Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $L$ be a real vector space of dimension 22 and $q$ a quadratic form on $L$ of signature $(3,19)$.

Let $V\subset L$ be a positive oriented subspace of dimension 2 and $G^{po}(2,L)$ be the Grassmannian of positive and oriented planes in $L$. I have read that the tangent space of $G^{po}(2,L)$ in $V$ is canonically identified with $Hom(V,V^\perp)$ (the orthogonality is intended with respect to $q$, of course).

I can not find a way to view this. I know that $Gr^o(2,L)$, the Grassmannian of oriented planes in $L$, is a double cover of $Gr(2,L)$ and is locally an isometry, so this two spaces have the same tangent spaces. Besides, i think there is no problem identifying $Hom(V,L/V)$ with $Hom(V,V^\perp)$. So my question now is: how does the positivity not change the tangent spaces?

share|improve this question
Why should positivity matter? –  Tom Goodwillie Jun 17 '13 at 11:53
i thought because i'm considering only the positive planes... but wait is it right to say that positivity is an open condition ($q(v)>0$) and so it doesn't change the tangent spaces? –  Filippo Amaducci Jun 17 '13 at 12:09

1 Answer 1

The catch is in the word "canonical". If $V$ is positive, then $V^\perp$ is transverse to $V$ and hence naturally isomorphic to $L/V$ (by means of the projection $L\to L/V$ restricted to $V^\perp$).

Without positivity, $V$ and $V^\perp$ are not always transverse, so the two spaces are isomorphic just because they have equal dimensions (not "canonically"). As a consequence, there may be no way to make the isomorphism depend continuously on $V$. This is essential if you want to figure out, for example, topological invariants of the respective fiber bundles.

share|improve this answer

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.