91Ó°ÊÓ

Let \(G\) be a topological group, and \(H \subset G\) a subgroup. Show that the closure \(\bar{H}\) of \(H\) is also a subgroup.

If \(G\) is a group, we define the opposite group \(G^{\circ}\) to be the same set with the multiplication \(+\) defined by \(a \cdot b=b \cdot a\). If \(g\) is a Lie algebra, with operation \([,\), , we define the opposite Lie algebra \(\mathfrak{g}^{\circ}\) to be the same set with the operation \([X, Y]^{\circ}=-[X, Y]\) (a) \(G^{\circ}\) is a group, and if \(\psi: G \rightarrow G\) is \(a \mapsto a^{-1}\), then \(\psi\) is an isomorphism from \(G\) to \(G^{\circ}\). (b) \(\Omega^{\circ}\) is a Lie algebra, and \(X \mapsto-X\) is an isomorphism of \(g\) onto \(\mathfrak{g}^{\circ}\). (c) \(\mathcal{L}\left(G^{\circ}\right)\) is isomorphic to \([\mathcal{L}(G)]^{\circ}=\mathfrak{q}^{\circ}\). (d) Let \([,\), be the operation on \(G_{e}\) obtained by using right invariant vector fields instead of left invariant ones. Then \((g,[,]\), is isomorphic to \(\mathcal{L}\left(G^{\circ}\right)\), and hence to \(\mathrm{g}^{\circ}\). (e) Use this to give another proof that \(\underline{g}\) is abelian when \(G\) is abelian.