetrical and yet have complex and
indeed chaotic geodesic curves on them; this is a novel addition
as the surfaces for which non-integrability of the geodesic flow
has been rigorously proved is small. Finally, we will also use
Poincar\'{e} sections to display the breakdown of regular motion
as well as to understand the various families of closed geodesics
on the surfaces; again this approach is novel in the case of
geodesic flows.

The layout of the paper shall be as follows: In Section 2 we
review the formulation of the geodesic equations and describe the
classes of surfaces which will be examined. In Section 3 we use
Poincar\'{e} sections to exhibit the breakdown in regular motion
suggesting non-integrability, as well as describe the main
families of closed geodesics on the surface. Finally in Section 4
we will review some necessary theory and prove the main result:
the non-integrability of the geodesic equations on sectoral
harmonic surfaces with $n>1$.

\input{psecsec}

\input{kovsec}

\section{Conclusions}

The application of Morales-Ramis theory and Poincar\'{e} sections,
two techniques most often seen in the realm of celestial mechanics
and mechanical systems, to study the integrability of geodesic
flow presents many interesting possibilities. By choosing a
specific class of surfaces to analyze we can rigorously prove the
non-integrability of the geodesic equations. It would be of
interest to extend the results of this work to examine further
surfaces; for example the tesseral harmonics mentioned in Section
2 again have (meridianal) planes of symmetry, and the NVE will
have coefficients that contain Legendre polynomials, and in
general, it would be very informative if we could examine
arbitrary surfaces with a plane of symmetry and try and arrive at
some obstructions to integrability. We note that the presence of a
plane of symmetry makes the decoupling of the NVE straightforward,
however Morales Ruiz \cite{morruiz} outlines a procedure in the
absence of an invariant plane. The completeness of the spherical
harmonics means that surfaces of a very broad class can be
decomposed into a set of spherical harmonics, and perhaps by
analyzing these individually we can make some conclusions about
their sum. Care must be taken however: the triaxial ellipsoid can
be decomposed into a set of (infinite) spherical harmonics, each
of which would likely have non-integrable geodesic flow, and yet
their recombination gives a surface which {\it is} integrable. A
further issue to consider is how the surface in question is
described; the parametric description of the harmonic surfaces led
easily to the calculation of the geodesic equations and in
particular the separation of the normal variational equation. If
the surface were defined in algebraic form, as is more suited to
the double torus for example, this separation may be more
problematic. Finally, this subject provides an interesting and
fruitful combination of techniques from dynamical systems theory,
differential geometry and differential Galois theory.

\section{Acknowledgements}

The author wishes to thank the School of Mathematics, National
University of Ireland, Galway, where some of this work was carried
out; Dr.\ Sergi Simon for useful conversations; and finally the
referees for their important and constructive comments.

redpajama_set_name: RedPajamaArXiv


text:
Bangladesh beat West Indies by 48 runs in first ODI
Providence, Guyana: A record double-century partnership between Tamim Iqbal and Shakib al Hasan paved the way as Bangladesh bounced back from their Test series drubbing with a 48-run win over West Indies in their opening one-day international on Sunday.
Man-of-the-Match Tamim blasted an unbeaten 130, the highest by a Bangladeshi in ODIs against the West Indies, while Shakib weighed in with 97 a second-wicket partnership of 207.
Captain Mashrafe Mortaza then led the way with the ball in taking four for 37 as the hosts were limited to 231 for nine in reply.
"It was important for us to make a strong statement here after the way we played in the Tests," Mashrafe said after the victory at Guyana's National Stadium.
"This ground has been a happy one for us over the years and we are looking now to clinch the series in the next match on Wednesday."
West Indies were on course to seriously challenge the target with Chris Gayle and Shimron Hetmyer at the crease.
However when Gayle was run out for 40 and Hetmyer fell to Mustafizur Rahman for a topscore of 52 the innings went into freefall and an entertaining unbroken last-wicket stand of 59 between Devendra Bishoo and Alzarri Joseph merely delayed the inevitable.
Smarting from the annihilation endured over the two Tests, both of which ended inside three days, Bangladesh immediately found the less challenging circumstances of ODI cricket more to their liking.
"For those of us who were involved in the Test series it mattered to us to come out here fighting and I am glad we did that," Tamim said.
Mushfiqur Rahim also belted 30 off just 11 deliveries and made a significant difference to his side's innings in the final overs.
His late assault contributed to 43 runs being taken off the last two overs -- the most productive ever for Bangladesh in their ODI history -- off the bowling of Jason Holder and Andre Russell.
For Holder it represented a turnaround from earlier in the morning, when he removed opening batsman Anamul Haque for a duck in his opening over after Mortaza chose to bat first.
Tamim then found a resolute partner in Shakib as they set about the task of preventing the sort of batting capitulation that was repeatedly evident in the Test matches.
Their cause was assisted considerably by sloppy West Indies catching. Tamim was missed on 17 and 20 while Shakib was let off on 15 and 85. They made the home side pay dearly for those lapses. Shakib's luck eventually ran out when three runs short of the landmark. Their partnership was the highest for any wicket in ODI's at Providence.
"We didn't help ourselves at all with poor work in the field and our death bowling wasn't the best either," Holder said. "We need to redouble our efforts and come back for what is now a must-win situation for us."
Shakib's dismissal and the swift demise of Sabbir Rahman turned out to be a blessing in disguise for Bangladesh as Mushfiqur launched into the bowling almost immediately while also stirring Tamim into life.
He struck consecutive sixes off Russell in the last over of the innings although it could not change the fact that his innings – 130 not out off 160 balls with ten fours and three sixes – was the slowest-ever hundred in ODI's by a Bangladeshi.
#TamimIqbal #Tamim #Bangladesh #WEstIndies #ODI #ShaqibAlHassan
Ozil's resignation sparks Germany racism storm as Ankara cheers
All Black magic as New Zealand retain World Cup Sevens crown
STILL IN A SPOT!
Battle lines drawn
Aussies register big victory
Afridi confirms for World Stars vs Qatar Stars tie
Australia set history, win big
Two-tier Test plan axed after India opposition
Australia script four-wicket win
Pakistan rout England in T20


Great day! We joined with Amnesty International's Write #4Rights campaign and wrote letters on behalf of people throughout the world who are in prison or being taken advantage of.
During this season of Advent we want to bring light to the injustices that we see, we want to help set the oppressed free, and see these captives released.


text:
Barakacity – francuska, muzułmańska organizacja pozarządowa o charakterze humanitarnym.

Historia
Organizacja z siedzibą w Courcouronnes pod Paryżem, powstała z inicjatywy Idrissa Sihamediego w 2008 choć oficjalnie została zarejestrowana dopiero w 2010. Prowadziła wówczas pierwszą dużą, międzynarodową akcję, nakierowaną na pomoc ofiarom powodzi w Pakistanie. W 2012 organizacja rozpoczęła realizację projektu Water is Life. Ta inicjatywa zrodziła się po podróży Idrissa Sihamediego do Togo, kraju cierpiącego na brak wody pitnej i infrastruktury ochrony zdrowia. W 2013 towarzystwo prowadziło akcję na rzecz mieszkańców Togo. W tym samym czasie działacze organizacji odbyli podróże do Birmy (lud Rohingja), a także do Syrii, z pomocą po zniszczeniach wojennych. Od 2014 organizacja dysponuje ciężarówką do wiercenia studni. Podejmowała też pierwsze działania w Republice Środkowoafrykańskiej, kraju wstrząsanym różnorakimi kryzysami. Od 2014 rozpoczęto też akcję pomocy birmańskim Rohingja, a także Palestyńczykom, którym poświęcona była kampania w czasie Ramadanu. W 2015 Barakacity działało w dwudziestu krajach i miało pięć biur zarządzanych przez lokalnych menedżerów. Deklarowało też działania w segmencie obrony praw człowieka.

Organizacja została rozwiązana w październiku 2020. Wcześniej oskarżana była przez rząd francuski o propagowanie idei radykalnego islamizmu, podżeganie do nienawiści, utrzymywanie stosunków w ramach radykalnego ruchu islamistycznego oraz usprawiedliwianie aktów terrorystycznych.

Przypisy

Organizacje pozarządowe we Francji
Hasła utworzone w ramach Tygodni Tematycznych 2019

redpajama_set_name: RedPajamaWikipedia


text:
Alzheimer's Disease, through the eyes of a hyperspectral camera
Imec's hyperspectral snapshot camera is used by researchers from KU Leuven and the university hospital to study protein accumulation in the retina of Alzheimer's patients.
What if Alzheimer's could be detected in a very early stage, only with an eye examination? Belgian biologists (KU Leuven), ophthalmologists (UZ Leuven) and engineers (VITO and imec) believe this is possible. Together, they aim to develop a rapid, user-friendly, and affordable test for diagnosing Alzheimer's.
Alzheimer's Disease (AD) is one of the three so-called trillion-dollar diseases in the world, next to cancer and diabetes. Today, there is still no adequate treatment, but scientists believe that diagnosing in an earlier stage could be the key to better therapy.
The eyes as a window to the brain, that's the starting point of a new series of studies worldwide to find an alternative diagnose method for Alzheimer's. As in the brain, the studies focus on finding the accumulation of two proteins - amyloid-beta (Aβ) and Tau – in the eyes. And that's where hyperspectral imaging (HSI) comes in. Every protein – when struck by light – reflects specific wavelengths. The hyperspectral camera captures this reflected light and analyses this 'signature'.
Researchers from the Biology Department at KU Leuven, the Ophthalmology research group at UZ Leuven, the Health Unit at VITO, and the hyperspectral group at imec collaborated to integrate some of imec's hyperspectral cameras with a microscope (for the biologists) and a fundus camera (for the ophthalmologists) to further investigate the hypothesis that Alzheimer's typical protein accumulations can be detected in the eyes, in a very early stage of disease development.
The results are encouraging and – if successful – the technique could be commercialized in a relatively short timeframe.
Read the full story on www.imechyperspectral.com.
Setup of combined hyperspectral and fundus camera at Stalmans' lab.
Get inspired by imec's hyperspectral imaging technology and contact our expert.
Read the paper 'Hyperspectral Imaging and the Retina:Worth theWave?' from Lemmens et al.
Learn more on the topic in the paper 'Combination of snapshot hyperspectral retinal imaging and optical coherence tomography to identify Alzheimer's disease patients' from Lemmens et al. This is the summary of the PhD thesis of Sophie Lemmens which can sent on request.
Discover more on the fundamental research in the paper 'The AppNL-G-F mouse retina is a site for preclinical Alzheimer's disease diagnosis and research' from Vandenabeele et al.
This research was done in the framework of the ADMIRE project - Alzheimer's Disease Detection Using Multimodal Imaging of the Retina – and funded through the Mission Lucidity initiative.
Part of this research work has been co-funded in the context of the HERALD project that was granted by the ATTRACT consortium, which received funding from the European Union's Horizon 2020 Research and Innovation Programme (2014-2020). Further, research is funded through a JNPD grant with the university of Melbourne and the Umea University and through a joint PhD program from the KU Leuven and the University of Melbourne.
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Let's read the Bible together every day! In one year we can read the entire Word of God. Just 15-20 minutes is all it takes to read or listen to a reading of the scriptures that will change your life. Click on the link below to take you to the page for today's reading. Join at any time and we will see how the power of God's Word transforms us together.


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KYAQ – Lincoln County's Public Radio Station at 91.7 on FM dial
On Friday, January 3, 2014.KYAQ(91.7 FM in Siletz, Oregon) went on the air and made way for the people of Lincoln County to shape their first local non-commercial radio station.
Bridget Wolfe, past Board President of Firebare, the station's sponsoring non-profit, said, "This is a gift to the community, a gathering spot for everyone on the Central Coast to share our wealth of intellectual and artistic resources. The airwaves are crucial to our area, often the only way people can know what is going on."
Wolfe said, "In the spring of 2007, my husband John Crawford and I decided to mark his birthday with a trip to Ashland to hear Amy Goodman speak. John and I had lived in Los Angeles before moving to the Oregon Coast and were avid KPFK listeners and supporters. We were frustrated by the lack of alternative voices in the area. We met people in Ashland who had similar concerns; listening to Amy inspired us, and we decided to make community radio here in Lincoln County a reality."
A local non-profit, Firebare, signed on as sponsor and, with the help and guidance of the Prometheus Radio Project, the group applied for a radio frequency. Firebare was initially rated second among eight competing applicants. In 2011, when the number one applicant withdrew, Firebare was issued its FCC construction permit to build a station, with a three-year deadline.
Community members, individuals and institutions, artists and musicians, belly dancers and a salsa DJ, all pitched in to raise money. Local artists collaborated to create a video for an Indiegogo campaign and designed the station's "Find Your Voice" posters, "to increase public awareness of KYAQ and how it will promote democracy, be a home to free expression, stimulate investigation of new ideas and become Lincoln County's premier creative outlet." National support came from Amy Goodman and Mike Malloy who called in during station fundraisers and David Barsamian of Alternative Radio who spoke at a local event. (Barsamian spoke again at a KYAQ event last fall.)
Some six to eight months into this process, Crawford met Alan Searle. Searle recalls, "I was standing in line at the feed store when I ran into Firebare Board Member John Crawford." Said Wolfe, "It was a once-in-a-lifetime opportunity. We could not have done it without Alan."
Searle, who had 30 years experience as a broadcast journalist, including positions with both Minnesota Public Radio and Free Speech Radio News, knew the ropes. He says, "From the onset, we demonstrated what an engaged community station can do." KYAQ operated as an internet station and broadcast live local music events, and also maintained a news, information and opinion website. During the early years, the website promoted the station project with construction updates, video commentaries from locals as well as national personalities.
Then, finally, came January 3, 2014; and, with some 14 hours of grace time before the FCC construction permit expired, KYAQ went on the air and began regular broadcast operations. Locally produced programs cover the gamut from environmental news and interviews, community affairs, political news, cooking, LGBTQ issues, Spanish language news Native American issues, and music, lots of music. KYAQ now has eight music programs created by local volunteers. Programming typically is talk, news, and educational shows during the day and music at night. Sunday afternoon features several Spanish language news and music programs, and Native American programming.
The station is completely volunteer run, and maintaining sufficient operating revenue has been challenging, as is typical with start-up community radio stations. After several years of operating losses, the station found itself $15,000 in debt. Over the last few years, the community stepped up with generous donations and the station has now retired most of its outstanding debt, but the financial challenge continues.
KYAQ's annual pledge and membership drive is scheduled for August 16th through the 19th, with two days of special live programming, followed by two days of broadcasting music live from the Toledo Wooden Boat Show.
Thursday and Friday, August 16th and 17th, KYAQ's local radio program hosts will be doing live broadcasts, interspersed with personal messages from community supporters. The goal is to add at least 20 new sustaining members, who will pledge a monthly donation of $10 or more.
On Saturday and Sunday, August 18th and 19th, the station will broadcast live from the Toledo Wooden Boat Show at Toledo's waterfront park, bringing great musical acts to the airwaves for the enjoyment of all. KYAQ will have a vendor booth at the Boat Show where they will take donations and memberships, and many of the local show producers will be on hand to chat with the public.
An exciting new feature this year is an online silent auction. Beginning August 16th and ending August 19th, the public can go to the website at www.KYAQ.org and bid on all the wonderful products and services that have been generously donated by local people and businesses. There are sure to be things in the silent auction that will appeal to every taste.
Lincoln County's only community radio station has grown to become a hub of information and education. The community calendar at www.KYAQ.org is filled with local events. Interviews with local people active in the environment, community service, politics and education abound. KYAQ has come a long way but it is just getting started. With the help of the community, the little radio station that could aims to fulfill its primary mission: " to strengthen the fabric of our community by weaving together vital threads of news, information, science, music, and art, creating an environment rich with surprise and discovery."


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Home / Drawing Tutorials / Comics / How to Draw Deadpool
How to Draw Deadpool
Hello everyone, today's lesson is made at the request of our readers who are not indifferent to the Marvel universe, and today we will draw the mysterious Deadpool!
Note, that Deadpool's costume is very similar to Spider-Man. But similarities between Spider-Man and Deadpool are ending on this. But similarities between Spider-Man and Deadpool are ending on this. He's not the classic hero, but rather mercenary. He often performs a variety of services to crime bosses. Also, Deadpool has a pronounced instability of the mental state and a penchant for very crazy things. In addition, Deadpool equipped with the latest technology fighter, widely known among fans of his jokes, "coolness" and black humor. Let's start the lesson and learn how to draw Deadpool!
We begin, as always, with stickman – man made of sticks and circles, which we draw to designate the posture of the character, the proportions of the body and the position on the paper. Speaking of proportions – because of the unusual perspective, drawing (we look at Deadpool from the top down), the proportions will appear very distorted, particularly during the first step.
First of all, draw a head in the shape of an oval. Draw the line of the back and shoulders, using simple lines. Then with simple lines mark the hands and feet. Just do not forget that in the initial stages you should draw using clear, but very light touches.
The stickman is ready. From this step we start adding volume, and in this step we will draw the lines of the facial symmetry. Draw the vertical line of symmetry of the face and the horizontal line of the eyes. Note, that these lines crossed in the center of the face. And these lines shows us the direction of view.
Because of the angle of view and posture of Deadpool we can not see his neck, so sketch out deltoids, using circles. And then draw the arms, which at this stage can be divided into three parts:
Deltoids;
Upper part of the arm;
Lower part of the arm;
Conditional fourth part is the hands. Sketch out the contours of the torso, that tapers to the waist. Draw the gun in the right hand and grenade on the left hand.
The upper part is ready. In this step we will start adding volume to the lower part of the Deadpool. Draw the pelvis area using the figure similar to the triangle. Draw the legs in the shapes of cylinders. About legs there is an interesting point in this lesson – the right of us is closer to us, so it looks bigger visually in spite of the fact that she strongly bent, and we can only see the hip.
Silhouette is ready, you can begin to detail. In this step, we draw the head of Deadpool. Erase the guidelines from the head of Deadpool. Draw the eyes, using the lines from the previous step. Draw elongated ovals around the eyes. Draw the shapes of the chin, cheekbones and ears under the mask. And do not forget to add wrinkles on the area of the bridge.
As we have said, Deadpool – an exotic and very interesting character. One of the interesting facts is that Deadpool is almost the only Marvel character, that realize that He – is the character of comics. So, continue drawing lesson, about how to draw Deadpool. Now draw the torso of Deadpool. First, draw the muscles, and then – patterns on the costume. The main muscles that are visible at this stage – is outlines of the pecs and abs.
In this step of the tutorial about how to draw Deadpool we draw the hands. Our hero is clothed in a suit from head to toe, but this suit is very fitting, therefore need to draw the muscles in great detail. Mark undulating mounds of muscles, passing through the whole arm – from the delta to the forearm. Using straight lines draw the gun in the right hand. And in the left hand draw the grenade. Don't forget to draw the lines of the costume and bracers.
In this step we finish the upper part of Deadpool. Erase the guidelines and all unnecessary guidelines, as we did in the example below. Circle the lines of the torso and hands (to learn more visit our drawing lesson about how to draw hands), using clean and sharp lines. Draw the belt, with a lot of pockets. Draw the logo of Deadpool on the center of the belt.
Note, that in this picture Deadpool appears to us without his most important attribute – the samurai sword.
So, continue to draw Marvel superhero. In this step we draw the legs of Deadpool. Erase the unnecessary lines of the previous steps and draw the vertical lines indicating the hilly and powerful thigh muscles. Actually, these lines are not so many, and they are all drawn from the waist to the knees. Draw the lines of the knees and sketch out the strap on the leg.
Here, our artists have used comics style of shadowing, which involves dark, contrasting shadows located on the large areas and consist of the usual thick hatching. In order to get such shadows, first draw the contours of the shadows, and then paint the areas of the shadows. To get halftones use thick hatching.
It was a lesson in which we told you about how to draw Deadpool. If you have comments about new lessons – feel free to write them in the comments, we will try to implement them. Good luck and see you at Drawingforall.net!
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ابوالفضل says:
it is very very beautifull
Oliver Salberg says:
This is awsome!!!!
Naaqruf says:
It looks awsome but I prefer the movie deadpool but still it looks mad
Farah N says:
this is cool.
It's good but it would be better if it had Deadpool's key weapons – the katanas
Dillan Braun says:
it is perfect! 🙂