# Graded Vector Spaces and the Interchange Law

I'm a little confused about how to correctly interchange factors in tensor products on graded vector spaces.

In particular let $V:= \bigoplus_{n \in \mathbb{N}} V_n$ be a $\mathbb{N}$-graded vector space and $\bigotimes^k V$ be the graded k-fold tensor product of $V$. Then one basic convention according to the interchange of factors in the tensor product is

$v_1 \otimes v_2 = (-1)^{|v_1||v_2|}v_2 \otimes v_1$.

But this is ambiguous, isn't it?

For example take $v_1 \otimes v_2 \otimes v_3 \in \bigotimes^3 V$. Then on the one side we could first interchange $v_1$ and $v_2$ to get

$v_1 \otimes v_2 \otimes v_3 = (-1)^{|v_1||v_2|}v_2 \otimes v_1 \otimes v_3$

and then interchange $v_3$ and $v_1$ to get

$v_1 \otimes v_2 \otimes v_3 = (-1)^{|v_1||v_2|+|v_3||v_1|}v_2 \otimes v_3 \otimes v_1$

But on the other side we could first interchange $v_2$ and $v_3$ to get

$v_1 \otimes v_2 \otimes v_3 = (-1)^{|v_2||v_3|}v_1 \otimes v_3 \otimes v_2$

and then interchange $v_1$ and $v_2$ to get

$v_1 \otimes v_2 \otimes v_3 = (-1)^{|v_1||v_2|+|v_2||v_3|}v_2 \otimes v_3 \otimes v_1$

From this we get

$(-1)^{|v_1||v_2|+|v_3||v_1|}v_2 \otimes v_3 \otimes v_1 = (-1)^{|v_1||v_2|+|v_2||v_3|}v_2 \otimes v_3 \otimes v_1$

and consequently

$(-1)^{|v_3||v_1|} = (-1)^{|v_2||v_3|}$ which is obviously not true in general.

What am I doing wrong here?

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The problem is on the line when you say "then interchange $v_1$ and $v_2$ to get"... This kind of manipulation, where you exchange two non-adjacent vectors, isn't covered by the sign rule you gave. What's in the middle matters.
ah! ok, that is because the tensor product of three factors is secretly $(V \otimes V)\otimes V$. And the interchange law is only defined on the 'basic' two fold tensor. Right? –  Mark Neuhaus Jun 15 '12 at 10:36