equivariant orientation Assume that $G$ is a compact Lie group and $M$ is a smooth oriented manifold on which $G$ acts freely. Then the orbit space $M/G$ is a smooth manifold with dimension $dimM-dimG$. In general $M/G$ may be unoriented, so when $M/G$ can be an oriented manifold?
 A: Choose a $G$-invariant Riemannian metric on $M$.  For any $x\in M$ let $U_x$ be the tangent space to the orbit $Gx$ at $x$, and let $V_x$ denote the orthogonal complement of $U_x$ in $T_xM$.  As the action is free, $U_x$ is canonically identified with the Lie algebra $LG$.  The spaces $V_x$ give a subbundle of $TM$.  Let $p:M\to M/G$ be the projection.  For $y\in M/G$, let $W_y$ be the set of equivariant sections of $V|{p^{-1}\{y\}}$.  As $p^{-1}\{y\}$ is a free $G$-orbit, we have natural identifications $W_y\simeq V_x$ for all $x\in p^{-1}\{y\}$.  It is also not hard to see that $W$ is the tangent bundle for $M/G$.  (You have to do most of this work to make $M/G$ a manifold in the first place.)  Put $n=\dim(M)$ and $d=\dim(G)$.  It now follows that sections of $\Lambda^{n-d}(T(M/G))$ biject with $G$-invariant sections of $\Lambda^{n-d}(V)$.  If we pick a nonzero element $u$ in the space $\Lambda^d(LG)\simeq\mathbb{R}$ then multiplication by $u$ gives a $G$-equivariant isomorphism 
$$ \Lambda^{n-d}(V) \to \Lambda^n(LG\oplus V) = \Lambda^n(TM). $$
Thus, orientations of $M/G$ biject with $G$-invariant orientations of $M$ (but the bijection depends on the sign of $u$, or in other words the orientation of $LG$).
A: Let's suppose that $M$ is connected. Then $M/G$ is orientable if and only if $G$ preserves orientations on $M$.
As the action of $G$ on the two possible orientations induces a continuous homomorphism $G\to \{\pm 1\}$, we infer that $M/G$ is orientable if $G$ is connected, for instance.
