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I have search the forum for answer to my question without much success. I am struggling putting together two definitions of the tensor product and tensors in general. I am restricting my question to only one vector space V.

Definition 1: Is the standard definition that a tensor is a multilinear map $f:V^*\times...\times V^*\times V\times...\times V \longrightarrow \mathbb{k}$. The tensor product is defined to be : $(f\otimes g)(...) = f(...)g(...)$

Definition 2: The tensor product is a new vector space where we want to define some product between vectors. We want to define the tensor product as a list of properties the product should have : \begin{align} \lambda(v\otimes w) = (\lambda v)\otimes w = v \otimes(\lambda w)\\ (v_1 + v_2)\otimes w = v_1\otimes w + v_2\otimes w\\ v\otimes(w_1 + w_2) = v\otimes w_1 + v\otimes w_2 \end{align} Then a tensor is a linear combination $\forall v \in V, w \in W$: \begin{align} \sum v_i\otimes w_i \end{align}

I know how to prove that the space $V\longrightarrow V$ is isomorphic to $V\otimes V^*$.

Here are my questions :

1) I don't know exactly to what property $(f\otimes g)(...) = f(...)g(...)$ refer to exactly in the second definition of tensor. For example if we have $f$ to be a (0,2) tensor and $g$ a (0,2) tensor. the product would be a (0,4) tensor so $(f\otimes g)(v_1,v_2,v_3,v_4) = f(v_1,v_2)g(v_3,v_4)$. To what kind of tensor in definition 2 would this be isomorphic ?

2) I am confused about what is a tensor component. As I understand tensor components are the scalars that form a linear combination of basis tensors. In definition 1, I see books defining the tensor components as $T^{a_1,...,a_k}_{b_1,...,b_k} = T(a_1,...,a_k,b_1,...,b_k)$. Where $a_i$ and $b_i$ are a basis of the vector space and covector space. For definition 2 I see tensor component written as : \begin{align} \sum\sum A_{ij}a_i\otimes b_j \end{align}

How do these components relate ?

3) I was trying to work out a basic example of an endomorphism $\mathbb{R}^2 \longrightarrow \mathbb{R}^2$ by using the two definitions but couldn't end up with the same set of components...

Thanks for any help !

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  • $\begingroup$ I'm writing a long answer for you. Please don't delete the question for whatever reason :P $\endgroup$ – Ivo Terek Sep 9 '16 at 19:26
  • $\begingroup$ Aha I so appreciate your time ! $\endgroup$ – user149705 Sep 9 '16 at 19:29
  • $\begingroup$ Note that $V^V$ is only isomorphic to $V \otimes V^*$ when $V$ is finite dimensional.... $\endgroup$ – Hurkyl Sep 9 '16 at 19:36
  • $\begingroup$ Yes sorry I skip that point. $\endgroup$ – user149705 Sep 9 '16 at 19:46
  • $\begingroup$ @user149705 I think I'm done $\endgroup$ – Ivo Terek Sep 9 '16 at 19:49
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Let's assume everything has finite dimension here.

1) I don't know exactly to what property $(f\otimes g)(...) = f(...)g(...)$ refer to exactly in the second definition of tensor. For example if we have $f$ to be a (0,2) tensor and $g$ a (0,2) tensor. the product would be a (0,4) tensor so $(f\otimes g)(v_1,v_2,v_3,v_4) = f(v_1,v_2)g(v_3,v_4)$. To what kind of tensor in definition 2 would this be isomorphic ?

A tensor product of two vector spaces $V$ and $W$ is a pair $(\mathsf{T} ,t)$, where $\mathsf{T} $ is a vector space and $t\colon V \times W \to \mathsf{T} $ is bilinear, such that if $\{{\bf v}_i\}$ and $\{{\bf w}_j\}$ are bases for $V$ and $W$, then $\{t({\bf v}_i,{\bf w}_j)\}$ spans $\mathsf{T} $, and if given $b\colon V \times W \to Z$ is any bilinear map (arriving at an arbitrary vector space $Z$), there is an unique linear map $\overline{b}\colon \mathsf{T} \to Z$ such that $\overline{b}\circ t = b$. Meaning that bilinear maps $b$ factor through $\mathsf{T} $ and we have all the information needed in a single linear map $\overline{b}$. One then proves that all tensor products of $V$ and $W$ are isomorphic, and so we put the usual notations $\mathsf{T} \equiv V \otimes W$, $t \equiv \otimes$, and write ${\bf v} \otimes {\bf w}$ for $t({\bf v},{\bf w})$.

An explicit construction is to take the free vector space with basis $V \times W$, and take its quotient by the subspace spanned by the elements of the form \begin{align} &({\bf v}_1+{\bf v}_2,{\bf w}) - ({\bf v}_1,{\bf w})-({\bf v}_2,{\bf w}), \\ & ({\bf v},{\bf w}_1+{\bf w}_2)-({\bf v},{\bf w}_1)-({\bf v},{\bf w}_2), \\ & (\lambda{\bf v},{\bf w}) - ({\bf v},\lambda{\bf w}).\end{align} We then denote the class of $({\bf v},{\bf w})$ by ${\bf v} \otimes {\bf w}$.

One generalizes all of this by considering more spaces, writing "multilinear maps" instead of "bilinear maps", and so on. The space $${\frak T}^{(r,s)}(V) = \{ f\colon (V^\ast)^r \times V^s \to \Bbb R \mid f \text{ is multilinear} \}$$is isomorphic to $V^{\otimes r}\otimes (V^\ast)^{\otimes s}$, and that isomorphism does not depend on a choice of basis (so it is better than your average run-of-the-mill isomorphism). Well, being more honest, we use a basis to define the isomorphism, but then we check that it would be same if we started with another basis. We say that $T \in {\frak T}^{(r,s)}(V)$ is a $r-$times contravariant and $s-$times covariant tensor. We of course have an operation $$\otimes \colon {\frak T}^{(r,s)}(V) \times {\frak T}^{(r',s')}(V) \to {\frak T}^{(r+r',s+s')}(V).$$

Now we hopefully understand a little better what a tensor product is, we can simply note that ${\frak T}^{(0,4)}(V) \cong (V^\ast)^{\otimes 4}$, and if $\{{\bf v}_i\}$ is a basis for $V$, and $\{{\bf v}^i\}$ is the dual basis, then $$f \otimes g = \sum_{i,j,k,\ell} f_{ijk\ell} {\bf v}^i \otimes {\bf v}^j\otimes {\bf v}^k \otimes {\bf v}^\ell,$$where $f_{ijkl} = f({\bf v}_i,{\bf v}_j,{\bf v}_k,{\bf v}_\ell)$ and ${\bf v}^i \otimes {\bf v}^j\otimes {\bf v}^k \otimes {\bf v}^\ell \in {\frak T}^{(0,4)}(V)$. It corresponds to that same expression seen as a linear combination of ${\bf v}^i \otimes {\bf v}^j \otimes {\bf v}^k \otimes {\bf v}^\ell$, the class of $({\bf v}^i,{\bf v}^j,{\bf v}^k,{\bf v}^\ell)$ in that quotient - that expression is an element of $(V^\ast)^{\otimes 4}$. Maybe I shouldn't have been lazy and used another notation for the classes until now - I'll gladly explain it all again if you have trouble following.

2) I am confused about what is a tensor component. As I understand tensor components are the scalars that form a linear combination of basis tensors. In definition 1, I see books defining the tensor components as $T^{a_1,...,a_k}_{b_1,...,b_k} = T(a_1,...,a_k,b_1,...,b_k)$. Where $a_i$ and $\{b_i\}$ are a basis of the vector space and covector space. For definition 2 I see tensor component written as : \begin{align} \sum\sum A_{ij}a_i\otimes b_j \end{align} How do these components relate ?

Components do depend on a choice of basis. The choice of notation used in your textbook was bad, we only keep indices on $T$, not the vectors. I mean one would write $$T^{i_1...i_r}_{\qquad j_1...j_s} \stackrel{\rm def.}{=} T({\bf v}^{i_1},...,{\bf v}^{i_r},{\bf v}_{j_1},...,{\bf v}_{j_s})$$instead. And with this notation, we'd have $$T = \sum_{i_1,...,i_r,j_1,...,j_s} T^{i_1...i_r}_{\qquad j_1...j_s} {\bf v}_{i_1}\otimes \cdots \otimes {\bf v}_{i_r}\otimes {\bf v}^{j_1}\otimes \cdots \otimes {\bf v}^{j_s}.$$With Einstein's summation convention, we'd only write $$T = T^{i_1...i_r}_{\qquad j_1...j_s} {\bf v}_{i_1}\otimes \cdots \otimes {\bf v}_{i_r}\otimes {\bf v}^{j_1}\otimes \cdots \otimes {\bf v}^{j_s}$$ with all the summations implied (and that's why index balance is good - if the same index appears twice, up and below, sum over it).

When you ask how these things relate, the reasonable thing to consider is another basis $\{{{\bf v}_i}'\}$, the corresponding dual basis $\{{{\bf v}^i}'\}$, write $${T^{i_1...i_r}_{\qquad j_1...j_s}}' = T({{\bf v}^{i_1}}',...,{{\bf v}^{i_r}}',{{\bf v}_{j_1}}',...,{{\bf v}_{j_s}}')$$and see how we can express this in terms of the "old" components $T^{i_1...i_r}_{\qquad j_1...j_s}$.

If you're still alive after all that index juggling, you'll certainly pardon me by ilustrating the relation only in the $(1,1)$ case. Write ${{\bf v}_j}' = \sum_i \alpha_{ij} v_i$. It is an easy linear algebra exercise to check that ${{\bf v}^j}' = \sum_i \beta_{ij} {\bf v}^i$, where $(\beta_{ij})$ is the inverse matrix of $(\alpha_{ij})$. Then $${T^i_{\hspace{1ex} j}}'=T({{\bf v}^i}',{{\bf v}_j}') = \sum_{k, \ell }\beta_{ki}\alpha_{\ell j}T({{\bf v}^k}',{{\bf v}^\ell}') = \sum_{k,\ell} \beta_{ki}\alpha_{\ell j}T^k_{\hspace{1ex}\ell}.$$

If we had more entries, then more $\alpha$'s and $\beta$'s would pop out. You can see that in any physics book, for instance. I like A Short Course in General Relativity, by Foster & Nightingale. By the way, it is costumary to denote the entries of the inverse matrix by writing indices upstairs. In this notation, and using Einstein's convention, we'd have simply $${T^i_{\hspace{1ex} j}}' = \alpha^{ki}\alpha_{\ell j}T^k_{\hspace{1ex} \ell}.$$

3) I was trying to work out a basic example of an endomorphism $\mathbb{R}^2 \longrightarrow \mathbb{R}^2$ by using the two definitions but couldn't end up with the same set of components...

When we write components in a basis, they're real numbers, and our discussion does not quite apply if the codomain of the bilinear map isn't $\Bbb R$.

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  • $\begingroup$ Thank you for the answer. I need more time to study it. Will comeback with further questions :) $\endgroup$ – user149705 Sep 9 '16 at 20:00
  • $\begingroup$ No worries! ${}{}$ $\endgroup$ – Ivo Terek Sep 9 '16 at 20:03
  • $\begingroup$ So basically the tensor product space linearize everything. Making an isomorphism between multi-linear maps and linear maps. Is the proof easy to derive for such a general result ? In my book the author only prove that a $V^* \otimes V$ is isomorphic to the set of maps of the type $ V \times V^* \rightarrow \mathbb{R}$, and then proceed to generalize to multilinear maps without any proof. $\endgroup$ – user149705 Sep 11 '16 at 7:39
  • $\begingroup$ Yes! Nice way to see it. The proof isn't too different from that initial case you said. Try to mimic it in the $(1,2)$ or $(2,3)$ case instead of the $(1,1)$ case to see the pattern. $\endgroup$ – Ivo Terek Sep 11 '16 at 11:21
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    $\begingroup$ Got the answer page 386. Thanks !! $\endgroup$ – user149705 Sep 19 '16 at 5:37

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