# Is $V\otimes V\otimes V^*$ and $V\otimes V^*\otimes V$ the same?

Are $V\otimes V\otimes V^*$ and $V\otimes V^*\otimes V$ the same?

I think tensor product is commutative and associative, so I think they are the same thing, then why is it necessary to use different notation $T_2^1$ and $T_{1\;1}^1$ do distinguish them?

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My commutative and associative means:

$M\otimes N \cong N\otimes M$

$(M\otimes N)\otimes P\cong M\otimes (N \otimes P)$

and my "same" means up to isomorphism.

Sorry for confusion..

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Actually, most definitions of the tensor product (on vector spaces) give a product that is merely symmetric and weakly associative. –  Hurkyl Nov 2 '12 at 8:59
@Hurkyl, I have edited my post. –  hxhxhx88 Nov 2 '12 at 9:03
No one has addressed the OP's question about notation yet. –  wj32 Nov 2 '12 at 9:05
@hxhxhx88 It's considered bad form to change your question after people have already posted answers to it. Neither of the two posted answers (as of this comment) make sense any more. Can you edit to make it clear what your original question was, and leave the new question as an addendum? –  Chris Taylor Nov 2 '12 at 9:07
@ChrisTaylor...oh sorry. I didn't consider it...I have make comments to those people who answered to look at my updated post..but .you are right..I will change it back.. –  hxhxhx88 Nov 2 '12 at 9:10
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The tensor product is commutative, as well as associative, only up-to-isomorphism. This explains why care is required, hence the notational distinction. It is important to realize that this is not just nit-picking. The tendency to just ignore such issues and pretend that one can get away with working as if the tensor product is actually commutative and associative is not without its dangers.

The situation is well understood in the context of monoidal categories. The Mac Lane coherence theorem says that under certain conditions one can indeed proceed as if the isomorphisms are indeed identities. There are two ways to state that precisely. One is the 'all diagrams commute' statement and the other is that every monoidal category is equivalent to a strict monoidal category. However, when going one (categorical) dimension higher such a general strictification result no longer holds. The relevance of this phenomenon (to topology) is discussed at length in Tom Leinster's book "Higher Operads, Higher Categories".

Edit: To consider a simpler case, we can look at the cross product of sets (and let's ignore commutativity and just look at associativity). For sets $A,B,C$ it is the case that $(A\times B)\times C \ne A\times (B\times C)$, simply because a typical element in the LHS is of the form $((a,b),c)$ while one in the RHS is of the form $(a,(b,c))$. Of course, the two sets are isomorphic, simply because they have the same cardinality. However (and it's a BIG however), among all bijective functions between the two sets there is one that is most natural, the one sending $((a,b),c)$ to $(a,(b,c))$. Let us call this bijection $a_{A,B,C}$ and call it the associator (of $A,B,C$). this is typically what one mentally does when pretending that $A\times (B\times C)=(A\times B)\times C$.

The nice, and not so trivial fact is that all of these associators mesh very nicely with each other. That means that if one starts now with any finite list of sets $A_1,\cdots, A_n$ and one places parenthesis around and forms a big repeated cartesian product, call it $C$, and then one moves parenthesis around in the same expression (but does not alter the order of the sets $A_i$) to get another big repeated cartesian product, call it $D$, then something very nice happens. One can use individual associators to obtain a function $C\to D$. To get such a function one chooses a particular order of rearranging parenthesis to get from the first arrangement (that gave rise to $C$) to the second arrangement (that gave rise to $D$). However, there is a lot of freedom as to the order in which parenthesis are rearranged. The nice thing is that no matter which order of rearrangement is chosen the resulting function $C\to D$ is the same.

This is essentially the reason why in this case (and more generally in any monoidal category, including thus the case of tensor products of vector spaces) nobody ever got into trouble for pretending (certain) non-equal things are equal. As said, the proof of this fact (basically Mac Lane's Coherence Theorem) is not trivial.

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...Your answer is too advanced to me.. but seems reasonable..at least I know it is danger to regard isomorphism as totally equivalent.. –  hxhxhx88 Nov 2 '12 at 9:18
Your updates makes me feel better:). So you mean that in set product, we can indeed regard isomorphism as identity. But in the tensor product case, it is dangerous? –  hxhxhx88 Nov 5 '12 at 12:06
Well, basically the same machinery that works for the cross product of sets works for the tensor product of vector spaces. I just used sets since it's simpler. The point is that given some conditions Mac Lane's coherence theorem will guarantee that as long as you do reasonable things you won't get in trouble if you pretend the associator isomorphisms are in fact identities. But to know what is reasonable and what is not it's a good idea not to treat isos as identities and instead keep track of the isos and actually use the coherence theorem. –  Ittay Weiss Nov 5 '12 at 20:41
OK. I got your point, thanks very much! –  hxhxhx88 Nov 6 '12 at 11:12
you're very welcome. –  Ittay Weiss Nov 6 '12 at 11:25
$V \otimes V \otimes V^*$ and $V \otimes V^* \otimes V$ are isomorphic, but not the equal. The isomorphism is given (on the basis vectors) by $v \otimes w \otimes \alpha \mapsto v \otimes \alpha \otimes w$ for $v,w \in V$ and $\alpha \in V^*$; but the elements on each side are not equal. $v \otimes \alpha \otimes w$ doesn't even lie in $V \otimes V \otimes V^*$!
@hxhxhx88: In that case, it really depends what you mean by 'the same'. They are the same up to isomorphism, but they are not the same in the sense of equality. (This is almost like the difference between intension and extension $-$ in some sense, they are the same in extension but not in intension.) –  Clive Newstead Nov 2 '12 at 18:54