Dold-Kan type theorems tell us that lots of categories are Morita-equivalent to the simplex category $\Delta$. In other words, there are a lot of stable $\infty$-categories which are secretly equivalent to simplicial objects in spectra. For instance, these include nonnegatively-increasingly-filtered spectra and nonnegatively-homologically-graded chain complexes in spectra. See Walde. This is some sort of statement about the non-injectivity of $Psh_{Spt}(-)$ on $\Pi_0$ of the space of categories (or maybe pointed categories or something). Here is a question about $\Pi_1$:

Question: What is the space of auto-equivalences of the category of simplicial spectra?

To start, would be nice to enumerate the auto-equivalences up to homotopy.

Example: For one example, note that the usual Dold Kan correspondence really comes in two dual flavors: from a simplicial spectrum, you can extract the spectral chain complex of "left normalized chains" or the chain complex of "right normalized chains". Composing one of these functors with its inverse gives an automorphism of simplicial spectra. Is this automorphism trivial?


2 Answers 2


Here is a proof that any autoequivalence of (the $\infty$-category of) filtered spectra is naturally equivalent to a suspension functor. Since filtered spectra are equivalent to simplicial spectra, this translates back. The argument takes some work but the following sketches it.

  1. Autoequivalences preserve categorical properties. Therefore, they preserve homotopy limits and colimits, hence the zero object, cofibers, sums, and so on.

  2. In particular, autoequivalences preserve compact objects. A filtered spectrum $X_0 \to X_1 \to \dots$ is compact if and only if each $X_i$ is compact and the maps $X_n \to X_{n+1}$ are equivalences for $n >> 0$. This condition is sufficient because such objects are built in finitely many stages from free objects $0 \to \dots \to 0 \to S^n \to S^n \to \dots$, and you can check necessity by mapping out to diagrams of objects of the form $Y \to Y \to \dots \to Y \to 0 \to 0 \to \dots$.

  3. Because filtered spectra are stable, they get mapping spectra as part of the categorical structure. If $X$ and $Y$ are compact then $map(X,Y)$ is a finite spectrum.

  4. Similarly because filtered spectra are a cocomplete, stable $\infty$-category, they also get a left-tensoring over spectra: $Z \otimes (X_n) \simeq (Z \otimes X_n)$. Any autoequivalence must preserve this left-tensoring.

  5. For any filtered spectrum $X$, we get a ring spectrum $end(X)$, and if $X$ is compact then by the above we find that for any ring spectrum $R$ we can calculate $end(X) \otimes R \simeq end_R(R \otimes X)$.

  6. Let's define the indicator $g(n,m)$ by $$ g(n,m)_k \simeq \begin{cases} \Bbb S &\text{if }n \leq k \leq m,\\0 &\text{otherwise,}\end{cases} $$ with all transition maps being identities. Note that $end(g(n,m)) \simeq \Bbb S$. (I'll allow $m=\infty$ in this definition.)

  7. For any field $k$, any filtered $Hk$-module $Y$ is a direct sum of shifts of indicators: $\Sigma^r Hk \otimes g(n,m)$. As a result, $end_{Hk}(Y) \simeq Hk$ if and only if if $Y$ has one such summand: it is a shift of an indicator.

  8. Here is a lemma: a compact filtered spectrum $X$ satisfies $end(X) \simeq \Bbb S$ if and only if it is equivalent to a suspension $\Sigma^r g(n,m)$ for some $r, n, m$. To check this, we note that for any field $k$, the filtered $Hk$-module $Hk \otimes X$ satisfies $end_{Hk}(Hk \otimes X) \simeq Hk$ by point 6, and hence it is an indicator by point 7. For this to be true, the universal coefficient theorem shows that $H\Bbb Z \otimes X$ must a shifted indicator $\Sigma^r H\Bbb Z \otimes g(n,m)$, and the Hurewicz / Whitehead theorems then let us construct an equivalence $g(n,m) \to X$.

  9. As a result, the family $\{\Sigma^r g(n,m)\}$ of (shifted) indicators is preserved (up to equivalence) by any autoequivalence.

  10. Now we note that $$ map(\Sigma^r g(n,m), \Sigma^s g(p,q)) \simeq \begin{cases} S^{s-r} &\text{if }p \leq n \leq q \leq m,\\ S^{s-r-1} &\text{if }n < p \leq m + 1, $m < q,\\ 0 &\text{otherwise.} \end{cases} $$

  11. I claim that property 10 allows us to identify $\Sigma^r g(0,\infty)$ among indicators, and therefore any autoequivalence must take $g(0,\infty)$ to something equivalent to $\Sigma^r g(0,\infty)$ for some $r$. In previous edits I've taken shortcuts with mixed success; here is a lengthier argument.

  • Let's define an essential map to be a map $f: g \to h$ between indicators such that $map(g,h) \simeq \Bbb S$, generated by $f$. By the above, essential maps are all equivalent to (shifts of) standard maps $g(n,m) \to g(p,q)$ for $p \leq n \leq q \leq m$ and maps $g(n,m) \to \Sigma (p,q)$ for $n < p \leq m+1$, $m < q$.
  • We will say that an essential map is bounded if there is an upper bound on the number of inequivalent ways to write it as a composite of essential maps. The maps $g(n,m) \to g(p,q)$ are unbounded if $q < m = \infty$, and otherwise they are bounded; the maps $g(n,m) \to \Sigma g(p,q)$ are all unbounded unless $q=\infty$.
  • Therefore, shifts of $g(n,\infty)$ are the only objects with no bounded maps out and only bounded maps in.
  • If there is an essential map $g(n,\infty) \to g(m,\infty)$, then $m \leq n$, and so shifts of $g(0,\infty)$ are uniquely determined among these objects.
  1. Similarly, for any other indicator $h$, $map(h, \Sigma^r g(0,\infty)) \simeq \Bbb S$ if and only if $h \simeq \Sigma^r g(n,\infty)$ for some $n$.

  2. If $h_1 \simeq \Sigma^r g(n_1,\infty)$ and $h_2 \simeq \Sigma^r g(n_2,\infty)$ are two such indicators as in point 11, we have that $n_1 \geq n_2$ if and only if $map(h_1,h_2) \simeq \Bbb S$.

  3. This lets us show that, up to equivalence, any autoequivalence of this category must, up to equivalence, take the tower $$ g(0,\infty) \leftarrow g(1,\infty) \leftarrow g(2,\infty) \leftarrow g(3,\infty) \leftarrow \dots $$ to a tower equivalent to $$ \Sigma^r g(0,\infty) \leftarrow \Sigma^r g(1,\infty) \leftarrow \Sigma^r g(2,\infty) \leftarrow \Sigma^r g(3,\infty) \leftarrow \dots $$

  4. This shows that, up to equivalence, the autoequivalence is equivalent to the shift $\Sigma^r$ on the subcategory spanned by $g(n,\infty)$. However, this is a set of generators for the category of filtered spectra: every other filtered spectrum is built from them by hocolims. Therefore, the autoequivalence must be equivalent to the shift $\Sigma^r$.

As a final note, there is an equivalence of categories between $Fun^L(Fil(Sp), Fil(Sp))$ and the functor category $Fun(\Bbb N \times \Bbb N^{op}, Sp)$ by a compact-generators argument. Therefore, the space of autoequivalences can be determined by looking at the space of self-maps of the object representing the identity functor in this category.

  • $\begingroup$ Just to clarify, in the last paragraph, \mathbb N denotes a downward pointing poset? I might have expected Z in place of N as these are unbounded filtrations, but perhaps I’m misundestanding this equivalence. $\endgroup$
    – Mori B.
    Nov 1, 2023 at 5:07
  • $\begingroup$ @MoriB. I usually use it to denote the upward-pointing poset $(0 \to 1 \to 2 \to \dots)$. In this case, the equivalence of categories is between simplicial spectra and nonnegatively filtered spectra, rather than unbounded ones, which is why $\Bbb N$ appears rather than $\Bbb Z$. $\endgroup$ Nov 1, 2023 at 13:18
  • $\begingroup$ thanks @Tyler, I missed that you meant non negatively filtered, tho I probably should have gathered that’s what you meant when you said “equivalent to simplicial spectra”. Is the bounded-belowness used somewhere in this argument? Do you expect unbounded filtrations to have more automorphisms? $\endgroup$
    – Mori B.
    Nov 1, 2023 at 13:33
  • $\begingroup$ @MoriB. The bounded-belowness is used to try and determine the indicator $g(0,\infty)$ as somehow preferred among indicators; but I believe that you can still get the result for filtered spectra. (I just want to hedge my bets because I believe that I'd need to think carefully about it.) Any compact object certainly has some bound below and so most of the same arguments immediately apply. $\endgroup$ Nov 1, 2023 at 16:57

Not an answer, but too long for a comment.

About your Example: If I understand right, you want to consider the self-equivalence of the category of simplicial spectra given by the automorphism of $\Delta$ that reverses the order in a finite ordered set. The corresponding self-equivalence of the category of nonnegative graded chain complexes just takes a chain complex and switches the sign of the boundary map in even degrees. That's naturally isomorphic to the identity. (At least, for ordinary chain complexes, and I suppose that that is basically true for spectral chain complexes as well.)

And here's a side comment: what if we look at spectral chain complexes of some definite dimension? For example, consider the category of 1-dimensional complexes $X_1\to X_0$ (that is, complexes with $X_n=0$ for all $n>1$). This is really just the category of maps of spectra. There is a self-equivalence of the category of these:

$F(X_1\to X_0)=(X_0\to X_0/X_1)$

We have $F^2(X_1\to X_0)=(X_0/X_1\to \Sigma X_1)$ and

$F^3(X_1\to X_0)=(\Sigma X_1\to \Sigma X_0)$.

So the group of (homotopy classes of) self-equivalences has in it not just the infinite cyclic group generated by suspension, but this larger group generated by a cube root of suspension. (Maybe that's all there is?)

What I just noticed is that something similar happens for higher-dimensional complexes. For two dimensions we are looking at objects $(X_2\to X_1\to X_0)$, where this really means three spectra, two maps, and a nullhomotopy of the composed map $X_2\to X_0$; or in other words a map $X_2\to X_1$ and a map $X_1/X_2\to X_0$.

Write $T(X)$ for the cofiber of that map $X_1/X_2\to X_0$; it is also the cofiber of the corresponding map $\Sigma X_2\to X_0/X_1$. I am thinking of it as the homology of the spectral chain complex. Here is a functor, an auto-equivalence of this category of $2$-dimensional spectral chain complexes:

$F(X_2\to X_1\to X_0)=(X_1\to X_0\to T(X))$

$F^2(X_2\to X_1\to X_0)=(X_0\to T(X)\to \Sigma^2 X_2)$

$F^3(X_2\to X_1\to X_0)=(T(X)\to \Sigma^2 X_2\to \Sigma^2 X_1)$

$F^4(X_2\to X_1\to X_0)=(\Sigma^2 X_2\to \Sigma^2 X_1\to \Sigma^2 X_0)$

So this time we have a fourth root of $\Sigma^2$.

It looks like in general for $d$-dimensional nonnegatively graded spectral chain complexes we have a self-equivalence that is a $(d+2)$nd root of $\Sigma^d$.

I don't know quite what to make of this, or whether it is useful at all in answering your question.


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