# Tensor product of vector bundles

So I have a problem that I encountered and I don't know what object I am looking at. So there are two manifolds $M_1$ and $M_2$ and a smooth map $f:M_1\rightarrow M_2$. We know that there is a vector bundle $\pi_2:E_2\rightarrow M_2$, and we also have a pullback vector bundle $\pi_1:f^\ast E_2\rightarrow M_1$. I want to know what exactly is the vector bundle with total space $T^\ast M_1\otimes f^\ast(E_2)$. I'm supposed to find a smooth section in this vector bundle, but I have no idea what the vector bundle is. Is it the map $T^\ast E_1\otimes f^\ast(E_2)\rightarrow M_1\otimes M_1$, this construction didn't make sense to me, so I thought it might be $T^\ast M_1\otimes f^\ast(E_2)\rightarrow T^\ast M_1$. Any thoughts on what it should be?

For reference this is the problem I'm looking at. It is part (c)

• Do you mean $f^*E_2$? You make no mention of a vector bundle $E_1$. – Michael Albanese Sep 7 '15 at 23:32
• Yes, that is what I mean. I have corrected the mistake. – Mark Sato Sep 7 '15 at 23:39
• Which book is this exercise from? – Michael Albanese Sep 8 '15 at 1:42
• It's an exercise that the professor gave to us, so the notation is a little bit ambiguous in some places(i.e. first draft). – Mark Sato Sep 8 '15 at 1:49

Given two vector bundles $V_1$, $V_2$ over $M$, we can form their tensor product bundle $V_1\otimes V_2$ which is a vector bundle over $M$. The fibre of the tensor product bundle is the tensor product of the fibres of the original two bundles. For more details on the construction of the tensor product bundle, see page $13$ of Hatcher's Vector Bundles and K-Theory.
In your question $T^*M_1$ and $f^*E_2$ are both vector bundles over $M_1$, so $T^*M_1\otimes f^*E_2$ is a vector bundle over $M_1$.
• Why do you think the base space is not $M_1$? – Michael Albanese Sep 8 '15 at 1:04
• I've read the question. The base space is $M_1$ as I said. I still don't see why you think it isn't. – Michael Albanese Sep 8 '15 at 1:08
• Why does it make sense to evaluate $Df$ at $(p,v)$, since $Df:M_1\rightarrow T^\ast M_1\otimes f^\ast TM_2$. – Mark Sato Sep 8 '15 at 1:10