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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.

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 (writing 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.

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.

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.

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.

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, $\mathrm{SO}(n)$. Now given $\delta x$ as above we have a ready $Z\in so(n)$ such that $\delta x=[Z,x]$, viz. $Z=J\delta x$. Hence (writing 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.

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.

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 (writing 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.

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 tired 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.

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.

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, $\mathrm{SO}(n)$. Now given $\delta x$ as above we have a ready $Z\in so(n)$ such that $\delta x=[Z,x]$, viz. $Z=J\delta x$. Hence (writing 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.