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..
 A: The tensor product is not commutative. (It's not even really associative.)
$V \otimes V \otimes V^*$ 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 $V \otimes V \otimes V^*$!
A: 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.
