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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.

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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. of Illusie, Complexe Cotangent et Déformations I, Springer LNM 239.

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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

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