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?


1 Answer 1


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?)

  • $\begingroup$ 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? $\endgroup$ Mar 31, 2014 at 20:54
  • $\begingroup$ I have no experience with functors but it seems very powerful. I'll check it out and try to absorb your answer. Thank you! $\endgroup$
    – AlanHarper
    Mar 31, 2014 at 22:00
  • $\begingroup$ @AlexNelson Yes, you are correct. I've made an edit. $\endgroup$ Apr 1, 2014 at 12:19
  • $\begingroup$ Dear @Robin, There is a natural map $\Gamma(E)\otimes\Gamma(F)\to\Gamma(E\otimes F)$, but is it really true that this is an isomorphism all the time? This is not such an easy thing in the category of, e.g., schemes. I actually asked a question about this (here: math.stackexchange.com/questions/492166/…) but never received an answer. $\endgroup$ Apr 1, 2014 at 19:22
  • $\begingroup$ I do not know if it is an isomorphism for schemes. But it is in the smooth category. You can find a proof in Manifolds and Differential Geometry by Jeffrey Lee, Proposition 7.32. Actually the first four sections of Chapter 7 are all relevant to this question. $\endgroup$ Apr 1, 2014 at 19:24

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