Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
    
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. –  dhagbert May 13 '13 at 10:03
    
Or edit it to ask for an explanation about "decalage" in general. –  dhagbert May 13 '13 at 10:25
    
Did you read the paper arxiv.org/abs/math/0601312 ? –  Nevermind May 20 '13 at 14:46
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.