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i hope my question is not too trivial. Let's suppose we have a vector space $V$ with a unimodular quadratic form $q$ of signature $(m,n)$.

My question is: which is the maximum dimension of an istropic subspace of $V$?

I would say it is $min\{m,n\}$. This is why. Suppose $min\{m,n\}=m$, if $min\{m,n\}=n$ the proof is the same. Given $U$, a positive subspace of dimension $m$, i choose $u_1,\cdots, u_m$ an ortonormal basis of $U$. Then i consider $U^\perp$ which is negative and i choose $w_1,\cdots,w_n$ an ortonormal basis. So the vectors $\{u_1+w_1,u_2+w_2,\cdots,u_m+w_m\}$ span an isotropic subspace of dimension $m$.

But could there be an isotropic subspace of dimension more than $min\{m,n\}$?

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    $\begingroup$ You could consider the projection map onto the positive definite subspace? $\endgroup$
    – Tom Lovering
    Commented Jun 26, 2013 at 11:25
  • $\begingroup$ sorry but i don't follow you $\endgroup$
    – michael waltz
    Commented Jun 26, 2013 at 11:36
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    $\begingroup$ @Tom Lovering: There is no such intrinsic thing as "the" positive-definite subspace (of maximal dimension, I suppose) in the indefinite case, so your suggestion is unclear. $\endgroup$
    – user61789
    Commented Jun 26, 2013 at 12:52
  • $\begingroup$ @michael: Please clarify in the question if you are working with vector spaces over $\mathbf{R}$ or not. It is a bit confusing, because you speak of $q$ being "unimodular", which suggests you have in mind a rational and even integral structure on $V$, raising the possibility that you are working over $\mathbf{Q}$ (briefly, why do you say "unimodular", and what do you mean by that when working over $\mathbf{R}$ if that is the intent?). $\endgroup$
    – user61789
    Commented Jun 26, 2013 at 12:54
  • $\begingroup$ Of course I meant $U$. So I was suggesting if you take the projection map $V \rightarrow U$, its kernel is negative definite, so in particular intersects any isotropic subspace trivially, so under this map, isotropic subspaces are embedded as subspaces of U. $\endgroup$
    – Tom Lovering
    Commented Jun 27, 2013 at 11:21

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This is the Witt index of the quadratic form. I'm assuming you're working over $\bf{R}$, in which case it is indeed $\min\{m,n\}$ (use Witt cancellation). It can be smaller over other fields (like $\bf{Q}$); for instance, you could have an indefinite quaternary form which is anisotropic. For more details see Chapter 1 of Milnor and Husemoller's "Symmetric Bilinear Forms" or Lam's "Introduction to Quadratic Forms over Fields".

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  • $\begingroup$ An example of an indefinite quadratic form over $\mathbb{Q}$ which is anisotropic (and hence has Witt index zero) is $q(x, y) = x^2 - 2y^2$. $\endgroup$
    – Michael Albanese
    Commented Nov 14, 2017 at 18:46

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