Automorphisms of a specific type of weighted projective space - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T03:06:34Z http://mathoverflow.net/feeds/question/107933 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/107933/automorphisms-of-a-specific-type-of-weighted-projective-space Automorphisms of a specific type of weighted projective space Stanley Yao Xiao 2012-09-23T21:04:16Z 2012-09-28T14:02:06Z <p>A question very close to this one was already asked: <a href="http://mathoverflow.net/questions/67363/automorphisms-of-a-weighted-projective-space" rel="nofollow">http://mathoverflow.net/questions/67363/automorphisms-of-a-weighted-projective-space</a></p> <p>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 $(1, 1, \cdots, 1, k)$. </p> <p>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 $\mathbb{P}(1, 1, \cdots, 1, k)$?</p> http://mathoverflow.net/questions/107933/automorphisms-of-a-specific-type-of-weighted-projective-space/107961#107961 Answer by rita for Automorphisms of a specific type of weighted projective space rita 2012-09-24T07:12:00Z 2012-09-28T14:02:06Z <p>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). </p> <p>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.</p> <p>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 $^t(1,0\dots 0)$.</p> http://mathoverflow.net/questions/107933/automorphisms-of-a-specific-type-of-weighted-projective-space/108337#108337 Answer by Jérémy Blanc for Automorphisms of a specific type of weighted projective space Jérémy Blanc 2012-09-28T12:20:15Z 2012-09-28T12:20:15Z <p>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.</p> <p>We can in fact give $G$ more explicitly (because there are a priori many extensions given two groups):</p> <p>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 $\mathbb{K}^{N+1}$.</p> <p>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:</p> <p>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.</p> <p>The action of $\mathrm{GL}(n,\mathbb{K})$ on $\mathbb{P}(1,\dots,1,k)$ is given by the action on $x_1,\dots,x_n$.</p> <p>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 $\{\lambda I| \lambda^k=1\}$.</p> <p>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 $L$. </p> <p>The surjective morphism $G\to \mathrm{PGL}(n,\mathbb{K})$ corresponds to the projection on $\mathrm{GL}(n,\mathbb{K})/I$ 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 $\mathbb{K}^{N+1}\rtimes\mathbb{K}^{*}/I$.</p>