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# Lesson 2

Congruent Parts, Part 2

### Lesson Narrative

In this lesson, students continue to apply the concept of corresponding parts. In a previous lesson, students justified that two figures being congruent guarantees that all pairs of corresponding parts are congruent. In this lesson, students explore a different direction. If any pair of corresponding parts is not congruent, then the two figures cannot be congruent. Students attend to precision (MP6) as they determine whether congruence statements about figures are true in the warm-up, and look for corresponding parts in the activities. Students continue working toward establishing triangle congruence shortcuts as they examine a set of congruent parts, Side-Side-Angle, that are not sufficient to establish the triangles as congruent.

Technology isn't required for this lesson, but there are opportunities for students to choose to use appropriate technology to solve problems. We recommend making technology available.

### Learning Goals

Teacher Facing

• Justify whether or not figures are congruent by reasoning about rigid transformations (in writing).
• Practice generating and comprehending congruence statements (orally and in writing) that establish corresponding parts.

### Student Facing

• Let’s name figures in ways that help us see the corresponding parts.

### Student Facing

• I can identify corresponding parts from a congruence statement.
• I can use rigid transformations to explain why figures are congruent.
• I can write a congruence statement.

Building On

Building Towards

### Glossary Entries

• corresponding

For a rigid transformation that takes one figure onto another, a part of the first figure and its image in the second figure are called corresponding parts. We also talk about corresponding parts when we are trying to prove two figures are congruent and set up a correspondence between the parts to see if the parts are congruent.

In the figure, segment $$AB$$ corresponds to segment $$DE$$, and angle $$BCA$$ corresponds to angle $$EFD$$.

### Print Formatted Materials

For access, consult one of our IM Certified Partners.


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Blog

# Linear recursions and geometric series

As I mentioned in a previous post, one way to look at a linear recursion is as a matrix. When you do this, it's natural to think about the eigenvalues and eigenvectors of that matrix, and the eigenvectors turn out to be the initial conditions for the recursion that get it to greate a geometric series. We can use this fact to create an nth order recursion that creates n different geometric series of our choice, with the particular series being created determined by the initial conditions of the recursion. So if we pick any two numbers, it's possible to find a single second-order linear recursion that we can make create a geometric series with our chosen numbers as the ration of consecutive terms. Which ratio we get is determined by the initial conditions for the recursion.

Suppose that we write the recursion

xn = ak-1 xn-1 + … + a1 xnk+1 + a0 xnk

as

 xn–k+1 … … xn

=

 0 1 … 0 1 a0 a1 … ak-1

 xn–k … … xn-1

If λ is a solution to

xkak-1 x k-1 – … – a1 xa0 = 0

then

 1 λ … λk-1

is an eigenvector that corresponds to the eigenvalue λ, and this eigenvector is also the initial state that turns the recursion into a geometric series: if we have

x0 = 1, x1 = λ, x2 = λ2, …, xk-1 = λk-1

then we’ll always have xn = λn.

This means that for an nth order recursion, we can pick any n values λi (which are exactly these eigenvalues), and find initial conditions for the recursion such that xn = λin. The eigenvectors are the initial conditions that do this, and they get the recursion to create a geometric series with the ratio between successive terms being the eigenvalue which corresponds to the eigenvector. It's also easy to write down the recursion that does this for us. In particular, if the eigenvalues are the roots of

xkak-1 x k-1 – … – a1 xa0 = 0

then the recursion

xn = ak-1 xn-1 + … + a1 xnk+1 + a0 xnk

does this for us.

An example

Suppose that we want to find a second-order recursion that will create either the series

xn = (-2)n

or

xn = 3n

This corresponds to having

λ1 = -2

and

λ2 = 3

for the zeroes of

(x – λ1) (x – λ2) = (x + 2) (x – 3)

= x2 x – 6

so that the recursion

xn+2 = xn+1 + 6 xn

will do this for us.

The initial conditions

x0 = 1

and

x1 = -2

give

xn = (-2)n

and the initial conditions

x0 = 1

and

x1 = 3

give

xn = 3n

format: plain


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# Raiden Glossary¶

available balance
The available balance $$B_a$$ of a channel participant is $$B_a = B_n - B_l$$.
balance proof
Balance proof is any kind of message we use in order to cryptographically prove to our counterparty (or them to us) that their balance has changed and that we have received a transfer.
channel capacity
A channel’s capacity is the sum of the total deposits of both its participants. It is also the sum of the channel participants net balance.
counterparty
The counterparty of a channel is the other participant of the channel that is not ourselves.
DirectTransfer
A direct transfer is a non-locked transfer, which means a transfer that does not rely on a lock to complete. Once they are sent they should be considered as completed.
hashlock
A hashlock is the hashed secret that accompanies a locked message: sha3(secret)
initiator
In a mediated transfer the initiator of the transfer is the raiden node which starts the transfer
lock expiration
The lock expiration is the highest block_number until which the transfer can be settled.
locked balance
The locked balance $$B_l$$ of a channel participant is the sum of the locked amount for all pending transfers $$T_p$$. So $$B_l = \sum_{k=0}^{N-1} T_p$$ where $$N$$ is the number of pending transfers.
MediatedTransfer
A mediated transfer is a hashlocked transfer between an initiator and a target propagated through nodes in the network.
merkletree root
locksroot
The root of the merkle tree which holds the hashes of all the locks in the channel.
net balance
The net balance $$B_n$$ of a channel participant $$P$$ is his total deposit $$P_d$$ along with the amount of token he received $$P_r$$ minus the amount $$P_s$$ of token he has sent. So $$B_n = P_d + P_r - P_s$$
netting channel
The on-chain payment channel between two raiden nodes.
payee
In a channel relationship between two raiden nodes the payee is the participant who receives a transfer
payer
In a channel relationship between two raiden nodes the payer is the participant who sends a transfer.
preimage
secret
The preimage, what we call the secret in raiden, is 32 bytes of random cryptographically secure data whose keccak hash ends up being the hashlock.
RefundTransfer
A refund transfer is a special type of MediatedTransfer that is used when a node can no longer propagate a transfer and a routing backtrack needs to be done.
reveal timeout
The number of blocks in a channel allowed for learning about a secret being reveal through the blockchain and acting on it.
RevealSecret
The reveal secret message is sent to a node that is known to have an interest to learn the secret.
secret message
The secret message is a message containing the secret and used for synchronization between mediated transfer participants.
SecretRequest
The secret request message is sent by the target of a mediated transfer to its initiator in order to request the secret to unlock the transfer.
settlement timeout
settlement window
The number of blocks after the closing of a channel within which the counterparty is able to call updateTransfer and show the transfers they received.
target
In a mediated transfer the target is the raiden node for which the transfer sent by the initiator is intended
transferred amount
The transferred amount is the total amount of token one participant of a netting channel has sent to his counterparty.


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# Are the given sum of abolute value functions increasing?

We have the set $$\{0,\cdots,2^K-1\}$$ in binary representation, $$\{00,01,10,11\}$$ for $$K=2$$. Then each $$0$$ is replaced by $$b_k$$ and each $$1$$ is replaced by $$1-b_k$$, This gives the set $$B=\{b_1b_2,b_1(1-b_2),(1-b_1)b_2,(1-b_1)(1-b_2)\}$$ The same process is done in reverse order for the $$a_k$$s i.e., by replacing each $$0$$ by $$1-a_k$$ and each $$1$$ by $$a_k$$. The result is then $$A=\{(1-a_1)(1-a_2),(1-a_1)a_2,a_1(1-a_2),a_1a_2\}$$ Then, we find the absolute difference sum of these two sets: $$f(a_1,b_1,a_2,b_2)=|b_1b_2-(1-a_1)(1-a_2)|+|b_1(1-b_2)-(1-a_1)a_2|+|(1-b_1)b_2-a_1(1-a_2)|+|(1-b_1)(1-b_2)-a_1a_2|$$

Here, each $$(a_k,b_k)$$ satisfies $$1>a_K>\cdots>a_1>0$$ and $$0 and $$a_k+b_k<1$$.

Given any set of pairs $$(a_k,b_k)$$ for $$k\in\{1,\ldots,K\}$$, the function $$f$$ is non-increasing in every $$b_k$$.

Example: $$(a_1,b_1)=(0.1,0.2)$$ and $$(a_2,b_2)=(0.2,0.1)$$. Then if we take $$b_1$$ as variable we expect that $$f(0.1,b_1,0.2,0.1)$$ is decreasing for $$0. I verified that this is true in Mathematica. If we take $$b_2$$ as variable we expect that $$f(0.1,0.2,0.2,b_2)$$ is decreasing for $$0. This is also true according to Mathematica.

I guess this should be general for any example. I dont have any idea how to prove it, however.

First, let us replace $$a_i$$ with $$1-a_i$$ to simplify notation. Note that we now have $$b_i < a_i$$. We denote $$b_i^0 = b_i$$ and $$b_i^1 = 1-b_i$$. Let $$f(\vec{a}, \vec{b}) = \sum_{v\in\{0, 1\}^n}|\prod_{i=1}^nb_i^{v_i} - \prod_{i=1}^na_i^{v_i}|$$. We wish to show that $$\frac{\partial f}{\partial{b_i}} \leq 0$$ for $$i\in [n]$$.
By symmetry it suffices to show that $$\frac{\partial f}{\partial{b_1}} \leq 0$$. The trick is to group the terms of $$f$$ in pairs, with each pair consisting of the $$v$$'s with $$v_0$$ either $$0$$ or $$1$$, and the remaining $$v_i$$'s fixed. For $$w\in \{0, 1\}^{n-1}$$, let $$f_w = |b_1\prod_{i=2}^nb_i^{w_{i-1}} - a_1\prod_{i=2}^na_i^{w_{i-1}}| + |(1-b_1)\prod_{i=2}^nb_i^{w_{i-1}} - (1-a_1)\prod_{i=2}^na_i^{w_{i-1}}|$$. Since $$f = \sum_w f_w$$, it suffices to show that for each $$w$$, that $$\frac{\partial f_w}{\partial{b_1}} \leq 0$$.
Fix $$w$$ and note that $$f_w$$ consists of two terms. For $$\frac{\partial f_w}{\partial{b_i}}$$ to be positive, we must have that the first term without the absolute value sign is positive and the second is negative. But suppose $$b_1\prod_{i=2}^nb_i^{w_{i-1}} > a_1\prod_{i=2}^na_i^{w_{i-1}}$$. Then since $$(1-b_1)/b_1 > (1-a_1)/a_1$$, we can multiply to get $$(1-b_1)\prod_{i=2}^nb_i^{w_{i-1}} > (1-a_1)\prod_{i=2}^na_i^{w_{i-1}}$$. Thus each $$\frac{\partial f_w}{\partial{b_i}} \leq 0$$ and we are done.
• So this is proof by contradiction right? I have some confusion with the construction of $w$. For $n=2$, we need to have $w=0$ and $w=1$ cases. Inside multiplications we have $w_1$, which should always be $1$? It is not very clear to me. Other point is the symmetry. I think we can just flip the things like instead of $b_1(1-b_2)$ we can say $(1-b_2)b_1$, do this for all other cases and the result that we find for $b_1$ is what we also need for $b_2$, since this is true for all $b_i$, then just a single $b_i$ is sufficient. Did I get you right? May 21 '21 at 23:30
• I am writing $w = (w_1, \ldots, w_{n-1})$. So for $n = 2$, $w_1 = 0$ and $w_1 = 1$ are the two cases. For $w_1 = 0$, we have $f_w = |b_1b_2 - a_1a_2| + |(1-b_1)b_2 - (1-a_1)a_2|$. For $w_1 = 1$, we have $f_w = |b_1(1-b_2) - a_1(1-a_2)| + |(1-b_1)(1-b_2) - (1-a_1)(1-a_2)|$. You can check that for each of these $f_w$'s the derivatives are at most $0$ using the argument I give. Regarding symmetry, the point is that we don't use anything special about $b_1$ in the above argument, so the same argument can be made for any $b_i$. May 22 '21 at 18:51
• Yes. Exactly what I said and thought. But this argument can be made by reordering the multiplicative elements of the function. The About $w$ I had understood just found not perfectly clear. Thank you for the explanation. May 23 '21 at 14:24


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# Custom POS tags with SpaCy for NER

Quite new to NLP and especially NER. I'm trying to train a NER model on a custom dataset. This is a dataset of houses for sale. As part of the entities I'm training the model to extract are reference numbers. These are of variable length (but usually between 4-9) and look like G55L7 or LPP01Z1-32.

How can I give these entities a new "POS tag", as from what I'm aware of, I can't find any in SpaCy's default list that would match these?

Ideally, I'd like to train this alongside a pre-existing NER model so that I can also extract ORGs which SpaCy already has support for.

For your first question, I would try to use Regex to identify the reference numbers since they seem unique comparing to normal words. I assume they are:

• All capital
• Start with one or more letters
• After follow numbers
• Any other pattern may follow (i.e. letters, numbers, hyphens)
• Do not include spaces

Could be something like this: \b[A-Z]+\d+([A-Z]|\d|-)*\b

I can imagine you could do it better...

Apart from the above answer, what you can do is create your own rule based entity matcher - https://spacy.io/usage/rule-based-matching

POS(part of speech) tagging won't help here.


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# Our world as a typical F-theory vacuum?

1. Nov 10, 2015

### mitchell porter

This paper presents an unusually concrete hypothesis about the string theory landscape - that it is numerically dominated by "flux vacua" arising from a single compactification manifold. And our world could be one of them.

http://arxiv.org/abs/1511.03209
The F-theory geometry with most flux vacua
Washington Taylor, Yi-Nan Wang
(Submitted on 10 Nov 2015)
Applying the Ashok-Denef-Douglas estimation method to elliptic Calabi-Yau fourfolds suggests that a single elliptic fourfold ${\cal M}_{\rm max}$ gives rise to ${\cal O} (10^{272,000})$ F-theory flux vacua, and that the sum total of the numbers of flux vacua from all other F-theory geometries is suppressed by a relative factor of ${\cal O} (10^{-3000})$. The fourfold ${\cal M}_{\rm max}$ arises from a generic elliptic fibration over a specific toric threefold base $B_{\rm max}$, and gives a geometrically non-Higgsable gauge group of $E_8^9 \times F_4^8 \times (G_2 \times SU(2))^{16}$, of which we expect some factors to be broken by G-flux to smaller groups. It is not possible to tune an $SU(5)$ GUT group on any further divisors in ${\cal M}_{\rm max}$, or even an $SU(2)$ or $SU(3)$, so the standard model gauge group appears to arise in this context only from a broken $E_8$ factor. The results of this paper can either be interpreted as providing a framework for predicting how the standard model arises most naturally in F-theory and the types of dark matter to be found in a typical F-theory compactification, or as a challenge to string theorists to explain why other choices of vacua are not exponentially unlikely compared to F-theory compactifications on ${\cal M}_{\rm max}$.

Last edited: Nov 10, 2015
2. Nov 11, 2015

### Staff: Mentor

I'm probably not the one whom you wanted to address since my knowledge here is basic at its best. Therefore my question: Did you post this as information for string theorists or as germ of a discussion? Do you want it to keep it under "unanswered"?
Am I right that I counted more than $10^{61}$ dimensions of this gauge group? That's a hell of a group.
If I'm allowed may I ask something general: Why are all gauge groups, except some small $SU(n)$ always exceptional Lie Groups? Is it because it has to be a Lie Group plus high coefficients, i.e. variability in the root system?

Last edited: Nov 11, 2015
3. Nov 20, 2015

### mitchell porter

It was meant for discussion...

Those Lie group factors in the "geometrically non-Higgsable gauge group" actually come from a geometric incarnation of the Dynkin diagrams. The "elliptic fourfold ${\cal M}_{\rm max}$" contains singularities which are what you would get if you started with a set of two-spheres touching each other with a topology encoded in a Dynkin diagram (point = sphere, edge = touching), and then shrank the sphere's volumes to zero. And I think the gauge field comes from D2-branes wrapping the two-spheres.

${\cal M}_{\rm max}$ is basically a Calabi-Yau space, such as you hear about in popular descriptions of string theory - a microscopic six-dimensional space which is to be understood as existing at each point in our macroscopic four-dimensional space-time - but filled with an "axio-dilaton field" that takes different values throughout its volume.

If I have understood correctly, the singularities I just mentioned pertain to the combination "metric + axio-dilaton". And at 33 separate locations in ${\cal M}_{\rm max}$, there is a point or a surface where this combination becomes singular in the way described, giving rise to a gauge field (or actually, an N=1 superfield) at that location.

All these gauge superfields lead separate lives - this is a type of braneworld model, where the 33 parallel "worlds" only interact gravitationally. So although that overall gauge group is big (not as big as you said - its dimension is more like $10^{3}$), in practice it's only the individual factors which are relevant, one for each braneworld.

If this does describe reality, we must be in one of the $E_8$ braneworlds.


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R functions to manage the Earth Engine Asset. The interface allows users to create and eliminate folders, move and copy assets, set and delete properties, handle access control lists, and manage and/or cancel tasks.

ee_manage_create(path_asset, asset_type = "Folder", quiet = FALSE)

ee_manage_delete(path_asset, quiet = FALSE, strict = TRUE)

ee_manage_assetlist(path_asset, quiet = FALSE, strict = TRUE)

ee_manage_quota(quiet = FALSE)

ee_manage_copy(path_asset, final_path, strict = TRUE, quiet = FALSE)

ee_manage_move(path_asset, final_path, strict = TRUE, quiet = FALSE)

ee_manage_delete_properties(path_asset, del_properties = "ALL", strict = TRUE)

ee_manage_asset_access(
path_asset,
owner = NULL,
editor = NULL,
viewer = NULL,
quiet = FALSE
)

ee_manage_asset_size(path_asset, quiet = FALSE)

## Arguments

path_asset

Character. Name of the EE asset (Table, Image, Folder or ImageCollection).

asset_type

Character. The asset type to create ('Folder' or 'ImageCollection').

quiet

Logical. Suppress info message.

strict

Character vector. If TRUE, the existence of the asset will be evaluated before performing the task.

final_path

Character. Output filename (e.g users/datacolecfbf/ic_moved)

List. Set of parameters to established as a property of an EE object. See details.

del_properties

Character. Names of properties to be deleted. See details.

owner

Character vector. Define owner user in the IAM Policy.

editor

Character vector. Define editor users in the IAM Policy.

viewer

Character vector. Define viewer users in the IAM Policy.

Logical. All users can see the asset element.

cache

Logical. If TRUE, the task report will be saved in the /temp directory and used when the function.

## Details

If the argument del_properties is 'ALL', ee_manage_delete_properties will delete all the properties.

## Author

Samapriya Roy, adapted to R and improved by csaybar.

## Examples

if (FALSE) {
library(rgee)

ee_Initialize()
ee_user_info()

# Change datacolecfbf by your EE user to be able to reproduce
user <- ee_get_assethome()
# 1. Create a folder or Image Collection
# Change path asset according to your specific user

# 1. List all the elements inside a folder or a ImageCollection

# 2. Create a Folder or a ImageCollection
ee_manage_create(
asset_type = "Folder"
)

ee_manage_create(
asset_type = "ImageCollection"
)

# 3. Shows Earth Engine quota
ee_manage_quota()

# 4. Move an EE object to another folder
ee_manage_move(
)

# 5. Set properties to an EE object.
ee_manage_set_properties(
add_properties = list(message = "hello-world", language = "R")
)

test_ic <- ee$ImageCollection(ic_id) test_ic$getInfo()

# 6. Delete properties
ee_manage_delete_properties(
del_properties = c("message", "language")
)
test_ic\$getInfo()

# 7. Create a report based on all the tasks
# that are running or have already been completed.


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# Mathenatucak Induction Problems in discrete math

1. Mar 2, 2010

### GoGoDancer12

1. The problem statement, all variables and given/known data

Prove that 3 divides n3 + 2n whenever n is a positive integer.

2. Relevant equations

3. The attempt at a solution

Basis Step :
P(1) : [13 + 2(1) ] /3
[1+2] /3
[3]/3
1
Since 3/3 =1, P(1) is true

Inductive Step:
[ k3 + 2k ] /3 = [(k +1)3 + 2(k+1) ] /3

[k(k2 + 2k)] /3 = [(k)3 + (3k)2 + 5k +3] /3

1. The problem statement, all variables and given/known data

Prove that f1 2 + f2 2 +...+ fn 2 when n is a positive integer.

2. Relevant equations
The Fibonacci numbers f0, f1, f2..., are defined by the equations f0 = 0, f1 = 1 and fn = fn-1 + fn-2 for n = 2,3,4...

3. The attempt at a solution

Basic Step:
P(1) : f1 2 = f1 * f2

12 = (1)*(1)
1= 1
Since f1 2 = f1 * f2 , P(1) is true.

Inductive Step:
fk 2 = fk * fk+1

fk 2 + fk+1 2
= f1+fk+1 + (fk+1)

Last edited: Mar 2, 2010
2. Mar 2, 2010

### Staff: Mentor

For the first problem, I'm having some trouble understanding what you're doing.
You have established the base case, P(1). IOW, when n = 1, n^3 + 2n = 1 + 2 = 3, so the proposition is true for n = 1.

Assume that the proposition is true when n = k. IOW, assume that k^3 + 2k is divisible by 3. Another way to say this is that k^3 + 2k = 3m for some integer m.

Now, when n = k + 1, you want to show/prove that (k + 1)^3 + 2(k + 1) is divisible by 3. You do NOT want to show that k^3 + 2k = (k + 1)^3 + 2(k + 1), which is essentially what you wrote in your post.

To show that (k + 1)^3 + 2(k + 1) is divisible by 3, expand the two terms. You will need to use the assumption that k^3 + 2k = 3m.

3. Mar 2, 2010

### GoGoDancer12

I'm lost about showing that (k + 1)^3 + 2(k + 1) is divisible by 3. I expanded the right side of the equation [k^3 + 2k)] /3 = [(k)^3 + (3k)^2 + 5k +3] /3. So, do I replace on the left side [k(k^2 + 2k)] /3 with 3m??

Last edited: Mar 2, 2010
4. Mar 2, 2010

### Staff: Mentor

This equation is bogus - [k^3 + 2k)] /3 = [(k)^3 + (3k)^2 + 5k +3] /3 - get rid of it, and also get rid of those /3 things that you have.

Take another look at what I said at the end of my previous post.

5. Mar 3, 2010

### GoGoDancer12

Ok, what about the second problem??

6. Mar 4, 2010

### Staff: Mentor

What is the second problem?
There's nothing there to prove. You have to have a statement of some kind in order to prove it.

Proof by induction, $(n!)^{2} \le (2n)!$. Mar 1, 2017


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Developing intuition for Lie groups and Lie algebras

Background: Until now, I've been able to motivate everything I've learned in mathematics, and develop some solid insights for everything. But I learned some Lie theory this summer, and while I have a good grasp of the elementary aspects and strong intuition for some or even most of what I've learned, there are some "holes" in my understanding of Lie algebras.

To give you an idea of what I'm looking for, I'll list some examples of things in Lie theory I DO understand and am able to motivate:

• The notion of a Lie group itself -- the idea comes from wanting to generalise what we know about discrete groups to more complicated contexts where the "manifold" structure of the group allows us to do so. Examples: compactness generalises finiteness, one-parameter groups generalise cyclic groups, etc.
• The exponential map -- For one-parameter groups to generalise cyclic groups, we need a "generalisation" of the group power to allow "real-index powers". The general way to define a real power is through the exponential map. Well, this real power stuff isn't always defined as it turns out (you need the exponential map to be surjective), but our motivation does explain why it "makes sense" that the exponential map is surjective in the connected abelian case (because then, the Lie algebra is basically a co-ordinate system on the Lie group -- I'm aware exponential co-ordinates are defined in more generality, but it's certainly more well-behaved here).
• The Lie algebra, i.e. "why is the logarithm/parameter space the tangent space?" We'd like to generalise the notion of a generator to a Lie group -- consider e.g. the circle group on the complex plane. An element near the identity generates a cyclic group, and as the element goes nearer to the identity -- as it becomes an infinitesimal generator, the cyclic group it approaches the entire group. Well, an element close to the identity is of the form $$1+\varepsilon t X$$, and generates a group element as $$(1+\varepsilon tX)^{1/\varepsilon}=e^{tX}$$. This is also intuition for the compound-interest limit, and for Euler's identity.
• The Lie bracket is the second-derivative of the commutator curve $$\gamma(t)=e^{tX}e^{tY}e^{-tX}e^{-tY}$$. Well, it's also the derivative of $$\gamma(\sqrt{t})$$, which proves closure under the Lie bracket.
• The real justification for the Lie bracket, however, comes from the fundamental fact that $$\mathrm{ad}:\mathfrak{g}\to\mathrm{Der}(\mathfrak{g}):=X\to[X,\cdot]$$ is the differential of the adjoint map $$\mathrm{Ad}:G\to\mathrm{Aut}(G):=g\mapsto\lambda x, gxg^{-1}$$, which is a group homomorphism. In particular, the preservation of the Lie Bracket by the differential of a group homomorphism is precisely the Jacobi identity: $$\mathrm{ad}([x,y])=[\mathrm{ad}(x),\mathrm{ad}(y)]$$. The basic point is that we are trying to reduce Lie group problems to Lie algebra ones as much as possible, and conjugation is an important idea that we'd like to see the map induced by on the Lie algebra -- we are seeing the result of the obvious fact that $$T\mathrm{Aut}(G)\subseteq\mathrm{Der}(TG)$$ (and also $$T\mathrm{Aut}(M)=\mathrm{Der}(M)$$ -- the fact that the automorphisms of an object form a group is equivalent to the derivations on an object forming a Lie algebra). Some more examples of the "study the Lie algebra approach":
• The uniqueness of the determinant as a map from $$G\to \mathbb{R}-\{0\}$$.
• An ideal is a subalgebra "induced" on the Lie algebra by a normal subgroup of the Lie group. This immediately provides the interpretation as "kernels of Lie algebra homomorphisms" as well as the condition $$[\mathfrak{g},\mathfrak{i}]\subseteq\mathfrak{i}$$.
• The idea behind the manifold-structure of a Lie group is that the flows are produced by left-multiplication by group elements, so those must be homeomorphisms. This motivation can be confirmed through various topological consequences, e.g.
• A neighbourhood of the identity generates the connected component. The idea behind the proof is this: if an entire open neighbourhood of the identity is contained in the subgroup, it means you can "flow in any direction" from the subgroup -- but to bring these flows to an arbitrary point of the manifold, you need left-multiplication to be a homeomorphism.
• The identity component is a (normal) subgroup. Because left-multiplication and inversion are continuous, they cannot tear the connected component apart (generalised "intermediate value theorem"), so it is closed under multiplication.
• Compact Lie groups -- How can a Lie group possibly "close in on itself"? Surely we keep "extending" an open neighbourhood $$W$$ of the identity by observing that $$xW$$ must be in the subgroup? The idea is that these translations of $$W$$ form an open cover of the group, if it has a finite subcover, then it makes sense for the group to close in on itself. By playing around with different open neighbourhoods $$W$$ and taking some suitable unions, one can see that this is equivalent to the condition that every open cover has a finite subcover, i.e. the group is compact.
• Characterisation of Abelian Lie groups -- "Compact Connected Abelian Lie Group is a torus" is a generalisation of "finite Abelian group is a product of cyclic groups" -- the idea is that the exponential map "wraps" the Lie algebra around into the Lie group -- this just gives the quotient of the Lie algebra by the kernel of the exponential map, which is topologically $$\mathbb{R}^n/\mathbb{Z}^n$$. The characterisation of a connected Abelian Lie group as a cylinder $$\mathbb{R}^{n+k}/\mathbb{Z}^k$$ follows similarly.

With that said, here are some stuff I DON'T (completely) understand, and would like to have a similar level of understanding for:

• Why is the structure of a Lie group characterised by its second-order structure? I know that this follows from the BCH formula, the local diffeomorphism nature of the exponential map and the fact that an open neighbourhood of 1 generates the group, but I have no intuition at all why the BCH formula "should" be true.
• What's the deal with simply-connected groups? I can certainly see why the Lie algebra cannot detect disconnectedness in a group -- I had expected that it could not detect compactness either (whether the group closes in on itself eventually), so the statement of Lie's third theorem would be "every Lie algebra has a corresponding unique connected, compact Lie group". Instead, the statement is "every Lie algebra has a corresponding simply connected Lie group".
• Non-surjectivity of the exponential map even in the connected case -- I'm not asking for counter-examples, I'm asking "what exactly goes wrong in groups like $$SL_\mathbb{R}(2)$$?", perhaps a hint about "what does the image of the exponential map look like?" (as an analogy, I would explain smooth functions failing to be analytic as "they are flatter than every polynomial at 0, and can be constructed as $$1/f(1/x)$$ where $$f$$ is any function that grows faster than every polynomial")
• Surjectivity when every element is contained in a maximal torus -- I read this here as a generalisation of "the exponential map is surjective in the connected compact case". Even if the generalisation isn't true, is there an intuitive way to understand why compactness makes the problem in the previous point go away.
• Characterisation of non-Abelian Lie groups -- Tell me if my understanding of simple and semisimple Lie algebras makes sense -- we want to classify non-Abelian Lie groups as products like we do Abelian Lie groups, and the only way to do so is as "semidirect products of simple Lie groups and Abelian Lie groups". A reductive Lie group is basically when this semidirect product is a direct product, and a semi-simple one is a reductive Lie group where there are no Abelian groups in the product. Is this right?
• Various abstract algebraic things -- I have no idea how to interpret things like nilpotent and solvable Lie algebras, radicals and so on in the context of Lie theory.
• At first when I heard of the Killing form, I presumed it would be some "natural" way to define a dot product on the Lie algebra -- but I honestly don't see how it is natural. Is it the only dot product that is invariant under Lie algebra automorphisms?

I've thought very hard about the theory, but I just can't seem to figure out how to fill these "holes". Am I missing some important central insight into Lie theory that are crucial to some of these questions?

• +1 for articulating what you know and don't know so well. But the “question” is very broad. Can you focus on one thing at a time? – Matthew Leingang Sep 24 '19 at 17:32
• The compactness can be detected. See negative definite Killing form. And solvable and nilpotent is not only restricted to Lie algebras. Groups, rings, algebras etc. have it, too, in particular also Lie groups. The Killing form is not a dot product. I suppose you will "miss several important central insights" without having worked with these topics. – Dietrich Burde Sep 24 '19 at 18:05
• Lie theory is a gigantic topic, there is tons and tons to say about it, and so it's not exactly surprising that some bits of it are still confusing at any particular stage. Try to have some patience! Isn't it wonderful that there's still so much more to learn? – Qiaochu Yuan Sep 24 '19 at 20:01
• Although this question surely falls in the category of too broad, I feel that if separated into multiple questions, they might go unanswered. Yes, MSE is platform for all the range of mathematical questions, but, somehow, something beyond common curriculum often doesn't attract as much as attention as it deserves. I would say that sheer amount of details that went into this question overweights the broadness, because if the OP included all of this context in any single question, it would be overkill. In the same spirit, I would count partial answers as good... and beneficial to community. – Ennar Sep 24 '19 at 22:36
• I really do encourage you, though, to ask your main questions separately, and then ask your follow-up questions in comments there. Not only will you be more likely to get coherent, organized answers (in ways that comments, mine included, are usually not), but it will make the site more useful for other people asking the same questions later. (In fact, I'd guess that at least some of the questions you're asking about already have answers on the site.) – Travis Willse Sep 25 '19 at 6:10


text:
# Galilean Transformation Question

1. Feb 13, 2005

### misogynisticfeminist

What happens if I have 2 frames,S and S', with S as my rest frame and S' moving in the +ve x direction (towards the right). Why is it that the equation for the galilean transformation for the x-coordinate is $$x'=x-vt$$ instead of $$x'=x+vt$$ ???

2. Feb 13, 2005

### Galileo

Think of an object lying on the positive x-axis (it therefore also lies on the x'-axis). At t=0 the x and x' coordinate of the object are the same, but since the origin in S' moves towards the object (as seen from S. In S' the object moves towards the origin.) the x' coordinate decreases with time. So x'=x-vt is the right equation.

Drawing a picture also helps.

3. Feb 13, 2005

### Staff: Mentor

The primed coordinates are the coordinates measured by the "moving" frame. Take a simple case: Imagine an object moving along with the origin of S'. After a time t, the S frame says the object's coordinates are (x,t) = (vt,t). Of course, the S' frame gives the coordinates as (x', t) = (0, t). Using $x'=x+vt$ gives the wrong answer.

4. Feb 13, 2005

### misogynisticfeminist

hey thanks alot, I've gottit, sorry if my questions are a little too simple, I'm mostly self-taught.

5. Feb 13, 2005

### dextercioby

There's no problem with us:it's essential for you to have these problems clear,especially if you want to go into studying SR,where things are a bit more difficult...

Daniel.


text:
# Clarifying the [sem] tag for simultaneous equation models?

The tag is described as being for structural equation modeling, but a proportion (my hunch: 25%) of its use is for (econometric) simultaneous equation models.

Is it worth clarifying, or splitting the tag?

• On the basis that, sadly, nobody reads the documentation splitting it seems more likely to achieve anything. – mdewey Jun 13 '16 at 21:06
• Clarifying never hurts. You could also go through the threads & remove it from those not about SEM. (A few at a time, during slow periods, etc., as usual.) – gung - Reinstate Monica Jun 14 '16 at 2:34
• @Gung - I've been doing that when the questions appear for a while. I feel a little bad that I don't have a tag to replace it with. – Jeremy Miles Jun 14 '16 at 16:50
• I think it's fine to just delete an inappropriate tag & replace it with nothing. – gung - Reinstate Monica Jun 14 '16 at 17:22
• Although it may not be used this way at the moment, it's also worth noting that SEM is commonly used as an abbreviation for "standard error of the mean" (e.g. Wikipedia does so) which is another potential source of confusion for people adding a tag. – Silverfish Jun 21 '16 at 0:12

With that said, I suppose it couldn't hurt to clarify the sem tag description to point out that some people using this tag will only have the special case of simultaneous equation models in mind.


text:
# Distance to the nearest pub

Building a matrix of the distance to the nearest pub using FSA and google data.

The plot below shows the distance to the nearest pub, from a grid of spots in Cambridge. Starting in the top left of the green box, I ran a script that calls the Google Distance Matrix API to ask where the nearest pub is. It then moves a little to the right, and asks again. When it hits the far side of the box, it drops down and makes a new row of calls. Distances are coloured based on the time Google says it will take to walk from that spot to the pub. I originally made this plot last year using Google’s data on Cambridge pubs, but the data Google has is very inaccurate. The Food Standards Agency though has a daily updated XML file, which has every pub nicely geo-coded. This updated plot should be far more accurate.

## The old plot

Google only allows 2,500 calls to the distance API a day, and there are 1,600 points in my grid. This means it would take days to find out the walking distance to every pub. When I first made this plot I simply went with the closest pub as the crow flies. When updating, I decided to look at the walking distance to the two closest pubs. The plot below is using the method from my old post, where I assumed the closest in a straight line was also the closest when walking.

## Where I went wrong

While the difference between the two plots above are a little hard to make out, the plot below shows all the points in my grid where the shortest as the crow flies is not the shortest when walking. I made this post in one day, and only had access to three API keys (I maxed out my own, Tina’s and the college IP address for day). This meant I could only check the three closest pubs in a straight line. I’m fairly confident the 316 (20%) points I found where the closest in a straight line, and walking, is different, covers all of the cases. The only potential problem is that shortcuts through colleges and greens aren’t currently accounted for. The plot below shows where the first post I made of distance to the nearest pub was wrong.

## Pub dense areas

The following plot simply shows where the nearest pub is.

## How to

Nathan Yau wrote a brilliant book called Datapoints1, where he introduces this style of map, and it is from there that based the code to handle the geo-coding.

Below are links to some snippets I used to make this plot

• Pull data from the online XML file
• Feed data to the Google Distance Matrix API

1. Nathan Yau (2013) Data Points: Visualization That Means Something, ISBN: 978-1-118-46219-5 ↩

##### James Black
###### PhD (Cantab)

James Black. Kiwi | Epidemiologist | Data Scientist | Engineering enthusiast.


text:
# If $a$ can be written as sum of squares of three rational numbers, prove that $a^m$ can be too.

Let a be a rational number that can be written as sum of squares of three rational numbers. Prove that $a^m$ can also can be written as sum of squares of three rational numbers for any positive integer $m$.

I tried to solve using Mathematical Induction Let a be a rational number that can be written as sum of squares of three rational numbers i.e. $$a=\left( p_1 / q_1\right)^2+\left( p_2 / q_2\right)^2+ \left( p_3 / q_3\right)^2\\$$ We prove $a^m$ is rational for every positive integer$\\$ For $n=1$, it is true. $\\$ Let it be true for $n=k$ i.e. $$a^m=\left( r_1 / s_1\right)^2+\left( r_2 / s_2\right)^2+ \left( r_3 / s_3\right)^2\\$$ Consider $$a^{m+1}=a^m *a$$ $$=(\left( p_1 / q_1\right)^2+\left( p_2 / q_2\right)^2+ \left( p_3 / q_3\right)^2) (\left( r_1 / s_1\right)^2+\left( r_2 / s_2\right)^2+ \left( r_3 / s_3\right)^2)\\$$

I am not sure how to proceed next to show that $a^{m+1}$ can be written as sum of squares of three numbers. Thanks in advance for any kind help.

• Based on (math.stackexchange.com/questions/1200794/…), A rational number $p/q$ is sum of three rational squares if and only if $pq$ is not of the form $2^{2a}(8b−1)$. Thus, by hypothesis, your number $a=p/q$ is such that $pq$ is not of the form $2^{2a}(8b−1)$. Now, What can you say about the powers $(pq)^m$? – MathChat Jan 7 '17 at 17:35
• Recall that any integer $n$ can be factored as $n=2^r\cdot s$, where $r\geq 0$, and $s$ is odd. So things pretty much reduce to compare the behavior of $s$ and $s^m$ $\mod 8.$ – MathChat Jan 7 '17 at 17:55
• Thanks for your kind response, But unfortunately I am not getting the idea, I need more assistance – yasir Jan 7 '17 at 18:13
• If $a=p/q$ is sum of three rational squares, then $pq=2^rs$, where either $s\equiv 1$ or $\pm 3 \mod 8$, or $r$ is odd and $s\equiv -1 \mod 8$. In the former case, the powers $s^m$ will never be $\equiv -1 \mod 8$. In the latter case, $s^m \equiv -1 \mod 8$ only if $m$ is odd, and then $rm$ is odd. Thus in either case, $(pq)^m$ will not be of the form $2^ab$, where $a$ is even and $b\equiv -1 \mod 8$. Therefore $a^m$ is the sum of three rational squares. – MathChat Jan 7 '17 at 18:15
• @ yasir: Please, let me know at which part of my response you are getting stuck. – MathChat Jan 7 '17 at 18:21

Based on (When is a rational number a sum of three squares?), a rational number $p/q$ is sum of three rational squares if and only if $pq$ is not of the form $2^{2a}(8b-1)$.
Note that any integer $n$ can be factored as $n=2^r\cdot s$, where $r\geq 0$ and $s$ is odd. So the result above is telling you that if a number $p/q$ is NOT a sum of three rational squares, then $pq=2^r\cdot s$, where $r$ is even and $s\equiv -1 \mod 8$. Equivalently, if a number $a=p/q$ is a sum of three rational squares, then $pq=2^r\cdot s$, where either $r$ is odd or $s\equiv 1, \pm 3 \mod 8$.
So things pretty much reduce to compare the behavior of $s$ and $s^m$ $\mod 8$.
Suppose $a=p/q$ i sa sum of three rational squares. That is, $pq=2^rs$, where either $s\equiv 1$ or $\pm 3 \mod 8$, or $r$ is odd. In the former case, the powers $s^m$ will never be $\equiv -1 \mod 8$. In the latter case, $s^m \equiv -1 \mod 8$ only if $m$ is odd. But then $(pq)^m=2^{rm}\cdot s^m$, with $rm$ odd.
Therefore, in either case, $(pq)^m$ will not be of the form $2^ab$, where $a$ is even and $b\equiv -1 \mod 8$. Hence $a^m$ is the sum of three rational squares.
There is a simple proof requiring no infection. If $m=2k$, $a^m = (a^k)^2+0^2+0^2$; if $m=2k+1$, $a^m=(a^k)^2a=(a^kr_1)^2+(a^kr_2)^2+(a^kr_3)^2$.