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Tensor fields and vector bundles

+ 3 like - 0 dislike
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Let $M$ be a differentiable manifold, $TM$ and $T^*M$ a tangent and cotangent bundle of $M$ and let $\Gamma (TM),\ \Gamma (T^*M)$ be spaces of smooth sections of $TM$ and $T^*M$. Let $T_s^r (M)$ denote a space of smooth tensor fields on $M$.

Since vector field on $M$ is a smooth section of $TM$, clearly
$$T_0^1 (M)=\Gamma (TM)$$ and similarly $$T_1^0 (M)=\Gamma (T^*M).$$

My question is related to something I saw in some lecture notes on differential geometry. There was a definition of a tensor field as a smooth section of $TM \otimes \dots \otimes TM \otimes T^*M \otimes \dots \otimes T^*M$. But since these bundles are not vector spaces, it seems a little suspicious to me. I think the product of (co)tangent bundles may be DEFINED "fiber-wise" in the following sense

$$ TM \otimes TM \equiv \bigcup_{p \in M} \{T_pM \otimes T_pM\} \ \tag{*}$$ (disjoint union). This would also correspond with the definition of tensor field I know. Since tensor field of some type is a map which assigns to every point $p \in M$ a tensor from $T_pM \otimes \dots \otimes T_pM \otimes T_p^*M \otimes \dots \otimes T_p^*M$ smoothly, one has

$$ T_s^r (M) = \Gamma \left( \bigcup_{p \in M} \{T_pM \otimes \dots \otimes T_pM \otimes T_p^*M \otimes \dots \otimes T_p^*M\}\right).$$

So my question is whether the following equalities are true:

$$ \bigotimes^r TM \otimes \bigotimes^s T^*M = \bigcup_{p \in M} \{\bigotimes^r T_pM \otimes \bigotimes^s T_p^*M\}\ \tag{1}$$

$$\Gamma \left( \bigotimes^r TM \otimes \bigotimes^s T^*M \right) = \bigotimes^r \Gamma (TM) \otimes \bigotimes^s \Gamma (T^*M) \ \tag{2}$$

and if the tensor product of (co)tangent bundles is not defined the way I proposed in $(*)$, what is the correct definition or from which more general concept it follows?

This post imported from StackExchange Physics at 2014-04-01 17:32 (UCT), posted by SE-user AlanHarper
asked Mar 31, 2014 in Mathematics by AlanHarper (15 points) [ no revision ]

1 Answer

+ 3 like - 0 dislike

I think you are looking for the concept of a smooth functor. Essentially, any functorial construct on vector spaces, such as taking duals, tensor products and direct sums, can be applied to a vector bundle, if for all vector spaces $V, W$ the map $$\mathcal F : \operatorname{Hom}(V,W) \to \operatorname{Hom}(\mathcal F(V), \mathcal F(W))$$ is smooth using the usual definition of what it means for a map between vector spaces to be smooth.

Thus if $E$ and $F$ both bundles over $M$ with fibers $V$ and $W$,, we can find a bundle denoted $E \otimes F$, also over $M$, the fiber of which is $V\otimes W$. So your first equality is true. In fact the proof is to do as you suggest and take the disjoint union of the fiber-wise tensor product. Then you only have to check that this gives a smooth bundle, but since the map $(A, B) \mapsto A\otimes B$ is smooth for linear transformations $A$ and $B$ between finite-dimensional vector spaces, this is the case.

On the level of sections, you also have $$\Gamma(E \otimes F) \cong \Gamma(E)\otimes\Gamma(F)$$ but the tensor product on the right is the tensor product of $C^\infty(M)$-modules. There is not strict equality here, but the spaces are naturally isomorphic. Here, naturally means that you can construct an isomorphism that doesn't depend on arbitrary choices. (To clarify: you probably know that if $V$ and $W$ are vector spaces of the same dimension, you can find an isomorphism $V \cong W$ by choosing bases for $V$ and $W$. But this isomorphism depends on which bases you choose, so they are not naturally isomorphic. However, $V \otimes W$ and $W \otimes V$ are naturally isomorphic. Can you think of why?)

This post imported from StackExchange Physics at 2014-04-01 17:32 (UCT), posted by SE-user Robin Ekman
answered Mar 31, 2014 by Robin Ekman (215 points) [ no revision ]
To be precise, on the level of sections, don't we have an isomorphism $\Gamma(E\otimes F)\cong\Gamma(E)\otimes\Gamma(F)$ and not a strict equality?

This post imported from StackExchange Physics at 2014-04-01 17:32 (UCT), posted by SE-user Alex Nelson
I have no experience with functors but it seems very powerful. I'll check it out and try to absorb your answer. Thank you!

This post imported from StackExchange Physics at 2014-04-01 17:32 (UCT), posted by SE-user AlanHarper
@AlexNelson Yes, you are correct. I've made an edit.

This post imported from StackExchange Physics at 2014-04-01 17:32 (UCT), posted by SE-user Robin Ekman

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