6
$\begingroup$

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.

$\endgroup$

2 Answers 2

8
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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, 2013 at 19:11
  • $\begingroup$ Seems like you are right. Unfortunately I don't speak French. Anyway the answer is ok. $\endgroup$
    – Nevermind
    May 25, 2013 at 6:17
2
$\begingroup$

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 $s^{-1}$.

$\endgroup$
5
  • $\begingroup$ 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. $\endgroup$ Jan 3, 2015 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, 2015 at 12:21
  • 2
    $\begingroup$ Yes. That is what I mean. I think that was the users problem after all. $\endgroup$ Jan 4, 2015 at 14:41
  • $\begingroup$ I don’t think the sign should be \sum |sx_i|. It should be \sum |x_i|. Let’s write (s(x_1),...,s(x_n)) as (s_1(x_1),...,s_n(x_n)), where the subscript of s_j only means it is applied to x_j. The we move these s_j to obtain (s_1,s_2,...,s_n)(x_1,x_2,...,x_n), which really produces a sign \sum |x_j|. The sign \sum |s(x_j)| that you claimed should come with (s_n,...,s_2,s_1)(x_1,x_2,...,x_n), which I don't think is the case. $\endgroup$
    – user188722
    Mar 25, 2021 at 3:26
  • $\begingroup$ In my last comment I forgot to write the (n-j) before |x_j| and |s(x_j)|. These two signs only differ by n(n-1)/2 on V^n, so they both work for the decalage isomorphism. But I was trying to say \sum (n-j)|x_j| is consistent with the convention (f,g)(v,w)=(-1)^{|g||v|}f(v)g(w), while \sum (n-j)|s(x_j)| is not. $\endgroup$
    – user188722
    Mar 25, 2021 at 3:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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