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

Let $V:=\oplus_{j\in\mathbb{Z}}V_j$ be a graded $\mathbb{F}$-vector space over the field $\mathbb{F}$. The graded tensor product of graded vector spaces is given by

$V \otimes W:= \oplus_{j\in \mathbb{Z}}\oplus_{p+q=j}V_p\otimes V_q$ and for the graded vector space $\mathbb{F}[j]$, which is $\mathbb{F}$ in degree $j$ and te zero vector space $\{0\}$ otherwise, the shift $V[j]$ is given by

$V[j]:=\mathbb{F}[j]\otimes V$

We then define a monoidal structure on the category of graded vector space, (more or less) given by the rule on homogeneous elements

$v\otimes w= (-1)^{deg(v)deg(w)}w\otimes v$

Then there is the decalage isomorphism

$ dec: V_1[1]\otimes \cdots \otimes V_n[1] \to (V_1 \otimes \cdots \otimes V_n)[n] $

given by $dec(v_1[1]\otimes \cdots \otimes v_n[1])= (-1)^{\sum_{j=1}^n(n-j)deg(v_j)}(v_1\otimes \cdots \otimes v_n)[n]$.

Now in work on graded (stuff), it is frequently said, that this isomorphism defines a natural isomorphism of the symmetric graded tensor-algebra of $V[1]$ and the antisymmetric graded tensor algebra, that is

$S(V[1])\simeq (\bigwedge V)[n]$

*The question is: How does the decalage induces such an algebra isomorphism? Or What is the natural isomorphism? *

If $dec$ itself would be the isomorphism, then

$dec(v[1] \vee w[1])= (dec(v_1)\wedge dec(w))[2]$ should hold, but this isn't true in general.

share|improve this question

1 Answer 1

up vote 3 down vote accepted

You have a detailed proof (in a much more general context but easy to read) in Proposition I.4.3.2.1 of Illusie, Complexe Cotangent et Déformations I, Springer LNM 239.

share|improve this answer
    
Maybe I should have left this as a comment instead of as an answer, but I have no choice (few points?). –  jjms May 23 '13 at 19:11
    
Seems like you are right. Unfortunately I don't speak French. Anyway the answer is ok. –  Nevermind May 25 '13 at 6:17

Your Answer

 
discard

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

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