Show that \(\Aut \PP^n\slice{k} = \PGL_{n+1}\slice{k} = \GL_{n+1}\slice{k} /\GG_m\).
Solution
Assume \(n\ge1\) and write \[ V=k^{n+1}, \qquad \PP^n_k=\PP(V). \] For \(n=0\) both sides are the trivial group.
1. Every element of \(\GL(V)\) induces an automorphism of \(\PP(V)\).
For \[ A\in\GL(V), \] define \[ \overline A:\PP(V)\longrightarrow\PP(V), \qquad [v]\longmapsto[Av]. \] This is a projective linear morphism, and \[ \overline{A^{-1}} \] is its inverse. Thus there is a group homomorphism \[ \GL(V)\longrightarrow\Aut_k(\PP(V)). \]
2. The kernel of\[ \GL(V)\longrightarrow\Aut_k(\PP(V)) \]is the scalar subgroup\[ k^\times I\cong\GG_m(k). \]
Every scalar matrix acts trivially on one-dimensional subspaces, so \[ k^\times I \] lies in the kernel.
Conversely, suppose \(A\in\GL(V)\) acts trivially on \(\PP(V)\). Then every nonzero vector spans an \(A\)-stable line. For a basis \[ e_0,\ldots,e_n \] write \[ Ae_i=\lambda_i e_i. \] Since the line spanned by \(e_i+e_j\) is also fixed, \[ \lambda_i e_i+\lambda_j e_j \] must be a scalar multiple of \[ e_i+e_j. \] Hence \[ \lambda_i=\lambda_j \] for every \(i,j\). Thus \(A=\lambda I\) is scalar.
3. Therefore the projective linear action gives an injectivehomomorphism\[ \PGL(V)=\GL(V)/k^\times \injects \Aut_k(\PP(V)). \]
This is the first isomorphism theorem applied to steps <1>1–<1>2.
4. Let\[ f\in\Aut_k(\PP^n). \]Then\[ \boxed{f^*\OO_{\PP^n}(1)\cong\OO_{\PP^n}(1)}. \]
The Picard group is \[ \Pic(\PP^n_k)\cong\ZZ, \] generated by \(\OO(1)\) D-CB9XS. Since \(f\) is an isomorphism, \[ f^*:\Pic(\PP^n)\longrightarrow\Pic(\PP^n) \] is a group automorphism. Therefore \[ f^*\OO(1)\cong\OO(\varepsilon), \qquad \varepsilon\in\{1,-1\}. \]
The pullback of an ample line bundle by an isomorphism is ample. The bundle \(\OO(1)\) is ample, whereas \(\OO(-1)\) is not ample. Hence \[ \varepsilon=1. \]
5. After choosing an isomorphism\[ \alpha:f^*\OO(1)\xrightarrow{\sim}\OO(1), \]pullback of sections gives an invertible linear map\[ T: H^0(\PP^n,\OO(1)) \longrightarrow H^0(\PP^n,\OO(1)). \]
Pullback along \(f\) gives an isomorphism \[ H^0(\PP^n,\OO(1)) \xrightarrow{\sim} H^0(\PP^n,f^*\OO(1)). \] Composing with the isomorphism on global sections induced by \(\alpha\) gives \(T\). Both maps are invertible, so \(T\) is an invertible \(k\)-linear transformation.
6. The automorphism \(f\) is induced by an element of \(\GL_{n+1}(k)\).
Let \[ x_0,\ldots,x_n \] be the standard basis of \[ H^0(\PP^n,\OO(1)). \] Put \[ \ell_i=T(x_i). \] Since \(T\) is invertible, the \(\ell_i\) form a basis of the space of linear forms. Hence there is an invertible matrix \[ A\in\GL_{n+1}(k) \] whose coordinate linear forms are \(\ell_0,\ldots,\ell_n\).
The morphism defined by this basis is \[ \overline A: [v]\longmapsto [\ell_0(v):\cdots:\ell_n(v)]. \] For each homogeneous coordinate section \(x_i\), its pullback along \(f\), after the chosen identification \(\alpha\), is exactly \(\ell_i\). On the open set where \(\ell_j\ne0\), both \(f\) and \(\overline A\) land in the standard chart \(D_+(x_j)\), and the pullback of its affine coordinate \(x_i/x_j\) is \[ \frac{\ell_i}{\ell_j} \] for both maps. The opens \(\{\ell_j\ne0\}\) cover \(\PP^n\) because the \(\ell_j\) form a basis and hence have no common projective zero. Therefore the two morphisms agree on an open cover: \[ f=\overline A. \] Therefore every automorphism of \(\PP^n\) is projective linear.
7. Consequently,\[ \boxed{ \Aut_k(\PP^n) \cong \PGL_{n+1}(k) \cong \GL_{n+1}(k)/k^\times. } \]
Step 3 gives injectivity of the projective linear action, and step 6 gives surjectivity. The quotient description is the definition of \(\PGL_{n+1}(k)\).
8. Q.E.D.
Step 7 is the required identification.