We know that $U(N)$ can be embedded into $SU(n+1)$ and that the quotient is isomorphic to complex projective space: $$ SU(n+1)/U(n) \simeq {\mathbb CP}^{n}. $$ We can split this process into two stages, first quotient $SU(n+1)$ by $SU(n)$ giving the sphere $$ SU(n+1)/SU(n) \simeq S^{2n+1}, $$ and then quotient $S^{2+1}$ by $U(1)$ to get ${\mathbb CP}^{n}$.

I am curious about the analogue for the compact symplectic groups $Sp(n)$. According to the Hermitian Symmetric Space Wikipedia page, $Sp(n)/U(n)$ is the space of complex structures on $\mathbb{H}^n$ compatible with the inner product. If it helps, note that like ${\mathbb CP}^{n}$, this is an Hermitian Symmetric Space, and a flag manifold. One can again split this quotient into a quotient by $SU(N)$, and a quotient by $U(1)$.

Question: What is the space $Sp(n)/SU(N)$? Does it have a name?

  • $\begingroup$ The 1st question: clearly YES with $K=SU(n)$. $\endgroup$ – Mikhail Borovoi Nov 6 '15 at 12:43
  • $\begingroup$ Question edited accordingly. $\endgroup$ – Ago Szekeres Nov 6 '15 at 12:48
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    $\begingroup$ This space is just (a covering of) the unit circle bundle of the canonical line bundle of the complex manifold $\mathrm{Sp}(n)/\mathrm{U}(n)$ in its canonical Kähler metric. $\endgroup$ – Robert Bryant Nov 6 '15 at 13:47
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    $\begingroup$ @AgoS: In general, any irreducible Hermitian symmetric space is of the form $U/H$, where $K$ has a nontrivial homomorphism $h:H\to S^1$ whose kernel $H$ is a simple Lie group. In this situation, the quotient $U/K$ is (a covering of) the bundle of unit complex volume forms (where 'unit' is defined in terms of the Kähler metric on $U/H$. $\endgroup$ – Robert Bryant Nov 6 '15 at 15:17
  • $\begingroup$ As another analogous situation for the symplectic groups $Sp(n)$, which is dear to twistor theorists, you could also consider $Sp(n+1)/(Sp(n) \times Sp(1)) \cong \mathbb{H}P^n$ and $Sp(n+1)/(Sp(n) \times U(1)) \cong \mathbb{C}P^{2n+1}$. The fibers of the map from the second to the first are of course in this case $2$-spheres. $\endgroup$ – Malkoun Nov 26 '15 at 6:35

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