## Frobenius Semi-linear Algebra 2

Recall our setup. We have an algebraically closed field ${k}$ of characteristic ${p>0}$. We let ${V}$ be a finite dimensional ${k}$-vector space and ${\phi: V\rightarrow V}$ a ${p}$-linear map. Last time we left unfinished the Jordan decomposition that says that ${V=V_s\oplus V_n}$ where the two components are stable under ${\phi}$ and ${\phi}$ acts bijectively on ${V_s}$ and nilpotently on ${V_n}$.

We then considered a strange consequence of what happens on the part on which it acts bijectively. If ${\phi}$ is bijective, then there always exists a full basis ${v_1, \ldots, v_n}$ that are fixed by ${\phi}$, i.e. ${\phi(v_i)=v_i}$. This is strange indeed, because in linear algebra this would force our operator to be the identity.

There is one more slightly more disturbing consequence of this. If ${\phi}$ is bijective, then ${\phi-Id}$ is always surjective. This is a trivial consequence of having a fixed basis. Let ${w\in V}$. We want to find some ${z}$ such that ${\phi(z)=w}$. Well, we just construct the coefficients in the fixed basis by hand. We know ${w=\sum c_i v_i}$ for some ${c_i\in k}$. If ${z=\sum a_i v_i}$ really satisfies ${\phi(z)-z=w}$, then by comparing coefficients such an element exists if and only if we can solve ${a_i^p-a_i=c_i}$. These are just polynomial equations, so we can solve this over our algebraically closed field to get our coefficients.

Strangely enough we really require algebraically closed and not merely perfect again, but the papers I’ve been reading explicitly require these facts over finite fields. Since they don’t give any references at all and just call these things “standard facts about ${p}$-linear algebra,” I’m not sure if there is a less stupid way to prove these things which work for arbitrary perfect fields. This is why you should give citations for things you don’t prove!!

Why do I call this disturbing? Well, these maps really do appear when doing long exact sequences in cohomology. Last time we saw that we could prove that ${E[p]\simeq \mathbb{Z}/p}$ for an ordinary elliptic curve from computing the kernel of ${C-Id}$ where ${C}$ was the Cartier operator. But we have to be really, really careful to avoid linear algebra tricks when these maps come up, because in this situation we have ${\phi -Id}$ is always a surjective map between finite dimensional vector spaces of the same dimension, but also always has a non-trivial kernel isomorphic to ${\mathbb{Z}/p\oplus \cdots \oplus \mathbb{Z}/p}$ where the number of factors is the dimension of ${V}$. Even though we have a surjective map in the long exact sequence between vector spaces of the same dimension, we cannot conclude that it is bijective!

Since everything we keep considering as real-life examples of semi-linear algebra has automatically been bijective (i.e. no nilpotent part), I haven’t actually been too concerned with the Jordan decomposition. But we may as well discuss it to round out the theory since people who work with ${p}$-Lie algebras care … I think?

The idea of the proof is simple and related to what we did last time. We look at iterates ${\phi^j}$ of our map. We get a descending chain ${\phi^j(V)\supset \phi^{j+1}(V)}$ and hence it stabilizes somewhere, since even though ${\phi}$ is not a linear map, the image is still a vector subspace of ${V}$. Let ${r}$ be the smallest integer such that ${\phi^r(V)=\phi^{r+1}(V)}$. This means that ${r}$ is also the smallest integer such that ${\ker\phi^r=\ker \phi^{r+1}}$.

Now we just take as our definition ${V_s=\phi^r(V)}$ and ${V_n=\ker \phi^r}$. Now by definition we get everything we want. It is just the kernel/image decomposition and hence a direct sum. By the choice of ${r}$ we certainly get that ${\phi}$ maps ${V_s}$ to ${V_s}$ and ${V_n}$ to ${V_n}$. Also, ${\phi|_{V_s}}$ is bijective by construction. Lastly, if ${v\in V_n}$, then ${\phi^j(v)=0}$ for some ${0\leq j\leq r}$ and hence ${\phi}$ is nilpotent on ${V_n}$. This is what we wanted to show.

Here’s how this comes up for ${p}$-Lie algebras. Suppose you have some Lie group ${G/k}$ with Lie algebra ${\mathfrak{g}}$. You have the standard ${p}$-power map which is ${p}$-linear on ${\mathfrak{g}}$. By the structure theorem above ${\mathfrak{g}\simeq \mathfrak{h}\oplus \mathfrak{f}}$. The Lie subalgebra ${\mathfrak{h}}$ is the part the ${p}$-power map acts bijectively on and is called the core of the Lie algebra.

Let ${X_1, \ldots, X_d}$ be a fixed basis of the core. We get a nice combinatorial classification of the Lie subalgebras of ${\mathfrak{h}}$. Let ${V=Span_{\mathbb{F}_p}\langle X_1, \ldots, X_d\rangle}$. The Lie subalgebras of ${\mathfrak{h}}$ are in bijective correspondence with the vector subspaces of ${V}$. In particular, the number of Lie subalgebras is finite and each occurs as a direct summand. The proof of this fact is to just repeat the argument of the Jordan decomposition for a Lie subalgebra and look at coefficients of the fixed basis.

## 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.

## 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.

## Fundamental Theorem on Lie Algebra Actions

All we’re going to do now is try to take what we did in the last couple of posts and generalize. So instead of working on Lie groups where we have nice left invariance, and we had all our flows were complete, we’re going to try to get things for arbitrary vector fields on a manifold where the flow is not necessarily guaranteed to be complete.

First off, I’m going to want to think of right actions instead of left now. This is because in the last post I showed that flowing is the same thing as right multiplication by $exp$. From now on, I’m assuming we have a Lie group acting smoothly on a manifold on the right $\theta(p,g)=p\cdot g$. We want a global flow action of $\mathbb{R}$ on our manifold $M$, so let $X\in Lie(G)$. Define the action $\mathbb{R}\times M \to M$ by $t\cdot p=p\cdot exp(tX)$ (note: the two dots are two different actions, the one on the right comes from the Lie group).

We have an infinitesimal generator for this flow, say $\widehat{X}\in\frak{X}(M)$. i.e. $\widehat{X}_p=\frac{d}{dt}\big|_{t=0}p\cdot exp(tX)$. Thus we have a map $\widehat{\theta}:Lie(G)\to\frak{X}(M)$ by $\widehat{\theta}(X)=\widehat(X)$.

Let’s break down what this map really is. For any $p\in M$, examine $\theta^p:G\to M$ by $\theta^p(g)=p\cdot g$. This is a smooth, since we can identify $G\cong \{p\}\times G$ and then it is just inclusion followed by the smooth action. Thus this is the orbit map of the action. You get everything in the orbit of $p$ by the action of $G$. Thus $\widehat{X}_p=d(\theta^p)_e(X_e)$.

Let’s go one step further and show that $X$ and $\hat{X}$ are $\theta^p$ related (note now that $X$, $p$, and the action are completely arbitrary, so this is really a very general statement).

By the group law $p\cdot gh=(p\cdot g)\cdot h$, we actually have $\theta^p\circ L_g(h)=\theta^{p\cdot g}(h)$. Now we just check:

$\widehat{X}_{p\cdot g}=d(\theta^{p\cdot g})_e(X_e)$
$= d(\theta^p)_g\circ d(L_g)_e(X_e)$
$= d(\theta^p)_g(X_g)$.

Which shows $\theta^p$ related.

Now we easily can get that $\widehat{\theta}: Lie(G)\to \frak{X}(M)$ is a Lie algebra hom. Using linearity of $\widehat{\theta}$ for a fixed p, and the previous statement, we get that $[\widehat{X}, \widehat{Y}]_p=d(\theta^p)_e([X, Y]_e)=\widehat{[X, Y]}_p$. Thus $[\widehat{\theta}, \widehat{\theta}]=\widehat{\theta}([X, Y])$.

Now we are ready to state the “Fundamental Theorem on Lie Algebra Actions”. A quick term, we say a $\frak{g}$-action $\widehat{\theta}$ is complete if $\widehat{\theta}(X)$ is complete for all $X$.

The FT on LAA says that given any complete $Lie(G)$-action $\widehat{\theta}:Lie(G)\to\frak{X}(M)$, there is a unique smooth right $G$-action on M whose infinitesimal generator is $\widehat{\theta}$.

I won’t prove this, but it was nice to state and a good ending place for the day.

## More Exponential Properties

I’ll just do a quick finish up on proving basic properties about the exponential map for today, since I’ll be moving on to a different topic after this.

The exponential map actually restricts to a diffeomorphism from some neighborhood of 0 in $\frak{g}$ to some neighborhood of $e\in G$. Well, if you’ve mastered your basic theorems, then this is no surprise. Last time we showed $dexp_0(X)=X$. The differential is the identity map and hence non-singular. Thus by the Inverse Function Theorem, there exists neighborhoods about 0 and $e$ such that there is a smooth inverse, and hence $exp$ restricted to these is a diffeo.

Probably the most interesting and non-obvious of these facts is the following one. Given another Lie group $H$ (and Lie algebra $\frak{h}$), and a Lie group homomorphism $F:G\to H$, we get a commutative diagram (which I’ve yet to figure out how to make on wordpress). Anyway it just says that $exp(F_*X)=F(exp(X))$. So pushing forward a vector field and then exponentiating it is the same thing as exponentiating first and then doing the group hom.

Let’s do it by showing that $exp(tF_*X)=F(exp(tX))$ for all $t\in\mathbb{R}$. Recall that the LHS is just the one-parameter subgroup generated by $F_*X$. Thus by uniqueness, we just need to show that $\gamma(t)=F(exp(tX))$ is a hom that satisfies $\gamma'(0)=(F_*X)_e$. It is a composition of homs, so is itself one. Now we compute the derivative: $\gamma'(0)=\frac{d}{dt}\big|_{t=0} F(exp(tX))$
$=dF_0\left(\frac{d}{dt}\big|_{t=0}exp(tX)\right)$
$=dF_0(X_e)$
$=(F_*X)_e$.

The last property for today is used quite a bit as well. It says that flowing by a vector field $X$ is the same thing as right multiplying by $exp(tX)$. In other words, $(\theta_X)t=R_{exp(tX)}$.

First note that for any $g\in G$, the map $\sigma(t)=L_g(exp(tX))$, satisfies $\sigma '(0)=dL_g (\frac{d}{dt}\big|_{t=0} exp(tX))=dL_g (X_e)=X_g$ by left invariance. Also, left multiplication takes integral curves to integral curves, so this is actually the integral curve of $X$ starting at $g$. Thus $L_g(exp(tX))=\theta^g_X(t)$.

Now we check what we wanted to show. $R_{exp(tX)}(g)=g\cdot exp(tX)=L_g(exp(tX))=\theta^g_X(t)=(\theta_X)_t(g)$. And we’re done!

Next I’ll talk about infinitesimal generators of Lie group actions, so that I can pull in some stuff I did from the Frobenius theorem.

Update: There is some sort of weird bug where when I hit preview it tells me to save the draft. Then when I hit save draft, it posts. So, sorry if you’ve been getting unedited versions of my posts.

## The Exponential Map

Notice from last time that the matrix exponential maps from the Lie algebra to the Lie group. It took an arbitrary matrix (i.e. a tangent vector at the identity) and mapped it to a non-singular matrix (i.e. a matrix in $GL_n(\mathbb{R})$). Now we want a map $exp: \frak{g}\to G$ that does the same basic idea. Let’s unravel what that idea was.

It was sort of hidden, but we said that the one-parameter subgroup of $GL_n(\mathbb{R})$ generated by A, is $F(t)=e^{tA}$. So it took a line through the origin: $t\mapsto tA$, and sent it to a one-parameter subgroup $F(t)$.

Let’s sort of work backwards. We have a one-parameter subgroup generated by $X$, say $F(t)$ from the last post (the integral curve starting at the identity). From our matrix example, we also saw that if we flowed along this for 1 time unit, $F(1)=e^X$. Thus we’ll define $exp(X)=F(1)$, where $F$ is the one-parameter subgroup generated by $X$. Thus $exp$ will be mapping from the Lie algebra to the Lie group.

Now we need to check the line through the origin property. i.e. Is $F(s)=exp(sX)$ the one-parameter subgroup generated by $X$? Well, it is really a simple matter of rescaling. Let $\overline{F}(t)=F(st)$ by rescaling. Then $\overline{F}(t)$ is the integral curve of $sX$ starting at $e$, so $exp(sX)=\overline{F}(1)=F(s)$.

Now there are lots of properties we should check. I’ll go through a few this time and the rest tomorrow before moving on. First off, we hope that $exp$ is a smooth map. In other words, in terms of the flow, we need $\theta_X^e(1)$ to depend smoothly on $X$.

Define a vector field on $G\times\frak{g}$ by $Y_{(g, X)}=(X_g, 0)$. Note we made the natural identification $T_{(g, X)}(G\times \frak{g})\cong T_gG\oplus T_X\frak{g}$. Let $X_1, \ldots, X_k$ be a basis for $\frak{g}$ and $(x^i)$ the coordinates for $\frak{g}$, $x^iX_i$. Also, let $(w^i)$ be smooth coordinates for $G$.

Let $f\in C^{\infty}(G\times \frak{g})$. Then in coordinates $Y(f(w^i, x^i))=\sum_j x^jX_jf(w^i, x^i)$. Each $X_j$ is a derivative in the $w^j$ direction, so the expression depends smoothly on $(w^i, x^i)$. Hence $Y$ is a smooth vector field.

Now the flow of $Y$, say $\Theta$, is $\Theta_t(g, X)=(\theta_X(t, g), X)$. Now the fact that $\Theta$ is a flow of a smooth map shows that it is smooth. Let $\pi_G:G\times \frak{g}\to G$ be projection. Then $exp(X)=\pi_G(\Theta_1(e, X))$ and hence is a composition of smooth maps so is itself smooth.

That was sort of exhausting, so I’ll just do one quick other property before quitting: $exp((s+t)X)=exp(sX)exp(tX)$. This is just because $t\mapsto exp(tX)$ is a one-parameter subgroup and hence homomorphism. The group structure on $\mathbb{R}$ is additive and the Lie group operation is multiplicative: $exp((s+t)X)=F(s+t)=F(s)F(t)=exp(sX)exp(tX)$.

Alright, one more quick one, since that one shouldn’t count. This one is a big one, in that it corresponds with a very characteristic property of the complex valued exponential. If we identify $T_0\frak{g}$ and $T_eG$ with $\frak{g}$, then $d\exp_0:T_0\frak{g}\to T_eG$ is the identity map!

Let $\gamma(t)=tX$. Then $\gamma'(0)=X$. Thus $dexp_0(X)=dexp_0(\gamma'(0))=(exp\circ\gamma)'(0)=\frac{d}{dt}\big|_{t=0}exp(tX)=X$ (the equality comes from the fact that exp is the flow).

I guess this ran a bit long. Oh well, time is running out.