2
$\begingroup$

In Hatcher's book on Vector Bundles, the tensor product of two vector bundles is defined through the gluing functions. But I need an example to understand it. So I think of the simplest case, the tensor product of two trivial line bundles.

Let $V_1$ and $V_2$ be two copies of $\mathbb R\times \mathbb R$, and $B = \mathbb R$. Let $\pi_i: V_i \rightarrow B, (x,y) \mapsto x$ for $i=1,2$. Are the gluing functions all identity maps? What is the tensor product of $(V_1,\pi_1)$ and $(V_2,\pi_2)$ as $B$-bundles and why? I guess that up to isomorphism, the result line bundle is still the trivial one, but I think what I need is something concrete.

Thanks.

$\endgroup$
  • 1
    $\begingroup$ For line bundles, the transition functions are "really" $1\times1$-matrices, and the tensor product corresponds to multiplication of numbers (which coincides with the Kronecker product). If the bundles are trivial, you may as well take the transition functions to be constant, i.e., $[1]$. Haven't read Hatcher's book, though, so perhaps this doesn't align perfectly with his definitions.... $\endgroup$ – Andrew D. Hwang Jan 1 '15 at 14:32
2
$\begingroup$

If $E,F$ are vector bundles over a common base space $B$ of rank $r_E$ and $r_F$ respectively, such that $E$ has $p\leq r_E$ everywhere linearly independent sections $s_1,\dots,s_p$, and $F$ has $q\leq r_F$ everywhere linearly independent sections $t_1,\dots,t_q$, then the tensor product $E\otimes F$ has (at least) $pq$ everywhere linearly independent sections, namely the $s_i\otimes t_j$.

This implies that the tensor product of two trivial vector bundles is itself trivial.

$\endgroup$
  • $\begingroup$ Thank you very much for the answer. It is happy to know that the tensor product is still trivial as I gussed. But I think I still need something concrete to know what is going on in the procedure of taking the tensor product. $\endgroup$ – sunkist Jan 1 '15 at 14:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.