Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm reading a book on multilinear algebra, and the author first establishes this easy isomorphism: if $V_1,\dots,V_k$ are vector spaces over the field $K$ and if $\sigma\in S_k$, then there is an isomorphism $f_\sigma : V_1\otimes\cdots\otimes V_k\to V_{\sigma(1)}\otimes\cdots \otimes V_{\sigma(k)}.$

That's fine, but then, he starts to define the tensor algebra of a vector space $V$. He does that in the following way: he defines $T^r_s(V)=V^{\otimes r}\otimes (V^\ast)^{\otimes s}$ and then defines the tensor algebra as

$$T(V)=\bigoplus_{r,s=0}^\infty T^r_s(V).$$

Then he comes to define the multiplication of this algebra. He first interpret tensor as multilinear mapping, then tensor multiplication is much simpler. He then goes to define multiplication when we really interpret tensors as elements of the tensor product of vector space He then says the following:

If tensors are not interpreted as multilinear mappings, then tensor multiplication can be defined with the help of the permutation operations, taking into account associativity, as the mapping $$f_\sigma : \underset{p}{V^\ast\otimes\cdots\otimes V^\ast} \otimes \underset{q}{V\otimes \cdots\otimes V}\otimes \underset{p'}{V^\ast\otimes \cdots\otimes V^\ast} \otimes \underset{q'}{V\otimes \cdots \otimes V}\to\\ \to \underset{p+p'}{V^\ast\otimes \cdots \otimes V^\ast}\otimes \underset{q+q'}{V\otimes \cdots \otimes V}$$ where $\sigma$ permutes the third group of $p'$ indices into the location after the first group of $p$ indices, preserving their relative order as well as the relative order of the remianing indices. In this variant, the bilinearity of tensor multiplication is equally obvious, and its associativity becomes and identity between permutations.

Well I've read this lots of times, but I simply didn't get how this defines a multiplication in $T(V)$. Elements of $T(V)$ are sequences of tensors with just finitely many nonzero terms. I can't see how this defines a multiplication. I can't see also how the map $f_\sigma$ comes into play, or even how this multiplication was defined.

What is the author doing there? How this defines multiplication in $T(V)$?

Thanks very much in advance!

share|improve this question
1  
The elements of the tensor algebra are arranged as direct sums of tensors of $V^*$ and $V$. The need for the permutation is mostly, in my opinion, book-keeping. To see the multiplication is into the tensor algebra you must again produce something with $V^*$ and $V$ in order. This is what the permutation does. That said, I like to think of a multilinear mapping on $V \times V^* \times V$ as a tensor, despite it's failure to fit into the standard book-keeping... the associativity and bilinear require further analysis, but as he says, they reduce to "identities amongst permutations" –  James S. Cook Nov 8 '13 at 13:37

2 Answers 2

up vote 3 down vote accepted

Let $T(V)$ be given by

$$T(V)=\bigoplus_{r,s=0}^\infty T^r_s(V).$$

Let us call the indices $r$ and $s$ the "weights". The idea is to define the associative product on the bi-homogeneous components $T^r_s(V)$ of $T(V)$, preserving the bi-weighting. To make things more complicated, please note that if $V$ was a $\mathbb Z$-graded vector space, then we would have 2 weights and a grading. Here we consider the ungraded case

  • Ungraded case: flip

Following the analyis in http://math.stackexchange.com/questions/557981/existence-of-isomorphism-between-tensor-products/558001 we introduce the flip operator

$$\tau: V_1\otimes V_2\rightarrow V_2\otimes V_1$$

with $\tau(v_1\otimes v_2):=f_{\sigma}(v_1\otimes v_2)=v_2\otimes v_1$, where $\sigma$ is just the exchange permutation. In our context $V_1=V$ and $V_2=V^{*}$.

The multiplication in $T(V)$ is then the associative map

$$\Gamma^{p,s}_{q,r}:=1^{\otimes p}\otimes \sigma_{q,r}\otimes 1^{\otimes s} : T^p_q(V)\otimes T^r_s(V)\rightarrow T^{p+r}_{q+s}(V) $$

with $$(1^{\otimes p}\otimes \sigma_{q,r}\otimes 1^{\otimes s})((v_1\otimes\dots\otimes v_p\otimes w_1\otimes\dots\otimes w_q)\otimes (z_1\otimes\dots\otimes z_r\otimes u_1\otimes\dots\otimes u_s)):= (v_1\otimes\dots\otimes v_p\otimes z_1\otimes\dots\otimes z_r) \otimes (w_1\otimes\dots\otimes w_q\otimes u_1\otimes\dots\otimes u_s)$$

on the bi-homogeneous components of $T(V)$, where $\sigma_{q,r}$ denotes the order -preserving permutation described in the OP, built by suitably applying the flip operator on the above strings. Associativity follows by considering the compositions

$$\Gamma^{p+r,m}_{q+s,n}\circ(\Gamma^{p,r}_{q,s}\otimes 1^{\otimes m+n}): (T^p_q(V)\otimes T^r_s(V))\otimes T^m_n(V)\rightarrow T^{p+r+m}_{q+s+n}(V)$$

and

$$\Gamma^{p,r+m}_{q,s+n}\circ(1^{\otimes p+q}\otimes\Gamma^{r,m}_{s,n}): T^p_q(V)\otimes (T^r_s(V))\otimes T^m_n(V))\rightarrow T^{p+r+m}_{q+s+n}(V)$$

and checking the strings, accordingly.

  • Ungraded case: antiflip

It is possible to introduce non trivial signs in the ungraded case. To do so we define the multiplication in $T(V)$ as the associative map

$$\Gamma^{p,s}_{q,r}:=1^{\otimes p}\otimes \sigma_{q,r}\otimes 1^{\otimes s} : T^p_q(V)\otimes T^r_s(V)\rightarrow T^{p+r}_{q+s}(V) $$

with $$(1^{\otimes p}\otimes \sigma_{q,r}\otimes 1^{\otimes s})((v_1\otimes\dots\otimes v_p\otimes w_1\otimes\dots\otimes w_q)\otimes (z_1\otimes\dots\otimes z_r\otimes u_1\otimes\dots\otimes u_s)):=(-1)^{qr} (v_1\otimes\dots\otimes v_p\otimes z_1\otimes\dots\otimes z_r) \otimes (w_1\otimes\dots\otimes w_q\otimes u_1\otimes\dots\otimes u_s)$$

on the bi-homogeneous components of $T(V)$, where $\sigma_{q,r}$ denotes the order -preserving permutation described in the OP. The sign $(-1)^{qr}$ reflects the move of $r$ factors through $q$ factors, without changing the order. This is equivalent to construct $\sigma_{q,r}$ by considering the anti flip

$$\tau_{-}: V_1\otimes V_2\rightarrow V_2\otimes V_1$$

with $\tau_{-}(v_1\otimes v_2):=-f_{\sigma}(v_1\otimes v_2)=-v_2\otimes v_1.$

Studying associativity, i.e. the compositions $\Gamma^{p+r,m}_{q+s,n}\circ(\Gamma^{p,r}_{q,s}\otimes 1^{\otimes m+n})$ and $\Gamma^{p,r+m}_{q,s+n}\circ(1^{\otimes p+q}\otimes\Gamma^{r,m}_{s,n})$ the signs we produce are $(-1)^{qr}(-1)^{(q+s)m}$ and $(-1)^{sm}(-1)^{(r+m)q}$, which are equal. Apart from the sign check, the proof of associativity is just a bit long but straightforward.

share|improve this answer
1  
Thanks for the answer @Avitus! Comparing to your answer on my question about the tensor algebra, it seems your map $\Gamma^{p,s}_{q,r}$ is just the concatenation you told me there. Is this right? And, why this factor $(-1)^{qr}$ appears? I thought this kind of thing would appear just when dealing with the exterior product. Also, this defines the product for the factorizable tensors, for the general case we extend by bilinearity? Thanks very much again. –  user1620696 Nov 8 '13 at 18:29
    
you are welcome! In the $\Gamma$ maps the concatenation is understood. The permutation exchanges the position of elements in $V^{\otimes q}$ and ${V^{*}}^{\otimes r}$ generating a sign. If you prefer, you can consider $\Gamma^{1,1} _{1,1}:(V\otimes V^{*})\otimes (V\otimes V^{*})\rightarrow (V\otimes V) \otimes (V^{*}\otimes V^{*})$ to fix ideas. The general case is extended linearly, you are right. –  Avitus Nov 8 '13 at 21:11
    
sorry for asking again, but I'm still a little confused. As I see you are defining the $\Gamma$ maps as tensor product of linear maps. The map $1^{\otimes p}$ and $1^{\otimes s}$ I understood as identity while the map $\sigma_{q,r}$ is the permutation I told about. The map $\sigma_{q,r}$ still confuses me, because in the author's proof of the existence of such permutations this sign didn't appear. Could you tell where this sign comes from? Thanks very much again for your help. –  user1620696 Nov 9 '13 at 13:35
    
no problem; your understanding of $\Gamma$ is correct (identities etc...). On signs: I believe there exists more than one choice; mine is one.I should have a look at your reference: could you please show it to me? If there exists no non trivial sign in your reference, I believe that the "permutation" is just a composition of "flips". I will edit my answer including this scenario, if you want :) –  Avitus Nov 9 '13 at 18:18
1  
you are welcome. I edited my answer, using both notations. –  Avitus Nov 9 '13 at 19:01

This situation combines two or three different issues in a confusing way.

First, "the tensor algebra" admits a clearer description, in terms of universal mapping properties, as follows, without overly talking about "tensors" at all: over a fixed ground field $k$ the universal algebra $A(V)$ of a $k$-vectorspace $V$ is an associative $k$-algebra and $k$-vectorspace map $V\to A(V)$ so that for any $k$-vectorspace map from $V$ to a $k$-algebra $B$, there is a unique $k$-algebra map $A(V)\to B$ such that the original $V\to B$ factors through it. For purely categorical reasons $A(V)$ is unique up to unique isomorphism, if it exists.

Second, the usual discussion of "tensor algebra(s)" is really just proof of existence by construction: take $A(V)=k\oplus V\oplus \bigoplus_{n\ge 2}\bigotimes^n V$, with obvious inclusion of $V$ and obvious $k$-vectorspace structure. The multiplication is defined (and the canonicalness can be discussed further) by distributivity and $$ (v_1\otimes \ldots \otimes v_m)\cdot (w_1\otimes \ldots \otimes w_n) = v_1\otimes\ldots\otimes v_m\otimes w_1\otimes\ldots\otimes w_n $$ Yes, this is induced from bilinear maps, etc.

Third, in the question, really the algebra tensor product of $A(V)$ and $A(V^*)$ is being considered, but (at least for the moment) no use is made of the fact that the second vectorspace is the dual of the first. For $x,x'\in A(V)$ and $y,y'\in A(W)$ for two vector spaces $V,W$, without referring to the internal structure at all, $(x\otimes y)\cdot (x'\otimes y')=xx'\otimes yy'$.

Now, some caution is in order, because in some scenarios there is a $\mathbb Z/2$ grading on things, with sign flips depending on the number of interchanges. But, even then, the algebras need merely be broken into odd and even parts, not so much expanded into long "words" in terms of the underlying vectors.

share|improve this answer
    
+1 very smooth! –  Avitus Nov 8 '13 at 14:15

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.