# Hulk Flatten Puny Octahedral Axiom

Alexander Ellis has just put up a very cool post over at Concrete Nonsense, called Wood, Glue, and the Octahedral Axiom.   For those of you who don’t know, the Octahedral Axiom is the Fourth and Most Annoying Axiom for triangulated categories.  It’s a nuisance for two reasons.  First, it’s a bit aesthetically unsatisfactory.   One of the triangulated category axioms asserts that for any map $f: A \to B$, there exists an exact triangle

$A \to B \to X \to A[1]$

The “cokernel” $X$ isn’t unique up to unique isomorphism — this is the aesthetically unsatisfying bit — so we’re forced to introduce the Octahedral Axiom, which asserts that $X$, while not canonical, is at least somewhat functorial.   Second, the Octahedral Axiom is usually written as (the projection onto the plane/page) of a certain partially commutative three-dimensional octahedron diagram.    It’s sort of a pain to see what the various parts of this diagram mean.

Anyways, Ellis has gotten crafty and put together a miniature  model of the Octahedral Axiom.   (The real one is kept in Paris, in the basment of IHES, and is quite large, but Ellis’ model fits in the palm of his hand.)  You should check it out.   It’s pretty sweet.

I think it might be worth mentioning, however, that it’s not necessary to use three-dimensional commutative diagrams to understand the octahedral axiom.    There’s a pretty reasonable way of writing it using just 2d diagrams.

Octahedral Axiom

If we have the following three exact triangles

then there exists a 4th “vertical” exact triangle $X \to Y \to Z$ such that the following diagram commutes.

Obviously, you can compress this more, since some of the commutative squares are trivial, but I think the content is clearer when the original three triangles are left in their original form.   What this axiom says, basically, is that if we identify $X = B/A$, $Y = C/A$, and $Z = C/B$, then we have $C/B = (C/A)/(B/A)$.

I think I’ll leave it to the reader to figure out what you have to do to this diagram to make it into an octahedron.

TeXnical Aside:   If anyone knows how to make wordpress support xypic I’d appreciate it.   At the moment I make commutative diagrams by typing xypic code into the BLANK in

http://www.codecogs.com/eq.latex?BLANK

This produces a .gif which I upload.

## 8 thoughts on “Hulk Flatten Puny Octahedral Axiom”

1. the link has an extra “http//” in it =)

2. That’s a great 2D diagram. If I had seen that before, perhaps I could’ve saved some cash at Rite Aid…

Anyway, xypic in WordPress would be a great holiday gift from the WordPress Gods.

3. Scott Carnahan says:

I seem to remember you giving a talk about two years ago about how stable infinity-categories are better than triangulated categories (in the sense that almost all useful triangulated categories are just homotopy categories of stable infinity-categories, and the latter are both more memorable and better-behaved). Have you changed your mind?

4. What is a good resource to read about the triangulated homotopy categories of stable infinity-categories?

5. A.J. Tolland says:

Scott,

That was an expository talk. I agree with the sentiment — how could it not be better? Stability is a property of infinity categories, rather than additional data — but I don’t want to give anyone the impression it was original. I learned it from Lurie’s thesis.

Urs,

The theorem I was explaining in the talk Scott mentioned — that the homotopy category of any stable infinity-category is triangulated — can be found in Chapter 1 of Jacob‘s thesis.

6. I’ve got to do a little sticking up for the octahedral picture; not that there’s anything wrong with trying to unpack it, but the octahedral axiom answers in the most elegant way possible the rather obvious question in a triangulated category: how are the cones of two morphisms $f$ and $g$ related to the cone of the composition $g\circ f$?

The answer: there’s an exact triangle $C_f\to C_{g\circ f} \to C_{g}\overset{[1]}\to$. That’s simple enough, isn’t it?

7. anon says:

I too prefer a 2D method of picturing the octahedral axiom, but a different one! Basically, you start with an arbitrary commutative square, and then the axiom says that there exists a way to embed it as the upper-left corner of a large 4×4 diagram where all rows and columns are distinguished, the right column is given by applying the shift functor to the left column, the bottom row is gotten by applying the shift functor to the top row, and all the squares commute except for the lower-right one, which anti-commutes. One recovers the usual octahedral axiom by taking, say, the top-most initial map to be the identity (which also makes the anti-commutativity a moot point, since it forces one of the objects in question to be zero).

But of course, we want to know: why does the bottom-right square anti-commute in general, instead of commuting? Well, there are pictures and mnemonics for this sort of thing, but in the end the whole issue seems to me to be something of a red herring.

Indeed, basically, the above version of the octahedral axiom is just the shadow (in the homotopy category) of a very basic infinity-categorical fact, itself a version of a familiar normal-category fact: that we can often break up limits or colimits over large diagrams in terms of iterated limits or colimits over smaller diagrams (cones in triangulated categories being instances of limits or colimits over certain very simple diagrams). This is one reason why enhanced versions of triangulated categories (especially, in my opinion, Lurie’s stable infinity categories) are preferable: the axioms and results become much simpler. In fact, they become so simple that one can actually state them! For instance, we have the nice statement that a stable infinity category will admit arbitrary finite limits and colimits (a perfect analog of the fact that an abelian category admits arbitrary finite limits and colimits); but the shadows of this fact in the level of the homotopy (triangulated) category have probably not been fully teased out, seem very complicated to state and understand, and, even for modest diagrams, are probably not a consequence of Verdier’s axioms. For instance, one would end up with a higher dimensional version of the octahedral axiom where one starts with a commutative n-cube instead of a commutative square. (And it’s not just higher axioms either; one has higher structure than just distinguished triangles, in particular a way to “totalize” in a certain sense an arbitrary finite complex; though understanding exactly in what sense is very delicate, or at least I have trouble). Nonetheless, it seems that the triangulated categories one really wants to work with do admit such stable infinity enhancements, and therefore carry all this great extra structure and these great extra properties, which are easily stated in the infinity-categorical language. Great news, as far as I’m concerned!