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How does Lie group help to solve ode's?

by ayan849
Tags: differential eqns, lie group
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ayan849
#1
Apr19-12, 01:03 PM
P: 22
Being not an expert, my question might sound naive to students of mahematics. My question is how on earth a Lie group helps to solve an ode. Can anyone explain me in simple terms?
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Alesak
#2
Apr25-12, 02:43 AM
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Suposedly if the ODE is somehow symmetric, the solutions are limited to representations of corresponding Lie group. Iīve seen this in Hallīs book on Lie groups, section "Why study representations", but I donīt understand it either.

Also I find it wierd noone here understands it. (or why else there is no response?)
ayan849
#3
Apr26-12, 07:14 PM
P: 22
Many thanks Alesak for your reply and the cited reference. It would be nice if someone could post a geometrical answer...

Stephen Tashi
#4
May2-12, 06:51 PM
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P: 3,287
How does Lie group help to solve ode's?

Quote Quote by ayan849 View Post
My question is how on earth a Lie group helps to solve an ode. Can anyone explain me in simple terms?

I've often wondered about it myself. I even have a modern book on the subject: "Solution of Ordinary Differential Equations by Continuous Groups", by Geore Emanuel, Champion and Hall/CRC. I've glanced at it from time to time, but never studied it throroughly.

I'd be glad to discuss it here and that might motivate me to study it more. I think you need at least a facility with the calculus of several variables in order to understand the material. After introducing continouous gropus, Emanuel discusses "The Method Characteristics", so understanding the geometric interpretation of that technique is probably necessary.

How big a non-expert are you? How serious are you about findng an answer?
lavinia
#5
May2-12, 08:11 PM
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Quote Quote by Stephen Tashi View Post

I'd be glad to discuss it here and that might motivate me to study it more. I think you need at least a facility with the calculus of several variables in order to understand the material. After introducing continouous gropus, Emanuel discusses "The Method Characteristics", so understanding the geometric interpretation of that technique is probably necessary.

How big a non-expert are you? How serious are you about findng an answer?
I would love to have a discussion of this topic - which I know zero about.
Stephen Tashi
#6
May2-12, 09:32 PM
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Do you already know something about Lie goups?

And I should have written "The Method Of Characteristics".
lavinia
#7
May3-12, 06:21 AM
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Quote Quote by Stephen Tashi View Post
Do you already know something about Lie goups?

And I should have written "The Method Of Characteristics".
I have read most Chevalley's book.
Stephen Tashi
#8
May4-12, 12:29 AM
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Quote Quote by lavinia View Post
I have read most Chevalley's book.
Then you know more than I do about Lie groups.

How shall we go about this? It might be boring to go through Emanuel's book in order. I'll jump to a crucial point and we can go back and pick up the background. Perhaps you can explain what a "once extended group" is.

Summarizing the first section of Chaper 5 with some excerpts:

The general form for a first-order ODE is:
[itex] f(x,y,y') = 0 \ [/itex] (5.1)
(I don't know how to make LaTex leave more spaces between the equations and the numbers, which follow them.)

"We assume this ODE is invariant under a [one parameter] group whose symbol is [itex] U f [/itex].
That sounds like old-fashioned jargon. What would the "symbol" of the group be? Is it the so-called "infinitesimal transformation"? I'll explain what he says about it below.



In a previous chapter he discusses "function invariance" under groups. Of "function invariance", he begins by saying:

We start with a function [itex] f(x,y) [/itex]. The condition under which it is invariant with respect to the the group is determined next. By this we mean that

[itex] f_1 = f(x_1, y_1) = f(x,y) [/itex] (4.1)
I have to go back another chapter to make sense of that notation. In the previous chapters he has defined the group element corresponding to the value of the parameter [itex] \alpha [/itex] as the transformation that sends [itex] (x,y) [/itex] to [itex] (x_1,y_1) [/itex] by the formulae:

[itex] x_1 = \phi(x,y,\alpha) [/itex] (2.3)
[itex] y_1 = \psi(x,y,\alpha) [/itex]

Where [itex] \phi [/itex] and [itex]\psi [/itex] are known functions.

In terms of those functions, he defines:

[itex] \zeta = \frac{\partial \phi}{\partial \alpha}|_0 [/itex]
[itex] \eta = \frac{\partial \psi}{\partial \alpha}|_0 [/itex]

The differential operator [itex] U [/itex] is defined by
[itex] U f = \zeta f_x + \eta f_y [/itex], which is (I think)
[itex] U f = \zeta \frac{\partial f}{\partial x} + \eta \frac{\partial f}{\partial y} [/itex].

Returning to his treatment of function invariance, he says

[tex] f(x_1,y_1) = f(x,y) + \alpha U f + \frac{\alpha^2}{2} U^2 f + .... [/tex]
we observe that a necessary a sufficient condition for [itex] f [/itex] to be an invariant function of the group, is for

[itex] Uf = 0 [/itex] (4.2)
I'll end this post with that note. To formulate the definition of invariance for [itex] f(x,y,y') [/itex] he has to introduce the notion of the "once extended" group.

Anyone have any geometric insight about the above concepts?
Sina
#9
May4-12, 09:09 AM
P: 120
One idea is that if the ODE admits enough 1-parameter group of symmetries (that are transformations that take solutions to solutions, where solutions are the curves in the phase space) which are Lie groups, one by one you find change of coordinates that takes each of these transformations to their normal form (i.e they become translations along a coordinate [call this coordinate s], if you are in R^n with sufficient smoothness conditions such transformations always exist where basically 1-parameter curves become one of your coordinates) and this change of coordinates reduces the order of your equation by 1. Ofcourse the order in which you choose these symmetries is an important detail. Moreover for an nth order ODE you have to have atleast n symmetries. To avoid locality issues I assume this was done on R^n. I don't how this is generalized to ODE on manifolds. Geometrically this means that solutions in these new coordinates do not essentially depend on the redundant coordinate s but only on its derivatives. So in the new form your ODE lacks the variable s but depends on its derivatives, reducing the equation by one order.

An accesible introduction (to a physics student) is Stephani ODE and Symmetries. However there are an immense amount of geometrical ideas hidden in that book (which are not explained for simplicity and mostly everything is introduced as a metodology) so it is very instructive to read or return back to that book after you had a proper course on differential geometry covering Lie groups and algebras.
Sina
#10
May4-12, 09:34 AM
P: 120
So really what you care about is the Lie group of transformations on the phase space which take solution curves to solution curves. Intiutively, existance of a symmetry may mean redundancy which can be used to lower the degree of the equation.
Stephen Tashi
#11
May4-12, 11:46 AM
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Quote Quote by Sina View Post
So really what you care about is the Lie group of transformations on the phase space which take solution curves to solution curves.
It will be interesting to square this with Emmauel's statements. The invariance I gave in my post is what he calls "function invariance". He defines another type of invariance (which he says is less important) called "curve invariance".

Another type of invariance is now discusses. Our goal will be to find the conditions wherein one curve of a one-parameter family of curves transforms into itself or into another curve of the family.

Let
[itex] f = f(x,y) = c [/itex] (4.8)
be a one-parameter family of curves. We select one of these curves, which is written as
[itex] f_1 = f(x_1,y_1) = c. [/itex]
Under the Transformation (2.3) this becomes
[itex] f_1 = f(\phi(x,y,\alpha),\psi(x,y,\alpha)) = \omega(x,y,\alpha)[/itex] (4.9)
or
[itex] \omega(x,y,\alpha) = c_1 [/itex].
By letting [itex] \alpha [/itex] vary with [itex] c_1 [/itex] fixed this relation becomes the other members of the family of curves.
Alesak
#12
May5-12, 04:54 AM
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P: 134
Hi guys,

I found this awesome book: http://www.physics.drexel.edu/~bob/LieGroups.html.

Itīs spiritual succesor of his earlier book (Lie Groups, Lie Algebras, and Some of Their Applications) and it seems to have the best motivation Iīve seen in a book about lie groups. Definitely worth checking out.
Sina
#13
May5-12, 07:08 AM
P: 120
A word of caution, Gilmore is not for the mathematically twitchy :p by meaning which you should be able to withstand some amount of informal arguements or should enjoy trying to make them rigorous. The contents of the book looks like a miracle which should also tell you that not everything is trated in as much detail as Knapp's Lie groups.

It is also a nice reference book for lie groups and algebras but many properties are given without proof.
Alesak
#14
May6-12, 04:45 AM
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P: 134
Iīm trying to go through Knapp for a school course, and I think he has the highest "proof/other text" ratio Iīve seen in a long time. The proofs seem form a dense set in his book even. It is really helpful to read about lie groups from other perspectives as well.

One good companion seems Quantum Mechanics. Symmetries, too bad I donīt know any quantum mechanics!
Sina
#15
May6-12, 07:24 AM
P: 120
yea I meant gilmore's book when I said many properties are given without proof
Stephen Tashi
#16
May6-12, 02:05 PM
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P: 3,287
My own interest in this is simply to obtain a geometric intuition for what's going on. I think that is in line with ayan849's original post.

Let's start with a simple question. What is the geometric interpretation of the operator U?.

As I recall, the convention in Lie groups is that when the parameter(s) are zero then the particular transformation they define is the identity. So, for a 1 parameter group, is U f(x,y) something like a directional derivative of f(x,y) taken along the direction that the group is moving the point (x,y)?

I say "something like" since there have been other threads about whether a "direction derivative" must be taken with respect to a direction defined by a unit vector. It was mentioned that Apostol, in a book, defines a derivative of a function "with respect to a vector" without requiring that the vector be a unit vector. However, he does not call this a "directional derivative". Is it correct to say that U is a "derivative with respect to a vector" that may not be a unit vector?
ayan849
#17
May17-12, 01:22 PM
P: 22
Can anybody care to give a geometrical interpretation? I can't understand a bit of what is going on here :(
Stephen Tashi
#18
May17-12, 11:30 PM
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P: 3,287
While we wait to see if anybody cares, could you explain whether your mathematical background includes directional derivatives and differential operators?

Currently, I can't answer the question ("In geometric terms, how does Lie theory helps solve differential equations?") and I only get a hazy idea from the post that said that it amounts to changing coordinates so every transformation amounts to a translation. I'll flatter myself by thinking that I am capable of getting geometric picture if I wade through some of the books that I have. (I like the Gilmore text.) However, I probably don't have the self discipline to do this without some motivation ( being a very busy person - retired, you know. It takes up all your time.) We might figure this out if you want to have a long discussion about it, which will keep me interested.

If you were just asking a casual question and wanted a quick answer without getting into a lot of posts, I can't help you.


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