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Let $X$ be a real finite dimensional linear space. Let $B:X\times X \rightarrow \mathbf{R}$ be a bilinear symmetric non-degenerated form. Let $M$, $N$ be totally isotropic subspaces with the same dimension $r$ such that $M\cap N=\{0\}$.

Assume that $e_1, ...,e_s$ are linearly indedendent in $M$, $f_1,...,f_s$ are linearly independent in $N$ (where $s< r$) and $B(e_i,f_j)=\delta_{i,j}$ for $i,j=1,...,s$.

Is it possible to extend $e_1,...,e_s$ to a basis $e_1,...,e_r$ in $M$, $f_1,...,f_s$ to a basis $f_1,...,f_r$ in $N$ in such a way that $B(e_i,f_j)=\delta_{i,j}$ for $i,j=1,...,r$?

I know only that for each basis $e_1,...,e_r$ in $M$ there exists a basis $f_1,...,f_r$ in $N$ such that $B(e_i,f_j)=\delta_{i,j}$ for $i,j=1,...,r$ (Bourbaki, Algebra, Chap.11, $\S$4,2, Prop.2)

Thanks.

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1 Answer 1

I think you need some additional assumption such as $X = M \oplus N$.

For example, if $s=0$, you're just asking whether any 2 totally isotropic subspaces with intersection $\{0\}$ have bases satisfying $B(e_i, f_j) = \delta_{i,j}$, and this will be false if you take some totally isotropic subspace $U = \langle e_1, e_2 \rangle $ and split it into $M = \langle e_1 \rangle$ and $N = \langle e_2 \rangle$. The whole space could still be big enough to be nondegenerate.

If you assume $X = M \oplus N$ (so $X$ has dimension $n=2r$), you can do it. It suffices to show how to do a single step from $s$ to $s+1$. Let $M_s$ be the span of $e_1, \ldots, e_s$ and let $N_s$ be the span of $f_1, \ldots, f_s$. Pick some $e = e_{s+1}$ not in $M_s$. By nondegeneracy (and the assumption $X = M \oplus N$), $N \cap e^\perp$ is a proper subspace of $N$. Therefore there exists a vector in $N$ outside both $N \cap e^\perp$ and $N_s$ (the union of two proper subspaces of $N$ cannot be all of $N$). Choose such a vector and make it $f_{s+1}$; by construction, it lies in $N$, and $B(e_{s+1}, f_{s+1})$ is nonzero and therefore we can scale $f_{s+1}$ to make the inner product equal to 1.

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Thanks. Could you write yet why $B(e_i,f_{s+1})=0$ for $i=1,...,s$. –  Richard Aug 28 '11 at 7:18
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The following almost works. After doing the above argument, Write $e'_{s+1} = \sum c_i e_i$ and $f'_{s+1} = \sum d_i e_i$ summing over $i=1, \ldots, s+1$. If we set $B(e_i, f'_{s+1}) = 0$ and $B(f_i, e'_{s+1})=0$ for $i = 1, \ldots, s$, then we conclude $c_i = -c_{s+1} B(e_{s+1}, f_i)$ and $d_i = -d_{s+1} B(f_{s+1}, e_i)$. The only problem is to make $B(e'_{s+1}, f'_{s+1}) = c_{s+1} d_{s+1} \sum_i B(e_{s+1}, f_i) B(f_{s+1}, e_i)$ nonzero. Unless that last sum is 0, we're done. And it would have to be 0 for all choices of $e_{s+1}$ and $f_{s+1}$ for a counterexample... –  Ted Aug 28 '11 at 20:18

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