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.