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Symmetric Matrices as Submfld. of M_n. Prelim

  1. Sep 2, 2007 #1


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    Hi, everyone.

    I am preparing for a prelim. in Diff. Geometry, and here is a question

    I have not been able to figure out:

    I am trying to show that Sym(n) , the set of all symmetric matrices in M_n

    = all nxn matrices, is a mfld. under inclusion.

    I see two possible approaches here, none of which I have been able to
    get an answer from:

    i) Showing Sym(n) is an open subset of M_n .

    This does not seem true.

    every open set containing a member ( in the topology as a subset of R^(n^2))

    contains both symmetric as well as non-symmetric matrices.

    ii) Figuring out directly if Sym(n) is a subspace of M_n, i.e, showing that every

    open set in Sym(n) equals W/\Sym(n) , with W open in M_n ( same topology

    as above).

    I don't know how to do this, because I cannot figure out the open sets in Sym(n).

    I don't see how to use any reasonably nice map like, e.g, the Det. to express

    Sym(n) as a subset of M_n. Some matrices in Sym(n) are invertible, others are

    not, so I don't see how this would work. Nor do I see how to express Sym(n)

    as the inverse image under a map of constant rank.

    I am out of ideas. Any suggestions?.

    Thanks For Any Help.

  2. jcsd
  3. Sep 2, 2007 #2


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    Which are you trying to do?

    Problem A. Sym(n) to be a subset of M(n). It therefore inherits a topology from M(n). Show that Sym(n), with this topology, is a manifold.

    Problem B. You have (previously) defined a topology on Sym(n). Show that Sym(n) is a manifold. Furthermore, show that the topology you have previously defined on Sym(n) is the same as the topology it inherits from M(n).
  4. Sep 3, 2007 #3
    Since this is a question about 'submanifolds', I guess, you're using the inherited topology.

    You should think about Sym(n) as a linear subspace of M(n). Then it's not hard to construct a (global!) parametrisation ...
  5. Sep 3, 2007 #4


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    use the implicit function theorem. i.e. find a function or functions on matrices such that symmetric ones are the common zero locus. then check that the derivatives of your functions, i.e. gradients, are independent.
  6. Sep 5, 2007 #5


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    Thanks to all for your replies. I ended up using MathWonk's idea (f(A)=A-A^T , and
    everything fell into place. Sym(n) is then a regular submanifold of the Lie group M_n,
    and so it is a submanifold. )because it was closest to my knowledge base at the moment. With the exam coming up, just got to be practical.

    Thanks Again.
  7. Sep 6, 2007 #6
    Sorry, but now I'm a little confused.

    From where to where is your map [itex]f(A):=A-A^\top[/itex] going? From M(n) to M(n)? But then the diagonal funktions [itex]f_{ii}(A) = A_{ii}-A_{ii} = 0[/itex] vanish, so their 'gradients' are 0 and so not independent of the others.

    In general: Is a submanifold [itex]M[/itex] defined as [itex]M=f^{-1}(0)[/itex] for some function [itex]f[/itex], then the directional derivative along [itex]M[/itex] always vanishes, so the 'gradients' cannot be independent.
    Mathwonk, what do you mean with your suggestion?

    I learned, that you have to show that the derivative of your function [itex]f(A):=A-A^\top[/itex] is onto. But therefore you have to restrict the codomain to the antisymmetric matrices and you have to know that these form a vector subspace. But then you know, that the antisymmetric matrices form a submanifold of M(n). This is the same reasoning as you can apply to the symmetric matrices, and it's much easier.

    So if you want to use the implicit function theorem you have to know, that the antisymmetric matrices form a submanifold of M(n), but that's the same hard to show than the original question.
    Last edited: Sep 6, 2007
  8. Sep 6, 2007 #7


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    if a smooth map from R^n to R^k has value zero precisely on the set S, and if at every point of S the derivative is a surjective linear map, then the set S is a submanifold of R^n of dimension n-k.

    in the present case the map taking A to A-At, goes from Mn to Mn, oopoos I guess i need the strong version of the implicit function theorem, the rank theorem. so here we assume the map goes from R^n to R^m and has locally constant rank equal to k along the set S where it vanishes.

    then the set S is amnifold of dimension n-k.

    so here the map not only has locally constant rank, it is inear! gosh this is trivial, since the symmetric matrices are actually a vector subspace, hence also a manifold.

    try this on a more interesting subset of matrices like O(n).
  9. Sep 6, 2007 #8


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    the rank theorem, which i think of as a version of the implicit function theorem, says that near a point where a smooth map has locally constant rank, there are smooth coordinates for domain and target in which the map looks exactly like the derivative.

    in particular the level set through that point is locally a manifold whose tangent space there is the kernel of the derivative for this stronger version of the IFT see Dieudonne, Foundations of modern analysis, pp273-276.

    (he separates this from the IFT, since he proves the weaker IFT also for banach spaces. this was the text for sophomore advanced calculus at harvard when i was an undergrad.)
  10. Sep 7, 2007 #9


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    i pick on rudin a lot as not too useful for learning, but he does have the rank theorem. maybe page 228, 3rd edition.

    i still hate his utilitarian discussion of differential forms, probably the worst available for a geometer.
  11. Sep 8, 2007 #10


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    quidamschwartz, are you less confused?
  12. Sep 11, 2007 #11
    Sorry for not answering yet. There was a lot of work to do.

    My confusion was primarily that I posted the 'linear subspace solution' in my first answer and I couldn't see how the problem could be solved with the implicit function theorem (or the rank-version of it) without using this linearity.
    And since you also used this, I'm not confused any more.

    I also was confused that you demanded the gradients should be independent, which cannot be true as I stated. But as you corrected the full-rank (resp. constant rank) condition is the right one.
    So thanks for your detailed explanations.
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