4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
9
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.