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 have a doubt on the associativity of the tensor product. I know that the tensor product of vector spaces is an associative operation up to a linear isomorphism and I'm just trying to prove that.

My idea is: Let $V_1, \dots ,V_p$ be vector spaces over the same field $\mathbb{K}$ and let $k, r \in \mathbb{N}$ such that $1 < k < r < p$. Define then the spaces:

$$T_1=\bigotimes_{i=1}^{k}V_i$$

$$T_2=\bigotimes_{i=k+1}^{r}V_i$$

$$T_3=\bigotimes_{i=r+1}^{n}V_i$$

Then, if I proove that $T_1 \otimes (T_2 \otimes T_3) = (T_1 \otimes T_2) \otimes T_3$ I'll be proving the general case, so the entire proof is reduced to prove that the tensor product of three vector spaces is associative up to a linear isomorphism, so that we can use this with $T_1$, $T_2$ and $T_3$.

Is this idea correct ? This approach really gives a proof of this fact, or there's some problem approaching it this way ?

Thanks a lot in advance!

share|improve this question

1 Answer 1

up vote 3 down vote accepted

Yes, that's enough.

For a general binary operation $*$ to be associative, is already defined as $$(x*y)*z=x*(y*z)$$ holds for all elements $x,y,z$. This basically enables us to write simply $x*y*z$ for $(x*y)*z=x*(y*z)$, i.e. naturally introduces a ternary operation. By induction, we can see that for any $n$, any parentheses put among $*$ of elements $x_1,x_2,..,x_n$ will result in the same element, i.e., naturally leading to the $n$-ary operation $x_1*x_2*...*x_n$.

This is almost the same for tensor product of vector spaces (or more generally, of modules over a commutative ring), we can just replace $=$ by $\cong$ everywhere above. That is, instead of equality, we have specific isomorphisms $\iota_{V_1,V_2,V_3}$ between $(V_1\otimes V_2)\otimes V_3$ and $V_1\otimes(V_2\otimes V_3)$, but the idea can still be applied: any parentheses put among the tensors of $V_1,V_2,..,V_n$ can be reordered using these specific isomorphisms to $(..((V_1\otimes V_2)\otimes V_3)...)\otimes V_n$. This proves that any two parentheses of a tensor are isomorphic to each other.

However, to ensure that, composing these corresponding $\iota$ isomorphisms, we always arrive to the same isomorphism, we need to check that for $n=4\ $ (and then by MacLane's theorem it also follows for all $n>4$): $$((V_1\otimes V_2)\otimes V_3)\otimes V_4 \to (V_1\otimes(V_2\otimes V_3))\otimes V_4\to V_1\otimes((V_2\otimes V_3)\otimes V_4) \to V_1\otimes(V_2\otimes (V_3\otimes V_4)) \ = \\ ((V_1\otimes V_2)\otimes V_3)\otimes V_4 \to (V_1\otimes V_2)\otimes (V_3\otimes V_4)\to V_1\otimes(V_2\otimes (V_3\otimes V_4)) $$ This is basically the concept of a monoidal category.

An alternative way is to explicitly introduce the $n$-ary tensor operation (just the same way as binary tensor is defined), and then we deal with canonical isomorphisms $V_1\otimes V_2\otimes...\otimes V_n\to V_1\otimes ..(V_i\otimes..\otimes V_j)..\otimes V_n$. This approach leads to the unbiased notion of monoidal category..

share|improve this answer

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.