# Naturality of Flows

This is something I always forget exists and has a name, so I end up reproving it. Since this sequence of posts is a hodge-podge of things to help me take a differential geometry test, hopefully this will lodge the result in my brain and save me time if it comes up.

I’m not sure whether to call it a lemma or not, but the setup is you have a smooth map ${F:M\rightarrow N}$ and a vector field on ${M}$, say ${X}$ and a vector field on ${N}$ say ${Y}$ such that ${X}$ and ${Y}$ are ${F}$-related. Define ${M_t}$ and ${N_t}$ to be the image of flowing for time ${t}$ and let ${\theta}$ and ${\eta}$ be the flows of ${X}$ and ${Y}$ respectively. Then the lemma says for all ${t}$ we have ${F(M_t)\subset N_t}$ and ${\eta_t\circ F=F\circ \theta_t}$ on ${M_t}$.

This is a “naturality” condition because all it really says is that the following diagram commutes:

${\begin{matrix} M_t & \stackrel{F}{\longrightarrow} & N_t \\ \theta_t \downarrow & & \downarrow \eta_t \\ M_{-t} & \stackrel{\longrightarrow}{F} & N_{-t} \end{matrix}}$

Proof: Let ${p\in M}$, then ${F\circ \theta^p: \mathbb{R}\rightarrow N}$ is a curve that satisfies the property $\displaystyle {\frac{d}{dt}\Big|_{t=t_0}(F\circ \theta^p)(t)=DF_{\theta^p(t_0)}(\frac{d}{dt}\theta^p (t)\Big|_{t=t_0})=DF_{\theta^p(t_0)}(X_{\theta^p(t_0)})=Y_{F\circ \theta^p(t_0)}}$. Since ${F\circ \theta^p(0)=F(p)}$, and integral curves are unique, we get that ${F\circ\theta^p(t)=\eta^{F(p)}(t)}$ at least on the domain of ${\theta^p}$.

Thus if ${p\in M_t}$ then ${F(p)\in N_t}$, or equivalently ${F(M_t)\subset N_t}$. But we just wrote that ${F(\theta^p(t))=\eta^{F(p)}(t)}$ where defined, which is just a different form of the equation ${\eta_t\circ F=F\circ \theta_t(p)}$.

We get a nice corollary out of this. If our function ${F:M\rightarrow N}$ was actually a diffeo, then take ${Y=F_*X}$ the pushforward, and we get that the flow of the pushforward is ${\eta_t=F\circ \theta_t\circ F^{-1}}$ and the flow domain is actually equal ${N_t=F(M_t)}$.

In algebraic geometry we care a lot about families of things. In the differentiable world, the nicest case of this would be when you have a smooth submersion: ${F: M\rightarrow N}$, where ${M}$ is compact and both are connected. Then since all values are regular, ${F^{-1}(n_0)}$ is smooth embedded submanifold. If ${N}$ were say ${\mathbb{R}}$ (of course, ${M}$ couldn’t be compact in this case), then we would have a nice 1-dimensional family of manifolds that are parametrized in a nice way.

It turns out to be quite easy to prove that in the above circumstance all fibers are diffeomorphic. In AG we often call this an “iso-trivial” family, although I’m not sure that is the best analogy. The proof basically comes down to the naturality of flows. Given any vector field ${Y}$ on ${N}$, we can lift it to a vector field ${X}$ on ${M}$ that is ${F}$-related. I won’t do the details, but it can be done clearly in nice choice of coordinates ${(x^1, \ldots, x^n)\mapsto (x^1, \ldots, x^{n-k})}$ and then just patch together with a partition of unity.

Let ${M_x}$ be the notation for ${F^{-1}(x)}$. Fix an ${x\in N}$, then by the above naturality lemma ${\theta_t\Big|_{M_x} : M_x\rightarrow M_{\eta_t(x)}}$ is well-defined and hence a diffeomorphism since it has smooth inverse ${\theta_{-t}}$. Let ${y\in N}$. Then as long as there is a vector field on ${N}$ which flows ${x}$ to ${y}$, then we’ve shown that ${M_x\simeq M_y}$, so since ${x}$, ${y}$ were arbitrary, all fibers are diffeomorphic. But there is such a vector field, since ${N}$ is connected.

# Lie groups have abelian fundamental group

Last year I wrote up how to prove that the fundamental group of a (connected) topological group was abelian. Since Lie groups are topological groups, they also have abelian fundamental groups, but I think there is a much neater way to prove this fact using smooth things. Here it is:

Lemma 1: A connected Lie group that acts smoothly on a discrete space is must be a trivial action.

Proof: Suppose our action is ${\theta: G\times M \rightarrow M}$ by ${\theta(g,x)=g\cdot x}$. Consider a point ${q}$ with non-trivial orbit. Then ${\text{im}\theta_{q}=\{q\}\cup A}$ where ${A}$ is non-empty. Thus ${G=\theta_{q}^{-1}(q)\cup \theta_q^{-1}(A)}$, a disconnection of ${G}$. Thus all orbits are trivial.

Lemma 2: A discrete normal subgroup of a Lie group (from now on always assumed connected) is central.

Proof: Denote the subgroup ${H}$. Then ${\theta(g,h)=ghg^{-1}}$ is a smooth action on ${H}$ (a discrete set). Thus by Lemma 1, ${ghg^{-1}=h}$ for all ${g\in G}$. Thus ${H}$ is central, and in particular abelian.

Now for the main theorem. Let ${G}$ be a connected Lie group. Let ${U}$ be the universal cover of ${G}$. Then the covering map ${p: U\rightarrow G}$ is a group homomorphism. Since ${U}$ is simply connected, the covering is normal and hence ${Aut_p(U)\simeq \pi_1(G, e)}$. By virtue of being normal, we also get that ${Aut_p(U)}$ acts transitively on the fibers of ${p}$. In particular, on the set ${p^{-1}(e)}$, which is discrete being the fiber of a discrete bundle. But this set is ${\ker p}$, which is a normal subgroup. I.e. a discrete normal subgroup, which by Lemma 2 is abelian.

Fix ${q\in \ker p}$. Then we get an ismorphism ${\ker p \simeq Aut_p(U)}$ by ${x\mapsto \phi_x}$ where ${\phi_x}$ is the unique covering automorphism that takes ${q}$ to ${x}$. Thus ${Aut_p(U)}$ is abelian which means ${\pi_1(G,e)}$ is abelian.

# PDE’s and Frobenius Theorem

I’ve started many blog posts on algebra/algebraic geometry, but they won’t get finished and posted for a little while. I’ve been studying for a test I have to take in a few weeks in differential geometry-esque things. So I’ll do a few posts on things that I think are usually considered pretty easy and obvious to most people, but are just things I never sat down and figured out. Hopefully this set of posts will help others who are confused as I recently was.

My first topic is about the Frobenius Theorem. I’ve posted about it before. Here’s the general idea of it: If ${M}$ is a smooth manifold and ${D}$ is a smooth distribution on it, then ${D}$ is involutive if and only if it is completely integrable (i.e. there is are local flat charts for the distribution).

What does this have to do with being able to solve partial differential equations? I’ve always heard that it does, but other than the symbol ${\displaystyle\frac{\partial}{\partial x}}$ appearing in the defining of a distribution or of the flat chart, I’ve never figured it out.

Let’s go through this with some examples. Are there any non-constant solutions ${f\in C^\infty (\mathbb{R}^3)}$ to the systems of equations: ${\displaystyle \frac{\partial f}{\partial x}-y\frac{\partial f}{\partial z}=0}$ and ${\displaystyle \frac{\partial f}{\partial y}+x\frac{\partial f}{\partial z}=0}$?

Until a few days ago, I would have never thought we could use the Frobenius Theorem to do this. Suppose ${f}$ were such a solution. Define the vector fields ${\displaystyle X=\frac{\partial}{\partial x}-y\frac{\partial}{\partial z}}$ and ${\displaystyle Y=\frac{\partial}{\partial y}+x\frac{\partial}{\partial z}}$ and define the distribution ${D_p=\text{span} \{X_p, Y_p\}}$.

Choose a regular value of ${f}$, say ${C}$ (one exists by say Sard’s Theorem). Then ${f=C}$ is a 2-dimensional submanifold ${M\subset \mathbb{R}^3}$, and since ${f}$ is a defining function ${T_pM=ker(Df_p)}$. But the very fact that ${f}$ satisfies, by assumption, ${X(f)=0}$ and ${Y(f)=0}$, we have ${T_pM=\text{span} \{X_p, Y_p\}}$. I.e. ${M}$ is an integral manifold for the distribution ${D}$. Thus ${D}$ must be involutive.

Just check now. ${\displaystyle [X,Y]=2\frac{\partial}{\partial z}}$, so in particular at the origin ${\displaystyle X_0=\frac{\partial}{\partial x}}$ and ${\displaystyle Y_0=\frac{\partial}{\partial y}}$ it is not in the span, and hence not involutive. Thus no such ${f}$ exists. This didn’t even use Frobenius.

Now let’s spice up the language and difficulty. Is it possible to find a function ${z=f(x,y)}$, ${C^\infty}$ in a neighborhood of ${(0,0)}$, such that ${f(0,0)=0}$ and ${\displaystyle df=(ye^{-(x+y)}-f)dx+(xe^{-(x+y)}-f)dy}$? Alright, the ${d}$ phrasing is just asking there is a local solution to the system ${\displaystyle \frac{\partial f}{\partial x}=ye^{-(x+y)}-f}$ and ${\displaystyle \frac{\partial f}{\partial y}=x^{-(x+y)}-f}$. Uh oh. The above method fails us now since it isn’t homogeneous.

Alright, so let’s extrapolate a little. We have a system of the form ${\displaystyle \frac{\partial f}{\partial x}=\alpha(x,y,f)}$ and ${\displaystyle \frac{\partial f}{\partial y}=\beta(x,y,f)}$. The claim is that necessary and sufficient conditions to have a local solution to this system is ${\displaystyle \frac{\partial \alpha}{\partial y}+\beta\frac{\partial \alpha}{\partial z}=\frac{\partial \beta}{\partial x}+\alpha \frac{\partial \beta}{\partial z}}$.

I won’t go through the details of the proof, but the main idea is not bad. Define the distribution spanned by ${\displaystyle X=\frac{\partial}{\partial x}+\alpha\frac{\partial}{\partial z}}$ and ${\displaystyle Y=\frac{\partial}{\partial y}+\beta\frac{\partial}{\partial z}}$.

Then use that assumption to see that ${[X,Y]=0}$ and hence the distribution is involutive and hence there is an integral manifold for the distribution by the Frobenius Theorem. If ${g}$ is a local defining function to that integral manifold, then we can hit that with the Implicit Function Theorem and get that ${z=f(x,y)}$ (the implicit function) is a local solution.

If we go back to that original problem, we can easily check that the sufficient condition is met and hence that local solution exists.

I had one other neat little problem, but it doesn’t really fit in here other than the fact that solutions to PDEs are involved.

# 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 $X$.