MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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|cite|improve this question
up vote 7 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.

share|cite|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

Let $V$ be a $\mathbb Z$ graded vector space over a field $\mathbb k$ of characteristic $0$ (for simplicity). The suspension $V[1]$ of $V$ is the graded vector space $V[1]:= V\otimes_{\mathbb k} \mathbb k[1],$ with $\mathbb k[1]$ concentrated in degree $-1$, and $\mathbb k_{-1}=\mathbb k$. With $s$ we denote the "suspension" morphism $s:V\rightarrow V[1]$, $x\mapsto sx:=x$ of degree $-1$; in other words, $|sx|=|x|-1$, where $|\cdot|$ denotes the degree of any homogeneous element in $V$ (or any other graded vector space). In general, we write $s^n: V\rightarrow V[n]$, for any integer $n$. We introduce the graded symmetric resp. antisymmetric algebras over $V$:

$$S(V ) = T (V )/ \langle x \otimes y − (−1)^{|x||y|} y \otimes x \rangle,~~ \text{resp.}~~\Lambda (V ) = T (V )/ \langle x \otimes y + (−1)^{|x||y|} y \otimes x \rangle, $$

For any $n\geq 0$ the decalage is a canonical isomorphism of graded vector spaces (not of algebras)

$$\Phi_n: S_n(V[1] )\rightarrow \Lambda(V)[n], $$

where $$\Phi_n(sx_1,\cdots, sx_n):= (-1)^{\sum_{i=1}^n(n-i)|sx_i|}s^{n}(x_1\wedge\cdots\wedge x_n). $$

Why that sign? The sign follows from the Koszul rule, once one removes the suspensions $s$ from the string $sx_1,\cdots, sx_n$ one by one applying $n$-times the desuspension morphism $s^{-1}$.

share|cite|improve this answer
So you say, that the map dec which the user Nevermind posted above is not correct? --Because he uses the wrong degree. ($|v_i|$ instead of $|v_i[1]|$) Indeed the version you presented here works. – Mark.Neuhaus Jan 3 '15 at 22:18
In Nevermind's formula for the decalage I would correct the sign: instead of $deg(v_j)$ one really has $deg(v_j)-1:=deg(s v_j)$, as the map picks up elements in $n$-copies of $V[1]$ (and not $V$). – Avitus Jan 4 '15 at 12:21
Yes. That is what I mean. I think that was the users problem after all. – Mark.Neuhaus Jan 4 '15 at 14:41

Your Answer


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.