# What does an endomorphism in a triangulated category give rise to?

Let $D\xrightarrow[]\varphi D\xrightarrow[]kE\xrightarrow[]j\Sigma D$ be an exact triangle in a triangulated category. I am trying to figure out what structure emerges from this on the base of the axioms. For example, the octahedron axiom gives an exact triangle $E\to E^{(2)}\to E\xrightarrow[]{\Sigma k\circ j}\Sigma E$, where $E^{(2)}$ is from the exact triangle $D\xrightarrow[]{\varphi\circ\varphi}D\xrightarrow[]{}E^{(2)}\to\Sigma D$. All this resembles a beginning of a spectral sequence obtained from an exact couple (e. g. obviously the composite $d:E\xrightarrow[]j\Sigma D\xrightarrow[]{\Sigma k}\Sigma E$ satisfies $\Sigma d\circ d=0$) but it quickly becomes too entangled for me.

Can anyone give me a reference for analyzing such situations? At least in simpler situations when e. g. $\varphi\circ\varphi$ is $\varphi$ or zero or invertible? Or on the other hand in more general situations when $\varphi$ has nonzero degree (i. e. $\varphi:\Sigma^nD\to D$ for any $n\in\mathbb Z$)?

I have vague feeling that something like this takes place in connection with the Devinatz-Hopkins-Smith theorem, and that Bondal-Kapranov's twisted complexes might be relevant, but cannot come up with anything definite...

• No matter what the degree of $\phi$ is, it gives rise to a filtered object. If you apply a homological functor $H$ to it (e.g. map another object into it), you get two spectral sequences, potentially converging to $H$ of the limit and colimit of the filtered object: ncatlab.org/nlab/show/… – Marc Hoyois Nov 17 '14 at 12:01
• @MarcHoyois thank you for the explanation, and for the link. But could you please write (in an answer) how precisely one obtains a filtered object, and whether one needs just triangulated structure to get anything from it? In the link you provided seemingly one needs some enhancements. – მამუკა ჯიბლაძე Nov 17 '14 at 19:27
• @MarcHoyois Also, colimit/limit seemingly amount to universally turning $\varphi$ into an isomorphism (either "from the left" or "from the right"), but then one loses e. g. the case when $\varphi$ is nilpotent. Cannot one obtain something meaningful in this case too? – მამუკა ჯიბლაძე Nov 17 '14 at 19:30

Here's an expanded version of my comment, addressing the spectral sequence part of the question. A filtered object in a triangulated category $T$ is simply a sequence

$$\dots \to X_{n-1} \to X_n \to X_{n+1} \to \dots$$

When you apply a homological functor $H : T\to A$, where $A$ is an abelian category, filtered objects in $T$ become bigraded spectral sequences in $A$. The construction of these spectral sequences uses nothing more than the triangulated structure (but you need more structure to make them functorial in the filtered object). In fact, the spectral sequence arises from the exact couple

$$H_*(X_*) \to H_*(X_*) \to H_*(C_*) \to H_*(X_*),$$

where $C_n=\operatorname{cofib}(X_{n-1}\to X_n)$ and $H_*=H\circ \Sigma^{-*}$. All this is described in much detail here, where some references are given.

The usual convergence result is the following. Suppose that $T$ and $A$ have enough sequential (homotopy) colimits and that:

• for every $r$, $H$ preserves the colimit of $n\mapsto \operatorname{cofib}(X_r\to X_{r+n})$,

• for every $r$, $H_r(X_n)=0$ for $n\ll 0$.

Then the spectral sequence converges strongly to $H_*(\operatorname{hocolim}_n X_n)$. Note that this applies to $H^{op}: T^{op}\to A^{op}$ as well, which gives a dual statement with limits.

Now, given an endomorphism $\phi\colon D\to\Sigma^d D$, I can think of two ways to get a filtered object:

(1) Just repeat the map $\phi$, with $X_n= \Sigma^{nd}D$,

(2) Let $X_0=0$, $X_{1}=\operatorname{fib}(\phi)$, $X_2=\operatorname{fib}(\phi^2)$, etc.

The potential targets are:

(1) $H_*$ of the colim-inversion of $\phi$,

(2) $H_*$ of the "$\phi$-primary torsion" in $D$.

In the dual story, we'll have potential targets:

(1') $H_*$ of the lim-inversion of $\phi$,

(2') $H_*$ of the "$\phi$-completion" of $D$.

I don't know if the spectral sequence (1) is useful. It converges strongly if $H$ preserves sequential colimits, $d<0$, and $H_*(D)$ is bounded below. It also converges trivially if $\phi$ has order $n$, the $(n+1)$st page being zero. But if $\phi$ is idempotent, the second page is zero and convergence fails completely.

The spectral sequence (2) is a Bockstein spectral sequence.

• Thank you for the extended answer! One question - I somehow thought that lim-inversion and $\phi$-completion are more or less the same thing? – მამუკა ჯიბლაძე Nov 23 '14 at 6:48
• What I mean by $\phi$-completion here is the inverse limit of the $cofib(\phi^n)$. For example, if $\phi$ is multiplication by $p$ on $\mathbb{Z}$, you get $\mathbb{Z}_p$. – Marc Hoyois Nov 23 '14 at 16:54