Let \(G\) be a topological group. Show that \(H_*(G)\) is an algebra. Show that \(G\actson H_*(G)\), which factors through the homomorphism \(G \into \pi_0(G)\) yielding a trivial action if \(G\) is path-connected.
Solution
1. Let \(m:G\times G\to G\) be multiplication and \(e\in G\) the identity. Define the Pontryagin product\[ \star:H_p(G)\otimes H_q(G)\longrightarrow H_{p+q}(G) \]by\[ a\star b=m_*(a\times b), \]where \(a\times b\) is the homology cross product.
The cross product sends \(H_p(G)\otimes H_q(G)\) naturally to \(H_{p+q}(G\times G)\), and multiplication then induces a map back to \(H_{p+q}(G)\).
2. This product makes \(H_*(G)\) a graded algebra, with unit the class \([e]\in H_0(G)\).
Associativity follows from associativity of the group multiplication together with naturality and associativity of the homology cross product: \[ m\circ(m\times\operatorname{id})=m\circ(\operatorname{id}\times m). \] The inclusion of the identity element is a two-sided unit for \(m\), so its degree-zero homology class is a two-sided unit for \(\star\).
3. The topological group \(G\) acts on itself by left translations\[ L_g(x)=gx, \]and hence acts on homology by\[ g\cdot a=(L_g)_*(a). \]
Since \(L_{gh}=L_g\circ L_h\) and \(L_e=\operatorname{id}\), functoriality of homology gives a genuine group action on \(H_*(G)\).
4. If \(g\) and \(h\) lie in the same path component of \(G\), then \((L_g)_*=(L_h)_*\).
Choose a path \(\gamma:I\to G\) from \(g\) to \(h\). Then \[ H(x,t)=\gamma(t)x \] defines a homotopy from \(L_g\) to \(L_h\). Homotopic maps induce the same map on homology.
5. Thus the action factors through the component group \(\pi_0(G)\); if \(G\) is path-connected, the action is trivial.
By 4 the action depends only on the path component of \(g\), so it factors through \(G\to\pi_0(G)\). If \(G\) is path-connected, every \(g\) lies in the component of \(e\), so \((L_g)_*=(L_e)_*=\operatorname{id}\).