# The Cohomology Computation

Alright, I’m in a sort of tough spot. Yesterday I started typing this up, but I just don’t have the motivation. There are lots of tedious details that no one is going to read and will not come up in our study after this. It is all incredibly standard chasing Fourier coefficients around, so I’m not going to do it. This post will be an outline in how one would go about doing it, and I even may provide quick ideas behind it, but if it will be weeks before I continue on if I don’t just get through this. Someday, if it seems important, I’ll come back and fill it in. Or if someone comments and really wants to see one particular part, at least I’ll have motivation that someone is going to read it.

Here goes. Recall some of my conventions. $X$ is a compact complex Lie group. We let $A=\Gamma(X, \mathcal{C})$ the global sections of the sheaf of $C^\infty$-functions on $X$. This was important to the Dolbeaut resolution. $\overline{T}$ are the $\mathbb{C}$-antilinear functionals. We use $\bigwedge^q$ to mean $\bigwedge^q(\overline{T})$.

We’ve shown that $A\otimes_\mathbb{C}\bigwedge^q$ is isomorphic to $\Gamma(X, \mathcal{C}^{0,q})$ as complexes and hence we have the iso $H^q(X, \mathcal{O}_X)\simeq H^q(A\otimes_\mathbb{C}\bigwedge)$.

Now we want to show that we actually have an induced isomorphism on cohomology from the inclusion map $i: \bigwedge \hookrightarrow A\otimes \bigwedge$. We’ll do this by comparing Fourier series. So we set up a normalized measure on $X$ say $\mu$. By integrating functions against this measure we get linear function $\mu_{\wedge}$ (I called this something different last time).

Now we need the lemma that for all $\omega\in A\otimes \bigwedge^q$ we have $\mu_\wedge (\overline{\partial} \omega)=0$. This follows from periodicity and translation invariance of the vector field that comes up when you go to write it down.

The next step is to define the “Fourier coefficients” of a function. Now if we choose any integer valued function from the lattice defining $X$, say $\lambda$. Then it extends to an $\mathbb{R}$-linear function on the tangent space at the identity, $V$. We then have the exponentiation map $\displaystyle v\mapsto e^{2\pi i \lambda(v)}$. This respects the lattice and hence descends to $\latex X$. Call this map $c_\lambda$.

Define the $\mathbb{C}$-linear function $A\to \mathbb{C}$ by $Q_\lambda(f)=\int_X c_{-\lambda}fd\mu$. Don’t be intimidated here. This is just the standard Fourier coefficient when your in a familiar situation. i.e. $f=\sum e_\lambda \otimes Q_\lambda(f)$.

Now choose a Hermitian inner product, which gives us a norm to work with. Here things will become less detailed. One can next prove a lemma that the map $f\to \{Q_\lambda(f)\}_\lambda$ is an isomorphism $A\to$ the space of maps decreasing at infinity faster than $\|\lambda \|^{-n}$ for all $n$.

Then do a computation to see that $Q_\lambda (\overline{\partial}\omega)=(-1)^p 2\pi i \left(Q_\lambda(\omega)\wedge \overline{C}(\lambda)\right)$.

Lastly we want to get back to showing that the inclusion is a homotopy equivalence. Thus use the Hermitian inner product to define a map $\lambda^*\in Hom_\mathbb{C}(\overline{T}, \mathbb{C})$ for every $\lambda$ by $\displaystyle \lambda^*(x)=\frac{\langle x, \overline{C}(\lambda)\rangle}{2\pi i \|\overline{C}(\lambda)\|^2}$.

We need to define for $\omega\in A\otimes \bigwedge^p$ a map $k(\omega)$ which will only be defined in terms of its Fourier coefficients. We define $Q_\lambda(k(\omega))=(-1)^p \lambda^* \neg Q_\lambda(\omega)$ if $\lambda\neq 0$ and the coefficient is 0 if lambda is 0. Now it has all been set up so that comparing Fourier coefficients on $\overline{\partial} k + k \overline{\partial}$ we get exactly the same ones as in $id_{A\otimes\wedge} - i \mu_\wedge$. Thus we are done by the magic of Fourier coefficients being unique.

That last computation I left out requires use of things such as “Cartan’s magic formula” and breaking it into two cases. Anyway, for not doing any details, I think this is a pretty thorough outline and filling any of the details you don’t believe or would like to know shouldn’t be too hard.

# Cohomology of Abelian Varieties II

Two posts in the same week! Before we get started today, we need to introduce one more piece of new notation. Let ${\overline{T}}$ be the ${\mathbb{C}}$-antilinear maps ${V\rightarrow \mathbb{C}}$. Our goal is to prove that ${H^q(X, \mathcal{O}_X)\simeq \bigwedge^q\overline{T}}$ and ${H^q(X, \Omega^p)\simeq \bigwedge^pT\otimes \bigwedge^q\overline{T}}$.

To do the calculation we will use the Dolbeault resolution: ${0\rightarrow \mathcal{O}_X\rightarrow \mathcal{C}^{0,0}\rightarrow \mathcal{C}^{0,1}\rightarrow \mathcal{C}^{0,2}\rightarrow\cdots}$. This is an acyclic resolution of the structure sheaf, and so is fine to use for the calculation of cohomology. The first upper index of ${\mathcal{C}}$ refers to the degree of the ${\mathbb{C}}$-linear part and the second upper index refers to the degree of the ${\mathbb{C}}$-antilinear part. The map of the chain complex is ${\overline{\partial}}$.

Let’s examine the complex a little more closely. Define ${\phi_{p,q}: \mathcal{C}\otimes (\bigwedge^pT\otimes \bigwedge^q\overline{T})\rightarrow \mathcal{C}^{p,q}}$ by ${\sum f_i\otimes \alpha_i\mapsto \sum f_i\omega_{\alpha_i}}$. Where we define ${\omega_{\alpha}}$ to be the natural translation invariant ${(p,q)}$-form associated to ${\alpha\in \wedge^pT\otimes \wedge^q\overline{T}}$ by left-invariantizing.

Note that ${\omega_{\alpha\wedge\beta}=\omega_\alpha\wedge \omega_\beta}$. Thus to prove that all ${\omega_\alpha}$ are ${\overline{\partial}}$-closed (which we’ll denote ${d}$ from now on for simplicity), we only need to check this for ${(1,0)}$ and ${(0,1)}$ forms. Now ${exp: V\rightarrow X}$ is a local iso, so we also only need to check ${d(exp^*(\omega_\alpha))=0}$. But ${exp^*(\omega_\alpha)=d\alpha}$, so ${d}$ of this expression, is ${(d\circ d)(\alpha)=0}$.

This gives us that our map ${\phi_{0,q}}$ is an iso ${\Gamma(X, \mathcal{C})\otimes_\mathbb{C} \bigwedge^q\overline{T} \rightarrow \Gamma(X, \mathcal{C}^{0,q})}$. The map ${\overline{\partial}}$ is defined to be ${\overline{\partial}(f\otimes \alpha)=\overline{\partial}(f)\otimes \alpha}$. So since the ${\omega_\alpha}$ are closed, the iso commutes with the differential and we get these are actually isomorphic as chain complexes. Thus computing cohomology of one is equivalent to computing cohomology of the other.

Explicitly, we know that ${H^q(X, \mathcal{O}_X)\simeq H^q(X, \Gamma(X,\mathcal{C})\otimes_\mathbb{C} \bigwedge^q\overline{T})}$.

Since this notation is cumbersome, let ${A=\Gamma(X, \mathcal{C})}$ and ${\bigwedge^*=\bigwedge^*\overline{T}}$.

Let ${i: \bigwedge \rightarrow A\otimes_\mathbb{C} \bigwedge}$ be the inclusion. We want to show this gives an iso ${\bigwedge^q\stackrel{\sim}{\rightarrow} H^q(X, A\otimes \bigwedge^*)}$. This is precisely the goal given at the start of the post.

Now ${V}$ is a vector space so we have a natural Euclidean measure. Let ${\mu}$ be the measure on ${X}$ induced from this that is normalized so that ${\mu(X)=1}$. (For those familiar, this is just the unique translation invariant Haar measure, and ${X}$ is compact, so is finite and can be normalized). Define the linear map represented by ${\mu}$ by ${S: A\rightarrow \mathbb{C}}$. It is just ${S(f)=\int_X fd\mu}$. If ${W}$ is any ${\mathbb{C}}$-vector space, denote by ${S_W}$ the map ${A\otimes_\mathbb{C} W\rightarrow W}$. In particular, we have ${S_\wedge: A\otimes \bigwedge^*\rightarrow \bigwedge^*}$ so ${S_\wedge\circ i=id_\wedge}$.

We have a good ways to go yet, so next time we’ll pick up with the lemma that ${S_\wedge(\overline{\partial}\omega)=0}$ for all ${\omega\in A\otimes \bigwedge}$.

# Cohomology of Abelian Varieties

Hopefully I’ll start updating more than once a month. Since it’s been awhile and the previous post was tangent to what we’re actually doing, I’ll recap some notation. ${X}$ will be a compact complex (connected) Lie group of dimension ${g}$. We showed that we have an analytic isomorphism ${X\simeq (S^1)^{2g}\simeq (\mathbb{R}/\mathbb{Z})^{2g}}$. Let ${V=T_0X}$ (note that I’ll assume ${0}$ is the identity).

Under the exponential map ${exp: V\rightarrow X}$ (which we showed was a local isomorphism), we have that ${V}$ is the universal covering space of ${X}$. We showed that ${ker(exp)=U}$ is a lattice. Now the title of this post will seem a little silly to experts out there in cohomology, since we know that topologically these things are all tori. We’ll go through the details anyway.

First, we’ll show that ${H^r(X, \mathbb{Z})\simeq }$ the group of alternating ${r}$-forms ${U\times \cdots \times U\rightarrow \mathbb{Z}}$. We proceed by induction.

By general covering space theory ${exp^{-1}(0)=U=\pi_1(X, 0)}$. Thus we get the base case ${H^1(X,\mathbb{Z})\simeq Hom(U=\pi_1(X), \mathbb{Z})}$. Now we’ll want to show that the cup product induces the isomorphism ${\bigwedge^r\left(H^1(X,\mathbb{Z})\right)\rightarrow H^r(X,\mathbb{Z})}$. By the ${r=1}$ case this proves the statement.

We first reduce to the case of showing it is true for ${S^1}$. Since ${X}$ is just a product of ${S^1}$‘s, if we show that if the statement is true for ${X_1}$ and ${X_2}$, then it is also true for the product we can make the reduction. (For simplicitly, coefficients are in ${\mathbb{Z}}$, but I’ll omit that). Since we only need to apply this in the case where ${X_1}$ or ${X_2}$ is finite product of ${S^1}$‘s, we can also assume that the cohomologies are finitely generated and that they two spaces are connected for simplicity.

First, by the K\”{u}nneth formula: ${H^1(X_1\times X_2)\simeq \left(H^1(X_1)\otimes_\mathbb{Z}H^0(X_2)\right)\bigoplus \left(H^0(X_1)\otimes_\mathbb{Z}H^1(X_2)\right)}$, but the spaces are connected, so ${H^0(X_i)=\mathbb{Z}}$. Thus ${\bigwedge^r(H^1(X_1\times X_2))\simeq \bigwedge^r(H^1(X_1)\oplus H^1(X_2))}$
${\simeq \displaystyle\sum_{p+q=r}\bigwedge^p(H^1(X_1))\otimes \bigwedge^q(H^1(X_2))}$.

But now our inductive hypothesis is that for all $p$ less than $r$, ${\bigwedge^p(H^1(X_i))\simeq H^p(X_i)}$. Thus we get ${\displaystyle \sum_{p+q=r}\bigwedge^p(H^1(X_1))\otimes \bigwedge^q(H^1(X_2))}$
${\displaystyle \simeq \sum_{p+q=r} H^p(X_1)\otimes H^q(X_2)}$
${\simeq H^r(X_1\times X_2)}$. In other words, stringing all these isos together we get the iso we wanted. So we’ve reduced to the case of showing the statement for ${S^1}$, which follows immediately since ${H^n(S^1)=0}$ for ${n>1}$.

Note that if you have basic facts about singular cohomology at your disposal, this isn’t at all surprising. But let’s look at sheaf cohomology instead. This will require us to look at the Hodge structure which could be interesting. We won’t go very far today, but let’s at least get a few things out of the way.

Let ${\Omega^p}$ be the sheaf of holomorphic ${p}$-forms on ${X}$. We’d like to compute ${H^r(X, \Omega^p)}$. Let ${T=Hom(V, \mathbb{C})}$, i.e. the (complex) cotangent space at the identity to ${X}$. As with vectors and vector fields, every ${p}$-covector, i.e. element of ${\bigwedge^pT}$ can be extended uniquely to a left invariant ${p}$-form by pulling back along the left multiplication by ${-x}$ map. We’ll denote the correspondence ${\alpha\mapsto \omega_\alpha}$. This map defines an isomorphism of sheaves ${\mathcal{O}_X\otimes_\mathbb{C} \bigwedge^pT\stackrel{\sim}{\rightarrow} \Omega^p}$.

This says that ${\Omega^p}$ is a free sheaf of ${\mathcal{O}_X}$-modules. Now take global sections to get that ${\Gamma(X, \Omega^p)\simeq \Gamma(X, \mathcal{O}_X\otimes \bigwedge^pT)\simeq \bigwedge^pT}$, since the global sections of ${\mathcal{O}_X}$ are constants. Thus the only global sections of ${\Omega^p}$ are the ${p}$-forms that are invariant under left translation. Thus this isomorphism reduces our calculation to ${H^r(X, \Omega^p)\simeq H^r(X, \mathcal{O}_X)\otimes_\mathbb{C} \bigwedge^pT}$. So we’ll start in on that next time.

# Complex Lie Group Properties

Today we’ll do two more properties of compact complex Lie groups. The property we’ve already done is that they are always abelian groups. We go back to the notation from before and let $X$ be a compact complex Lie group and $V=T_eX$.

Property 1: $X$ is abelian.

Property 2: $X$ is a complex torus.

Proposition: $exp: \mathcal{L}(X)\simeq V\to X$, the exponential map, is a surjective homomorphism with kernel a lattice.

Proof: Fix $x,y\in X$. Note that the map $\psi: \mathbb{C}\to X$ by $t\mapsto (exp(tx))(exp(ty))$ is holomorphic since it is the composition of multiplication (holomorphic by being a Lie group) and the fact that $\phi_x(t)= exp(tx)$ which was checked to be holomorphic two posts ago. This is a homomorphism since $X$ is abelian.

Note that $d\psi_0\left(\frac{\partial}{\partial t}\Big|_0\right)=x+y$. By the uniqueness property of flows and exp just being a flow, $t\mapsto exp(tz)$ is the unique map with the property that the differential maps $\frac{\partial}{\partial t}\Big|_0\mapsto z$. Thus $\psi(t)=exp(t(x+y))$. Let $t=1$ and we get $exp(x)exp(y)=exp(x+y)$. i.e. $exp$ is a homomorphism.

Just as before, since $X$ is connected and $exp$ maps onto a neighborhood of the origin, the image is all of $X$. Let $U=ker(exp)$. We also saw two posts ago that there is a neighborhood of zero on which $exp$ is a diffeo and in particular is injective. Thus the $U$ is a discrete subgroup of $V$. But the only discrete subgroups of a vector space are lattices. This proves the proposition.

Corollary: $X$ is a complex torus.

Proof: We can holomorphically pass to the quotient and hence get a holomorphic isomorphism of groups $V/U\simeq X$.

Property 3: As a group $X$ is divisible and the $n$-torsion is isomorphic to $(\mathbb{Z}/n\mathbb{Z})^{2g}$ (recall that $g=dim_\mathbb{C}(X)$).

Proof: By property 2 we have that as a real Lie group $X\simeq (\mathbb{R}/\mathbb{Z})^{2g}=(S^1)^{2g}$. This proves both parts of property 3.

This is a good stopping point, since next time we’ll start thinking about the cohomology of $X$.

# Notes Regarding the Complex Structure

I know I said I wasn’t going to do this, but it isn’t that hard, and for completeness I should actually explain how the posts on real Lie groups/algebras relate to the complex case.

Let $V$ be a finite-dim real vector space. Then we call $J$ a complex structure of $V$ if $J\in End_{\mathbb{R}}(V)$ and $J^2=-id_V$. Given such a pair $(V, J)$, we can turn $V$ into a complex vector space $\widehat{V}$ by defining scalar multiplication by $(a+bi)v=av+J(bv)$. Likewise, we get a complex Lie algebra out of a real one if the complex structure also satisfies $[u, J(v)]=J([u,v])$ for all $u,v\in\frak{g}$. This is the associated complex Lie algebra to $\frak{g}$ denoted $\widehat{\frak{g}}$.

We can of course go the other direction much more easily. Given a complex Lie algebra $\frak{g}$, then restricting the $\mathbb{C}$-action to $\mathbb{R}$ gives us a real Lie algebra $\frak{g}_\mathbb{R}$. Note that $\widehat{\frak{g}_\mathbb{R}}=\frak{g}$ under the complex structure $J: \frak{g}_\mathbb{R}\to\frak{g}_\mathbb{R}$ by $u\mapsto iu$.

Suppose that $G$ and $H$ are complex Lie groups. Then if we have a morphism $\phi: G\to H$ in the category of complex Lie groups, i.e. a holomorphic map that is also a group homomorphism, then we can regard $d\phi: \mathcal{L}(G)\to\mathcal{L}(H)$ as a morphism of complex Lie algebras. Then $\phi$ can be regarded as a map of real Lie groups whose differential commutes with the complex structure on Lie algebras. The other way works as well. Given a real Lie group map whose differential is $\mathbb{C}$-linear, we have that $\phi$ is actually a map of complex Lie groups.

So far this is a fast overview. I don’t want to spend more than one post on this, but if you want to see more about any of these things, just comment.

The main purpose of bringing this up is that if we have a complex lie group $G$, then we’ll denote the underlying real Lie group as $G_\mathbb{R}$. By the posts I’ve already done, we can explicitly construct $exp_{G_\mathbb{R}}: \frak{g}\to G$. Note this is only a real holomorphic map. By the above statements, if $d(exp_{G_\mathbb{R}})$ is $\mathbb{C}$-linear, then it is actually complex holomorphic.

Just as in the exponential map post, we define the $\mathbb{C}$-linear map $\alpha: \mathbb{C}\to\frak{g}$ by $\alpha(z)=zv$ where $v\in T_eG$. Since $\mathbb{C}_\mathbb{R}$ is simply connected, there is a unique lift of this map to all of $\mathbb{C}$ which we call $\phi_v: \mathbb{C}\to G$ such that $d\phi_v=\alpha$. Which means that $\phi_v=exp_{G_\mathbb{R}}\circ \alpha$. i.e. $\phi_v(z)=exp_{G_\mathbb{R}}(zv)$.

Thus $d\phi_v=\alpha$ is complex holomorphic giving the 1-parameter subgroup in $G$ satisfying $d\phi_v(1)=v$ and exponential map $exp_G(v)=\phi_v(1)$. We have essentially proved the theorem that $exp_G=exp_{G_\mathbb{R}}$ which means the previous posts on the subject still apply.

As motivation for later, we’ll now do the example. Let $\mathbb{C}^n$ be the (only) simply connected complex Lie group of dimension $n$. We have global coordinates $z_1, \ldots, z_n$ since this is a vector space. Thus $\frac{\partial}{\partial z_1}\Big|_0, \ldots, \frac{\partial}{\partial z_n}\Big|_0$ forms a basis for $\mathcal{L}(\mathbb{C}^n)$.

We have that $[\frac{\partial}{\partial z_i}, \frac{\partial}{\partial z_j}]=0$, so the Lie algebra is abelian. i.e. we can identify $\mathcal{\mathbb{C}^n}$ with $\mathbb{C}^n$. The exponential map is just the identity.

The other example is to take a real basis $e_1, \ldots, e_{2n}$ of $\mathbb{C}^{n}$ and let $\displaystyle D=\{\sum_{i=1}^{2n}m_ie_i : m_i\in \mathbb{Z}\}$ and quotient $\mathbb{C}^n/D$. This is a real torus and will be incredibly important in when we return to compact complex Lie groups.

# Analytic Theory of Abelian Varieties I

It will be useful to have past posts as reference: one-parameter subgroups, exponential map, exponential properties, and lie algebra actions.

As a summary and way to get notation going I’ll just list some important things that are proved there and that we’ll use. First off, everything in those posts used smooth manifolds, but we’ll be using complex manifolds. This just means transition maps must be holomorphic rather than smooth. We can still do all of those things with $\mathbb{C}$ in place of $\mathbb{R}$ just making sure that everything is actually holomorphic rather than only smooth.

Let $X$ be a connected complex manifold of dimension $g$ with a group structure such that inversion $X\to X$ by $x\mapsto x^{-1}$ and multiplication $X\times X\to X$ by $(x,y)\mapsto xy$ is holomorphic. This is called a connected complex Lie group.

Let $V=T_eX$ be the tangent space at the identity. This is a complex vector space of dimension $g$. If $v\in V$, then we get an entire left-invariant vector field on $X$ by defining $Y_x=(dL_x)_e(v)$ where $L_x$ is left multiplication by $x$. Any left-invariant vector field turns out to automatically be holomorphic. The set of all left-invariant vector fields on $X$ is denoted $\mathcal{L}(X)$ and it called the Lie algebra associated to the Lie group $X$.

Thus we get an integral curve of the flow of this vector field associated to $v$. Since the vector field is complete, we get the one-parameter subgroup $\phi_v: \mathbb{C}\to X$. See the post on this for more rigor. This map satisfies $d\phi_v(\frac{d}{dz})=v$.

Note that we can always identify $\mathcal{L}(\mathbb{C})$ with $\mathbb{C}$ by the isomorphism $w\mapsto w\frac{d}{dz}$. Under this identification, we have $d\phi_v(1)=v$. The map $\phi_v(t): \mathbb{C}\times V\to X$ is holomorphic. Define $exp: V\simeq \mathcal{L}(X)\to X$ by $exp(v)=\phi_v(1)$.

See the exponential map post to see why we get some nice properties such as $\phi_{sv}(t)=\phi_v(st)$. If we identify the tangent space to $V$ at 0 with $V$ itself, then we get that $(dexp)_0(v)=v$. Lastly, given any homomorphism of complex Lie groups $T: X_1\to X_2$ we get that $T(exp_{X_1}y)=exp_{X_2}((dT)_ey)$.

I should probably be explaining the subtleties going on between considering the tangent space as complex versus real. Basically, if you write down all the maps and identifications carefully, all of these things actually respect the $\mathbb{R}$-structure. But we won’t really worry about that unless it comes up later.

We’ll do one theorem: If $X$ is a compact connected complex Lie group, then $X$ is abelian.

Consider the conjugation map $C_x: X\to X$ by $C_x(y)=xyx^{-1}$. Then $(dC_x)_e: V\to V$ is an automorphism. We also have that $x\mapsto (dC_x)_e$ a holomorphic map $\psi: X\to Aut(V)$ which is a subspace of $End(V)$. Since $X$ is compact and $\psi$ holomorphic, it must be constant. i.e. $(dC_x)_e=(dC_e)_e=id_V$.

Since $C_x$ is a homomorphism of complex Lie groups $X\to X$, we get the exponential property mentioned above: $C_x(exp y)=exp((dC_x)_ey)=exp(id_V(y))=exp(y)$. This tells us that $exp(V)=\{exp(v) : v\in V\}$ is in the center of $X$. But $(dexp)_0$ is the identity and in particular has full rank, so by the Implicit Function Theorem $exp$ is a homeomorphism from a neighborhood of $0$ in $V$ to a neighborhood of the identity in $X$.

But $X$ is connected, so any neighborhood of the identity generates all of $X$. Thus $exp(V)$ generates the whole group $X$, and hence the center of $X$ is all of $X$ meaning $X$ is abelian.

# Handle Decomposition

Today I’ll just prove that a Morse function will give a handle decomposition of a closed manifold. Let’s use all the notation already set up (meaning critical points, values, attaching maps, dimension, Morse function, gradient-like vector field, etc).

We just induct on the subscripts of critical points. We’ve already done the base case (it is a min and hence a 0-handle from here). So we just need to show that if $M_{t}$ is a handlebody for $t\in (c_{i-1}, c_i)$, then $M_{c_i+\varepsilon}$ is a handlebody with the appropriate handle attached.

So we’ve assumed that we have some decomposition $M_{c_{i-1}+\varepsilon}\cong \mathcal{H}(D^m;\phi_1, \ldots , \phi_{i-1})$. We also know that we attach a handle of index $\lambda_i$ when crossing $c_i$, so we do have a diffeo to a manifold $M_{c_i-\varepsilon}$ with a $\lambda_i$-handle attached with attaching map $\phi: \partial D^{\lambda_i}\times D^{m-\lambda_i}\to \partial M_{c_i-\varepsilon}$.

Note that $[c_{i-1}+\varepsilon, c_i-\varepsilon]$ contains no critical values, so by flowing along $X$ we get a diffeo $M_{c_{i-1}+\varepsilon}\cong M_{c_i-\varepsilon}$. Let $\psi:M_{c_{i-1}+\varepsilon}\to M_{c_i-\varepsilon}$ be this diffeo.

So by inductive hypothesis, $M_{c_i-\varepsilon}\cong \mathcal{H}(D^m;\phi_1, \ldots , \phi_{i-1})$, so we can assume $\psi$ actually maps from the handlebody to $M_{c_i-\varepsilon}$. Now by composing we get our actual attaching map (note that before now the handle was attached to $M_{c_i-\varepsilon}$ and not the handlebody itself).

i.e. $\psi^{-1}\circ \phi : \partial D^{\lambda_i}\times D^{m-\lambda_i}\to \partial (\mathcal{H}(D^m;\phi_1, \ldots , \phi_{i-1}))$. So let $\phi_i=\psi^{-1}\circ \phi$, and we get that $M_{c_i+\varepsilon}\cong \mathcal{H}(D^m;\phi_1, \ldots , \phi_{i-1}, \phi_i)$, so we are done.

So I sort of dragged on longer than probably necessary there, since there was essentially nothing new. It was just being pedantic about the diffeo of the manifold and the handlebody.

There are some subtleties that should be pointed out, though. The index of the critical point did determine the index of the handle, and we went in “ascending” order. The other much more important and also more subtle point is that the choice of gradient-like vector field was how we constructed the attaching map. So even the same Morse function with a different choice of gradient-like vector field could actually give a “different” handle decomposition when considering attaching maps as part of the data.

# Handlebodies III

I keep naming my posts “handlebodies”, so I think it is officially time to define what one is. A handlebody is a manifold obtained from $D^m$ by attaching various $\lambda$-handles successively. Thus a general handlebody will look like $D^m\cup D^{\lambda_1}\times D^{m-\lambda_1}\cup \cdots \cup D^{\lambda_n}\times D^{m-\lambda_n}$.

If you’re familiar with how to construct a CW-complex, this is pretty similar. You just inductively attach the handles using smooth maps, and then smooth out the manifold so that at each step we have a legitimate smooth manifold. It may be useful to introduce a notation for this. The first attaching $D^m\cup_{\phi_1} D^{\lambda_1}\times D^{m-\lambda_1}$ with attaching map $\phi_1: \partial D^{\lambda_1}\times D^{m-\lambda_1}\to \partial D^m$ will be denoted $\mathcal{H}(D^m; \phi_1)$. So inductively denote the i-th attaching by $\mathcal{H}(D^m ; \phi_1, \ldots , \phi_{i-1}, \phi_i)$.

After i steps, we will always have i attaching maps even if some are formally meaningless (attaching a 0-handle is a disjoint union, so there is no attaching).

If we express M as a handlebody we call that a handle decomposition of the manifold.

Next time I’ll prove the result that everyone that has been reading the posts will have already guessed. Given a Morse function $f: M\to \mathbb{R}$ on a closed manifold, $f$ determines a handle decomposition of $M$. Moreover the handles of this handlebody correspond to the critical points of $f$, and the indices of the handles coincide with the indices of the corresponding critical points.

I’m short on time today, so I’m going to put off proving it.

I’m not sure how much to say about my other news, since it is still sort of up in the air. I passed two of my qualifying exams (which was all that was necessary for now), and I may officially lock myself into the path of algebraic geometry as my field in the next couple of days. Before I say too much about this, I’ll just say that I should have more information on Monday about what I’m officially doing.

# Handlebodies II

Let’s think back to our example to model our $\lambda$-handle (where $\lambda$ is not a max or min). Well, it was a “saddle point”. So it consisted of a both a downward arc and upward arc. If you got close enough, it would probably look like $D^1\times D^1$.

Well, generally this will fit with our scheme. An n-handle looked like $D^n$ … or better yet $D^n\times 0$, and a 0-handle looked like $0\times D^n$, so maybe it is the case that a $\lambda$-handle looks like $D^\lambda\times D^{n-\lambda}$. Let’s call $D^\lambda\times 0$ the core of the handle, and $D^{n-\lambda}$ the co-core.

By doing the same trick of writing out what our function looks like at a critical point of index $\lambda$ in some small enough neighborhood using the Morse lemma, we could actually prove this, but we’re actually more interested now in how to figure out what happens with $M_t$ as $t$ crosses this point.

By that I mean, it is time to figure out what exactly it is to “attach a $\lambda$-handle” to the manifold.

Suppose as in the last post that $c_i$ is a critical value of index $\lambda$. Then I propose that $M_{c_i+\varepsilon}$ is diffeomorphic to $M_{c_i-\varepsilon}\cup D^\lambda\times D^{m-\lambda}$ (sorry again, recall my manifold is actually m-dimensional with n critical values).

I wish I had a good way of making pictures to get some of the intuition behind this across. I’ll try in words. A 1-handle for a 3-manifold, will be $D^1\times D^2$, i.e. a solid cylinder. So we can think of this as literally a handle that we will bend the cylinder into, and attach those two ends to the existing manifold. This illustration is quite useful in bringing up a concern we should have. Attaching in this manner is going to create “corners” and we want a smooth manifold, so we need to make sure to smooth it out. But we won’t worry about that now, and we’ll just call the smoothed out $M_{c_i-\varepsilon}\cup D^\lambda\times D^{m-\lambda}$, say $M'$.

Let’s use our gradient-like vector field again. Let’s choose $\varepsilon$ small enough so that we are in a coordinate chart centered at $p_i$ such that $f=-x_1^2-\cdots - x_\lambda^2 + x_{\lambda +1}^2+\cdots + x_m^2$ is in standard Morse lemma form.

Let’s see what happens on the core $D^\lambda\times 0$. At the center, it takes the critical value $c_i$ and it decreases everywhere from there (as we move from 0, only the first $\lambda$ coordinates change). This decreasing goes all the way to the boundary where it is $c_i-\varepsilon$. Thus it is the upside down bowl (of dimension $\lambda$). Likewise, the co-core goes from the critical value and increases (as in the right side up bowl) to the boundary of a $m-\lambda$ disk at a value $c_i+\delta$ (where $0<\delta<\varepsilon$).

Let's carefully figure out the attaching procedure now. If we think of our 3-manifold for intuition, we want to attach $D^\lambda\times D^{m-\lambda}$ to $M_{c_i-\varepsilon}$ by pasting $\partial D^\lambda\times D^{m-\lambda}$ along $\partial M_{c_i-\varepsilon}$.

So I haven't talked about attaching procedures in this blog, but basically we want a map $\phi: \partial D^\lambda\times D^{m-\lambda}\to \partial M_{c_i-\varepsilon}$ and then forming the quotient space of the disjoint union under the relation of identifying $p\in \partial D^\lambda\times D^{m-\lambda}$ with $\phi (p)$. Sometimes this is called an adjunction space.

So really $\phi$ is a smooth embedding of a thickened sphere $S^{\lambda - 1}$, since $\partial D^\lambda=S^{\lambda-1}$. And the dimensions in which it was thickened is $m-\lambda$. Think about the "handle" in the 3-dimensional 1-handle case. We gave the two endpoints of line segment (two points = $S^0$) a 2-dimensional thickening by a disk.

Now it is the same old trick to get the diffeo. The gradient-like vector field, $X$, flows from $\partial M'$ to $\partial M_{c_i+\varepsilon}$, so just multiply $X$ by a smooth function that will make $M'$ match $M_{c_i+\varepsilon}$ after some time. This is our diffoemorphism and we are done.

# Handlebodies I

We now come to the main point of all these Morse theory posts. We want to somehow figure out what a closed manifold looks like based a Morse function that it admits (who knows how long I’ll develop this theory, maybe we’ll even get to how Smale proved the Poincare Conjecture in dimensions greater than or equal to 5).

Suppose $M$ is closed and $f:M\to\mathbb{R}$ a Morse function. We’ll use the convenient notation $M_t=\{p\in M : f(p)\leq t\}$. So again, with the height analogy, as t increases, we will be looking at the entire manifold up to that height. Since M is compact, there is some finite interval $[a,b]$ such that $M_a=\emptyset$ and $M_b=M$.

Note that with essentially no modification, we have already proved the Theorem that if $[c,d]$ contains no critical values, then $M_c\cong M_d$. So really, the point is to now figure out what happens as we pass through the critical values.

First off, there are only finitely many critical points, and we can assume that each of these has distinct critical values by raising and lowering critical values. So if $p_0, \ldots, p_n$ are the critical points and $c_k=f(p_k)$, we can order the indices so that $c_0 < c_1 < \cdots < c_n$.

To be explicit, $c_0$ is the min, so $M_t=\emptyset$ for $t < c_0$ and $M_t=M$ for t greater than $c_n$, since $c_n$ is the max (also, wordpress hates inequalities, or me, I haven't decided yet, but it always cuts out lots of stuff and I just have to write the inequality in words).

These two critical points would be a nice place to start our examination. By the Morse lemma and the fact that a min has index 0, we know that there exists a neighborhood of $p_0$ on which $f=x_1^2+\cdots + x_m^2+c_0$ (Alright, I’m sorry about that, but I just realized I have n critical points, so the dimension of my manifold is now m).

More explicitly there is some $\varepsilon>0$ such that $M_{c_0+\varepsilon}=\{(x_1, \ldots , x_m) : x_1^2+\cdots + x_m^2\leq \varepsilon\}\cong B^m$. So if we are thinking of height (of a 2-dim manifold), we’ll want to visualize this as a “bowl” where you have the bottom of the bowl the min and then it slopes upward along a sphere, and then you have the boundary circle at height $c_0+\varepsilon$.

So note that the only thing we used about this critical point is that it had index 0. This shape is called a (m-dimensional) 0-handle.

The reverse happens at our max. We have $M_{c_n-\varepsilon}=\{(x_1, \ldots , x_m) : x_1^2+\cdots +x_m^2\geq \varepsilon\}$, since the critical point has index m. This is an $m$-handle and thinking in 2-d height, it is a downward facing bowl.

Again, there is nothing special about being the absolute max, any index m critical point will locally be an $m$-handle.

Index k critical points where $k\neq 0,m$ are more complicated so I’ll leave those for next time.

Now we have a nice overview of how this will work. We just need to figure out what a $k$-handle looks like, then as t increases through a critical value with index k, $M_t$ will “attach a k-handle”. When we are not near a critical value, the $M_t$ will not change diffeomorphism-type. We just need to make this a little more precise next time (or maybe even the time after).