Show that a topological group with a finite-dimensional CW structure is an orientable manifold. [Consider the homeomorphisms \(x \mapsto gx\) or \(x \mapsto xg\) for fixed \(g\) and varying \(x\) in the group.]
Solution
Let \(G\) be a topological group with a finite-dimensional CW structure, and let \(d\) be the largest dimension of a cell of \(G\).
Choose a \(d\)-cell \(e^d\). Since there are no cells of dimension larger than \(d\), the union of all cells other than \(e^d\) is a subcomplex, hence closed. Therefore the open cell \(e^d\) is an open subset of \(G\), homeomorphic to \(\mathbb R^d\). Choose a point \(x\in e^d\).
For any \(y\in G\), left multiplication by \[ g=yx^{-1} \] is a homeomorphism \[ L_g:G\to G, \qquad z\mapsto gz, \] carrying \(x\) to \(y\). Hence \[ L_g(e^d) \] is an open neighborhood of \(y\) homeomorphic to \(\mathbb R^d\). Thus every point of \(G\) has a Euclidean \(d\)-neighborhood, so \(G\) is a \(d\)-manifold.
It remains to orient it. Since a CW complex is locally path-connected, each connected component of \(G\) is path-connected. First orient the identity component \(G_0\). Choose an orientation at one point, say the identity \(e\), and transport it to \(g\in G_0\) by the local homeomorphism \(L_g\). This is well-defined coherently: if \(g,h\in G_0\), then the transition between the translated charts is a left translation \[ L_hL_g^{-1}=L_{hg^{-1}}. \] Because \(hg^{-1}\in G_0\), choose a path \(\gamma\) from \(e\) to \(hg^{-1}\). The family \[ L_{\gamma(t)} \] is an isotopy from the identity homeomorphism to \(L_{hg^{-1}}\). An isotopy cannot reverse the local orientation class, so all transition maps preserve the transported orientations. Hence \(G_0\) is orientable.
Every other component is a left coset \(aG_0\) and is homeomorphic to \(G_0\) by \(L_a\), so transport the orientation of \(G_0\) to each component. Therefore the whole manifold \(G\) is orientable.
Thus \[ \boxed{\text{every topological group with a finite-dimensional CW structure is an orientable manifold}.} \]