# Trouble understanding Tensor product in context of Torsion Tensor

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??

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.

• 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. Nov 15, 2013 at 16:08
• 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. Nov 15, 2013 at 23:33
• 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)? Nov 16, 2013 at 4:54
• I don't think there is an analogue of the notation for the case of general tensor products. Nov 16, 2013 at 6:44
• Aah Ok..Thanks again. Nov 16, 2013 at 6:44