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Let $T(\mathbb{R}P^{n})$ be a tangent bundle to a $n$-dimensional projective space and let also denote $\gamma_{1}$ to be a trivial bundle of rank $1$.

Moreover, let $E \rightarrow \mathbb{R}P^{n}$ be a universal (tautological) bundle.

I would like to prove that $$T(\mathbb{R} P^{n}) \oplus \gamma_{1} = E \oplus E \oplus \ldots E = \bigoplus_{i=1}^{n} E$$

On the one hand, it may looks as if it follows from the existence of the Euler exact sequence of sheaves, written in the following way: $$ 0 \rightarrow \mathcal{O}_{\mathbb{P}^{n}} \rightarrow \mathcal{O}(1)^{\oplus (n+1)} \rightarrow \mathcal{T}_{\mathbb{P}^{n}} \rightarrow 0$$ One can show that this sequence splits (though not canonically???).

Are there any ways to derive the statement from the propositon above or are there any easier ways to obtain the desired result (maybe, by pulling back the $T(\mathbb{R}P^{n})$ via $f: S^{n} \rightarrow RP^{n}$?)

hyperkahler
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1 Answers1

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It's well-known (and not hard to prove) that $T\mathbb{R}P^n \cong \text{Hom}(\gamma, \gamma^{\perp})$ where $\gamma$ is the tautological line bundle. Observe $\text{Hom}(\gamma, \gamma)$ is a trivial line bundle, and summing it with $T\mathbb{R}P^n$ we get $\text{Hom}(\gamma, \gamma \oplus \gamma^\perp)$. Since $\gamma \oplus \gamma^\perp$ is trivial, the result follows from the fact that, for a trivial line bundle $\epsilon$, there is an isomorphism $\text{Hom}(\gamma, \epsilon) \cong \gamma$ (this is just saying $\gamma$ is isomorphic to its dual, which is easy to see since e.g. it admits a metric).

Pedro
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