Posted by: Alexander Ellis | December 22, 2008

## Wood, glue, and the Octahedral Axiom

Arts and crafts and mathematics have met before. For instance, mathematical knitting isn’t new. I’d like to talk about my humble (read: shoddy) contribution.

An important structure in homological algebra, especially when talking about things up to (co)chain homotopy,  is that of a triangulated category.  If you haven’t seen triangulated categories before, Wikipedia is decent, but Chapter 10 of Weibel’s Introduction to Homological Algebra is better. Here, I’m just going to be concerned with the octahedral axiom, whose statement is notoriously hard to picture. (Throughout this post, I’ll be using the formulations and notations of Weibel.)

Let $\mathbf{K}$ be an additive category and let $T:\mathbf{K}\to\mathbf{K}$ be an automorphism of this category. A triangle in $\mathbf{K}$ is an ordered triple $(A,B,C)$ of objects along with three morphisms: $u:A\to B$, $v:B\to C$, and $w:C\to TA$. We write $\lbrace(A,B,C),(u,v,w)\rbrace$ for short. A triangulated cateogry is an additive category $\mathbf{K}$ equipped with an automorphism $T$ and a distinguished family of triangles which are called the exact triangles. We usually draw an exact triangle as shown in the diagram to the right. These data must satisfy four axioms, the last and most confusing of which is the following.

Octahedral Axiom: Let $\mathbf{K}$ be a category equipped with an automorphism $T$. Suppose we have three exact triangles: $\lbrace(A,B,Z),(u,j,\partial)\rbrace$, $\lbrace(B,C,X),(v,x,i)\rbrace$, and $\lbrace(A,C,Y),(vu,y,\delta)\rbrace$. Then there is a fourth exact triangle $\lbrace(Z,Y,X),(f,g,(Tj)i)\rbrace$ such that $\partial=\delta f$, $x=gy$, $yv=fj$, and $u\delta=ig$.

This is usually pictured as an octahedron not unlike the image on the right, which I’ve shamelessly re-LaTexed from a diagram in section 10.2 of Weibel. (The picture looks better there.) In this view, the equalities of morphisms in the Axiom take on geometric significance: the equalities $\partial=\delta f$ and $x=gy$ mean that the four non-exact faces all commute (two did automatically), and the equalities $yv=fj$ and $u\delta=ig$ say that the two shortest paths between $B$ and $Y$ in either direction agree. (There is an asymmetry here: the other two pairs of antipodal vertices do not have any such paths to compare.)

There is another common visualization, a planar one involving line segments consisting of two morphisms each. See Weibel or Wikipedia.

There are a few annoying drawbacks to the octahedral diagram above:

• At a glance it’s not immediately clear which triangles commute and which are exact.
• At a glance it’s not immediately clear which three exact triangles were given and which one was implied by the Axiom.
• Some arrows involve the objects $A$, $B$, and $Z$, while some involve $TA$, $TB$, and $TZ$. But in the diagram, we only have room for one or the other, so we omit the $T$. Since the convention for drawing an exact triangle already takes this into account, this no problem when considering exact triangles; but when we view the commuting triangles, we have to keep in mind this convention of writing $B$ in place of $TB$.

I like to see things clearly, immediately. I don’t to have to think about which triangles commute and which are exact, or whether an arrow is really hitting $B$ or $TB$. All the problems above are fixed by my small pet project for this past weekend. Check out the eye candy (in the second picture, you can kind of see six of the eight sides):

Other than the shoddy workmanship, here are some things to notice about the model and why it improves on the diagram above:

• It’s 3D! You can hold it (but please, be gentle)!
• Blue sides are exact triangles; yellow sides are commutative triangles. The “output” exact triangle has been labeled (though not in the pictures above). So the meanings of the triangles are obvious at a glance.
• If a map goes to or from $TB$ instead of $B$, then the vertex corresponding to $B$ is labeled as $B$ when part of an exact triangle (following the usual convention), but as $TB$ on the yellow, commuting faces. This is my favorite feature of the model.