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.

  • $\begingroup$ You mean 2|n. (I have no clue what three vertical dots mean) $\endgroup$ – M.B. Feb 6 '14 at 17:15
  • $\begingroup$ One way to prove it is to use study the first Stiefel-Whitney class. $\endgroup$ – M.B. Feb 6 '14 at 17:19

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.

  • $\begingroup$ This deserves both an upvote and OP's attention! Never seen this proof before, but yet again, I know (too) little about Lie groups/algebras. $\endgroup$ – M.B. Feb 7 '14 at 19:05
  • $\begingroup$ @M.B. Glad you like it - I just made it up on the fly! $\endgroup$ – Jason DeVito Feb 7 '14 at 19:33
  • $\begingroup$ @Jason: I just stumbled on this. Very nice. Of course, your argument can be mirrored by working with the Maurer-Cartan forms of $SO(n)$ to give a well-defined volume element $\bigwedge\limits_{1\le\alpha\le k<\mu\le n} \omega_{\alpha\mu}$. $\endgroup$ – Ted Shifrin Apr 18 '16 at 23:28
  • $\begingroup$ @Ted: I'm not familiar with those forms (at least by that name). Are those the bi-invariant differential forms on $SO(n)$? And thanks for fixing my typos! $\endgroup$ – Jason DeVito Apr 18 '16 at 23:43
  • $\begingroup$ Yup, that's what they are. This is the standard coframing for any homogeneous space. You're looking at how the frames $e_1,\dots,e_k$ for the $k$-plane are twisting into the normal directions. $\endgroup$ – Ted Shifrin Apr 19 '16 at 0:00

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.


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