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.


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.

  • $\begingroup$ Maybe I should have left this as a comment instead of as an answer, but I have no choice (few points?). $\endgroup$ – jjms May 23 '13 at 19:11
  • $\begingroup$ Seems like you are right. Unfortunately I don't speak French. Anyway the answer is ok. $\endgroup$ – 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 $s^{-1}$.

  • $\begingroup$ 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. $\endgroup$ – Mark.Neuhaus Jan 3 '15 at 22:18
  • $\begingroup$ 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 $V$). $\endgroup$ – Avitus Jan 4 '15 at 12:21
  • 2
    $\begingroup$ Yes. That is what I mean. I think that was the users problem after all. $\endgroup$ – Mark.Neuhaus Jan 4 '15 at 14:41

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