Geometrically quantizing real Grassmannians

Question

It seems that the Grassmannian of oriented 2-dimensional planes in $\mathbb{R}^n$

$$ \mathrm{Gr}(n,2) = \frac{\mathrm{SO}(n)}{\mathrm{SO}(n-2) \times \mathrm{SO}(2)} $$

has a symplectic structure that's invariant under the action of $\mathrm{SO}(n)$. My argument for this is as follows. Start with the cotangent bundle of $T^*\mathrm{S}^{n-1}$ and put the usual kinetic Hamiltonian

$$ H(x,p) = \frac{1}{2} \|p\|^2 , \qquad x \in \mathrm{S}^{n-1}, p \in T^*_x \mathrm{S}^{n-1}$$

on this, getting $H \colon T^*\mathrm{S}^{n-1} \to \mathbb{R}$. This generates a flow on $T^*\mathrm{S}^{n-1}$ which corresponds to geodesic motion on the sphere. Now do symplectic reduction: form the submanifold $\{H = E\}$ for some constant $E > 0$, and mod out by the flow generated by $H$. This gives the space of oriented great circles in $\mathrm{S}^{n-1}$, and thanks to the theory of symplectic reduction this should have a symplectic structure.

But any oriented great circle in $\mathrm{S}^{n-1}$ arises by intersecting this sphere with unique oriented 2d subspace of $\mathbb{R}^n$. So, the space of oriented great circles in $\mathrm{S}^{n-1}$ should be diffeomorphic to $\mathrm{Gr}(n,2)$.

So, $\mathrm{Gr}(n,2)$ should have a $\mathrm{O}(n)$-invariant symplectic struture depending on $E > 0$. I guess changing the energy $E$ just rescales this symplectic structure.

Question 1. Is this correct so far? If so, it must be known. Do you know references?

Question 2. What's a nice simple formula for this symplectic structure on the real Grassmannian?

Question 3. Is this symplectic structure the imaginary part of an $\mathrm{SO}(n)$-invariant Kähler structure?

Question 4. If so, has someone geometrically quantized $\mathrm{Gr}(n,2)$ for values of $E$ making the symplectic structure integral? What Hilbert spaces do we get?

If it works, we should get some nice finite-dimensional unitary representations of $\mathrm{SO}(n)$.

Answer 1

Write the left invariant Maurer-Cartan 1-form $\omega$ on $SO(n)$ as $$ \begin{pmatrix} \omega^i_j & \omega^i_J \\ \omega^I_j & \omega^I_J \end{pmatrix}. $$ The structure equations of Cartan are $d\omega=-\omega\wedge \omega$. I claim that the 1-form $\Omega = \sum_I \omega^I_1 \wedge \omega^I_2$ is a Kaehler form for the metric $g=\sum_{I,j} \omega^I_j \omega^I_j$. Clearly the linear algebra is ok: the 1-forms $\omega^I_j$ are semibasic for the map $SO(n) \to Gr(2,n)=SO(n)/SO(2)\times SO(n-2)$, and transform by orthogonal transformation under the action of the structure group, so $g$ is a metric on $Gr(2,n)$. The structure equations force $\Omega$ to be closed. The structure group action preserves the almost complex structure for which $\omega^I_1 + i\omega^I_2$ are complex linear, as (under right action $r_g$ for $g \in SO(2) \times SO(n-2)$, $r_g^* \omega = Ad(g)^{-1}\omega$, so if we write $$ g=\begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix} $$ then $r_g^*\omega = b^{-1}ga$, so $$ r_g^*(\omega^I_1+i\omega^I_2)=\sum_J b^J_I(\omega^J_1+i\omega^J_2)(a_1 + i a_2). $$ This almost complex structure is complex, as $$ d(\omega^I_1+i\omega^I_2)=-(\omega^I_J-i\omega^1_2)\wedge(\omega^J_1+i\omega^J_2) $$ (by the structure equations) so there are no $(\omega^I_1+i\omega^I_2)\wedge(\omega^J_1-i\omega^J_2)$ terms, i.e. no torsion, i.e. the Nijenhuis tensor vanishes, so a complex structure.

As abx points out, if you take a 2-plane, say the span of an orthonormal basis $u,v \in \mathbb{R}^n$, then the vector $z=u+iv$ is null for the quadratic form $\sum z_i^2$, clearly. Conversely, every null vector arises this way uniquely up to a complex scalar. The 2-plane determines $z$ up to rescaling by a unit complex number, so $Gr(2,n)$ is a quadric hypersurface in $\mathbb{P}_{\mathbb{C}}^{n-1}$.

Answer 2

Another way to view this: think of $T^*S^{n-1}$ as $\lbrace (q,p)\in\mathbb{R}^{2n}\,|\,\Vert q\Vert = 1, q.p=0\rbrace$, with symplectic structure $\omega((v_q,v_p),(w_q,w_p)) := v_q\cdot w_p - v_p\cdot w_q$. The $\operatorname{SO}(n)$-action is Hamiltonian, with $\operatorname{SO}(n)$-equivariant momentum map $J:T^*S^{n-1}\rightarrow \mathfrak{so}(n)^*$ given by $$ \langle J(q,p),\xi\rangle = p(\xi\cdot q)\qquad\textrm{for}\qquad\xi\in\mathfrak{so}(n). $$

Since the $\operatorname{SO}(n)$-action preserves the Hamiltonian, it drops to the reduced space $M_{\textrm{red}} = H^{-1}(E)/{\lbrace\textrm{Hamiltonian flow}\rbrace}$. Since the unreduced action acts transitively on $H^{-1}(E)$, the reduced action acts transitively on $M_{\textrm{red}}$. The $\operatorname{SO}(n)$-equivariance of the corresponding reduced momentum map $J_{\textrm{red}}:M_{\textrm{red}}\rightarrow \mathfrak{so}(n)^*$ implies that $J_{\textrm{red}}$ maps $M_{\textrm{red}}$ to a coadjoint orbit $\mathcal{O}\subset\mathfrak{so}(n)^*$. Kostant's coadjoint orbit covering theorem then implies that $J_{\textrm{red}}$ is a symplectic covering map of $M_{\textrm{red}}$ onto $\mathcal{O}$ , equipped with the (positive) Kostant-Kirillov-Souriau form $$ \omega^+_\mu(-\operatorname{ad}_\zeta^*\mu,-\operatorname{ad}_\chi^*\mu) := \langle\mu,[\zeta,\chi]\rangle. $$ In fact, in this case $J_{\textrm{red}}$ is a symplectomorphism.

To see things more concretely: identifying $\mathfrak{so}(n)$ with its dual via the trace form, we get the (still equivariant) Lie-algebra-valued momentum map $j:T^*S^{n-1}\rightarrow \mathfrak{so}(n)$ given by $$ j(q,p) = \frac{1}{2}(q p^\top - p q^\top)\in\mathfrak{so}(n). $$ Then for example taking the generic point $(q,p) = (e_1,\sqrt{2E} e_2)\in H^{-1}(E)$, this gives $$ j(q,p) = \begin{bmatrix} 0 & \sqrt{\frac{E}{2}} & \ldots & 0\\ -\sqrt{\frac{E}{2}} & 0 &\ldots & 0 \\ \vdots \\ 0 & 0 &\ldots & 0\end{bmatrix} $$ Clearly the stabiliser of this element under the adjoint action of $\operatorname{SO}(n)$ is $\operatorname{SO}(2)\times\operatorname{SO}(n-2)$. Viewed in $T^*S^{n-1}$, the second factor acts trivially on $(q,p)$, while the first factor produces the great circles corresponding to the Hamiltonian flow. Hence the fibres of the quotient map $\pi:H^{-1}(E)\rightarrow M_{\textrm{red}}$ and the momentum map $j:H^{-1}(E)\rightarrow \mathcal{O}$ agree, which implies that the covering $j_{\textrm{red}}:M_{\textrm{red}}\rightarrow \mathcal{O}$ is actually a diffeomorphism.

It's straightforward now to put a complex structure on the coadjoint/adjoint orbit (the same construction works for the orbits of any compact Lie group). I will use adjoint orbits here. At any point $\xi\in \mathcal{O}\subset \mathfrak{so}(n)$, the operator $\operatorname{ad}_\xi:\operatorname{so}(n)\rightarrow\operatorname{so}(n)$ is skew-adjoint with respect to the trace form, and so has pure imaginary eigenvalues. Let $$ \mathfrak{n}_\xi^+ := \lbrace \zeta\in\mathfrak{so}(n)\,|\, \operatorname{ad}_\xi\zeta = i\lambda\, \zeta \quad\textrm{for some }\lambda>0\rbrace, $$ and define the polarization $F$ on $\mathcal{O}$ by $$ F_\xi := \lbrace \operatorname{ad}_\zeta \xi\,|\,\zeta\in\mathfrak{n}_\xi^+\rbrace. $$ At any point $\xi\in\mathcal{O}$, we have $\ker(\operatorname{ad}_\xi) = \mathfrak{so}(n)_\xi$ (the adjoint stabiliser algebra), and we can choose a Cartan subalgebra $\mathfrak{h}\subset\mathfrak{so}(n)_\xi$, from which we can introduce an ordering on weights. As Ben McKay mentioned, the usual Borel-Weil construction implies that the holomorphic sections with respect to the polarization $F$ give an irreducible representation with highest weight related to $\xi$. I would have to check my signs and conventions carefully, but I think in this case you get the dual to the irrep with highest weight $-\frac{i}{\hbar}\xi^\flat$, where $\cdot^\flat:\mathfrak{so}(n)\rightarrow \mathfrak{so}(n)^*$ is the duality with respect to the trace form (don't quote me on that, though :)).

Answer 3

For question 1 (references): this is covered in Besse, Einstein Manifolds, pp. 230-233 for even and then odd $n$.

For question 2 (nice formula), I think you want to do something like what works for all complex grassmannians $X$ as coadjoint orbits of $\mathrm U(n)$: identify a subspace with the self-adjoint projector $x$ onto it. Then the 2-form maps tangent vectors $\delta x$, $\delta'x\in T_xX\subset\operatorname{End}(\mathbf C^n$)$ $ to the number $\operatorname{Tr}(\delta'x J\delta x)$ where $\smash{J\delta x=\frac1i[x,\delta x]}$. I am too lazy to spoil your fun figuring out the analogue :-)

For question 3 (Kähler structure): bingo, the $J$ you’ll have found in step 2 does it.

For question 4 (geometric quantization): Your resulting $x$’s can be regarded as a coadjoint orbit. (Adapt the $\mathrm U(n)$ case where we pair self-adjoint $x$’s with skew adjoint $Z\in\mathfrak u(n)$ by $\smash{\frac1i\operatorname{Tr}(xZ)}$.) It intersects your choice of dominant Weyl chamber in a single point (viz. in my example, a diagonal of 1's followed by 0's). Borel-Weil says that geometric quantization gives the representation with this highest weight — in the example, an exterior power of $\mathbf C^n$. Rescaling the coadjoint orbit (or symplectic form) by an integer factor, you get symmetric powers of that instead.

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Edit (for question 2):

I sorted out the analogue I had in mind. To an oriented 2-plane in $\mathbf R^n$, attach $x:=$ the symmetric projector onto it followed by 90º rotation in it. (As it’s oriented, this is well-defined.) Writing bar for transpose, we are therefore representing the oriented 2-plane with oriented orthonormal basis $(u,v)$ by the composition of $u\bar u + v\bar v$ with $(u,v)\mapsto(v, -u)$, i.e. by $x=v\bar u-u\bar v$. Those $x$’s clearly belong to $$ X:=\left\{x\in so(n): x^3 + x = 0 \quad\text{and}\quad \mathrm{rank}(x)=2\right\}; \tag1 $$ conversely they exhaust $X$ as one sees by observing that members of $X$ have minimal polynomial $t(t+i)(t-i)$, so they are diagonalizable over $\mathbf C$, which gets us many pairs $(u,v)$ giving $x$. Now the point is that the following is a complex structure on $X$: $$ J\delta x := [x, \delta x] \qquad(\delta x \in T_xX). \tag2 $$ Indeed, deriving the equation in (1) gives $\delta x.x^2+x.\delta x.x+x^2.\delta x + \delta x=0$; and using our map $(u,v)\mapsto x$ (Stiefel $\to X$), one actually checks the first and hence both of: $$ x.\delta x.x=0, \qquad \delta x.x^2+x^2.\delta x + \delta x=0 \qquad\forall\,\delta x\in T_xX. \tag3 $$ From this, verifying $J^2\delta x=-\delta x$ is straightforward. And then the second point is that (using minus the trace form to identify $so(n)$ with its dual), (1) is clearly homogeneous under, hence a (coadjoint) orbit of, $\mathrm{SO}(n)$, hence indeed symplectic. Moreover given $\delta x\in T_xX$, the complex structure gives a ready $Z\in so(n)$ such that $\delta x=[Z,x]$, viz. $Z=J\delta x$. So (with also $\delta'x=[Z',x]$ if you like), \begin{align} \omega(\delta x,\delta'x)&=\omega([Z,x],\delta'x)&&\text{by construction}\\ &=-\langle\delta' x,Z\rangle&&\text{by definition of the KKS 2-form}\\ &=\mathrm{Tr}(\delta'x.Z)&&\text{by choice of }\langle,\rangle\\ &=\mathrm{Tr}(\delta'x\,J\delta x)&&\text{by choice of }Z. \tag4 \end{align} Together with (2), this is my preferred “nice simple formula” for the symplectic and indeed Kähler structure of this oriented grassmannian.