tensor product of two vector bundles

Question

Let $\xi$ and $\eta$ be two vector bundles over the same base space $B$. Then $\xi\otimes \eta$ is orientable if and only if $\xi$ and $\eta$ are both orientable. How to prove this true or not true? Thanks.

Is the tensor product of a mobius band with itself orientable ?

Answer 1

You can't prove that result it because it is false!
For example the tautological bundle $\xi$ on $\mathbb P^n(\mathbb R)$ is non-orientable but its tensor square $\xi^{\otimes 2}=\theta$ is the trivial bundle (like all squares of line bundles on a compact topological space !) $\theta=\mathbb P^n(\mathbb R) \times \mathbb R$ and is thus certainly orientable.

The assertion that $\xi$ is non orientable can be proved in an elementary way but I recommend learning to use Stiefel-Whitney classes, which allow you to obtain such results quite mechanically:
a line bundle $\eta$ on a topological space $M$is orientable iff its first Stiefel-Whitney class $w_1(\eta)\in H^2(M,\mathbb Z/2)$ is zero and in our case $w_1(\xi)\neq 0\in H^2(M,\mathbb Z/2)$ is the generator of the group $H^1(\mathbb P^n(\mathbb R),\mathbb Z/2)=\mathbb Z/2$.
The best reference hands-down for Stiefel-Whitney classes (and much, much more) is Milnor-Stasheff's Characteristic Classes (of which PDF versions float online...).

Answer 2

As Georges points out the result isn't true: $\xi \otimes \eta$ being orientable doesn't imply $\xi$ and $\eta$ orientable. Trivially you can take $\xi$ to be the zero vector bundle...

In the other direction though: A vector bundle $\xi$ (of rank $k$) is orientable iff $\Lambda^k \xi$ is a trivial line bundle iff $\Lambda^k \xi$ has a nowhere vanishing section. Let's assume that both $\xi$ and $\eta$ are orientable.

Let $s \in \Gamma(B, \Lambda^k \xi)$ and $t \in \Gamma(B, \Lambda^l \eta)$ be nowhere vanishing sections. Then you can construct a section $u$ of $\Lambda^{kl} (\xi \otimes \eta)$ by using the following construction on vector spaces: $$\begin{align} \Lambda^k V \otimes \Lambda^l W & \to \Lambda^{kl} (V \otimes W) \\ (v_1 \ldots v_k) \otimes (w_1 \ldots w_l) & \mapsto (v_1 \otimes w_1) \ldots (v_1 \otimes w_l) (v_2 \otimes w_1) \ldots (v_k \otimes w_l) \end{align}$$

This gives a continuous functor on vector spaces, so it yields a functor on vector bundles. And you can check that this is injective for vector spaces (it's tedious). Therefore $u$ is a nowhere vanishing section of $\Lambda^{kl}(\xi \otimes \eta)$, what we wanted.

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Of course if you're okay with higher level machinery, then $w_1(\xi \otimes \eta)$ can be expressed as a linear polynomial in $w_1(\xi)$ and $w_1(\eta)$ (use the splitting principle and the computation for line bundle $w_1(\ell_1 \otimes \ell_2) = w_1(\ell_1) + w_1(\ell_2)$). So if both are zero (ie $\xi, \eta$ are orientable) then $\xi \otimes \eta$ is orientable.

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Yet another proof: a vector bundle is orientable if you can find charts and transition functions that preserve orientation, ie. take values in $GL^+_k(\mathbb{R})$. If both $\xi$ and $\eta$ are orientable, you can find such charts, restrict them, and you get charts for $\xi \otimes \eta$ with transition functions expressed in terms of those of $\xi$ and $\eta$. If the transition functions for $\xi$ and $\eta$ have positive determinant, then so do those of $\xi \otimes \eta$ you constructed that way.