A question very close to this one was already asked: Automorphisms of a weighted projective space

But the answer given does not satisfy my needs. So avoiding having two questions that are identical, I am interested in a specific type of weighted projective space, namely $\mathbb{P}(1, 1, \cdots, 1, k)$ for some natural number $k$. The case $k = 1$ gives rise the usual projective space. To be more precise, I am considering a real weighted projective space with weight vector $(1, 1, \cdots, 1, k)$.

So the question is can one characterize, in general, the automorphism group of such a space. If $k = 1$ then we get the projective linear group, namely the group $\textbf{GL}(\mathbb{R}^n)/\mathbb{R}^\ast \cong \textbf{PGL}(\mathbb{R}^n)$. In general, can one characterize the automorphism group of weighted projective spaces of the type $\mathbb{P}(1, 1, \cdots, 1, k)$?


2 Answers 2


I'm not sure how one deals with the general case, but your example is the cone in $\mathbb P^{N+1}$ over the $k$-th Veronese embedding of $\mathbb P^{n-1}(\mathbb K)$ in $\mathbb P^N$ ($n=$ number of 1's, $N=$ the appropriate dimension for the Veronese embedding).

So for $k>1$ the automorphism group $G$ is an extension $$1\to \mathbb K^*\times {\mathbb K}^{N+1}\to G\to \mathbb PGL(n)\to 1.$$ EDIT: as J\'er\'emy points out below and in his answer, the kernel in the sequence above is not a direct product, but a more complicated group.

This is easily seen by taking homogeneous coordinates in $\mathbb P^{N+1}$ such that the vertex of the cone is, say, $[1,0\dots 0]$ and representing an automorphism by a matrix such that the first column is $^t(1,0\dots 0)$.

  • 2
    $\begingroup$ Pay attention that the kernel of the map $G\to \mathrm{PGL}(n,\mathbb{K})$ is not equal to $\mathbb{K}^*\times \mathbb{K}^{N+1}$. In fact is it not abelian: for example, the elements $(x_1:\dots:x_n:y)\mapsto(x_1:\dots:x_n:\alpha y)$ and $(x_1:\dots:x_n:y)\mapsto (x_1:\dots:x_n:y+(x_1)^k)$ do not commute in general. See below. $\endgroup$ Sep 28, 2012 at 12:25
  • $\begingroup$ You are right, I had been too hasty. Thanks for pointing this out. I'll edit my answer. $\endgroup$
    – rita
    Sep 28, 2012 at 14:00
  • $\begingroup$ Thanks for editing, anyway the good point was to describe the isomorphism and action on $\mathbb{P}^{n-1}$. But it is funny that the kernel is quite strange and depends in fact on the roots of unity in $\mathbb{K}^*$. $\endgroup$ Sep 28, 2012 at 16:58

As rita said $\mathbb{P}(1,\dots,1,k)$ is naturally isomorphic to the cone in $\mathbb{P}^{N+1}$ over the $k$-th embedding (Take the map which sends $(x_1:\dots:x_n:y)$ onto $((x_1)^k:(x_1)^{k-1}x_2:...:(x_n)^k:y)$ where the $N+1$ first coordinates are the monomials of degree $k$ in $x_1,\dots,x_n$), so there is a natural morphism from the group $G=\mathrm{Aut}(\mathbb{P}(1,\dots,1,k))$ to $\mathrm{PGL}(n,\mathbb{K})$. However, the kernel is not the one which was described in the above answer.

We can in fact give $G$ more explicitly (because there are a priori many extensions given two groups):

We choose $k>1$ (otherwise the description is different and obvious). We identify $\mathbb{K}^{N+1}$ with the set of homogeneous polynomials of degree $k$ in $n$ variables. The group $\mathrm{GL}(n,\mathbb{K})$ naturally acts on $\mathbb{K}^{N+1}$.

Let $H$ be the semi-direct product $\mathbb{K}^{N+1}\rtimes \mathrm{GL}(n,\mathbb{K})$. There is a natural surjective map $H\to G$, that we describe now:

The action of $\mathbb{K}^{N+1}$ on $\mathbb{P}(1,\dots,1,k)$ is given by $(x_1:\dots:x_n:y)\mapsto (x_1:\dots:x_n:y+P(x_1,\dots,x_n))$ where $P\in\mathbb{K}^{N+1}$ is the corresponding polynomial.

The action of $\mathrm{GL}(n,\mathbb{K})$ on $\mathbb{P}(1,\dots,1,k)$ is given by the action on $x_1,\dots,x_n$.

It yields thus a morphism $H\to G$ whose kernel is the subgroup $L$ of $\mathrm{GL}(n,\mathbb{K})$ consisting of diagonal matrices of the form $\{\lambda I| \lambda^k=1\}$.

The group $G=\mathrm{Aut}(\mathbb{P}(1,\dots,1,k))$ is thus equal to the quotient of $\mathbb{K}^{N+1}\rtimes \mathrm{GL}(n,\mathbb{K})$ by the subgroup $L$.

The surjective morphism $G\to \mathrm{PGL}(n,\mathbb{K})$ corresponds to the projection on $\mathrm{GL}(n,\mathbb{K})/L$ followed by the quotient by the image of all diagonal matrices (we have first killed only finitely many and then kill all others). The kernel of this map is thus equal to $\mathbb{K}^{N+1}\rtimes\mathbb{K}^{*}/L$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.