Maps between shifted complexes

Question

I am studying the higher stack of perfect complexes $E$ of fixed length $n=(b-a)+1$ (i.e. their homology groups are zero outside the interval $[a,b]$) and it seems that the higher homotopy groups (which express higher automorphims of the stacky points) at a point $E$ are:

$\pi_{i}\left(\operatorname{Perf}^{\leq n} ,E\right) \simeq E x t^{1-i}(E,E)\: n > i>1$

According to the following defnition:

Let $\mathcal{A}$ be an abelian category. Let $i \in \mathbf{Z}$. Let $X, Y$ be objects of $D(\mathcal{A})$. The ith extension group of $X$ by $Y$ is the group $$ \begin{array}{r} \operatorname{Ext}{A}^{i}(X, Y)=\operatorname{Hom}_{D(\mathcal{A})}(X, Y[i])=\operatorname{Hom}_{D(\mathcal{A})}(X[-i], Y) \\ \end{array} $$

I think we get $E x t^{j}(E,E)=\operatorname {Hom}\left ( E , E[j] \right ), -1>j>-n$. This should correspond to higher automorphisms of the complex $E$ but I don't see why.

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Let us be fairly concrete and let our $E$ be of length $4$ chain complex of vector spaces:

$E:=V_{1} \stackrel{d_{2}}{\leftarrow} V_{2} \stackrel{d_{3}}{\leftarrow} V_{3} \stackrel{d_{4}}{\leftarrow} V_{4}; \; d^{2}=0$

So we will have the groups

$E x t^{-1}(E,E)=\operatorname {Hom}\left ( E , E[-1] \right )$, $E x t^{-2}(E,E)=\operatorname {Hom}\left ( E , E[-2] \right ), $ and $E x t^{-3}(E,E)=\operatorname {Hom}\left ( E , E[-3] \right ), $

As the complex $E$ has cohomology zero outside $[1,4]$ we get that $E[-1], E[-2]. E[-3]$ have only $3,2$ and $1$ non-zero terms respectively (also the differential operators will be shifted).

QUESTION: If this is true can anybody tell me why these maps $\operatorname {Hom}\left ( E , E[j] \right )$ encode higher automorphims of $E$?

Answer 1

As we discussed in the comments below my answer here, a morphism $E\to E[j]$ in the derived category (say $E$ is a complex of projectives, which is always the case over a field) is a homotopy class of homotopies from the $0$ homotopy to itself (this $0$ homotopy itself being the $0$ homotopy between the $0$ homotopy between the $0$ homotopy ... and so forth $j-1$ times), so it can be interpreted as a homotopy group.

A statement that works more generally is the following : suppose $C$ is a pointed $\infty$-category with pullbacks, then for any $A,B\in C, j\in \mathbb N$, $map_C(A,\Omega^j_C B) = \Omega^j(map_C(A,B),0)$ where $map_C$ is the mapping space in $C$, and $\Omega^j_C$ is the $j$-th iterated loop in $C$ ($\Omega_C X := 0\times_X 0$, where $0$ is the $0$ object in $C$) and $\Omega^jY$ on the right is the usual loop space, namely $map_*(S^j, Y)$, where $map_*$ denotes the space of pointed maps; and $0\in map_C(A,B)$ is the $0$ morphism (yes, the notation is a bit overloaded)

Recall that $\pi_0\Omega^jY \cong \pi_j Y$ for a pointed space $Y$ so that this formula tells you in particular that $\pi_j(map_C(A,B),0) \cong \pi_0(map_C(A,\Omega^j_C,B)) = [A,\Omega^j_CB]$, where $[-,-]$ is "homotopy classes of maps". Note that on the left, $map_C(A,B)$ is pointed at $0$. In a (homotopically) additive setting, this doesn't really matter because all components are equivalent, but in general it can be important to note this restriction.

In any case, you want to apply this to $C= D(\mathcal A)$, the derived $\infty$-category of $\mathcal A$, which is indeed pointed and has pullbacks (and is additive, in fact much more, it's stable, but we don't need to know that). Then there is a further claim in this setting which is:

Let $E$ be a chain complex in $\mathcal A$, then $E[j]$ is a model for $\Omega^jE$ in $D(\mathcal A)$.

With this in mind, it follows that $[E,E[j]] \cong \pi_j(map(E,E),0)\cong \pi_j(map(E,E),id_E)$.

Here's a slight caveat : $[E,E[j]]$ means homotopy classes of maps in $D(\mathcal A)$, which don't necessarily correspond to homotopy classes of maps in the naive "chain homotopy" sense. They do correspond to those if $E$ is a complex of projectives, which as I said above, will hold over a field.

But if you forget about chain homotopies, you can still interpret morphisms $E\to E[j]$ as elements of $\pi_j$, as long as you're careful about what you mean when you say "morphisms $E\to E[j]$" (they need not come from morphisms of chain complexes, and even if they do, might be equivalent without being related by an honest chain homotopy without doing some replacement first)

Finally, up to a sign, you can compute $[E,E[j]]$ as Ext groups: $[E,E[j]]\cong \mathrm{Ext}^{-j}(E,E)$

(the $1-i$ in your first paragraph comes from the fact that you have a shift because you're looking at Perf, and not only at $E$)