Orientability of Grassmannians How can I understand when $Gr(n,k)$ is orientable and when not? I found that answer is yes if and only if $n \vdots 2$, but I do not know how to prove it.
 A: A manifold $M$ is orientable if and only if $w_1(TM) = 0$.
Using the splitting principle, one can show that $w_1(E\otimes F) = \operatorname{rank}(F)w_1(E) + \operatorname{rank}(E)w_1(F)$. Now recall that 
$$TGr(n, k) \cong \operatorname{Hom}(\gamma, \gamma^{\perp}) \cong \gamma^*\otimes\gamma^{\perp} \cong \gamma\otimes\gamma^{\perp}$$
where $\gamma \to Gr(n, k)$ is the tautological bundle which has rank $k$, and $\gamma^{\perp}$ is the orthogonal complement of $\gamma \subset \varepsilon^n$ and therefore has rank $n - k$. So
$$w_1(TGr(n, k)) = w_1(\gamma\otimes\gamma^{\perp}) = \operatorname{rank}(\gamma^{\perp})w_1(\gamma) + \operatorname{rank}(\gamma)w_1(\gamma^{\perp}) = (n-k)w_1(\gamma) + kw_1(\gamma^{\perp}).$$
It follows from $\gamma\oplus\gamma^{\perp} \cong \varepsilon^n$ that $w_1(\gamma^{\perp}) = w_1(\gamma)$ so 
$$w_1(TGr(n, k)) = (n-k)w_1(\gamma) + kw_1(\gamma^{\perp}) = (n - k)w_1(\gamma) + kw_1(\gamma) = nw_1(\gamma).$$
As $w_1(\gamma) \neq 0$, we see that $w_1(TGr(n, k)) = 0$ if and only if $n$ is even.
A: Here's a sequence of propositions which will prove the theorem.
Proposition 1  $Gr(n,k)$ is diffeomorphic to the homogeneous space $G/H = SO(n)/S(O(k)\times O(n-k))$, where $S(O(k)\times O(n-k))$ means all $n\times n$ matrices whose top left $k\times k$ and bottom right $n-k\times n-k$ are orthogonal, all other entries are $0$, and the overall determinant is 1.
Proposition 2  If $G$ is compact, then $G$ admits a bi-invariant metric.  In particular, one has a splitting $\mathfrak{g} = \mathfrak{h}\oplus \mathfrak{p}$ where both $\mathfrak{h}$ and $\mathfrak{p}$ are preserved by the $Ad_H$ action.
Proposition 3  $G/H$ is orientable iff the $Ad_H$ action on $\mathfrak{p}$ is by orientation-preserving isometries.
Believing these, here's a proof.
By Propositions 1 and 2, $\mathfrak{so}(n) = \mathfrak{so}(k)\oplus\mathfrak{so}(n-k)\oplus\mathfrak{p}$ and $Ad_H$ preserves $\mathfrak{p}$.
A straightforward calculation shows that $\mathfrak{p}$ consists of block $n\times n$ matrices in $\mathfrak{so}(n)$ with the top left $k\times k$ and bottom right $n-k\times n-k$ blocks both $0$.  Hence, by making a single really long column out of the individual columns of $\mathfrak{p}$, we may identify $\mathfrak{p}$ with $\mathbb{R}^{k(n-k)}$.
Now, $H$ has two path components which are distinguished by the determinant of the upper left $k\times k$ column.  Because the identity acts in an orientation- preserving manor, so too does everything in the connected component of the identity.  So, we need only figure out if $Ad_A$ is orientation preserving for a single $A\in H$ which is not in the identity component.
Let's use $A = \operatorname{diag}(-1,1,...,1,-1,1,...,1)$ where the two $-1$s are in the top left of the $k\times k$ and $n-k\times n-k$ blocks.
Then $Ad_A(X)$ for $X\in \mathfrak{p}$ negates the first row and the first column of $X$ (so, in particular, does nothing to the top left entry of $X$).  Thinking about this as a map on $\mathbb{R}^{k(n-k)}$, $Ad_A$ has the matrix representation a diagonal matrix with a $-1$ for each entry of $X$ which is flipped.  In particular, it has determinant equal to $(-1)^{\text{number of }-1s}$.
But, there are $k -1 + n-k-1 = n-2$ entries equal to $-1$, so the is orientation preserving iff $n$ is even.  By Proposition 3, this implies $Gr(n,k)$ is orientable iff $n-2$ is even iff $n$ is even.
If you'd like, I can prove some or all of the propositions, but this post is already getting kind of lengthy.
