Michi’s blog » Triangulated categories

## Triangulated categories

• July 17th, 2006
• 3:47 pm

One predominant tendency in the algebra/category theory camp is to seek out the minimal set of conditions needed to be able to perform a certain technique, and then codifying this into a specific axiomatic system. Thus, you only need to verify the axioms later on in order to get everything else for free.

One such system is the theory of triangulated categories. This pops up in homological algebra; where you like to work with Tor and Ext – both of which turn out to be derived functors, generalizing the tensor product and the homomorphism set respectively. With the construction of the derived category, we can find a category, in which the tensor product in that category is our Tor, and the hom sets is our Ext.

Once that entire yoga is worked through, you could start backtracking, and pulling out all properties you used. Minimizing this set of properties in one specific way leads to the concept of a triangulated category, and in this post, I intend to retrace, using the definition of a triangulated category, and looking more specifically on the case of the derived category of the category of chain complexes of modules over a fixed ring for a canonical example.

## Triangulated category

Weibel gives a definition, due to Verdier, of a triangulated category. It is an additive category – meaning that each Hom(A,B) is an abelian group, and the group operation distributes over composition of morphisms – equipped with an automorphism T called the translation functor, and with a distinguished family of triangles (u,v,w) – by which we mean , and . We further expect these to fulfill the following axioms

1. Every morphism u can be embedded into an exact triangle (u,v,w). Furthermore, (1,0,0) is exact and exactness is closed under isomorphisms of triangles.
2. If (u,v,w) is an exact triangle, then so is (v,w,-Tu) and (-T-1w,u,v).
3. If (u,v,w) and (u’,v’,w’) are exact triangles with f and g such that gu=u’f, then there is a morphism h such that (f,g,h) is a morphism of triangles. In clear text, this means that the following diagram has the dotted arrow, such that everything commutes:

Note that the leftmost square commutes by the condition on f and g.
4. Suppose we have exact triangles through the triples A,B,C’, A,C,B’ and B,C,A’. Then these determine an exact triangle on C’,B’,A’. This is, due to one of the ways to visualize it, called the octahedral axiom.

Now, for an exact triangle (u,v,w), we can form the following diagram

where the morphism exists due to the axiom (3). Thus, since the diagram commutes, vu=0.
By forming similar diagrams for all the rotations, we also get wv=0 and (Tu)w=0. Thus, exact triangles behave as expected, if we view them as representatives for long exact sequences.

## Chain complexes form a triangulated category

Proposition: For the category of chain complexes of R-modules, with homotopy equivalence classes of chain maps as morphism, we have a triangulated category structure.

Proof:
We set TA=A[-1], so that TAi=Ai+1. We further call a triangle (u,v,w) exact if it is isomorphic to a triangle (u’,v’,d) on the complex triple (A’,B’,cone(u)). That is, the exact triangles are (up to isomorphism) the triangles that go through a mapping cone.

### Digression: The cone of a chain map

Suppose is a chain map. We define the mapping cone cone(f) as the complex , with degree n part and with differential .

Claim: The cone of the identity map cone(1) is split exact, with splitting map .
Proof: In order to verify this claim, we need to check that dsd=d and that the complex is exact. For exactness, note that

and thus it is the set of pairs (dc,c) for c in Cn

so this is also a set of pairs (dc,de-c); and as we keep c fixed and let e run through all possible values, we still will get pairs that are already represented in our description of the kernel. So these sets are equal, and thus the cone is exact.
For split exactness, we calculate
dsd(c,c’)=ds(-dc,dc’-c)=d(c-dc’,0)=(dc,dc’-c)=d(c,c’)
and thus everything behaves as it should, and we can conclude that cone(1) is split exact.

Claim: is null-homotopic if and only if f extends to a map .
Proof: Suppose, first, that f is null-homotopic. Then there is some such that f=ds+sd. Then the map (c,c’)\mapsto f(c’)-sc is a chain map [Unparseable or potentially dangerous latex formula. Error 2 ] which restricts to f on the second summand . This holds, because and will be the two corresponding paths through the corresponding commutative square. Now, it remains to show that df(c’)-dsc=sdc-f(c)+fd(c’). But fd=df since f is a chain map, and sd+ds=f implies that -ds=sd-f, so -dsc=sdc-f(c). This shows one direction.

For the other direction, suppose we can extend f to a chain map . Then the calculation above shows that thus df(c’)-fd(c’)+f(c)-dsc-sdc=0. But f still is a chain map, so the df(c’)-fd(c’) portions disappears. And the remaining equation says that for any c in C, we need f(c)=(ds+sd)(c). Thus the s used in extending f to a chain map from the cone is precisely a contracting homotopy.

Given a map , we can construct a short exact sequence

with the maps and . This is a s.e.s. of chain complexes, since the calculations and , for the first map and and for the second map show that all relevant squares commute. Remember, thereby, that the differential on A[-1] gains a sign, due to the shift.

This last section gives the prototype for our exact triangles. Anything isomorphic to this is exact. Suppose is a chain morphism. Then this fits into the triangle for and . Furthermore, since the mapping cone of the identity map is split exact, this splitting map gives a null homotopy of any map to that cone. So the cone is isomorphic to the zero object in this category, and thus the identity map has the required embedding. By defining exact triangles as all triangles isomorphic to a triangle with a cone, it’s clear that isomorphism of triangles preserves exactness.

Rotation of triangles follows after working out that the maps appearing when matching the rotation to what it should be are chain homotopy equivalences. I’m not going to do this here.

The existence of morphisms follows since the construction of a mapping cone is natural.

And right about now, I’m slowly growing bored by the verification of axioms. The octahedral axiom seems (by a quick readthrough) to be a diagram chase to find the right homotopies and the right maps to get to the end.

## The funky stuff

So, what are these things good for? The first meatier construction we’ll see is that of a long exact sequence in cohomology. For a triangulated category K and an abelian category A, we say that an additive functor H from K to A is a cohomological functor whenever all exact triangles have long exact sequences in cohomology. The cohomology functor is the canonical example.

Having this machinery, we will – in future posts – strive to construct a derived category, in which the quasi-isomorphisms are inverted by a localisation process; and thus the objects are isomorphic when they have induced isomorphisms on cohomology. Work in the derived category is very reminiscent of work with derived functors – with Ext and Tor.

### 1 Person had this to say...

• Ulf
• July 20th, 2006
• 19:07

Hej !

Såg din website… som jag inte visste fanns. Själv har jag http://ulfpolitik.blogspot.com/