Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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)


share|cite|improve this question

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.

share|cite|improve this answer
Thanks. Could you write yet why $B(e_i,f_{s+1})=0$ for $i=1,...,s$. – Richard Aug 28 '11 at 7:18
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.