Let us (again) consider the bilinear form $\beta(A,B)=\operatorname{Tr}(AB)$ for $A,B \in \mathbb{F}^{n,n}$ (quadratic matrices over a field $\mathbb{F}$). I am interested in finding the biggest subspace $U \subset \mathbb{F}^{n,n}$ such that for all $A \in U: \beta(A,A)=\operatorname{Tr}(A^2)=0$.
Investigations about the trace form
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0What do you mean by "closed under addition and multiplication"? – 2011-06-28
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0That you can add linear combinations of matrices in $U$ and for the resulting matrix $B$ the fact $\beta(B,B)=0$ is still valid. – 2011-06-28
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0So if I understand correctly, you are looking for a generating set? – 2011-06-28
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0Yes I am looking for a basis of $U$. But if its not possible to give one, just the dimension of the maximal $U$ would already be great, too. – 2011-06-28
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0I took the liberty of making some minor modifications in formatting. Nitpick: the Killing form should be capitalized, as it bears the name of [Wilhelm Killing](http://en.wikipedia.org/wiki/Wilhelm_Killing). – 2011-06-28
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0Is this the Killing form? It doesn't seem to arise from the adjoint representation of any Lie algebra. I'd just call it the trace form. – 2011-06-28
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0Ok I did not know the expression is only used in that context. I read on wikipedia that the bilinear form that is defined that way is called Killing form on the article about the trace. I fixed the expression in case it confuses someone. – 2011-06-28
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0In what sense is your example closed under addition? If you add two matrices of this form, the $(2,1)$ entry will be $-1$, not $-1/2$. – 2011-06-28
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0In the sense that the trace of the sum will remain 0 (if I did not make an error in my calculation). – 2011-06-28
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1$Tr(A+B)$ will be $0$, because trace is additive, but $Tr((A+B)^2)$ will not be. Try basically any example. I'm hesitating to write out an answer, because this sort of thing is making it hard for me to understand what you're looking for. – 2011-06-28
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0Yes I made an error in the calculation, sorry for that. However Pete seems to have understood what I am looking for and already pretty much answered my question :) – 2011-06-28
1 Answers
By coincidence, I learned very recently that the trace form on $M_n(k)$ is nondegenerate iff the characteristic of $k$ does not divide $n$. This was a mistake: it is in fact easy to see that the bilinear form $x,y \in M_n(k) \mapsto \operatorname{Trace}(xy)$ is nondegenerate for all $n \in \mathbb{Z}^+$ and all fields $k$. Indeed, if $x \neq 0$, then it has some $(i,j)$ entry nonzero, and then if you multiply on the right by the matrix $E_{ji}$ which has $(j,i)$ entry equal to $1$ and all other entries $0$, you get a matrix with nonzero trace. I was thinking of the definition of a strongly separable algebra, which is a separable $k$-algebra with nondegenerate trace form. But the trace form on an arbitrary finite dimensional algebra is the "unreduced trace", i.e, the trace of $x \bullet$ acting $k$-linearly on $A$. When $A$ is a matrix algebra (or more generally a central simple algebra), this unreduced trace is precisely $\sqrt{[A:k]}$ times the
reduced trace, so when $A = M_n(k)$ it is $n$ times the usual matrix trace. Of course, when the characteristic of $k$ is divisible by $n$, this makes the unreduced trace identically zero, so $M_n(k)$ is not "strongly separable" (but the definition looks a little strange to me now, since it seems like we are focusing our attention on the wrong trace form).
I will assume throughout that the characteristic of $k$ is not $2$ so that the standard algebraic theory of quadratic forms can be applied.
You are asking for the maximal dimension of a totally isotropic subspace. If your quadratic form is nondegenerate, every totally isotropic subspace $U$ has an "isotropic supplement" $U'$ such that $U \cap U' = 0$, $\dim U = \dim U'$ and the quadratic form restricted to $U + U'$ is an orthogonal direct sum of $\dim U$ copies of the hyperbolic plane $\mathbb{H} = \langle 1, - 1 \rangle$. (See e.g. $\S 6$ of these notes on quadratic forms.) Therefore the dimension of a maximal totally isotropic subspace is equal to the number, say $r$, such that the Witt Decomposition of $q$ is
$q \cong \bigoplus_{i=1}^r \mathbb{H} \oplus q'$,
where $q'$ is anisotropic, i.e., $q'(x) = 0 \implies x = 0$.
So we want to know the Witt Decomposition of the trace form. When $k = \mathbb{R}$, joriki's answer to your previous question shows that
$q \cong \left(\frac{n^2+n}{2} \right) \langle 1 \rangle \oplus \left( \frac{n^2-n}{2} \right) \langle -1 \rangle \cong \left( \frac{n^2-n}{2} \right) \mathbb{H} + \left(n \right) \langle 1 \rangle$,
so the number $r$ is $\frac{n^2-n}{2}$. This is different from the formula you have given -- in fact, eventually smaller -- so if I have not made a mistake then you have: you should check first of all that the subspace you have in mind is really totally isotropic.
The next order of business is to compute the Witt Decomposition for the trace form over a more general field. Looking at the "matrix units" $E_{ij}$ as joriki did when $k = \mathbb{R}$ seems like a good start, but I haven't done this computation myself.
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1@Listing: no problem. Let me emphasize that such a maximal subspace is not unique: there will be infinitely many such subspaces, all conjugate under the orthogonal group of the quadratic form. – 2011-06-28
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2For an example, I think the subspace of upper triangular matrices with $0$ on the diagonal fits the bill. – 2011-06-28
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0Yes I deleted the comments. Now everything is clear, thank you. – 2011-06-30