# Origin of the sign convention in the Tensor product of graded vector spaces

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?

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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. –  Mark Grant May 10 '12 at 11:15
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 $S^{n+k}$. –  Mike Shulman May 10 '12 at 20:39

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:

For a recent discussion see MathOverflow question: Tensor product of linear mappings versus chain complexes

In particular Simon Letner and Qiaochu Yauns' answers provide a nice conceptual backdrop.

I also found Theo Johnson-Freyd's answer in this MathOverflow question quite illuminating: References for sign conventions in homological algebra

Theo Johnson-Freyd's answer in this thread is also instructive: Sign convention for derivations in CDGAs

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1+. This answer basically shows that this question is a duplicate. –  Martin Brandenburg May 10 '12 at 14:02

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 $f(1,1)=\pm 1$. 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 K\"unneth formula and say $H^\*(M\times N)=H^\*(M)\otimes H^\*(N)$. In what sense these two are isomorphic as algebras? To define a 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.

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