Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,047 questions , 2,200 unanswered
5,345 answers , 22,709 comments
1,470 users with positive rep
816 active unimported users
More ...

  Tensor fields and vector bundles

+ 3 like - 0 dislike
952 views

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

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ys$\varnothing$csOverflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...