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# I SU(2) representations

#### Silviu

Hello! I just started reading about SU(2) (the book is Lie Algebras in Particle Physics by Howard Georgi) and I am confused about something - I attached a screenshot of those parts. So, for what I understood by now, the SU(2) are 2x2 matrices whose generators are Pauli matrices and they act on a vector space. Also you can have any n x n representation of SU(2) (here I am a bit confused why is it called SU(2) if you can represent it as any n x n matrix).
So I assumed that this means they act (in the 2 x 2 case) on 2x1 column matrices. But, from what i read in the attached screenshots, I understand that the Hilbert space where these matrices act is the one that undergoes a representation, as they say "to completely reduce the Hilbert space of the world to block diagonal form". And also the (j, m) notation I thought it was for eigenvectors, but they say that they use that notation for the representations of the Hilbert space. From what I read (I just started all this stuff with representation theory so I really need help to understand this), I understood that the matrices that act on the Hilbert space can have different representations, but the Hilbert space itself (formed of vectors) stays the same.
So, to resume, my question is why you need to have a representation of the Hilbert space itself and not just of the operators (the Pauli matrices in this case), or in case I misunderstood what I read, what does the author means by the parts I attached? Thank you!

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#### MisterX

Maybe the question can be boiled down to "Why do we need abstract groups and can we describe everything about a group with a specific set of generator transformations?" Some points
• There may be multiple choices of the field type ($\mathbb{R}$ or $\mathbb{C}$) and dimension of linear transformations which obey the abstract group algebra. As you mention $SU(2)$ can be represented by $n\times n$ matrices as long as $n\geq2$.It may be useful to see matrix representation as a separate notion from the group itself.
• Pauli matrices have an element of arbitrary choice. Even with field and dimension fixed there is an element of arbitrary parameter-ization which doesn't reflect any new group structure.
• Not all Lie groups are simply connected and can be expressed as exponential expansion. The Lorentz group is a common example of a Lie group that is not simply connected.
As for the statement about $(j,m)$ - they are looking at representations that consist of operators on a subspace of the $\mid j,m\rangle$ states. They label the representations by $j$, which is the highest $J_3$ value in that subspace. The dimension of a subspace of fixed $j$ is $2j+1$.

So a $j=\frac{1}{2}$ representation would be $2\frac{1}{2} + 1 = 2$ dimensional and thus we would use $2\times2$ matrices.

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#### lpetrich

A good place to look for stuff on representations of SU(2) is discussions of quantum-mechanical angular momentum. Each possible value of angular momentum j corresponds to a representation (rep) of SU(2), an irreducible representation (irrep) with dimension 2j+1. Each of the dimensions in it corresponds to a value of the projected angular value or magnetic quantum number m. The matrices of the rep are given by the Wigner D-matrices, matrices for doing 3D rotations in that rep. Addition of angular-momentum values corresponds to a product representation constructed from each value's rep. The resulting angular-momentum values correspond to irreps in that rep.

"SU(2) representations"

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# Motion in two dimention problem

1. Nov 15, 2012

### Ammar w

1. The problem statement, all variables and given/known data
A projectile is fired from the top of a cliff of height h above the ocean below. The projectile is fired at an angle θ above the horizontal and with an initial speed vi. (a) Find a symbolic expression in terms of the variables vi, g, and θ for the time at which the projectile reaches its maximum height. (b) Using the result of part (a), find an expression for the maximum height hmax above the ocean attained by the projectile in terms of h, vi, g, and θ.

2. Relevant equations

vfy = viy - gt = visinθi - gt

the maximum hight = $\frac{vi^2 (sinθi)^2}{2g}$

3. The attempt at a solution

a)
vfy = viy - gt = visinθi - gt

it reaches the maximum hight when vfy = 0.

=> visinθi - gt = 0

t = $\frac{ visinθi}{g}$

right??

b) hmax = viyt - $\frac{1}{2}$gt2

hmax = visinθi ($\frac{ visinθi}{g}$) - $\frac{1}{2}$g ($\frac{ visinθi}{g}$)2

hmax = ..... how to simplify it??

2. Nov 15, 2012

### tiny-tim

Hi Ammar w!

(btw, there's no need to write θi

there's only one θ, so just write θ ! )

just expand that second bracket!

(and then get some sleep! :zzz:)

btw, if you know the standard constant acceleration equations, you should be able to find one that solves the problem straight away

3. Nov 15, 2012

### SammyS

Staff Emeritus
Your solution to part a) looks fine.

For part b, what is the height (above the ocean) of the projectile at time t = 0 ? It's h, correct?

So you need to modify the equation, hmax = viyt - $\frac{1}{2}$gt2 to reflect that.

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Here’s a really absurd question: “What’s the expected number of games I’d have to play against Magnus Carlsen before I win one game?”. In the spirit of What if?, let’s try to give a serious scientific answer to this absurd hypothetical.

My Elo is 1420. Magnus’ is 2835. That’s a pretty big gap.

According to the Elo formula1 my expected score against Magnus is 0.0003 (ha!). Let’s call this $$p$$. In case it’s not clear, your expected score is the probability you win + 1/2 the probability you draw. So, if there were no draws in chess, then E[#games until I win] = 1/p ~= 3450 games. However, there are draws, which is obviously terrible news for my expected #games. Let’s say that if Magnus doesn’t win outright, he’s 500x more likely to draw against me than to lose to me.

Given that:

\begin{align*} \mathbb{P}[win] + \frac{1}{2} \mathbb{P}[draw] &= p = 0.0003 \\ \mathbb{P}[win] + \frac{1}{2} 500 \mathbb{P}[win] &= p \\ 256 \mathbb{P}[win] &= p \\ \mathbb{P}[win] &= \frac{p}{256} \\ \mathbb{E}[\text{#games until win}] = \frac{1}{\mathbb{P}[win]} &= \frac{1}{\frac{p}{256}} \approx \text{850,000} \end{align*}

I’m sure Elo breaks down when the rating difference is this large, but I’m not actually sure which way it breaks down2, so I’m just going to stick w/this guess - let’s call it an even million.

Now, another thing to consider is the fact that after playing Magnus in 100,000 games I might actually improve quite a bit! But I’ll ignore that for now…

1. If Player A has a rating of $$R_{A}$$ and Player B a rating of $$R_{B}$$, the exact formula for the expected score of Player A is $$\frac{1}{1 + 10^{(R_B - R_A)/400}}$$

2. This question motivated a later post. Turns out, Elo overpredicts the ability of the better player to win when the rating difference is extremely large.


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# zbMATH — the first resource for mathematics

Fragments of the word $$\Delta$$ in a braid group. (English. Russian original) Zbl 0599.20044
Math. Notes 36, 505-510 (1984); translation from Mat. Zametki 36, No. 1, 25-34 (1984).
The braid group $$B_{n+1}$$ is defined by generators $$a_1,\ldots,a_n$$ and relations $$a_ia_{i+1}a_i = a_{i+1}a_ia_{i+1}$$, $$i=1,\ldots,n-1$$, $$a _ia_j = a_ja_i$$, $$i<j-1$$. Let $$\Pi_{n+1}$$ be the semigroup with the same presentation. For $$1\le i\le j\le n)$$ let $$A_{i,j}$$ and $$B_{i,j}$$ denote the words $$a_ia_{i+1}\cdots a_j$$ and $$a_ja_{j-1}\cdots a_i$$, respectively: $$A_{i,i-1}$$ and $$B_{i,i+1}$$ are empty. Let $$\Delta$$ denote the so-called fundamental word $$A_{1,n}A_{1,n-1}\cdots A_{1,1}$$. Words of the form $$A_{\gamma_n,n}A_{\gamma_{n-1},n-1}\cdots A_{\gamma_1,1}$$, where $$\gamma_i\le i+1$$ for all $$i$$, are called right fragments of $$\Delta$$. Words of the form $$B_{1,\gamma_1}B_{2,\gamma_2}\cdots B_{n,\gamma_n}$$ are called left fragments of $$\Delta$$. It is proved that a word in the alphabet $$a_1,\ldots,a_n$$ is a divisor of $$\Delta$$ in $$\Pi_{n+1}$$ iff it is equal in $$\Pi_{n+1}$$ to some fragment of $$\Delta$$. F. Garside [Matematika, Moskva 14, No. 4, 116–132 (1970; Zbl 0211.34103)] proved that in $$B_{n+1}$$ every word is uniquely representable in the form $$\Delta^ mX$$, where $$X$$ is a positive word that is not divisible by $$\Delta$$ in $$\Pi_{n+1}$$. Using the concept of fragment the author introduces the notions of left and right normal form and proves that in $$B_{n+1}$$ every word is uniquely representable in left normal form and right normal form. A criterion is found for fragments of $$\Delta$$ to be divisible in $$\Pi_{n+1}$$ by a given letter $$a_i$$.

##### MSC:
 20F36 Braid groups; Artin groups 20F10 Word problems, other decision problems, connections with logic and automata (group-theoretic aspects) 20F05 Generators, relations, and presentations of groups
Full Text:
##### References:
 [1] F. L. Garsaid, ?On Braid groups and other groups,? Matematika,14, No. 4, 117-132 (1970). [2] E. Artin, ?Theory of braids,? Ann. Math.,48, 101-126 (1947). · Zbl 0030.17703 [3] G. G. Gurzo, ?Systems of generators for normalizers of some elements of braid groups,? Akad. Nauk SSSR, Ser. Mat.,48, No. 3, 476-519 (1984). · Zbl 0548.20022 [4] G. S. Makanin, ?On normalizers of braid groups,? Mat. Sb.,86, No. 2, 171-179 (1971). [5] J. McCool, ?On reducible braids,? in: Word Problems. II, North-Holland (1980), pp. 261-295. · Zbl 0434.20021 [6] V. B. Styshnev, ?Root extraction in braid groups,? Izv. Akad. Nauk SSSR, Ser. Mat.,42, No. 5, 1120-1131 (1978). · Zbl 0402.20029
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


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# zbMATH — the first resource for mathematics

Existence of infinitely many homoclinic orbits in Hamiltonian systems. (English) Zbl 0725.58017
We consider a Hamiltonian system in $${\mathbb{R}}^{2N}$$, $$z'=J\nabla_ zH(t,z)$$, H being 1-periodic in time, and 0 being a hyperbolic rest point. Under global assumptions on H, we prove that there are always infinitely many orbits homoclinic to 0, i.e. such that $$z(\pm \infty)=0$$. Those orbits are geometrically distinct, in the following sense: $(x,y\text{ are geometrically distinct}),\quad \Leftrightarrow \quad (\forall n\in {\mathbb{Z}}:\;x(.)\neq y(.-n)).$ The approach we use here is variational, and no transversality hypothesis is needed.
Reviewer: E.Séré (Paris)

##### MSC:
 37J99 Dynamical aspects of finite-dimensional Hamiltonian and Lagrangian systems
Full Text:
##### References:
 [1] Ambrosetti, A., Rabinowitz, P.: Dual variational methods in critical point theory and applications. J. Funct. Anal.14, 349–381 (1973) · Zbl 0273.49063 · doi:10.1016/0022-1236(73)90051-7 [2] Clarke, F.: Periodic solutions of Hamiltonian inclusions. J. Differ. Equations40, 1–6 (1951) · Zbl 0461.34030 · doi:10.1016/0022-0396(81)90007-3 [3] Coti Zelati, V., Ekeland, I., Séré, E.: A variational approach to homoclinic orbits in Hamiltonian systems. Math. Ann.288, 133–160 (1990) · Zbl 0731.34050 · doi:10.1007/BF01444526 [4] Ekeland, I.: Convexity Methods in Hamiltonian’ Systems. Berlin Heidelberg New York: Springer 1989 · Zbl 0742.58047 [5] Hofer, H., Wysocki, K.: First order elliptic systems and the existence of homoclinic orbits in Hamiltonian systems. Preprint Rutgers University 1989 · Zbl 0702.34039 [6] Lions, P.L.: Solutions of Hartree-Fock equations for Coulomb systems. Preperint CEREMADE n. 8607. Paris 1988 · Zbl 0693.35047 [7] Lions, P.L.: The concentration-compactness principle in the calculus of variations. Rev. Mat. Iberoam.1, 145–201 (1985) · Zbl 0704.49005 [8] Moser, J.: Stable and random motions in Dynamical Systems. Princeton: Princeton University Press 1973 · Zbl 0271.70009 [9] Poincaré, H.: Les Méthodes Nouvelles de la Mécanique Céleste. Paris: Gauthier-Villars 1899 [10] Rabinowitz, P.H.: Periodic and heteroclinic orbits for a periodic Hamiltonian system. Preprint University of Wisconsin-Madison 1988 · Zbl 0701.58023
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


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# LS3 6.2 V8 Ready To Mount

### Help Support HomeBuiltAirplanes.com:

#### TXFlyGuy

##### Well-Known Member
Just got back from Titan. They are almost ready to hang the V8 on the fuselage.

#### dcstrng

##### Well-Known Member
Impressive -- the sound alone should keep you smiling for a month or so...

#### ekimneirbo

##### Banned
Someone did some nice welding there. Are those just for looks rather than the actual exhaust for the engine ? The LS3 is an awesome engine.........

#### TXFlyGuy

##### Well-Known Member
Someone did some nice welding there. Are those just for looks rather than the actual exhaust for the engine ? The LS3 is an awesome engine.........
These are functional stacks. The first and last pipe are non-functional, the middle 4 are live exhaust ports. The sound is awesome!

#### mcrae0104

##### Well-Known Member
HBA Supporter
Log Member
Just got back from Titan. They are almost ready to hang the V8 on the fuselage.
Tex, I genuinely mean no offense, but is this truly a "homebuilt" airplane? You seem to post a lot about the paint job you're going to buy, the engine someone else is hanging, and the avionics someone else installed. (Disclaimer: I have not built an aircraft, so I don't have a leg to stand on.)

#### TXFlyGuy

##### Well-Known Member
Tex, I genuinely mean no offense, but is this truly a "homebuilt" airplane? You seem to post a lot about the paint job you're going to buy, the engine someone else is hanging, and the avionics someone else installed. (Disclaimer: I have not built an aircraft, so I don't have a leg to stand on.)
We built the center section, fuel tanks, many items in the fuselage, etc. It will meet the 51% rule. The FAA has no problem with items like avionics being professionally installed. They do not care who paints the plane, or if it gets painted.

We are spending time at the factory getting "hands on" time as well. There are a number of items that are beyond the skill set of most of us...like building the control surfaces, spar, motor mounts, etc.

This is simply Titan's Builder Assist Program that we are taking advantage of!

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#### cheapracer

##### Well-Known Member
Log Member
Tex, I genuinely mean no offense, but is this truly a "homebuilt" airplane? You seem to post a lot about the paint job you're going to buy, the engine someone else is hanging, and the avionics someone else installed. (Disclaimer: I have not built an aircraft, so I don't have a leg to stand on.)
I have no issue at all with that type of build other than jealousy. Add that some people genuinely have no time or skillz, or both.

As long as the person doesn't stand on a plinth and put down others who are struggling to achieve their goals who's bank account obviously isn't quite as flush.

Great build Tex, looking forward to the result.

#### TXFlyGuy

##### Well-Known Member
I have no issue at all with that type of build other than jealousy. Add that some people genuinely have no time or skillz, or both.

As long as the person doesn't stand on a plinth and put down others who are struggling to achieve their goals who's bank account obviously isn't quite as flush.

Great build Tex, looking forward to the result.
There are many "hands on" photos on the Project: Mustang FB page. Much of this project was built in a local school that has an aviation program. My skills in the building area are limited. Thus, we have recruited others to help along the way. All the time keeping a careful eye (and hand) on the situation.

Those who can undertake a project like this solo have earned my respect and admiration. Much of this requires two sets (or more) of hands and eyes.

Bank account? This has cost 50% more than what we originally estimated. Much of that being the gear doors and avionics. But worth every penny invested!

#### cheapracer

##### Well-Known Member
Log Member
This has cost 50% more than what we originally estimated.!
As they say, I hope when I die my wife doesn't sell my toys for what I told her I paid for them! :gig:

#### gtae07

##### Well-Known Member
We built the center section, fuel tanks, many items in the fuselage, etc. It will meet the 51% rule. The FAA has no problem with items like avionics being professionally installed. They do not care who paints the plane, or if it gets painted.

We are spending time at the factory getting "hands on" time as well. There are a number of items that are beyond the skill set of most of us...like building the control surfaces, spar, motor mounts, etc.

This is simply Titan's Builder Assist Program that we are taking advantage of!
The 51% rule is extremely heavily tilted towards primary structure and firewall-forward items like baffling and control brackets. The entire electrical, avionics, and fuel systems together comprise six steps on the kit checklist. You get more credit for building your elevators than you do for almost all of your systems work.

This kind of rankles me a bit as an aircraft systems engineer, but I can see about three different sides to it all. For one, the overwhelming majority of early homebuilts, and still a fair number of current ones, are very simple on the systems side and have little in the way of dependency on those beyond making sure the engine keeps getting fuel. You don't really need electrical and avionics to build a viable homebuilt, so the FAA sort of considers them "optional".

Then again, on aircraft that do have involved wiring and systems work, that stuff takes up a significant portion of the build time. It seems a bit of a slight that someone who legitimately wires up a full-glass IFR homebuilt with electronic injection himself gets less credit for that than a guy who pop-rivets together elevators from a pre-punched kit.

To the third point, though, it's comparatively rare for a homebuilt to crash due to a structural failure. Systems-related failures (including powerplant systems) comprise the majority of accidents with some kind of aircraft failure; the FAA may consider it a safety benefit to let the complex stuff be done by professional shops at minimal penalty to the builder (which is also why I suspect they've interpreted the rules more leniently than one might expect on structural items in kits and builder-assist programs).

#### Chris In Marshfield

##### Well-Known Member
HBA Supporter
Log Member
Direct drive or geared? Kinda tough to tell from the photo, but the image makes it look like there's an offset for a gearbox. Of course it's on an engine stand

Looks great!

#### TXFlyGuy

##### Well-Known Member
Direct drive or geared? Kinda tough to tell from the photo, but the image makes it look like there's an offset for a gearbox. Of course it's on an engine stand

Looks great!
The PSRU is 1.9 - 1, from Autoflight / Titan. Manufactured in New Zealand.

#### TXFlyGuy

##### Well-Known Member
This is what she will look like very soon...

#### BoeveP51

##### Well-Known Member
Really nice work, if I had a barrel of cash I would be up there with you. You don't want to know how long it has taken me to get where I am. It takes a lot of hands, effort and MONEY to get there. I only have two hands, taking me a bit longer... Will not get into the cash situation

#### TXFlyGuy

##### Well-Known Member
Really nice work, if I had a barrel of cash I would be up there with you. You don't want to know how long it has taken me to get where I am. It takes a lot of hands, effort and MONEY to get there. I only have two hands, taking me a bit longer... Will not get into the cash situation
We have spread the cost out over 3.5 years. That includes the time that Titan was building our kit at the factory prior to us taking delivery. While it is expensive, you can make it less painful.

Yes, it requires many hands. Those that think they can build this "solo" need to think again.

My most valuable asset in this whole affair has been my wife. Why? Because she has never complained about the \$ I have spent on this project. It is a big help when your spouse is on board...figuratively, and literally.

#### TXFlyGuy

##### Well-Known Member
The engine is mounted. The fuselage is now on the wing, with the hydraulics functioning. The landing gear and inner gear doors are fully operational. The photo is a month old, so many changes since then!

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Latest update:

#### TXFlyGuy

##### Well-Known Member
Actual exhaust installed. It has been modified from the first version.

#### TXFlyGuy

##### Well-Known Member
Ceramic coating of the exhaust. Performance Coatings of Arlington, Texas did the work. And a very nice job.

#### Attachments

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## Verification

Verify, whether $$\mathbf x=(4{,}0,-3{,}0,2)^T+p(-2{,}3,2{,}1,0)^T+q(1{,}2,-1{,}0,1)^T$$ is a solution of the system

\begin{alignat*}{11} x_1 &-& x_2 &+& 2 x_3 &+& x_4 &+& 3 x_5 &=\ & 4 \\ 2 x_1 &-& 2 x_2 &+& 5 x_3 & & &+& 7 x_5 &=\ & 7 \\ x_1 &-& x_2 &+& x_3 &+& 3 x_4 &+& x_5 &=\ & 3 \\ x_1 &-& x_2 & & &+& 5 x_4 &+& 3 x_5 &=\ & 9 \end{alignat*}

• #### Resolution

It suffices to parse the solution values into the original system – either any three independent points (the solution space is a plane), or the values including parameters:

$$(4-2p+q)-(3p+2q)+2(-3+2p-q)+(p)+3(2+q)=$$

$=(4-6+6)+p(-2-3+4+1)+q(1-2-2+3)=4$

$$2(4-2p+q)-2(3p+2q)+5(-3+2p-q)+0+7(2+q)=$$

$=(8-15+14)+p(-4-6+10)+q(2-4-5+7)=7$

$$(4-2p+q)-(3p+2q)+(-3+2p-q)+3(p)+1(2+q)=$$

$=(4-3+2)+p(-2-3+2+3)+q(1-2-1+1)=6+q$

The last equation does not hold for $$q \ne 0$$.

The test showed that $$\mathbf x$$ is not a solution of the system.


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# pre-cal

Find the roots of the function f(x)= x^2+2x+2

Determine f(x)
a) (x+1-i)(x+1+i)
b)(x+1-i sqrt of 2)(x+1+i sqrt of 2)
c)x-1+i)(x-1-i)
d)x-1+i sqrt of 2) (x-1-i sqrt of 2)

For all of these, use the rule
[f(x) + ai]*[(f(x) - ai} = [f(x)]^2 + a^2
where a is a real constant.
For example, the answer to (a) is
(x+1)^2 + 1 = x^2 + 2x + 2

1. 👍
2. 👎
3. 👁

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text:
Laplacians on graphs vs. Laplacians on Riemannian manifolds: $\lambda_2$?

A graph $G$ is connected if and only if the second-largest eigenvalue $\lambda_2$ of the Laplacian of $G$ is greater than zero. (See, e.g., the Wikipedia article on algebraic connectivity.)

Is there an analogous statement for the eigenvalue $\lambda_2(M)$ of the Laplacian operator $\Delta$ for an $n$-dimensional connected, closed Riemannian manifold $M$?

($\Delta(f) = \nabla^2(f) = −\mathrm{div}(\mathrm{grad}(f))$.)

I am trying to understand the relationship between Laplacians on graphs and Laplacians on Riemannian manifolds. Pointers to help elucidate the connection would be greatly appreciated!

Addendum. See Richard Montgomery's interesting new comment on the Laplacian on the integer lattice.

-
I don't understand why you require that $M$ is connected. Isn't it true in both cases that the space of harmonic functions is precisely the space of locally constant functions, so its dimension is the number of connected components? –  Qiaochu Yuan Jan 22 '12 at 2:15
If the manifold has finitely many connected components, then L^2(M) decomposes as direct sum over the L^2 spaces of the separate connected components. Every connected component contribute the constant space $$\mathbb{C}\cdot 1$$ to be an eigenvector of eigenvalue $0$, hence the multiplicity of the the eigenvalue $0$ equals to the number of connected components. The connection between graphs and manifolds is more explicit by approximating the manifold with its 1 skeleton structure see in the book by Lubotzky - ma.huji.ac.il/~alexlub/BOOKS/On%20property/On%20property.pdf , p.34 –  Asaf Jan 22 '12 at 7:45
This is a general comment about the relation between the laplacian graph and the one on Riemannian manifold. Suppose $M$ compact and let $T$ be a triangulation of $M$. Given $n\in\mathbb N$ denote by $T_n$ the triangulation of $M$ obtained by $T$ doing $n$-times a barycentric subdivision of $T$. Denote by $\Delta_n$ the discrete Laplacian of the natural connected graph constructed on the 1-skeleton of $T_n$. It is possible that $\Delta_n$ converges in some suitable sense to $\Delta$ and this should give some transference principle. –  Valerio Capraro Jan 22 '12 at 13:10
This probably won't answer your question directly, but I got a lot of useful references in the responses to this question: mathoverflow.net/questions/66892/… –  Paul Siegel Jan 22 '12 at 23:42
Thanks Asaf, Valerio, and Paul--All very useful comments! –  Joseph O'Rourke Jan 23 '12 at 0:25

In my opinion, Colin de Verdiere's book "Spectre de Graphes" is the best place to start investigating the connection between the discrete Laplacian and the manifold Laplacian.

Recently, investigations in computer science (machine learning) lead to considerable progress.

Pick a "cloud" of points in a Riemann manifold. Consider the complete graph with vertices on these points. Next add weights to the edges that correlate with the geodesic distance between the corresponding points on the manifold. Then form a certain weighted Laplacian associated to this weighted graph. This operator converges with probability one to the the manifold Laplacian as the cloud is bigger and bigger and is chosen randomly with respect to the metric volume measure on the manifold.

I know I skipped many details, but you can get precise statements in this nice paper by M. Belkin and P. Niyogi:

Towards a Theoretical Foundation for Laplacian-Based Manifold Methods

M. Belkin's webpage has additional info.

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@Liviu: Very nice re cloud converging to manifold Laplacian! Thanks! –  Joseph O'Rourke Jan 22 '12 at 13:24
Belkin and Niyogi have a paper called "Convergence of Laplacian Eigenmaps" where they show that the spectrum of the above weighetd discrete Laplacian converges to the spectrum of the manifold Laplacian. –  Liviu Nicolaescu Jan 22 '12 at 13:30

Take a finite cover $\cal{U}$ of $M$ by open subsets, and view the nerve $N({\cal U})$ of this cover as a finite simplicial complex, carrying a combinatorial Laplace operator in each degree. By choosing $\cal{U}$ carefully, with a small enough mesh, in all degree $p$ you may approximate the $k$ first eigenvalues of the Laplace-Beltrami operator on $p$-forms of $M$, by the corresponding eigenvalues of the combinatorial Laplace operator in degree $p$ on $N({\cal U})$. Here $k=s_p({\cal U})-b_p(M)$, where $s_p({\cal U})$ is the number of $p$-simplices. See Theorem 3.1 in a paper by T. Mantuano http://arxiv.org/pdf/math/0609599.pdf Reference: Discretization of Riemannian manifolds applied to the Hodge Laplacian, Tatiana Mantuano, Source American Journal of Mathematics, Volume 130, Number 6, December 2008, pp. 1477-1508

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@Alain: Your reference is precisely on point. That paper's introduction gives a useful summary of discretizing a manifold to study the spectrum of its Laplacian. Thanks! –  Joseph O'Rourke Jan 23 '12 at 0:24

Hi,

1. Note that in either case (graph or manifold) several Laplace Operator discrtizations exists (with different properties). Some of these do not converge to the true (continuous) operator when mesh size is reduced. Higher order FEM approaches can yield fast convergence. Most operators are linear approximations.

2. On the manifold discretization (usually triangles for surfaces), you have geometric information, e.g. the angles between edges, which does not necessarily exist in the graph case.

3. For any operator on a manifold (mesh) that discretizes the contniuous Laplacian this holds: the eigenvalues are a diverging sequence of real positive numbers (including zero, and of course the number of eigenvalues is limited by the discretization). The first eigenvalue is zero if the manifold is closed, or if the Neumann boundary condition is applied at the boundary. It is larger zero for Dirichlet boundary condition. Iff you have n eigenvalues that are zero, you have n connected components.

There may be easier ways of testing connectedness (e.g. using the mesh representation, Euler characteristic or flooding algorithms).

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