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Consider the symmetric monoidal category of graded vector spaces in which the symmetric structure is given by the Koszul sign rule. Assume if necessary that the ground field is of characteristic zero. Then (at least for finite dimensional vector spaces?), people say that there is a canonical "decalage" isomorphism $Sym^{k}(V[1]) \simeq (\bigwedge^{k}V)[k]$, where $[1]$ means changing the degree of an element in $V$ by $1$.

I think that I understand this, but the isomorphism seems a bit curious to me in terms of the sum over all $k$, since it seems not to be multiplicative. For example, suppose $V$ is $1$-dimensional in degree $0$ with basis element $v$ and $1$-dimensional in degree $1$ with basis element $w$. Then in $Sym^{2}(V[1])$, $v \cdot w=w \cdot v$ while in $(\bigwedge^{2}V)[2]$, $v \cdot w=-w \cdot v$.

Is this correct? And if so, how are the graded commutative algebras $Sym^{\bullet}(V[1])$ and $\bigwedge^{\bullet}(V)$ related? I always assumed they would be isomorphic after forgetting gradings, but the above seems to say that the obvious map doesn't respect multiplication.

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  • $\begingroup$ I think the answer to this question is that the "decalage" map is not as obvious as I thought, but involves signs. One should interpret $V[1]$ as $V \otimes k[1]$ and then for example under the isomorphism $V[1] \otimes V[1] \simeq V[2]$, there is a sign. Perhaps I should just delete this question. $\endgroup$
    – dhagbert
    May 13, 2013 at 10:03
  • $\begingroup$ Or edit it to ask for an explanation about "decalage" in general. $\endgroup$
    – dhagbert
    May 13, 2013 at 10:25
  • $\begingroup$ Did you read the paper arxiv.org/abs/math/0601312 ? $\endgroup$
    – Nevermind
    May 20, 2013 at 14:46
  • $\begingroup$ Check the answers to this question: mathoverflow.net/questions/131579/… $\endgroup$
    – Avitus
    Sep 12, 2014 at 15:52

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