Derivatives of morphisms of linear algebraic groups

Question

I am currently trying to learn about linear algebraic groups and their lie algebra structure.

However, I am struggling to explicitly calculate the derivatives of morphisms between algebraic groups, as soon as they are not explicitly given by polynomial functions. (e.g. det: $GL_n \to G_m$ is clear for me as you can just differentiate and get the trace).

However, if you take e.g. quotient maps, I don't know how to find derivatives in a kind of "algorithmic" way.

Take for example Exercise 4.4.11(3) in Springer's book on Linear Algebraic groups. There, he gets the map $\varphi: SL_2 \to PSL_2$ by firstly taking the inclusion $SL_2 \to GL_2$ and then the quotient map $GL_2 \to GL_2/Z(GL_2)$. How would one go about calculating the derivative $d\varphi$ to show that it is an isomorphism in this case (for char(k) not 2)?

Answer 1

I mean I don't really think you can find them in an algorithmic way that doesn't require you giving a description of the map in coordinates. But, this is perhaps not that unexpected. We need something more abstract.

The following can be phrased more generally, with a slightly less desirable conclusion. But, this level of generality is maybe not really helpful for you right now. I am not sure what your background level is. Let me know if what I have written is not understandable.

Proposition: Suppose that $$1\to K\to G\to Q\to 1$$ is a short exact sequence of smooth algebraic groups over $F$. Then, there is an exact sequence $$0\to T_e(K)\to T_e(G)\to T_e(Q)\to 0.$$

I am not sure what your background is, but let me try to clarify some things:

• here by 'algebraic group', I mean finite type affine group $F$-scheme,

• by smooth I mean in the sense of algebraic geometry but, for groups, this is equivalent to being geometrically reduced, equivalently reduced if $F$ is perfect, and actually it's automatic if $F$ has characteristic $0$,

• the exactness means that $K$ maps isomorphically to $\ker(G\to Q)$, and that $G\to Q$ is faithfully flat, and as $Q$ is smooth over $F$ this is equivalent to $G\to Q$ being surjective.

Proof of Proposition: As this question is entirely unchanged after base changing to $\overline{F}$, we may assume without loss of generality that $F$ is algebraically closed.

Recall that one way to define $T_e(H)$ is as $\ker(H(F[\varepsilon])\to H(F))$, where $F[\varepsilon]:=F[x]/(x^2)$ is the dual number, and $F[\varepsilon]\to F$ is the map sending $\varepsilon$ to $0$. Moreover, if $H$ is smooth over $F$ then we have that $H(F[\varepsilon])\to H(F)$ is actually surjective -- this is just the infinitesimal lifting criterion!

So, we get a map of exact sequences

$$\begin{matrix}1 & \to & K(F[\varepsilon]) & \to & G(F[\varepsilon]) & \to & Q(F[\varepsilon]) & \to & 1 \\ & & \downarrow & & \downarrow & & \downarrow & & \\ 1 & \to & K(F) & \to & G(F) & \to & Q(F) & \to & 1\end{matrix}$$

where the rows are exact by the definition of exact sequence, except for exactness on the right. The fact that $G(F)\to Q(F)$ is surjective is clear. The fact that $G(F[\varepsilon])\to Q(F[\varepsilon])$ is surjective uses the smoothness of $K$. One way to say this is that $K$ being smooth actually implies that $G\to Q$ is smooth, and being smooth implies that that you can always lift a section after moving to an etale cover (if that means anything to you) -- but the connected etale covers of $F[\varepsilon]$ are just itself, so you can actually just lift the section.

So, from the Snake Lemma (see here for why this makes sense here) we deduce that there is an exact sequence

$$1\to T_e(K)\to T_e(G)\to T_e(Q)\to \mathrm{coker}(Q(F[\varepsilon])\to Q(F)).$$

But, as $Q$ is smooth, the map $Q(F[\varepsilon])\to Q(F)$ is surjective, and so this last cokernel is trivial. $\blacksquare$

So, let's apply this in your situation. So, as you said, we will assume that $F$ does not have characteristic $2$. Then, we have the short exact sequence

$$1\to \mu_2\to \mathrm{SL}_2\to \mathrm{PSL}_2\to 1,$$

where $\mu_2$ is the second roots of unity, embedded diagonally in $\mathrm{SL}_2$. All the groups here are smooth over $F$ (we are using that $F$ does not have characteristic $2$ here to say that $\mu_2$ is smooth). So, by our proposition we get a short exact sequence

$$0\to T_e(\mu_2)\to T_e(\mathrm{SL}_2)\to T_e(\mathrm{PSL}_2)\to 0.$$

But, note that $T_e(\mu_2)$ is trivial. Namely, $\mu_2$ is a finite (and so 'discrete') group, and so has trivial tangent space. More concretely, the map

$$\{\pm 1\}=\mu_2(F[\varepsilon])\to \mu_2(F)=\{\pm 1\},$$

is injective, so the kernel is trivial. Thus, we deduce that $T_e(\mathrm{SL}_2)\to T_e(\mathrm{PSL}_2)$ is an isomorphism as you desired.

In fact, inspecting the proof shows that if $\varphi\colon G\to H$ is a smooth isogeny of smooth algebraic groups (i.e. $\varphi$ is surjective with finite smooth kernel), then $d\varphi$ is an isomorphism. The intuition for this is clear, as such smooth isogenies are like 'finite covering spaces', and so like 'local isomorphisms' -- so of course they induce isomorphisms on tangent spaces.

Answer 2

Question:"However, if you take e.g. quotient maps, I don't know how to find derivatives in a kind of "algorithmic" way."

Answer: A "hint" on how to make an explicit and elementary calculation (I mentioned this in the comments but it was deleted).

Let $SL_2:=Spec(k[SL_2])$ and $PSL_2:=Spec(k[PSL_2])$. You define $PSL_2:=SL_2/\mathbb{G}_m$. There is an action

$$\sigma^*: \mathbb{G}_m \times SL_2 \rightarrow SL_2$$

inducing the "coaction"

$$ \sigma: k[SL_2] \rightarrow k[\mathbb{G}_m]\otimes k[SL_2]$$

and the coordinate ring of $PSL_2$ is the ring of coinvariants: $k[PSL_2]:=k[SL_2]^{\sigma}$, where $k[SL_2]^{\sigma}$ is the ring of elements $x$ with $\sigma(x)=1 \otimes x$. The ring $k[PSL_2]$ is calculated in the Springer book. Let $k[SL_2]:=k[x_1,x_2,x_3,x_4]/(D-1)$ where $D:=x_1x_4-x_2x_3$. There is a natural inclusion

$$i: k[PSL_2] \rightarrow k[SL_2]$$

and the ideal of the "identity" $e$ in $k[SL_2]$ is the ideal $I:=(x_1-1, x_2,x_3,x_4-1)$. You must calculate the inverse image maximal ideal $i^{-1}(I):=J$ and you get a canonical surjection

$$ j:J/J^2 \rightarrow I/I^2.$$

The dual map $j^*$ is the tangent map:

$$j^* :=d_e(\pi): T_e(SL_2) \rightarrow T_e(PSL_2).$$

Then you must calculate $d_e(\pi)$.

Hence you must understand what the coaction $\sigma$ is, calculate the invariant ring $k[PSL_2]:=k[SL_2]^{\sigma}$ and calculate the ideals $I,J$.

This gives an "algorithm":

1. Understand and calculate the invariant ring $k[SL_2]^{\sigma}$.
2. Calculate the ideals $I,J$
3. Calculate the map $J/J^2 \rightarrow I/I^2$ and the tangent map $d_e(\pi)$.