8
$\begingroup$

Suppose $V := \bigoplus_{i \in \mathbb{N}}V_i$ and $W := \bigoplus_{i \in \mathbb{N}}W_i$ are $\mathbb{N}$-graded vector spaces. Then their graded tensor power is defined by $V \bigotimes W := \oplus_{n \in \mathbb{N}} \oplus_{i+j=n} V_i \otimes W_j$ and there is a natural isomorphism $\sigma_{V,W}: V \bigotimes W \to W \bigotimes V$ defined on homogeneous elements by the 'sign convention'

$$\sigma_{V,W}(v \otimes w) = (-1)^{\deg(v)*\deg(w)}w \otimes v$$

What is the origin of this natural isomorphism? Is this the only one or are there others defined by other sign conventions?

$\endgroup$
2
  • 5
    $\begingroup$ This is known as the Koszul sign convention. My guess (and it is only a guess) is that it has its origins in geometry: for a wedge of differential forms we have $\eta\wedge\omega = (-1)^{\operatorname{deg}(\omega)\operatorname{deg}(\eta)}\omega\wedge\eta$, which makes the exterior derivative well-behaved and keeps track of orientations. $\endgroup$
    – Mark Grant
    May 10, 2012 at 11:15
  • 10
    $\begingroup$ Even more geometrically, the twist isomorphism $S^n \wedge S^k \cong S^k \wedge S^n$ has degree $(-1)^{n k}$ under the identification of both sides with $S^{n+k}$. $\endgroup$ May 10, 2012 at 20:39

2 Answers 2

8
$\begingroup$

Is this the only one or are there others defined by other sign conventions?

As has already been mentioned - there are essentially two but the Koszul one is amongst other things, the only one which makes the tensor product of complexes a complex.

Since I have been interested in this topic of late, I would point you towards the following internal (within MathOverflow) references:

$\endgroup$
1
  • $\begingroup$ 1+. This answer basically shows that this question is a duplicate. $\endgroup$ May 10, 2012 at 14:02
4
$\begingroup$

If you want to find another $\sigma'_{V,W}\colon V\otimes W\to W\otimes V$ so that $\sigma'_{U,V\otimes W}=\sigma'_{U,V}\sigma'_{U,W}$ and assume that on homogeneous elements $\sigma'_{V,W}(v\otimes w)=f(\deg(v),\deg(w))w\otimes v$ for some elements $f(\cdot,\cdot)\colon\mathbb{N}\times\mathbb{N}\to F$, then it is easy to deduce that $f(a,b)=f(1,1)^{ab}$. This, together with the requirement that $\sigma'_{V,W}\sigma'_{W,V}=Id_{V\otimes W}$, instantly shows that $f(1,1)=\pm1$. But of course there were many assumptions made along the way.

As for the origin, I think that in addition to what Mark Grant says, when you have the product on the cohomology $H^*(M)$ (over a field), you then can apply the Künneth formula and say $H^*(M\times N)=H^*(M)\otimes H^*(N)$. In what sense these two are isomorphic as algebras? To define aa product on the tensor product $H^*(M)\otimes H^*(N)$, you need those isomorphisms $\sigma$, so that you can do $H^*(M)\otimes H^*(N)\otimes H^*(M)\otimes H^*(N)\to H^*(M)\otimes H^*(M)\otimes H^*(N)\otimes H^*(N)$ and then compute the product in $H^*(M)$ and in $H^*(N)$. The Koszul convention is precisely the choice for which you obtain an algebra isomorphism.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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