Let $k$ be an algebraically closed field. I am looking for an easily quotable description of automorphism groups of $\mathrm{Spec} k[x]/(x^n)$. I could compute explicit matrix representations for several small $n$, but a more intrinsic description would be desirable.

*Smooth points are all alike*; *every fat point is happy in its own way.*

"Anna Karenina", Leo Tolstoy

Let $k$ be an algebraically closed field. Denote $A_n:=k[x]/x^{n+2}$, $G_n:=\text{Aut}A_n$, and let $N_n\subset G_n$ stand for the unipotent radical of $G_n$. Since there is no problem in finding a faithful matrix representation of $G_n$, I guess, Dima asks about a coordinate-free description of $G_n$.

We will mainly deal with $N_n$ because $G_n=N_n\rtimes T$ is a semidirect product, where $\text{G}_m\simeq T\subset G_n$ has no canonical choice (though all tori are conjugated in $G_n$).

**Answer.** We assume for simplicity $k$ of characteristic $0$. For sufficiently big $n$, $N_{n+1}$ is a central extension (with that very centre) of $N_n$. This extension is determined by a certain orbit of the group $\text{Out}N_n$ naturally acting on the projective plane
${\mathbb P}_kH^2(N_n,k)$. The orbits are: a dense one, two $1$-dimensional orbits $L'$ and $H'$ dense in the projective lines $L$ and $H$, and a couple of fixed points $f_0,f_1\in L$, where $f_0$ is the intersection $L\cap H$. Our **choice** determining the extension $0\to k\to N_{n+1}\to N_n\to1$ is $H'$. The action of $T$ can be easily defined on the way.

The interesting stuff appears at the end, when the point becomes really fat.

**Boring stuff.** Every element $g\in G_n$ is determined by the polynomial $f_g:=gx$, where $f_g(x)=f_0x+f_1x^2+\dots+f_nx^{n+1}$ with $f_i\in k$ and $f_0\ne0$. The composition in $G_n$ is expressed in these terms as
$f_{g_1g_2}(x)\equiv f_{g_2}\big(f_{g_1}(x)\big)\mod x^{n+2}$. Note that
$g\in N_n$ iff $f_0=1$ and $g\in T$ iff $f_1=\dots=f_n=0$. Sometimes, it is more convenient to use other coordinates: $f_g(x)=x\Big(h_g(x)+\displaystyle\sum_{i=0}^n(p_1x)^i\Big)$, where
$h_g(x)=\displaystyle\sum_{i=2}^np_ix^i$. So, $p_1=f_1$ and $p_i=f_i-f_1^i$ for $i\ge2$.

The Lie algebra $L_n$ of $G_n$ is formed by all the derivations from $A_n$ to $A_n$, i.e., by all the derivations from $k[x]$ to $k[x]$ that preserve $\text{Ideal}(x^{n+2})$. The formula $df(x)=f'(x)dx$ correctly defines a derivation $d\in L_n$ by its value $dx\in A_nx$. Taking $d_ix=x^{i+1}$ for $i=0,1,\dots,n$, we obtain a basis of $L_n$ subject to $[d_i,d_j]=(j-i)d_{i+j}$ for $i+j\le n$ and $[d_i,d_j]=0$ for $i+j>n$. The Lie algebra $M_n$ of $N_n$ is simply spanned by $d_1,\dots,d_n$.

**0.** $N_0=1$ and $G_0=\text{G}_m$.

**1.** $N_1=\text{G}_a$ and $G_1=N_1\rtimes\text{G}_m$ is the semidirect product with respect to $\text{G}_m=\text{Aut}\text{G}_a$.

**2.** $N_2=\text{G}_a\oplus\text{G}_a$ (the corresponding coordinates are, for instance, $p_1,p_2$) and $G_1=N_1\rtimes\text{G}_m$, where the action of $c_0\in\text{G}_m$ is given by the rule $c_0\cdot(p_1,p_2):=(c_0p_1,c_0^2p_2)$.

**Preliminaries.** The homomorphism $\pi:A_n\to A_{n-1}$ induces the surjective homomorphisms $\pi:G_n\to G_{n-1}$ and $\pi:N_n\to N_{n-1}$ whose kernel $\text{G}_a\simeq C_n\subset N_n$, formed by all $c\in N_n$ with $f_c=x(1+fx^n)$, $f\in k$, lies in the centre of $N_n$ and coincides with this centre unless $n=2$.

A few more well-known and easy facts. Under the assumption that there is a rational section of $\pi$ (which is valid in our case of $X_h=N_{n+1}$ and $X=N_n$), a central extension $0\to k\to X_h\overset\pi\to X\to1$ of algebraic groups is given by an (arbitrary) element $h\in H^2(X,k)$ of rational $2$-cohomologies of $X$ (the action of $X$ on $k=\text{G}_a$ is trivial). The groups $\text{Aut}X$ and $\text{G}_m$ act on $X$ and on $k$, hence, on $H^2(X,k)$. Assuming that $k$ coincides with the centre of $X_{h_i}$ for both $i=1,2$, the groups $X_{h_1}$ and $X_{h_2}$ are isomorphic iff $h_1$ and $h_2$ are in a same orbit of $\text{Out}X\times\text{G}_m$ (the inner automorphisms of $X$ act trivially on $H^2(X,k)$).

In our case, the central extension is nontrivial. Denote by
$0\ne[h_n]\in H^2(N_n,k)$ an element providing the extension $\pi:N_{n+1}\to N_n$. By the above, the group $N_{n+1}$ is in fact given by the choice of the orbit of $[h_n]$ with respect to the action of $\text{Out}N_n$ on
${\mathbb P}_kH^2(N_n,k)$. **Our task** is therefore to indicate this orbit and to explain how $T\simeq\text{G}_m$ acts on $N_{n+1}$ (already knowing how does $T$ act on $N_n$).

The homomorphism $\pi$ induces a $k$-linear map $\pi^*:H^2(N_{n-1},k)\to H^2(N_n,k)$ and, unless $n=2$, a homomorphism $\pi:\text{Out}N_n\to\text{Out}N_{n-1}$. Another elementary remark: an element $\alpha\in\text{Out}N_{n-1}$ belongs to the image $\pi\text{Out}N_n$ (i.e., is liftable) iff $[h_{n-1}]\in{\mathbb P}_kH^2(N_{n-1},k)$ is a fixed point of $\alpha$.

Using the Chevalley-Eilengebrg complex of $M_n$, one can calculate $H^2(M_n,k)\simeq H^2(N_n,k)$ for $n\ge2$ (the action of $M_n$ on $k$ is trivial), getting the following result. For any $n\ge4$, we have the cocycle $a:=d_2^*\wedge d_3^*$; for any $n\ge6$, we have the cocycle $b:=d_2^*\wedge d_5^*-3d_3^*\wedge d_4^*$; for any $n\ge2$, we have the cocycle $h_n:=\displaystyle\sum_{1\le i<j\le n\atop i+j=n+1}(j-i)d_i^*\wedge d_j^*$. Now, $1$-dimensional $H^2(M_2,k)$ and $H^2(M_3,k)$ are spanned respectively by $[h_2]$ and $[h_3]$, $2$-dimensional $H^2(M_4,k)$ and $H^2(M_5,k)$ are spanned respectively by $[a],[h_4]$ and $[a],[h_5]$, and a $3$-dimensional $H^2(M_n,k)$ is spanned by $[a],[b],[h_n]$ when $n\ge6$. Note also that (when it makes sense) $\pi^*:[a]\mapsto[a]$, $\pi^*:[b]\mapsto[b]$, and $\pi^*:[h_n]\mapsto0$.

**First steps.** I claim (without proof as it is bulky) that $\text{Out}N_n$ is $4$-dimensional.

Since $N_2$ is in fact a $2$-dimensional $k$-linear space, we conclude that $\text{Out}N_2=\text{GL}_2k$. In view of $\dim_kH^2(N_2,k)=1$, we have a unique possible extension, thus obtaining the famous Heisenberg group $N_3$. The group $\text{Out}N_2$ is entirely liftable, and $\text{Aut}N_3$ is formed by $3\times 3$-matrices $G:=\left[\begin{smallmatrix}g&0\\{*}&\det g\end{smallmatrix}\right]$, where $g\in\text{GL}_2k$. Taking $g=1$, we obtain the normal subgroup $N_2\subset\text{Aut}N_3$ of inner automorphisms of $N_3$, hence, $\text{Out}N_3=\text{GL}_2k$ is $4$-dimensional, as claimed. The action of $T$ on $N_3$ will be defined later.

Since $\dim_kH^2(N_3,k)=1$, we again get a unique possible extension $N_4$, the entirely liftable $\text{Out}N_3$, and $\text{Out}N_4=\text{GL}_2k$.

When $\dim_kH^2(N_4,k)=2$, we are stuck.

**Lemma.** *For $n\ge5$, there is a $1$-dimensional subgroup $\text{G}_a\simeq U_n\subset\text{Out}N_n$ such that
$\text{Out}N_n=U_n\cdot\pi\text{Out}N_{n+1}$, $U_n\cap\pi\text{Out}N_{n+1}=1$, $\pi^3U_n=1$, and $\pi^2U_n\ne1$.*

In place of a **proof**, we only define how $u\in k$ acts on $f(x)=x\Big(1+\displaystyle\sum_{i=1}^nf_ix^i\Big)\in N_n$. (Let me skip some reasons allowing to get rid of an unpleasant straightforward verification.) We use the coordinates $p_1:=f_1$ and $p_i:=f_i-f_1^i$ for $i\ge2$. Consider the "quadratic" polynomial $q_f(x):=q_{f,0}+q_{f,1}x+q_{f,2}x^2$, where
$$q_{f,0}:=p_2,\qquad q_{f,1}:=(n-3)p_3-2(n-4)p_2p_1,$$
$$q_{f,2}:=\frac{(n-2)(n-3)}2p_4-(n-3)(n-4)p_3p_1-\frac{(3n-14)(n-1)}4p_2^2+\frac{(n-4)(n-9)}2p_2p_1^2.$$
The action of $u\in k$ on $f$ is given by the rule $u\cdot f:=f+ux^{n-1}q_f$.

**Fat point fun.** Using Lemma and omitting the $\pi$'s, we can write $\text{Out}N_n=U_nU_{n+1}U_{n+2}T$. As $U_{n+1},U_{n+2},T$ are liftable, they fix the point $[h_n]\in H^2(N_n,k)$. As $U_n$ is not liftable, the orbit $(\text{Out}N_n)[h_n]=U_n[h_n]$ is $1$-dimensional. The action of $U_n,U_{n+1},U_{n+2}$ on $[a]$ and $[b]$ is trivial because it can be performed at the level of $H^2(N_{n-3},k)$, where $U_n,U_{n+1},U_{n+2}$ vanish. It is easy to see that $[a]$ and $[b]$ are fixed points of $T$ of different weights. Finally, we can get the picture of $\text{Out}N_n$-orbits in
${\mathbb P}_kH^2(N_n,k)$. There is a dense orbit which is the complement of two lines $L$ and $H$. The intersection $L\cap H$ is a fixed point $f_0$ (either $[a]$ or $[b]$; no idea which one). The line $L$ spanned by $[a]$ and $[b]$ contains therefore one more fixed point $f_1$. Our choice is the $1$-dimensional orbit $H'$ such that the compliment $H\setminus H'$ in its closure $H$ is a single point $f_0$.

I forgot to define the action of $T$ on $N_n$. For big $n$, it is easy: just take any torus $T\subset\text{Aut}N_n$. In order to get the action of $T$ for small $n$, simply apply a number of $\pi$'s.

**Conclusion.** When we killed $[h_n]$ by inserting it as a centre, a new $[h_{n+1}]$ was born, and the other cohomologies $[a]$ and $[b]$ survived. What if we kill $[a]$ or $[b]$, once or several times. How probable is that we obtain a group of automorphisms of some fat point? (The algebras $A_n$ provide possibly the simplest series of artinian local algebras. Of course, there is no hope to classify all local artinian algebras. Nevertheless, my question may have a tame nature.)

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