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Thread: The Matrix Lie Groups

  1. #1 The Matrix Lie Groups 
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    The Lie groups describe continuous symmetries.

    First notice well the word "describe", and now let's have a simple example.

    It is intuitively obvious that the 2-sphere is symmetrical under any and all rotations. What is slightly less obvious that the mapping that takes a point to the point on the 2-sphere is continuous. For now let's assume this is the case, and ask if there exists a Lie group that describes this symmetry.

    Simple answer - yes! But we have some way to go before this will become clear. But note this well: if we think of our 2-sphere as being some kind of "physical object" (whatever that means), then this a special and not very interesting case. Mathematicians and more especially physicists are interested in the symmetries of equations and theories, and I intended to give examples in due course.

    But for now we have some work to do. Suppose first that be vector spaces, and define the set of all linear transformations by . Then the set of all transformations becomes , and no serious ambiguity results if I write this as .

    Now suppose I insist that every element (a linear transformation, recall) in has an inverse. By this stipulation the identity transformation comes free, and we recover the structure of a GROUP.

    To celebrate, let's write this as and call it the General Linear Group. Notice that thus far I have not specified the dimension of nor the field over which is defined. We'll do that in a mo.....

    But first I need to show that this is a MATRIX group. I confess I had to think long and hard how best to explain this, so if what follows doesn't quite float your goat, I apologize.

    Well the easy way out is to wave my hands and say it is an elementary fact of operator theory (oops - I use the terms operator and transformation interchangeably -I am allowed!) that any linear operator can be written as a matrix. But that is a cop-out.

    So first some boring stuff. Almost any vector can be written - expanded - as where the are scalars chosen from whatever field we are working over, and the set are called basis vectors. I say "almost" because, in order to be an element in the set we require that this sum has trivial multiplicative coefficients - that is one or zero according to whether or not. One calls this "linear independence of the basis vectors"

    First let us assume that our vector space is defined over , the real numbers. Second let us assume this vector space has dimension

    So suppose the linear transformation and further suppose we allow this to act on basis vectors (as we must) .

    Notice first the trivial fact that for no very good reason that conventionally one does not parenthesize the argument of an operator. More importantly notice that if are basis vectors then the cannot possibly be.

    So we have that, for the non-basis vectors there must be non-trivial real coefficients on the basis set i.e in all so the following applies:

    If I say that the index denotes the the vector that I wish to expand, and the index denotes the element in the basis upon which it is being expanded, I am free to write

    and say that the superscript index as it roams over references the columns in some real-valued matrix, and the subscript index likewise references the rows in the same matrix.

    That is



    You think that is all? Dream on.......


    Last edited by Guitarist; October 28th, 2012 at 05:46 PM.
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  3. #2  
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    Well, you guys are no fun!

    Let me just point out that any time a physicist (or would-be physicist) mentions anything to do with post-Newtonian physics, say the Special or General Theories of Relativity, or the so-called Standard Model, they are implicitly invoking continuous symmetries.

    Hell, this is even true of para-Newtonian physics - like the Maxwell equations - or even pre-Newtonian physics like those of Galilei

    There are probably other examples, but, thanks to you guys, I cannot be bothered to give it any thought.......


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  4. #3  
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    Im really interested in this but I don't really know how to comment since I pretty much followed the entire first post.
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  5. #4  
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    Quote Originally Posted by Guitarist View Post
    Well, you guys are no fun!

    Let me just point out that any time a physicist (or would-be physicist) mentions anything to do with post-Newtonian physics, say the Special or General Theories of Relativity, or the so-called Standard Model, they are implicitly invoking continuous symmetries.

    Hell, this is even true of para-Newtonian physics - like the Maxwell equations - or even pre-Newtonian physics like those of Galilei

    There are probably other examples, but, thanks to you guys, I cannot be bothered to give it any thought.......
    Wow, easy Guitarist !
    I am definitely interested in what you have posted, just haven't had the time yet to sit down and think about it in detail, because I am flat out with work and family commitments. This is not easy stuff for a non-mathematician to get his head around, but it's on my "to-understand" list for the weekend
    I'll post my questions then.
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  6. #5  
    Moderator Moderator Markus Hanke's Avatar
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    Ok, having read this carefully I think I can follow you thus far. Basically it comes down to the fact that a linear operator "acting" on a vector can be expressed as a matrix - this is rather intuitive, since the multiplication of a matrix with a vector ( of same dimensionality ) yields another vector. If GL(V) is the group of all linear transformations V -> V, then its elements can thus be represented by matrices. It makes sense now
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  7. #6  
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    Right, so we all agree that a linear transformation can be represented as a matrix, and so we can use the terms "matrix", "linear transformation" and "linear operator" interchangeably, which I will certainly do from mow on (depending on my mood!).

    So, for the GROUP of all invertible linear transformations we write where is the dimension of our vector space and accept this refers to the group of all invertible real matrices.

    Note that the condition for a matrix to be invertible is that it has a non-zero determinant - one calls this a non-singular matrix. So the set is the group of all non-singular real matrices.

    I now need to convince you all this group is in fact a MANIFOLD.

    To be blunt first, the unhelpful claim that the group of all non-singular real matrices is a vector subspace of the vector space of ALL real matrices, and therefore inherits its topology and is therefore a manifold. Clear? No of course not! What follows is not a proof, think of it more as a "persuasion".....

    First I argue that every finite-dimensional vector space is a manifold. Consider for any non-zero . This is trivially a manifold - it "looks locally like itself". As a manifold we my assign to each and any point a vector space called the "tangent space" . But is "flat" in the sense of Euclid,, so is in fact ALL of , so is both a manifold and a vector space.

    Now there is an easy theorem from elementary linear algebra that proves a natural isomorphism between ANY -dimensional real vector space and , so every finite-dimensional vector space is a manifold. With me so far?

    Good. So consider the set of ALL linear transformations where I do not insist these are different. I called this set as . Now it is an early exercise in any linear algebra course to prove this is a vector space i.e. the arithmetic addition of operators and the scalar multiplication of operators is closed in . Let us assume we have proved this, thus, from the above this is a manifold.

    Oh well - a caution. The arithmetic additive inverse of the matrix whose entries are is of course the matrix whose entries are and whose entry-wise sum is the matrix whose entries are , the zero matrix - the killer transformation if you like. This is not, repeat NOT what is ordinarily meant by the matrix inverse, which is taken as the multiplicative inverse.

    So the set of non-singular real matrices is a subset of ALL real matrices - a vector space therefore a manifold - and, is therefore a vector subspace and therefore a manifold.

    So is a group that is also a manifold. It's a Lie Group.
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    Guitarist, I read most of your stuff with great interest. I am a layman and have nothing to contribute, but I am learning all of the time. Thank you.
    .
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  9. #8  
    Moderator Moderator Markus Hanke's Avatar
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    Does this extrapolate to vector spaces with infinitely many dimensions, e.g. Hilbert spaces ?

    Does that mean that every group which is also a manifold is a Lie Group ?
    Last edited by Guitarist; November 6th, 2012 at 05:44 PM.
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  10. #9  
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    Quote Originally Posted by Markus Hanke View Post
    Does this extrapolate to vector spaces with infinitely many dimensions,
    No
    e.g. Hilbert spaces ?
    Are you under the impression that every Hilbert space is infinite-dimensional? This is not so

    Does that mean that every group which is also a manifold is a Lie Group ?
    Yes - it is a definition
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  11. #10  
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    Quote Originally Posted by Guitarist View Post
    Are you under the impression that every Hilbert space is infinite-dimensional?
    Not at all. It was just an example that came to mind.
    Thanks for the answers !
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  12. #11  
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    Quote Originally Posted by Markus Hanke View Post
    Does that mean that every group which is also a manifold is a Lie Group ?
    Not exactly, there are a couple of compatibility issues that need to be sorted first. For starters you need a differentiable manifold and the group operations have to be compatible with the differentiable structure on the manifold.
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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