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Let $\mathcal{A}$ be a Differential graded $k$ category, where $k$ is a commutative ring. I want to extend the shift functor $[1]$ on chain complexes to dg modules over $\mathcal{A}$. Now a right dg module M is a DG functor $M:\mathcal{A}^{op}:\rightarrow \mathcal{C}_{dg}(k)$, where $\mathcal{C}_{dg}(k)$ is the category with chain complexes as objects, and for two objects $C$ and $D$ a chain complex $\mathcal{H}om(C,D)$ with n'th component being graded $k$ homomorphisms of degree $n$ (which doesn't necessarily commute with the differential), and with differential $d(f)=d_D\circ f -(-1)^nf\circ d_C$. The first logical step would therefore be to extend the shift functor to $\mathcal{C}_{dg}(k)$. Now you get a DG functor $[1]:\mathcal{C}_{dg}(k)\rightarrow \mathcal{C}_{dg}(k)$ by sending a chain complex $C$ to a chain complex $C[1]$ defined by $C[1]^n=C^{n+1}$ and $d_{C[1]}=-d_C$ as usual, and sending a map $f$ of degree $n$ to $(-1)^n\cdot f$. I would therefore expect the shift of $M$ to be the composition \begin{align*} [1]\circ M:\mathcal{A}^{op}:\rightarrow \mathcal{C}_{dg}(k) \end{align*} which would send an element $g\in \mathcal{A}(X,Y)^n$ to $(-1)^nM(g)$. However in section 2.1 of the paper "On the cyclic homology of exact categories" http://www.math.jussieu.fr/~keller/publ/cyex.pdf they state that the shift of $M$, denoted by $M[1]$, acts on $g$ by \begin{align*} M[1](g)^m=(-1)^{m\cdot n}M(g)^{m+1} \end{align*} where for a map $f:C\rightarrow D$ between chain complexes, $f^n$ denotes the restriction of $f$ to the component $C^n$. This is also a consequence if you extend the definition of the shift functor on modules over graded categories as defined in section 1.1 of the paper "Deriving DG Categories" http://www.math.jussieu.fr/~keller/publ/ddc.pdf

What is the reason behind the sign convention chosen in the papers? And what is wrong with my reasoning?

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  • $\begingroup$ Too many signs to keep track of (as usual in this subject), but are you sure your definition of $[1]:\mathcal{C}_{dg}(k)\rightarrow \mathcal{C}_{dg}(k)$ on morphisms is the right one? In particular I wonder if the differential applied to a composition $f[1]g[1]$ satisfies the graded Leibniz rule. $\endgroup$ – Dag Oskar Madsen Feb 23 '14 at 21:26
  • $\begingroup$ If I haven't made any mistakes in my computation, $[1]$ on morphisms should give a map on chain complexes (remember that we have to put a minus sign in the differential when shifting). So the composition should satisfy leibniz $\endgroup$ – Sondre Feb 23 '14 at 22:07
  • $\begingroup$ Actually, as written, the sign conventions in the two papers you refer to differ slightly from each other. Does $m$ refer to the degree in $M[1]$ or the degree in $M$? The second sounds most reasonable, but I don't know. $\endgroup$ – Dag Oskar Madsen Feb 24 '14 at 8:45

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