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Negative-definite inner products

  1. Dec 22, 2007 #1
    Let V be an inner product space whose inner product g is not positive-definite but simply nondegenerate (Minkowski spacetime is an example). Using Zorn's Lemma I've proven fairly easily that V has a maximal subspace on which g is negative-definite. Now I'm investigating:

    Conjecture: V has a subspace, of maximal dimension, on which g is negative-definite. If V is finite-dimensional, this is obviously true. Do you guys think it is true if V is of infinite dimension?
     
  2. jcsd
  3. Dec 22, 2007 #2
    I'm having difficulty proving this using Zorn's Lemma (if it is true at all). Let A be the collection of all subspaces of V on which g is negative-definite. Give A the partial order relation: U < W iff dim(U) < dim (W). Let B be a totally ordered subcollection of A. Now what is an upper bound of B in A??? The union of subspaces here is not a subspace because we are not dealing with a chain of subspaces.
     
    Last edited: Dec 22, 2007
  4. Dec 22, 2007 #3
    Idea: Given a totally ordered subcollection B of A, let W = the SUM of the subspaces in B. Consequently, the dimension of W will be greater than the dimension of each subspace, since W is the span of all the subspaces in B. Thus W is an upper bound of B.

    Is W in A? W will be a subspace even if the sum is infinite if I define elements in W to be the set of all finite sums of the elements of the subspaces, right?

    The negative-definite property of W: g(v_1+...+v_n, v_1+...+v_n) = oops, though all the g(v_i,v_i) terms are non-positive, there are many g(v_i, v_j) terms that may be positive.
     
    Last edited: Dec 23, 2007
  5. Dec 23, 2007 #4
    V has an orthonormal basis if it is finite-dimensional (where a unit vector v here is defined as g(v,v)= 1 or -1), but not necessarily if it is infinite-dimensional. It will always have a maximal orthonormal basis however, and thus there will be a maximal submatrix that is diagonalizable. But there is no matrix to represent g if the dimension of V is not countable.

    If V has an orthogonal basis then all the g(v_i,v_j) terms would be zero and the proof is complete. Perhaps that is the necessary and sufficient condition for the maximality of the dimension. Though the condition is sufficient, it may not be a necessary condition. My original conjecture is that there is no extra necessary condition on V.

    Note: This was a response to a post that was deleted.
     
    Last edited: Dec 23, 2007
  6. Dec 23, 2007 #5
    Proof complete.

    The conjecture is true if V has an orthogonal basis. What I wonder is if this condition on V can be weakened (or have no condition at all) and the conjecture still remain true.
     
    Last edited: Dec 23, 2007
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