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One of the most famous examples of higher Artin stacks is the stack of perfect complexes. I recall here the basic stuff:

We fix a function $b: \mathbb{Z} \rightarrow \mathbb{N}$ which is zero outside of finitely many values, and we let $\operatorname{Perf}^{\leq b} $ be the higher stack whose value on a scheme $X$ is the $\infty$ -category of perfect complexes $C$ over $X$ such that at any point $x \in X$, we have $h^{i}\left(C_{x}\right) \leq b(i) .$ So $\operatorname{Perf}^{\leq b} $ is an $n$ -stack, where $n$ is the length of the interval on which $b$ is nonzero.

It is proven that the Buchsbaum–Eisenbud schemes (or similarly the varieties of complexes) provide an atlas for it, so this is stack turns to be the quotient of these varieties of complexes by the equivalence relation identifying two complexes which are quasi-isomorphic.

My question is basically: what are geometrically the $n$-(auto)morphisms in this stack?

I guess that we have that chain complexes as objects, chain maps as $1$-morphisms, chain homotopies as $2$-morphisms, chain homotopies between chain homotopies as 3-morphisms (and so on...)

EDIT 1

It seems according to this, this and this paper, each point of the stack would be a complex $C$ with automorphism and higher automorphisms groups (in which the $n$-(auto)morphims that I wonder about live in) equal to:

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

which reflects the fact that "morphisms between complexes of amplitude in $[a, b]$ has homotopies and higher homotopies up to degree $n-1$, or equivalently that the $\infty$-category of complexes of amplitude in $[a, b]$ has $(n-1)$ -truncated mapping spaces."

Now according to the following defnition:

I think we get $E x t^{j}(C,C)=\operatorname {Hom}\left ( C , C[j] \right ), 0>j>-n$.

Why these maps between shifted complexes econde higher automorphisms of $E$? There should be some $-j$-simplex around there but I don't know how to see them (my knowledge of Homotopical Algebra is not big)

EDIT 2

A good explanation of the higher homotopy groups in terms of the $\operatorname{Ext}$ groups is already here. Anyway, it would be interesting to see how these homotopy groups reflect themselves in the higher automorphims of the objects that provide the atlas to the stack, i.e. the varieties of complexes.

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    $\begingroup$ Sorry, it was a typo. The ">" symbol that appears with the quote command mixed with the own equations symbols a couple of times. I think it is correct now $\endgroup$
    – Martin Hurtado
    Commented Apr 15, 2021 at 13:06
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    $\begingroup$ The key thing you seem to be missing in your intuition is the Dold-Kan correspondence that identifies chain complexes with abelian groups in simplicial sets (i.e. ``spaces''), and their homology groups with the homotopy groups of the underlying simplicial set. $\endgroup$
    – Denis Nardin
    Commented Apr 17, 2021 at 9:47
  • $\begingroup$ Sure, I have to think on that to understand the higher homotopy groups of the stack. I am going to rephrase a bit more the question (sorry it is a bit messy) $\endgroup$
    – Martin Hurtado
    Commented Apr 17, 2021 at 14:56
  • $\begingroup$ I guess the Dold-Kan correspondence allows you to pass from the homology groups of the complex that appear in the Ext's in to mapping spaces of the simplicial object but still I dont know the form of this mapping spaces so I cannot see the geometry there (Sorry if say any stupid thing, I am not an expert in homotopy theory/higher category theory) $\endgroup$
    – Martin Hurtado
    Commented Apr 17, 2021 at 15:53
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    $\begingroup$ The Dold-Kan correspondence tells you exactly how to build that Kan complex. Then there's a theorem by Quillen that basically tells you that a Kan complex is as good as a topological space for what concerns homotopy type (and in fact I and other homotopy theorists often refer to a Kan complex as a space, and for me the mapping space is defined as a Kan complex) $\endgroup$
    – Denis Nardin
    Commented Apr 17, 2021 at 20:19

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