Introduction to Groups and Geometries

We will show that \(L_g\) is invertible. Similar arguments show that \(R_g,C_g\) are invertible. We havefor all \(x\in G\text{.}\) A similar argument shows that \(L_g(L_{g^{-1}}(x))=x\) for all \(x\text{.}\) We conclude that \(L_g\) is invertible, with inverse \((L_g)^{-1}=L_{g^{-1}}\text{.}\) Thus \(L_g\) is an element of \(\Perm(G)\text{,}\) as desired.

You just have to check that \(L_{gh}=L_g\circ L_h\text{,}\) \(R_{gh}=R_g\circ R_h\) (here you see why the inverse is needed!), and \(C_{gh}=C_g\circ C_h\text{.}\) Here is the argument for \(L\text{.}\) Similar arguments work for \(R,C\text{.}\) Let \(g,h\in G\text{.}\) We have \(L_g(L_h(x))=g(hx) = (gh)x = L_{gh}(x)\) for all \(x\in G\text{,}\) so we have \(L_g\circ L_h = L_{gh}\text{.}\)

For injectivity, if \(L_g=L_h\text{,}\) then \(g=L_g(e)=L_h(e)=h\text{.}\) By the First Isomorphism Theorem, we conclude that \(G\approx L(G)\text{,}\) which is subgroup of \(\Perm(G)\text{.}\)

Apply Proposition 2.5.4.

Apply Fact 1.4.4 (check that the action of \(g\) is constant on projective equivalence classes of vectors).

Let \(\phi\colon GL(V)\to \Perm(\Proj(V))\) be the action defined by Equation (2.5.1). It is easy to check that \(K\subseteq \ker(\phi)\text{.}\) Containment in the other direction requires a short linear algebra argument. Suppose \(g\in \ker(\phi)\text{.}\) This means that for every \(v\in V\text{,}\) there is a nonzero scalar \(\lambda_v\) such that \(gv=\lambda_v v\) (that is, every vector in \(V\) is an eigenvector of the operator \(g\)). If \(\dim(V)=1\text{,}\) we are done. If \(\dim(V)\gt 1\text{,}\) choose independent vectors \(v,w\in V\text{.}\) We have \(\lambda_{v+w}(v+w)=g(v+w)=\lambda_v v+\lambda_w w\text{.}\) Because \(v,w\) are independent, it follows that \(\lambda_v=\lambda_w=\lambda_{v+w}\text{.}\) Thus it must be that there is a single \(\lambda\in \F^\ast\) such that \(gv=\lambda v\) for all \(v\in V\text{,}\) that is, \(g\) is an element of \(K\text{.}\)

Apply the First Isomorphism Theorem (Corollary 2.4.7) to \(\phi\colon GL(V)\to \Perm(\Proj(V))\) to obtain

Given \(\alpha,\beta\) in \(\F^\ast\) and \(T\) in \(GL(V)\text{,}\) the fact that \((\alpha \beta)T=\alpha(\beta T)\) verifies that the map \(\F^\ast \to \Perm(GL(V))\) given by \(\alpha\to [T\to \alpha T]\) is a homomorphism. The equivalence relation that corresponds to this action is given by \(T\sim_{\F^\ast} S\) if and only if \(S=\alpha T\) for some \(\alpha\in \F^\ast\text{.}\) Because \(S=\alpha T\) if and only if \(S=(\alpha \Id)T\text{,}\) the equivalence classes that define \(PGL(V)\) in the previous part of this problem are the same as the equivalence classes \(GL(V)/\F^\ast\text{.}\)

Factor the given map as \(\Proj(\C^2)\to \extC\to S^2\) given byLet \(\mu\) denote the map \([(\alpha,\beta)] \to \alpha/\beta \text{.}\) That \(\mu\) is well-defined follows from the observation that \((k\alpha)/(k\beta)=\alpha/\beta\) for all \(k\in \F^\ast\text{.}\) To see that \(\mu\) is bijective, we will exhibit an inverse map. Let \(\nu\colon \extC\to \Proj(\C^2)\) be given by \(\gamma \to [(\gamma,1)]\) for \(\gamma \in \C\) and \(\infty \to [(1,0)]\text{.}\) To check that \(\nu\circ\mu=\Id\text{,}\) for \(\beta\neq 0\text{,}\) we haveand for \(\beta=0\text{,}\) we haveReading the two displayed equations from right to left establishes that \(\mu\circ \nu=\Id\text{.}\) We conclude that \(\nu=\mu^{-1}\text{.}\) Using the fact that stereographic projection \(s\) is bijective, we have \(s^{-1}\circ \mu\) is bijective, as desired.