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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?

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    $\begingroup$ Why should positivity matter? $\endgroup$
    – Tom Goodwillie
    Commented Jun 17, 2013 at 11:53
  • $\begingroup$ 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? $\endgroup$
    – Filippo Amaducci
    Commented Jun 17, 2013 at 12:09

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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.

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