What is the proof that SO(2n+1) is non-orientable for any positive integer n?

Question

Inquiring minds want to know. :) I read somewhere that SO(3) is not orientable but I did read somewhere that for any odd dimension N>1, SO(N) is a non-orientable manifold. If such is true I'm eager to see how.

Answer 1

Since this has sat around on MSE for an eternity (=$22$ hours!), I'll go ahead and turn my comment into an answer.

For any $n$ at all, $SO(n)$ is orientable. More generally, for any $n$ at all, $SO(n)$ is a Lie group and all Lie groups are orientable.

Since you already seem to know $SO(n)$ is a manifold, let me briefly argue it's a Lie group, that is, that multiplication and inversion are smooth. The point is that $SO(n)$ naturally sits inside of $M(n,\mathbb{R})$ which is a vector space of dimension $n^2$. Elements of $M(n,\mathbb{R})$ can be multiplied in the usual way and the key fact is that each coordinate of $AB$ is given by a polynomial in the entries of $A$ and $B$. In particular, multiplication on $M(n,\mathbb{R})$ is smooth and so, by restriction it is smooth on $SO(n)$ as well.

Inversion, when defined, is given by a rational function of the entries of the matrix (Cramer's rule), so is smooth so long as it is defined. Hence, by restricting to $SO(n)$ (where everything is invertible), we see that inversion is also smooth.

Thus, we need only argue that all Lie groups are orientable. One definition of orientability that I like is that a manifold is orientable if it has a nonvanishing volume form at every point. (This can be shown to be equivalent to any of the other definitions of orientability).

To see that a Lie group has a non-vanishing volume form, note that $T_e G$ (where $e\in G$ is the identity) has one because it's just a vector space. Choose one and call it $\omega$. To define a non-vanishing volume form at every point, define $\Omega(g) = L_{g^{-1}}^\ast \omega$ where $L_g:G\rightarrow G$ denotes left multiplication: $L_g(h) = gh$. Since $L_{g^{-1}}$ is a diffeomorphism, it maps nonzero forms to nonzero forms, so $\Omega$ is non-vanishing.