# A topological group which is also a (not necessarily smooth) manifold is orientable

I am trying to show that a topological group which is also a (not necessarily smooth) manifold is automatically orientable. I know of a proof involving transition functions for smooth manifolds, in which case the object in question is a Lie group.

I am using Hatcher's definition of orientability: An $n$-manifold $M$ is orientable if it admits a local orientation $\eta_x$ at each $x\in M$ where $\eta_x$ is a generator of $H_n(M\mid x)\cong \mathbb{Z}$, with the following compatibility property: For each $x\in M$, there is an open ball $x\in B_x\cong \mathbb{R}^n$ so that for every $y\in B_x$, the local orientation $\eta_y$ is the isomorphic image (induced by inclusion of pairs) of the same generator $\eta_{B_x}$ of $H_n(M\mid B_x)$.

I have a clear candidate for such a local orientation, but I am having trouble showing the compatibility: Let $e$ be the identity of the topological group $M$. Choose any generator $\eta_e$ of $H_n(M\mid e)$, and for any $g\in M$, let $\eta_g = L^g_*(\eta_e)\in H_n(M\mid g)$ where $L^g:M\to M$ is left multiplication by $g$ ($L^g$ is a homeomorphism, so it certainly induces an isomorphism on homology).

To start showing the compatibility condition, given $x\in M$, let $B_x$ be any open neighborhood of $x$ homeomorphic to $\mathbb{R}^n$. We are required to show that the following diagram commutes: $\require{AMScd}$ $$\begin{CD} H_n(M\mid B_x) @>id>\cong> H_n(M\mid B_x) \\ @VV{\cong}V @V{\cong}VV \\ H_n(M\mid x) @>{\cong}>L^{(y^{}x^{-1})}_*> H_n(M\mid y) \end{CD}$$

where the vertical maps are induced by inclusion. Here is where I am stuck. The corresponding diagram on the level of topological spaces certainly does not commute. Any ideas, thoughts, hints, or full solutions are welcome!

• To clarify the nonstandard notation, $H_n(M\,|\,A)$ means $H_n(M,M-A)$. As a hint for an answer, try replacing the term $B_x$ in the upper right corner by $L^{(yx^{-1})}(B_x)$ and the upper map by $L^{(yx^{-1})}_*$. – Allen Hatcher Aug 20 '16 at 0:08
• Thanks for the clarification and the hint! Perhaps I am misunderstanding something, but if I do as you suggest, neither leg (top then right, nor left then bottom) is a map induced by inclusion. I could add the diagonal arrow from top left to bottom right, induced by inclusion, but here it seems I am back to the original problem: it is not clear to me that either triangle commutes (although the large square certainly does). – Doeke Aug 20 '16 at 3:13
• See here: math.stackexchange.com/q/348552 – Ramiro de la Vega Aug 20 '16 at 19:28
• Thanks for the link, Ramiro. Indeed I came across this post, but the question is slightly different (I do not want to assume smoothness) and the answer does not address my concern. However, for those looking for an answer, a duplicate has been answered on the stackexchange. – Doeke Aug 23 '16 at 23:55

In fact beside your question there is a beautiful theorem in homotopy theory. This theorem due to T. Bauer, N. Kitchloo, D. Notbohm and E. K. Pedersen guarantees that any loop space $X=\Omega B$ where $B$ is a $CW$-complex and such that $H_*(X)=\oplus H_i(X;\mathbb{Z})$ is a finitely generated abelian group is homotopy equivalent to a compact, smooth, parallelizable manifold ("Finite loop spaces are manifolds" Acta Math. 2004).
This theorem applies to your case where $X=G\simeq \Omega BG$.
The very first step in the proof of this result which is very closely related to your question is to prove that $X$, thus $G$ in your case, is a Poincaré duality space. The proof is completely algebraic and uses the fact that for any field $\mathbb{F}$ the cohomology algebra is a finitely generated, connected graded Hopf algebra. Then you use Borel's classification that tells you that $H^*(G;\mathbb{F})$ is a tensor product of exterior algebras and truncated polynomials algebras and get a family of top classes $[G]_p\in H_{n_p}(G;\mathbb{Z}/p\mathbb{Z})$ for an integer $n_p$. Then you show that all these $n_p$ are equal to a fixed integer $n$ and that $[G]_p$ is the reduction $mod(p)$ of an integral fundamental class whose cap product induces a Poincaré duality isomorphism. All details and references are given in the introduction of Bauer, Kitchloo, Notbohm, Pedersen's paper.