Tensor product of universal vector bundles

Question

Let $E\rightarrow RP^n$ be the universal (tautological) vector bundle. Is $E\otimes E$ trivial?

Answer 1

Viewing $\mathbb{RP}^n$ as $\operatorname{Gr}(1, n+1)$, we see that the tautological bundle $E \to \mathbb{RP}^n$ is a line bundle, and hence, so is $E\otimes E$. Recall that line bundles are determined up to isomorphism by their first Stiefel-Whitney class. As

$$w_1(E\otimes E) = w_1(E) + w_1(E) = 0,$$

we see that $E\otimes E$ is trivial for any line bundle $E$, in particular, for the tautological line bundle over $\mathbb{RP}^n$.

Another way to see that the tensor product of a line bundle with itself is trivial is as follows. First, after choosing a Riemannian metric $g$ on $E$, we obtain an isomorphism $E \to E^*$ given by $e \mapsto g(e, \cdot)$. Therefore $E\otimes E \cong E^*\otimes E \cong \operatorname{End}(E)$. Now note that $\operatorname{End}(E)$ is trivial for any line bundle $E$ as it has a nowhere-zero section, namely $\operatorname{id}_E$.