Is the square of a vector bundle trivial?

Question

Let $E$ be a smooth vector bundle with a metric. Then $E^*$ is isomorphic to $E$ by mapping $v \in E$ to $\langle v, \circ\rangle \in E^*$.

Because $E \otimes E^*$ is a trivial bundle, can we say that the square $E \otimes E \cong E \otimes E^*$ is also trivial ?

Answer 1

In Mihail's answer, they are taking the tensor product of complex line bundles as complex bundles, but you could also take the tensor product as real bundles. This gives rise to examples where $E\otimes E$ is non-trivial.

First, note that $E\otimes E \cong E^*\otimes E \cong \operatorname{End}(E)$. This bundle is always orientable, because $w_1(E\otimes F) = \operatorname{rank}(F)w_1(E) + \operatorname{rank}(E)w_1(F)$, so $E\otimes E$ has an Euler class. As $\operatorname{End}(E)$ has a nowhere-zero section, namely $\operatorname{id}_E$, we have $e(E\otimes E) = 0$. In particular, you cannot use the Euler class to demonstrate that such a bundle is non-trivial.

Now let $E$ be a real orientable rank two bundle. Choosing an orientation for $E$, we can view $E$ as a complex line bundle. Then

$$\operatorname{End}(E) = \operatorname{End}_{\mathbb{C}}(E)\oplus\overline{\operatorname{End}}_{\mathbb{C}}(E)$$

where the terms of the decomposition are complex linear and complex antilinear endomorphisms respectively. If $J$ denotes the almost complex structure on $E$, then the decomposition is given by $L \mapsto \frac{1}{2}(L - JLJ) + \frac{1}{2}(L+JLJ)$. Note that $\operatorname{id}_E$ defines a nowhere-zero section of $\operatorname{End}_{\mathbb{C}}(E)$, so $\operatorname{End}_{\mathbb{C}}(E) \cong \varepsilon_{\mathbb{C}}^1$ (alternatively, $\operatorname{End}_{\mathbb{C}}(E) \cong E^*\otimes_{\mathbb{C}} E \cong \varepsilon^1_{\mathbb{C}}$). On the other hand, a complex anti-linear endomorphism of $E$ can be viewed as a complex linear homomorphism $E \to \overline{E}$, so

$$\overline{\operatorname{End}}_{\mathbb{C}}(E) \cong \operatorname{Hom}_{\mathbb{C}}(E, \overline{E}) \cong E^*\otimes_{\mathbb{C}}\overline{E} \cong \overline{E}\otimes_{\mathbb{C}}\overline{E} \cong \overline{E}^2.$$

Therefore

\begin{align*} p_1(\operatorname{End}(E)) &= p_1(\varepsilon_{\mathbb{C}}^1\oplus\overline{E}^2)\\ &= p_1(\overline{E}^2)\\ &= -c_2(\overline{E}^2\otimes_{\mathbb{R}}\mathbb{C})\\ &= -c_2(\overline{E}^2\oplus E^2)\\ &= -c_1(\overline{E}^2)c_1(E^2)\\ &= -4c_1(\overline{E})c_1(E)\\ &= 4c_1(E)^2. \end{align*}

So, for example, if $E = \mathcal{O}(1)$ over $\mathbb{CP}^2$, then $E\otimes E$ is non-trivial as $p_1(E\otimes E) = 4\alpha^2 \neq 0$ where $\alpha = c_1(\mathcal{O}(1))$ is a generator of $H^2(\mathbb{CP}^2; \mathbb{Z})$.

Answer 2

My answer holds for the complex vector bundles rather than real (Thank you @ConnorMalin for this remark in the comments below)

You are correct in the case of (real) line bundles but in general $E \otimes E^*$ is not a trivial bundle.

It is known that there is a group isomorphism between $H^2(M,\mathbb Z)$ and the complex line bundles over $M$ given by the first Chern class. Pick $M$ with nontrivial $H^2$ for example consider $\mathbb{CP}^2$. Let $E$ be the tautological bundle over $\mathbb{CP}^2$. Then $E\otimes E$ is not trivial because $c_1(E)$ is a generator of $H^2(M,\mathbb Z)=\mathbb{Z}$ and hence $c_1 (E\otimes E)$ is twice the generator.

Answer 3

For any line bundle, real or complex, $E\otimes E^*$ is in fact trivial. In the real case, since $E^*\cong E$, as you asserted, we know that $E\otimes E$ is indeed trivial. The "universal" example is the Möbius line bundle $E$ over the circle. It is, of course, nontrivial (corresponding to the generator of $H^1(S^1,\Bbb Z/2)$). However, its square corresponds to take a Möbius strip with two half-twists, and this is, as you can easily check, isomorphic to the trivial bundle. In particular, it has a nowhere zero section.

EDIT: Mike Miller suggested to me in chat that you consider $E^*\otimes E$ with $E = \varepsilon^1 \oplus \Lambda^2 (T\Bbb RP^2)$ as a rank-$2$ bundle on $\Bbb RP^2$. A Stiefel-Whitney class computation shows that $w_2(E\otimes E)\ne 0$, and so the bundle must be nontrivial.

I would like to do the computation with $E$ the tautological bundle on $\tilde G(2,4)$ (the Grassmannian of oriented $2$-planes in $\Bbb R^4$). It shouldn't be too bad to get $p_1(E\otimes E)$.