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I am trying to show that $T\mathbb{RP}^n$ and $\text{Hom}(\gamma_1,\gamma_1^{\perp})$ are isomorphic bundles over $\mathbb{RP}^n$.

For $[x]\in\mathbb{RP}^n$, let $L_x$ be the line in $\mathbb{R}^{n+1}$ joining the two representatives $x$ and $-x$ of $[x]$ and let $L_x^{\perp}$ be its orthogonal complement in $\mathbb{RP}^n \times \mathbb{R}^{n+1}$. Given $(x,v)\in T\mathbb{RP}^n$ we have a linear transformation $T(x,v):L_x\rightarrow L_x^{\perp}$ defined by $T(x,v)(x)=v$. This is clearly well defined because $T(x,v)(-x)=-v$. Now, for each $[x]\in\mathbb{RP}^n$ we have an isomorphism on each fiber $T(x,*): T_{[x]}\mathbb{RP}^n \rightarrow \text{Hom}(L_x,L_x^{\perp})$ given by $T(x,*): (x,v) \mapsto T(x,v)$. However, finding an isomorphism for each fiber is not enough, how can I show these fiber isomorphisms actually induce a homeomorphism of the total spaces of $T\mathbb{RP}^n$ and $\text{Hom}(\gamma_1,\gamma_1^{\perp})$?

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Sorry, what's $\gamma_1$? – Dylan Moreland Aug 14 '11 at 19:17
$\gamma_1$ is the line bundle of over $\mathbb{RP}^n$, that is $\gamma_1=\{([x],v): [x]\in \mathbb{RP}^n, v\in [x]\}$. – Manuel Aug 14 '11 at 19:25
Ah, it's $\mathscr{O}(-1)$. Thanks! – Dylan Moreland Aug 14 '11 at 19:37
Oh well, I guess if a continuous map between total spaces is an isomorphism on each fiber then it is a homeomorphism... we could restrict to a local trivialization and construct an inverse there? – Manuel Aug 14 '11 at 19:39
You should be more explicit about how you are viewing the tangent space. Your map make it seem like you are already identifying $\gamma^{\perp}$ and $T\mathbb{RP}^n$. If $q:S^n\to \mathbb{RP}^n$ is the quotient map, you have that $q^*\gamma^{\perp}\cong q^* T\mathbb{RP}^n \cong TS^n$ and $q^* \gamma$ is the trivial bundle, but you need to be careful to do something which explains why you need to take the hom (because that reason disappears after you pull back). – Aaron Aug 14 '11 at 22:56
up vote 1 down vote accepted

Your comment is correct. You need to use the following lemma

(Milnor-Stasheff 2.3): Let $\xi$ and $\eta$ be vector bundles over $B$, and let $f:E(\xi) \to E(\eta)$ be a continuous function which maps each vector space $F_b(\xi)$ isomorphically onto $F_b(\eta)$. Then $f$ is necessarily a homeomorphism. Hence $\xi$ is isomorphic to $\eta$

The proof is, as you would expect, to think locally! (Let me know if you want me to put the proof up)

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This is also Lemma 1.1 here – Juan S Aug 14 '11 at 21:42
That's good. I should have thought a bit more before posting the question... Thanks! – Manuel Aug 15 '11 at 1:54

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