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I know that there are quite a few threads dealing with this question already. I have pored through them for quite some time and they have been informative. However there are still some clarifications that I seek. If this is too much of a duplicate, then I wouldnt mind if it is closed. I will sum up what I have understood.These are the definitions that I have seen:

  1. Tensor Product of two vector spaces $V,W$ over a field $\mathbb F$ is another vector space $T$ over $\mathbb F$ equipped with a bilinear mapping $\otimes:V \times W \to T$ ; $\otimes (v,w) \mapsto v\otimes w$ such that for any vector space $U$ over the same field and any bilinear map $f:V \times W \to U$, $ \exists !$ linear map $f^* :T \to U$ where $f^*\circ \otimes =f $.

  2. An (r,s) tensor is a multilinear function $B_{i_1 \ldots i_r}^{j_1 \ldots j_s}$ on $(V^*)^r \times (V)^s$ onto the field $\mathbb F$, denoted as $V \otimes \ldots \otimes V \otimes V^* \otimes \ldots \otimes V^*$ .

Now my question is the first is a definition of Tensor Product of Vector spaces, whereas the 2nd defines just a "tensor"??Also is the 2nd definition something that restricts itself to merely copies of a single vector space and its dual only??

Further a Torsion Tensor on a Riemannian Manifold M is defined as a vector valued two-form on $\chi(M) \times \chi (M) \to \chi(M) $ where $\chi (M)$ is the set of all $C^{\infty}$ vector fields on M. So in this case what kind of tensor is this??Its said to be (1,2) tensor , but I am not able to see that. The closest I would say is a (0,2) tensor. And does vector-valued mean this is the 1st definition in work??

Also am I right in saying the two definitions are similar only in the case of Finite Dimensional spaces whereas the 1st definition is the more general one??

Please help me differentiate between the two definitions and put the Torsion in proper context.

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That's a lot of question marks, Vishesh. I hope I can help.

In the two definitions above, the first defines the tensor product of vector spaces. The tensor product of the vector spaces $V$ and $W$ is written $V \otimes W$, where the underlying field is understood from the context. The second definition identifies tensors as multilinear maps whose domain is the vector space

$$\underbrace{V^{*} \times \cdots \times V^{*}}_{r} \times \underbrace{V \times \cdots \times V}_{s}$$

and whose codomain is $\mathbb{F}$. The type of the tensor allows us to identify its domain, and in this case we say the tensor is of type $(r,s)$.

The connection between these definitions can be seen by considering the simple case $V^{*} \otimes V$. An element $T \in V^{*} \otimes V$ is a $(1,1)$-tensor in the following way. Write $T = v^{*} \otimes v$, and let $(w,w^{*}) \in V \times V^{*}$. Then

$$T(w,w^{*}) = v^{*}(w)w^{*}(v)$$

That is, $T \in V^{*} \otimes V$ has a component in $V^{*}$ which "eats" elements of $V$, and a component in $V$ which is "eaten" by elements of $V^{*}$. Thus it makes sense to pass pairs of elements in $V \times V^{*}$ into $T$ and get an element of $\mathbb{F}$, and the function you get is easily seen to be multilinear. This example shows how to identify mutilinear maps on

$$\underbrace{V^{*} \times \cdots \times V^{*}}_{r} \times \underbrace{V \times \cdots \times V}_{s}$$

with elements of the tensor product space $$\underbrace{V \otimes \cdots \otimes V}_{r} \otimes \underbrace{V^{*} \otimes \cdots \otimes V^{*}}_{s}$$

explaining why we call those multilinear maps "tensors".

The general construction of a tensor product can involve more vector spaces than $V$ or $V^{*}$, but these are not the sort of tensors you will be interested in as you begin to study differential geometry.

The torsion tensor is a $(1,2)$-tensor because it takes in two vector fields and spits out a vector field. Thus it must have the capacity to "eat" two vector fields, so it has two components in $\chi(M)^{*}$. However, the end result is not an element of $\mathbb{F}$ but a vector field, so there must be an extra copy of $\chi(M)$ lying around, which is why the torsion tensor lives in $\chi(M) \otimes \chi(M)^{*} \otimes \chi(M)^{*}$.

In the infinite dimensional case, there are more multilinear maps than objects of the tensor product space, so you are correct that the correspondence breaks down there.

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  • $\begingroup$ Good God.That's perfect!!!!!!! I hope the exclamation marks equal the question marks. Seriously thanks a lot. I have spent an entire day trying to get my head around this.One small clarification, the 2nd definition also works for different vector spaces, rather than just copies of $V$ and $V^*$. I meant V and W and anything else or even their duals in tandem?? I think your answer says so, I just wanted to confirm. $\endgroup$ – Vishesh Nov 15 '13 at 16:08
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    $\begingroup$ Yes, you can look at multilinear maps out of products of spaces other than $V$ or $V^{*}$, so long as all the vector spaces are over the same field. $\endgroup$ – Elchanan Solomon Nov 15 '13 at 23:33
  • $\begingroup$ Sorry to bother, but one last doubt, if I have a multilinear map onto the underlying field $V^* \otimes W^* \otimes V \otimes W$ and combinations therein, what type of tensor(r,s??) would it be??Would I still call it using the number of dual spaces and vector spaces(even though they may not be copies of a single vector space)? $\endgroup$ – Vishesh Nov 16 '13 at 4:54
  • $\begingroup$ I don't think there is an analogue of the notation for the case of general tensor products. $\endgroup$ – Elchanan Solomon Nov 16 '13 at 6:44
  • $\begingroup$ Aah Ok..Thanks again. $\endgroup$ – Vishesh Nov 16 '13 at 6:44

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