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The Special Linear Group

  1. May 7, 2006 #1
    Hey, quick question... if I am considering the set of nxn matrices of determinant 1 (the special linear group), is it correct if I say that the set is a submanifold in the (n^2)-dimensional space of matrices because the determinant function is a constant function, so its derivative is zero everywhere? Also, I've been told that the tangent space to this set at the identity is just the space of trace 0 matrices, but I'm not seeing where this relationship comes from. Any help?
     
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  3. May 7, 2006 #2

    Hurkyl

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    I'm not sure what that has to do with anything.

    And note that the determinant map is constant on any subset of SL(n) -- even those subsets that are not manifolds.


    Well, what is the tangent space?

    Suppose I have a curve f(x) in SL(n) such that f(0) = I. What can you say about its derivative?
     
  4. May 7, 2006 #3
    Well the first part of my question doesn't really have much to do with anything, you're right, but it was more of just a verification.. as in is it enough if I say the derivitave map is constant everywhere to show that SL(n) is a submanifold of the nxn space.

    The trouble I'm having with the tangent space of SL(n) is that I am having difficulty with the curve f(x) in SL(n) itself. When you say we have a curve f(x) in SL(n) such that f(0) = I, is the derivative just going to be the derivative of I? And since the diagonal is composed of constant elements, we get their derivatives to be zero, therefore giving us a trace 0 matrix? I apologize if this is obvious.. I had trouble in Linear Algebra with this sort of thing. Thank you for the response though
     
  5. May 7, 2006 #4

    Hurkyl

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    No; you want the derivative of f, not the derivative of f(0).

    f is a function that takes values in an n²-dimensional vector space. Taking its derivative is no different than what you did in vector calculus when you worked in R² or R³... you just have many more components, and you arrange them in a square shape instead of a column.


    (Incidentally, I think what you want to do is a differential approximation)
     
  6. May 8, 2006 #5

    mathwonk

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    you seem to be referring to the fact that a level set of a smooth function is a manifold if the function has maximal rank at every point of that level set. this does indeed apply to the set of matrices of determinant one and does prove they are a manifold.

    the fact that the tangent space at the identity is the set of trace zero matricers can be checked directly, and is also the basis of the "frenet serret formulas".
     
  7. May 8, 2006 #6

    Hurkyl

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    I ought to work through this problem again. :smile: Last time I did it, I threw my hands up in the air, decreed that "dt" was an infinitessimal, and looked at the equation det(I + A dt) = 1, where A was a tangent vector.

    Actually, I still think about matrix Lie groups and algebras in that way. :frown: I suppose I ought to figure out how to justify it in terms of nonstandard analysis so I can feel better about myself. :smile:
     
  8. May 9, 2006 #7

    mathwonk

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    it is easy to see that det has rank one at an nxn matrix A where det(A) is not zero. just compose with a transverse parameter getting det(tA) = t^n det(A), and take the derivative at t=1.

    to get the tangent space to f=c at p, just take the orthogonal complement of the gradient of f at p. for the determinant polynomial, the rgadient is the vector whose entry in the i,j place is plus or minus the minor determinant associated to that entry.

    Hence when evaluated at the identity amtrix it has a zero in every entry except those indexed by i,i, and at those places it has a 1. the orthocomplement is thus matrices which dot to zero with this vector, i.e. those of trace zero.
     
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