PCMI/Graduate Summer School 2006/Introduction to Link Homology

The core of these notes were taken by Gabriel C. Drummond-Cole at the PCMI 2006 Graduate Summer School. A link to his original notes can be found in the references.

Lecure 1

It's time to begin. I'd like to welcome you here on behalf of half of the organizers. The other half is still in the air. I have a couple of announcements. First thing is that the problem sessions for today are cancelled because you guys don't know enough yet. The other thing is that there will be an organizational meeting today at 9:40 for the research program. There's a TA meeting at 7:45 tonight. Both of those are here. That's all the announcement. This is our first lecturer, M. Khovanov from Columbia.

I'd like to thank the organizers. I'll give six lectures. I plan to spend the first two discussing braid actions on categories. Most of the time will be spent on one particular action. I want to discuss invariants of braid cobordisms. The last four I'll talk about link homology.

I'm going to do some preparation that will be algebraic but will become more topological as we go on.

We'll start with path rings. Let $\mathrm{\Gamma }$ be an oriented graph, with finitely many vertices and edges. It may have loops. The path ring $\mathbb{Z}\mathrm{\Gamma }$ is the free Abelian group of with all paths as a basis.

So for the graph there are six paths, one of length two, two of length one and three of length zero. Multiplication is by concatenation of paths. Most products in this ring are zero because they don't compose.

We have the paths $a,b,c,\alpha ,\beta ,$ and $\alpha \beta .$ Then we have $(a),(b),(c)$ with $(a{)}^2=0,(b{)}^2=0,(c{)}^2=0$ and then $a(\alpha )=\alpha$ and $\alpha (a)=0$ is $0.$

Ex 1: Note that the sum of the zero length paths over all vertices is the unit.

Note that the path ring on a loop $\alpha$ is $\mathbb{Z}[\alpha ].$

I'd like to say a few words about modules over path rings. I'll talk about $mod-\mathbb{Z}\mathrm{\Gamma },$ the category of right $\mathbb{Z}\mathrm{\Gamma }$-modules. Then $M={\oplus }_{\mathrm{v}\mathrm{e}\mathrm{r}\mathrm{t}\mathrm{i}\mathrm{c}\mathrm{e}\mathrm{s}}M(a).$ Now if you multiply by $\alpha$ you get a map from $M(a)$ to $M(b).$

Ex 2: What are morphisms in this language?

Ex 3: Describe the analogous structure for left modules.

Another thing you can do is, people like to put a field instead of $\mathbb{Z},$ and then this algebra has homological dimension one so every module is projective.

Something about, say $k\mathrm{\Gamma }$ is a hereditary ring (homological dimension one)

We want also to quotient the path ring by relations. If there is more than one path between vertices, you can quotient by the relation ${\gamma }_1+\lambda {\gamma }_2=0.$ This is too general.

The case we care about is zigzag rings, $A_n:$

And then the relations we want to add are $(i|i+1|i+2)=0,(i|i-1|i-2|)=0,$ and $(i|i+1|i)=(i|i-1|i).$ We can write this as ${\delta }_1^2=0,{\delta }_2^2=0,$ and ${\delta }_1{\delta }_2={\delta }_2{\delta }_1$ where ${\delta }_1,{\delta }_2$ are arrows to the right, left, respctively. So $A_n$ modules are the same as bicomplexes, complexes with two commuting differentials. Note also that $A_n\cong A_n^{op}.$

Now $\mathbb{Z}\Gamma$ is graded by path length. The relations quotiented in $A_n=\mathbb{Z}\mathrm{\Gamma }/\sim$ is also graded since the relations are homogeneous.

Any length three path is zero. A subpath which contains two steps to the right or left it's zero, and ${\delta }_1{\delta }_2{\delta }_1={\delta }_2{\delta }_1^2=0.$ So $A_n$ is small. It's a free Abelian group on $(1),(2),\dots ,(n),$ the paths $(i|i+1)$ and $(i|i-1)$ and the paths $X_i=i|i-1|i.$

So from $A_n$ we will get a braid group action by means of the Temperley-Lieb algebra.

Take $P_i=A_n(i),$ which is spanned by paths that end at $i.$ When $i$ is in the middle, there are four such paths, one each of lengths zero and two and two of lengths one: $(i),(i-1|i),(i+1|i),$ and $X_i.$

This is left-projective, because they are direct summands, $A_n=\oplus A_n(i).$

Likewise we can define ${}_{i}P$ as $(i)A_n,$ and this is a right projective module. There is a similar decomposition. If you take ${}_{i}P{\otimes }_{A_n}{P}_j$ you get an Abelian group spanned by paths that start at $i$ and end at $j.$ Most of the time this is zero, that is if $i-j|>1.$ It's $\mathbb{Z}(i|j)$ if $j=i\pm 1.$ If $i=j$ there are two paths, the zero path and $X_i$ so the tensor product is $\mathbb{Z}(i)\oplus \mathbb{Z}{X}_i.$

Introduce $U_i=P_i{{\otimes }_{\mathbb{Z}}}_{i}P.$ This is an $A_n$-bimodule. If you tensor $U_i{\otimes }_{A_n}{U}_i.$ So you get $P_i{\otimes }_{i}P{\otimes }_{A_n}{P}_i{\otimes }_{i}P.\cong (P_i{\otimes }_{i}P)\oplus (P_i{\otimes }_{i}P)$ where these are for $(i)$ and $X_i.$ So this is $U_i\oplus U_i.$

Then $U_i\otimes U_{i+1}\otimes U_i$ decomposes into six terms and what's left will be $U_i.$ Similarly for $i-1.$ When you tensor $U_i$ and $U_j$ when they are far away from one another ($>1$) you get zero.

These relations are the Temporley-Lieb algebra with some additional relations.

This algebra is generated by diagrams $u_i$ where there are $n-1$ vertical lines and cups and caps in positions $i,i+1.$ You say a circle gives multiplication by $q+q^{-1}.$ The less interesting picture is to set $q=1$ and multiply by $2.$ So $u_i^2=2u_i.$ If you multiply $u_i{u}_{i+1}{u}_i$ you get $u_i$ isotopically. You have $u_i{u}_j=u_j{u}_i$ for $i,j$ far from one another, but you don't get that these are equal to zero.

This is just an algebra with generators and relations, but it has this nice geometric picture. Now you are taking tensor products instead of multiplications, and the equalities are isomorphisms. So we're lifting the equations of the Temperley-Lieb algebra.

The braid group has generators ${\sigma }_i$ which go to $0\to U_i=P_i{\otimes }_{i}P\to A_n\to 0.$ This middle map ${\beta }_i$ takes $a\otimes b$ to $ab\in A_n.$ I'll talk about this at length.

Let $A$ be any ring. Denote by $Kom(A)$ the category of complexes of $A$-modules. We take the quotient category, (the principle is that you have to modify morphisms) with the same objects, but where $f=g$ in the quotient category if the difference is nullhomotopic. What this means for complexes of modules is:

In this diagram, we have $t$ is nullhomotopic if there exists $h$ with $t=dh+hd$ Ex 4: Nullhomotopic morphisms form an ideal in $Kom(A).$

So $C(A)$ is the quotient category. $Ho{m}_{C(A)}(M,N)=Ho{m}_{Kom(A)}(M,N)/\sim .$ Ex 5: For any $K,0\to K\stackrel{1}{\to }K\to 0$ and $0\to 0\to 0\to 0$ are isomorphic in $C(A).$

Let $[1]$ be the left shift operator, namely let $M[1{]}^i=M^{i+1}$ and $d_{M[1]}=d_M.$

Then let the cone $C(f)$ for $f:M\to N$ be $M[1]\oplus N.$ with differential $D=-d_M+f+d_N.$

For $N$ a bimodule and $M$ a module we have $N{\otimes }_{A}M$ a module, so $N\otimes$ is a functor from $A-mod$ to $A-mod.$ If $N$ and $M$ are complexes of bimodules, modules, resepectively, then $N\otimes M$ is a complex of modules via

just taking the total differential.

Theorem: There are isomorphisms $R_i{\otimes }_{A_n}{R}_{i+1}\otimes R_i\cong R_{i+1}\otimes R_i\otimes R_{i+1}.$ \\ $R_i\otimes R_j\cong R_j\otimes R_i$ for $|i-j|>1.6,$ and \\ $R_i\otimes {R_i}^{\prime }\cong A_n\cong {R_i}^{\prime }\otimes R_i.$

Ex 6: We need ${R_i}^{\prime }$ to be the complex $0\to A_n\stackrel{{\gamma }_i}{\to }{U}_i\to 0$ with ${\gamma }_i(1)=(i-1|i)\otimes (i|i-1)+(i+1|i)\otimes (i|i+1)+X_i\otimes (i)+(i)\otimes X_i$ representing the inverse of ${\sigma }_i.$

The theorem fails in $Kom(A-bimodules)$ but holds in the homotopy category of complexes of bimodules. \\ This action is faithful.

The tensor product can be viewed at a functor $C(A_n)\to C(A_n),$ and so equivalences in the braid group become isomorphisms of functors.

This is striking, why would we have an action and why would it be faithful? This is related to tricomplexes. We have $A_n$ with ${\delta }_1,{\delta }_2,$ and when we deal with complexes we have the additional $d.$ So we have tricomplexes up to homotopy and in this category we have a faithful action of the braid group.

So far beyond this, in four-complexes, I have seen nothing interesting. Bicomplexes correspond to spectral sequences and tricomplexes to the braid group action. I don't know about four.

References

• Notes
  • Gabriel C. Drummond-Cole's Lecture Notes

• Articles
  • A categorification of the Jones polynomial, Duke Math. J. 101 (2000), no. 3, 359--426 (Mikhail Khovanov)

• • Quivers, Floer cohomology, and braid group actions (Mikhail Khovanov and Paul Seidel), J. Amer. Math. Soc. 15 (2002), no. 1, 203--271 (electronic)

• • Matrix factorizations and link homology (Mikhail Khovanov and Lev Rozansky)

• • Matrix factorizations and link homology II (Mikhail Khovanov and Lev Rozansky)