## 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.

## The Grothendieck Spectral Sequence

Well, I meant to do lots more examples building up some more motivation for how powerful spectral sequences can be in some simple cases. But I’m just running out of steam on posting about them. Since we’ve done spectral sequences associated to a double complex, we may as well do the Grothendieck Spectral Sequence, then I might move on to another topic for a bit (I admit it is sort of sad to not prove the Kunneth formula using a SS).

I haven’t scoured the blogs to see whether these topics have been done yet, but I’m thinking about either basics on abelian varieties a la Mumford, or some curve theory, possibly building slowly to and culminating in Riemann-Roch.

In any case, we have the tools to do the Grothendieck Spectral Sequence (GSS) quite easily. Let $\mathcal{A}, \mathcal{B}, \mathcal{C}$ be abelian categories with enough injectives. Let $\mathcal{A}\stackrel{G}{\to}\mathcal{B}\stackrel{F}{\to}\mathcal{C}$ be functors (and $FG:\mathcal{A}\to\mathcal{C}$ the composition). Suppose that $F$ and $G$ are left exact and for every injective $J\in\mathcal{A}$ we have $G(J)$ is acyclic. This just means that $R^iF(J)=0$ for all positive $i$.

Then there exists a spectral sequence (the GSS) with $E_2^{pq}\simeq (R^pF)(R^qG)(X)\Rightarrow R^{p+q}(FG)(X)$ with differential $d_{r}:E_r^{pq}\to E_r^{p+r, q-r+q}$.

The proof of this is just to resolve $X$ using the injectives that we know exist. This gives us a double complex. From a double complex, the way to get the $E_2$ term is to take vertical then horizontal homology, or horizontal and then vertical. Both of these will converge to the same thing. One way completed collapses to the “0 row” due to the fact that the exact sequence remained exact after applying the functor except at the 0 spot. Thus it stabilizes at this term and writing it out, you see that it is exactly $R^{p+q}(FG)(X)$. Taking homology in the other order gives us exactly $(R^pF)(R^qG)(X)$ by definition of a derived functor. This completes the proof.

I probably should write the diagram out for clarity, but really they are quite a pain to make and import into wordpress. The entire outline of the proof is here, so if you’re curious about the details, just carefully fill in what everything is from the previous posts.

This is quite a neat spectral sequence. It is saying that just by knowing the derived functors of $F$ and $G$ you can get to the derived functors of the composition of them. There are two important spectral sequence consequences of this one. They are the Leray SS and the Lyndon-Hochschild-Serre SS. The later computes group cohomolgy.

I promised early on to do the Leray SS for all the algebraic geometers out there. The Leray SS gives a way to compute sheaf cohomology. Let $\mathcal{A}=Sh(X)$ and $\mathcal{B}=Sh(Y)$ be the category of sheaves of abelian groups on X and Y. Let $\mathcal{C}=Ab$ the category of abelian groups. Let $f:X\to Y$ be a continuous map, then we have the functor $F=f_*$ and the two global section functors $\Gamma_X$ and $\Gamma_Y$.

Applying the GSS to these functors, we get that $H^p(Y, R^qf_*\mathcal{F})\Rightarrow H^{p+q}(X, \mathcal{F})$.

There are a few things to verify to make sure that the GSS applies, and we need the fact that $\Gamma_Y\circ f_*=\Gamma_X$. It would also be nice to have an example to see that this is useful. So maybe I’ll do those two things next time.