Tensor product for vector bundles is commutative, associative, and has an identity element?

Question

How do I see that the tensor product operation for vector bundles over a fixed base space is commutative, associative, and has an identity element?

Answer 1

Suppose from the start that I have three vector bundles $E_1, E_2, E_3$ and an open cover of the base space on which all three of these trivialize. Then if the transition functions for $E_i$ are denoted $\phi_{i,U}: U \to GL_n$, then the tensor product $E_1 \otimes E_2$ is the vector bundle given by transition functions $\phi_1 \otimes \phi_2$, where I mean the tensor product of matrices. That these satisfy the cocycle condition follows from the fact that $(A \otimes B)(C\otimes D) = (AC \otimes BD)$. If this is not your definition of tensor product, I leave it to you to verify that it agrees with your definition.

Then your question amounts to...

1) $\phi_1 \otimes \phi_2$ is conjugate to $\phi_2 \otimes \phi_1$ by a specific permutation matrix.

2) $(\phi_1 \otimes \phi_2) \otimes \phi_3 = \phi_1 \otimes (\phi_2 \otimes \phi_3)$.

3) $\phi_1 \otimes \bf 1 = \phi_1$, where by $\bf 1$ I mean the number 1 as a linear map $\Bbb R \to \Bbb R$ (aka, the identity map). So the vector bundle defined by the constant transition functions (which obviously satisfy the cocycle condition) is an identity element for the tensor product operation; this is the trivial line bundle $M \times \Bbb R \to M$.

All of these facts are easily checkable, either yourself or in any linear algebra book that talks about the tensor product.