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\author{Gabriel C. Drummond-Cole}
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\title{Dennis Seminar\\February 14, 2007}
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\maketitle
I want to talk about some nonlinear structures on chain complexes, e.g., $C\otimes C\to C$ or $C\to C\otimes C.$  Multilinear maps like that, not really linear as a function of $C.$  There's an interesting thing in analysys, you take two functions, you can think of them as tensors, and then multiply them, formally you have the idea that the derivative satisfies
$$
D(f\cdot g)= Df\cdot g + f\cdot Dg
$$
Suppose these are measurable functions and I can multiply them.  If $f,g,Df,Dg$ are bounded, well $f\cdot g$ is bounded and then $D(f\cdot g)$ is bounded.  So the space of almost everywhere bounded functions with almost everywhere bounded derivatives is an algebra.

Suppose we modify this a little bit, we could do $f$ and $g$ bounded and $Df$ in $L^2.$  That would be another algebra.  If you put like $f,g\in L^2$ and $Df\in L^2,$ you would have $f\cdot g\in L^1$ only.  So a good one would be $f,g$ bounded and $Df\in L^p.$  Certain spaces in analysis form algebras, and when you're studying a nonlinear PDE like 
$$
\frac{d}{dt}U(x,t)=U(x,t)\frac{d U}{dx},
$$
then the right side might not be in the same space for all $t.$

Let's look at a geometric example, in cell complexes.  These nonlinear structures that are models of this analysis, if you put grids down and take averages of your continuum objects over these grids, you get discrete objects that approximate objects, you can always multiply your discrete objects but then new features appear because you can't really multiply pointwise.  This whole world of nonlinear structures on the chain complex can have bearing on the analysis.

You have chains and the chain complex $C_.$ with $\boundary,$ and you always have the Hom dual picture, $C^.,\delta$ which is $Hom_R(C_., R).$  There are restricted Homs, if its infinite dimensional you can take formal linear combinations in terms of a dual basis and look at restricted Homs.  I'm thinking of things that are finite now so this difference doesn't matter.  

Then you have the pair subdivision, which is a nice amalgamation of the barycentric subdivision.  It's, it looks like this.  The combinatorics and the shapes of the pieces is bounded.  On a vertex, how many things come into it, and how many things are around an edge?

It has this nice notation, if the cells are $\eta,\eta',\eta'',$ I'm assuming this is regular so that the boundary of each cell is a closed ball.  Then if this list is of the cells of $C,$ the pair subdivision $(P_.,\boundary)$ are pairs $(\eta,\eta')$ such that $\eta'\subset \eta$ and the codimension of $\eta'$ is the dimension of the pair.  Then the boundary of the pair is 
$$\boundary (\eta,\eta')=(\delta \eta,\eta')+(-1)^{|\eta|}(\eta,\boundary \eta')$$$$
\delta (\eta,\eta')=(\boundary \eta,\eta')+(-1)^{|\eta|}(\eta,\delta \eta')$$

When you take the cell decomposition of a manifold and then the dual cell decomposition, which gives you a decompositon transversal to the first one, if you intersect these two, this is the smallest simultaneous refinement of both of them, it's the pair subdivision of both of the decompositions.  That makes it somehow canonical.  A cell $\sigma$ has dual $(\sigma, K).$

[I wanted to ask a slightly naive question, here in the formula for the boundary of the pair complex I need the coboundary, that's already identifying with the dual.]

In a cell complex one often identifies chains and cochains, except in the infinite case you have to say some words.

[Are you implicitly using an orientation on edges?  You could just take these geometrically.]

The choices of orientation are all canonically isomorphic.  You take the free Abelian group on pairs of cells and orientations subject to $(C,o)=-(C,o^{op}).$  The boundary operator is $\boundary (C,o)=(\text{face},\text{induced orientation}).$  All of the choices are canonically isomorphic to this construction.

[What about geometric pairs?  Then boundary is geometric boundary.  Every triangle has three pieces.]

But then boundary boundary won't be zero.  Okay, so what is homology?  It's different for each triangulation.  So a homology element assigns to each triangulation an element of its homology so that given any two triangulations the cells are related by the canonical isomorphism.

For me, the cell decomposition is like choosing a Riemannian metric.  Given that choice you want to have canonical constructions.  There are many choices here.  There is a canonical vector space and a canonical map.  These vector spaces don't have bases.  Since you're over the integers, you have a natural integer structure but no basis element.  You go and choose a basis and then when you write out the canonical linear operator in terms of this basis, you get the oriented chain complex.

You can do exactly the same thing with the coboundary.

[Is this analagous to saying that if you have a basis, you have the dual basis?]

Sometimes you say basis up to sign, which are given by the cells.  You could refine that to an actual basis.  You can let the dual guy be one or minus one on your basis element.

All right.  Mahmoud provoked that by asking what I meant by these equations, and the answer is that they are all canonically isomorphic.

There is another feature that when you take a subdivision you can induce orientations.  

So Steenrod opened my eyes to say that there is an actual set which is the universal homology, not just the isomorphism class of a group.  So say you have a construction that has choices, you have an output that is logically different but canonically isomorphic, then you take invariant sections. 

By the way, there is no way to do this for spinors.  There is no canonical set of spinors.  That's one reason they're kind of deep.

Okay, so let's move on to the product structure.  There's a nice formula for the product.  You define $$(\eta,\alpha)(\beta,\nu')=\left\{\begin{array}{l}0\text{ if }\alpha\ne \beta\\(\eta,\nu)\text{ if } \alpha=\beta\end{array}\right.$$
[Picture.]

If you just work with the singular complex there's another formula.  The singular complex is great for algebraists.  You have coordinates and you just have one extra coordinate.  So $\langle \nu_0,\ldots \nu_k\rangle$ tells you what a simplex is.  Then a linear function on that is a singular cochain.  Then $$
\varphi_k'\cup \varphi_\ell''\langle \nu_0,\ldots, \nu_k,\nu_{k+1},\ldots,\nu_{k+\ell}\rangle
=
\varphi'\langle \nu_0,\ldots,\nu_k\rangle\varphi''\langle nu_k,\ldots, \nu_{k+\ell}\rangle.$$

This is called the Alexander Whitney product and is the only concrete formula that people use.  You can see they're kind of similar.  I use a particular vertex twice.

An interesting question is to ask when you can write down a formula for the cup product.  If you write down all the vertices, you can do it.  There's some precise combinatorial data you need.

These are violently noncommutative.  When you write forms you symmetrize somehow to get something graded commutative.  This is like one of the terms in the multiplication of forms.

This is compatible with the coboundary but not with the boundary.  There's another formula, a similar formula, a map $P_.\to P_.\otimes P_.,$ call it $\Delta,$ and 
$$\Delta(\eta,\nu)=\sum_{\eta'}(\eta,\eta')\otimes (\eta',\nu).$$  This is a chain mapping for $\boundary.$  These two things, $\cdot$ and $\Delta$ are transposes of one another.  One is the dual of the other.  This type of formula often comes up.

[Discussion of differential geometry.]

Now the next thing to note is that $\ker \delta$ is a subalgebra and the image of $\delta$ is an ideal.  So the cohomology is an algebra.  Unfortunately, in the homology case, you get
$$
\xymatrix{\ker (P_.\otimes P_.)\\
\ker\ P_. \otimes \ker\ P_.\ar[u]&&\ker\ P_.\ar@{.>}[rr]\ar[urr]}$$
So you don't have the necessary map for homology to be a coalgebra.

I'm ending early.  Next time I want to discuss the well-definedness of this, of the cohomology product, and the commutativity of the cohomology product, the graded commutativity.  This has a lot of interesting ramifications.  One thing you might think about, there's no class next week.  We're going to have a quantum field theory day.  Next week there are talks from 1-2 (Terilla), 2:30-4:15 (Kontsevich) and 5-6 (Traedler).  There is lunch (11:30-12:45) and dinner (6:30 to whenever).

[I would like to understand the business of having a commutative associative product at the chain level over $\zZ$ and $\qQ.$  I would also like to understand the Massey products, higher order products, and Steenrod operations on cohomology.]

[What special features does this have when your space is a manifold? How about a Lie group?] 

\section{John Terilla}

So I just want to remind you that if you have a differential graded Lie algebra $(V,d,[\,,\,])$ and a parameter ring $B$ then you can look at the following set: $$\{\gamma\in (V\otimes m_B)^1 :d\gamma+\frac{1}{2}[\gamma,\gamma]=0\}/\sim$$
You could replace $m_B$ with $B$ but keep in mind the following.  If $d\gamma+[\gamma,\gamma]=0$ then $d_\gamma=d+[\gamma,\,]$ is a differential from $V\otimes B$ to itself, $d_\gamma^2=0,$ and $d_\gamma$ is a derivation of the bracket.  If you set the parameters equal to zero, this should return to $d.$  That's equivalent to saying that $\gamma$ comes from the maximal ideal, since the augmentation exactly kills the maximal ideal.  What I was reminding you of was that given the differential graded Lie algebra $L$ one gets a functor $Def_L$ from the category of parameter rings to the category of sets.  You look for the set of solutions given that parameter ring.  In general $Def_L(B)$ is only a set but $Def_L(k[t]/t^2)$ is a vector space, identified with $H^1(L)$ (assume $|t|=0$)

Suppose $\epsilon\in k[t]/t^2$ and $m_\epsilon=tk[t]/t^2,$ with the degree of $t$ equal to $-i.$  Then $\gamma\in (V\otimes m_\epsilon)^1$ is o form $\alpha\otimes t=\alpha t\in V^{1+i}.$  Then $$d\gamma+\frac{1}{2}[\gamma,\gamma]=d\alpha t + \frac{1}{2}[\alpha,\alpha]t^2=d\alpha t$$
So then $Def_L(\epsilon_i)=H^{i+1}(L)$  The isomorphism is $[\gamma]=[\alpha t]\to [\alpha].$

Okay, so here you get this functor from parameter rings to sets, and what I wanted to say was, one, there is this pattern of extending the parameter rings.  Algebraically, this seems harmless.  But there are great repercussions.  One is to go from these Artin algebras $\Aa$ to $\Aa^\zZ,$ graded Artin algebras.  This can make a big difference.  Doing this for deformations of associative algebras give deformations in the categories of associative algebras and $A_\infty$ algebras, respectively.  Now let's look at $d\Aa^\zZ.$

For deformations of a complex structure, first you take sections of exterior powers of the tangent bundle tensor exterior powers of the antiholomorphic part of the cotangent bundle.  So these are like $\frac{d}{dz}\otimes d\bar{z}.$  This bracket is the product of the bracket on the first half and the product on the second half.  The differential is $\bar{\partial}.$  So I wanted to discuss two things.  There's a pattern of extending the parameter ring.  You can pass to graded or differential graded parameter rings.  There are two things to say about that.  In general, there may be obstructions to solving the Maurer-Cartan equation in general.  But adding a differential structure to the parameter rings allows you to bypass the obstructions.  You can encode the obstructions in a differential to go on.

Let me do an example of that.  Let $B\to B'$ be a surjection of differential graded parameter rings.  Call the map $\tau.$  Then we call $\tau$ small or a small extension if for $J=\ker\ \tau$ we have $Jm_B=0.$  So this is an example: $t^2 k[t]\to k[t]/t^3\to k[t]/t^2.$  The problem then is, is the map $Def_L(B)\to Def_L(B')$ surjective?  Can you extend a solution over $t^2$ to a solution over $t^3?$  This amounts to extending a solution to the next level.  In general, if the answer is yes, if $\tau_*$ is surjective for all small extensions $\tau$, then $Def_L$ is called smooth.  In this case, you can represent $Def_L$ which is equivalent to saying that there exists a universal solution to the Maurer Cartan equation, namely a $\Gamma\in L\otimes m_{SH^*}.$  You can access this property just by looking at little extensions.  This is because every finite dimensional Artin algebra can be obtained from a trivial one in this way.

It turns out that $Def_L$ is not smooth in general but is quasismooth.  This is called smooth because the moduli space of solutions is smooth.  The moduli space is a 	``formal pointed space.'' The points are maximal ideals.  Think of affine functions on the line.  Polynomials in one variable are smooth.  So I think, you have a point, there may be a direction in which you can't move.  I'm sort of replacing a space by a functor and the tangent space with the tangent space to the functor, its value on the ring of dual numbers.  This is the right idea of a smooth space.

So the structure of $\Gamma$ is
$$t^\alpha x_\alpha+t^\alpha t^\beta x_{\alpha\beta}+t^\alpha t^\beta t^\gamma x_{\alpha\beta\gamma}+\ldots$$
where $\{[x_\alpha]\}$ is a basis for $H(L)$ and $t^\alpha$ a dual basis for $H^*.$  Then $[x_\alpha]\to x_\alpha, [x_\alpha]\wedge [x_\beta]\to x_{\alpha\beta}$ and so on are an $L_\infty$ morphism $(H,0)\to L=(V,d,[\,,\,])$

Recall that an $L_\infty$ morphism is a pair $(V,D)$ where $D$ is a coderivation on the free symmetric algebra on $V[1].$  So $S^n V[1]\cong \wedge^n V[n].$

A coderivation is determined by $d:V\to V,$ and then $[\,,\,]:S^2 V\to V,$ and then lift to $SV.$  Then $\tilde{d}+\widetilde{[\,,\,]}$ is a coderivation, and it squaring to zero means that $d^2=0,$ the bracket satisfies Jacobi, and $d$ is a derivation of the bracket.  A morphism of $L_\infty$ algebras $(V,D)\to (V',D')$ is a coalgebra map $\sigma:SV[1]\to SV'[1]$ that commutes with $D,D',$ so $\sigma D=D'\sigma,$ of degree zero.  This is actually a nice, there's a nice thing to say.  Let $B$ be a parameter ring.  If $\gamma\in V\otimes m_B,$ this is the same as, this gives a map $\gamma:m_B^*\to V$ by linear algebra.  I'm going to recharacterize solutions.  Now in the finite dimensional case $m_B^*$ is a coalgebra, and this is a linear map of degree one of vector spaces. By the universal property of coalgebras, this lifts to a degree zero map to $SV[1].$  As a free construction it comes from the filler $V[1]\gets SV[1].$  Any linear map lifts to a coalgebra map to $SV.$  So $\gamma$ gives a coalgebra map $\tilde{\gamma}:m_B^*\to SV[1].$  I didn't say anything about it satisfying the Maurer-Cartan equation.  Then $\gamma$ satisfies the Maurer-Cartan equation in $V$ if and only if $\tilde{\gamma}$ is a map of differential coalgebras $(m_B,0)\to (SV[1], D=\tilde{d}+\widetilde{[\,,\,]}),\ D\tilde{\gamma}=0.$  So solutions are like sets of differential coalgebra maps.  

In the case of the universal solution, then the parameter ring is $SH^*$ which is completed, a free commutative coalgebra.  Then $\Gamma$ gives $\tilde{\Gamma}$ which is a map $(SH[1],0)\to SV[1],D)$ which by definition is an $L_\infty$ morphism.  The use of these rings, the duals of the maximal ideals are like finite dimensional approximations of $L_\infty$ algebras.  Compare an $L_\infty$ algebra with the dual of an Artin algebra.  One is finite dimensional, the other is free.

Let me say, this gives a very simple way to extend the parameters to differential parameters.  Let $\Aa^\zZ_d$ be the category of differential graded parameter rings.  Then $Def_L:\Aa^\zZ_d\to Sets$ is given by $B\mapsto \{\text{differential coalgebra maps }m_B^*,d\to SV,D\}$ up to equivalence.  I can define equivalence with homotopies of coalgebra maps.  These are not $L_\infty$ morphisms.

[I give an algorithm for finding the formula for the homotopy, in Infinitessimal Computations.]

This equivalence is interesting, but if I say this, then there is the following analogy, I said what it was for the functor to be smooth.  Here I can say what it is to be quasismooth.  We call a small extension $\tau:(B,d)\to (B',d')$ acyclic if $H(\ker\ \tau)=0.$  It's not necessary to go to this point of view.  This is useful when you go beyond $L_\infty$ algebras.  Then the functor $Def_L$ is quasismooth if and only if $Def_L(B)\to Def_L(B')$ is surjective for all small acyclic extensions.

Let me put something down that will illuminate.  Suppose that $dx=0$ and that $[x,x]$ is not exact.  Then you cannot extend the solution to the Maurer Cartan equation.  So $tx$ is a solution in $k[t]/t^2$ but there exists no solution $tx+t^2y$ in $k[t]/t^3.$  You don't have that.  Here's a way of getting around that.  Let $B=k[s,t]/s^2,st,t^3,$ and let $ds=-\frac{1}{2}t^2,dt=0.$  Then $B'=k[t]/t^2.$  Then there is the natural surjection $B\to B'.$  The kernel is $sk\oplus t^2k.$  The closed elements are $t^2,$ but that's exact.  So what I'm doing is, I have only $t$ in $B',$ and in $B$ I have $s,t,t^2.$  So then $tx+s[x,x]$ is a solution to the Maurer Cartan equation in $V\otimes B.$   I've added a differential, so the equation is $$d_V\gamma+d_B\gamma+\frac{1}{2}[\gamma,\gamma]=0$$
So for this $\gamma$ what you get is 
$$d_V(tx+s[x,x])+d_B(tx+s[x,x])+\frac{1}{2}[tx+s[x,x],tx+s[x,x]]=0.$$  You cook up, if you had an obstruction, you add a new variable and encode the obstruction into the differential.  The theorem is, if you have a functor which is smooth, then it's represented by a ring with no differential.  If it's quasi-smooth, then it's represented by a ring with a differential.

If you want to go all the way to the universal solution, you take $x_\alpha$ a basis, and you have $[x_\alpha,x_\beta]=c_{\alpha,\beta}^\delta x_\delta +dx_{\alpha\beta}.$  Then define $d:SH^*\to SH^*$ starting with $t^\delta\mapsto c_{\alpha\beta}^\delta t^\alpha t^\beta+\ldots,$ and the further terms will be determined by the Massey products.  Because this is a Lie bracket, this $d$ squares to zero.  So the universal solution will be $t^\alpha x_\alpha+t^\alpha t^\beta x_{\alpha\beta}+\ldots$ and it will satisfy $$d\Gamma+\frac{1}{2}[\Gamma,\Gamma]+d_H\Gamma=0.$$
I guess I will stop.
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vector space with an inner product, you have two ring structures on the exterior algebra.


( a==> n ) ---> na
 (-a =>-n) ---> na 


ab cd --> hopf thing
ax xb cy yd
axcy xbyd
b=y, x=c
ab cd if c<b

ab cd --> D*
ax xd if b=c

ab cd --> *D
ax xd if b=c, x<=b


ab --> *D
#N ab
