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Consider the homogeneous space $SU(n)/K$, where $K$ is a sub-group of $SU(n)$ and the bi-invariant metric on $SU(n)$.

What is the appropriate quotient metric on the homogeneous space and what are the geodesics? Clearly, for example, $K=U(n-1)$ gives $\mathbb{C}P^{n-1}$ and the Fubini-Study geodesics which can be expressed in terms of the original geodesics on $SU(n)$. Is any similar formula available for the more general case and is the resulting metric unique (upto constant positive multiple) as in the FS case?

I'm not seeking homogeneous geodesics specifically.

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  • $\begingroup$ In the case when $K$ is maximal torus $T$ or Borel group, then we have such homogeneous spaces as coadjoint orbits , see arxiv.org/abs/math-ph/0602016 $\endgroup$
    – user21574
    Commented Jan 24, 2017 at 21:13
  • $\begingroup$ Kirillov-Kostant-Souriau metric, when $K$ is maximal torus $T$ or Borel group, note that such metrics on coadjoint orbits are K\"ahler and complex coadjoint orbit is projective in general. see pages 150, 151 for more precise information, emis.de/proceedings/Varna/vol9/Bernatska.pdf $\endgroup$
    – user21574
    Commented Jan 24, 2017 at 21:19

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I believe the answers to your and related questions are covered in Ch.X of

[1] Foundations of Differential Geometry, vol. II, Kobayashi, Nomizu

More specifically, let $\mathfrak g:=Lie(SU(n))$ and $\mathfrak k:=Lie(K)$ be Lie algebras of the corresponding groups. Since $SU(n)$ is a compact simple Lie group there is unique (up to multiple) invariant metric $\kappa$ on $\mathfrak g$ coming from the Killing form.

Let us fix a point $p=[eK]\in M=SU(n)/K$; any invariant tensor on $M$ is completely determined by its value at $p$. In particular, invariant metric on $M$ is determined by a metric on $T_p M\simeq \mathfrak g/\mathfrak k$, where isomorphism is given by the infinitesimal action along $X\in\mathfrak g$. One of the ways to equip $\mathfrak g/\mathfrak k$ with a metric is to identify it with a $\kappa$-orthogonal complement of $\mathfrak k$. This way you get so-called naturally reductive homogeneous space. For $K=SU(n-1)$ this gives Fubini-Study metric on $\mathbb P(n-1)$; for $K=S(U(k)\times U(n-k))$ this metric makes $Gr(n,k)$ a symmetric space.

There are two important connections on $M$ with a naturally reductive metric. One is the standard Levi-Civita connection $\nabla^{LC}$ corresponding to the metric defined above, another one is canonical connection $\nabla$ (see Ch X.2 of [1]). Important facts are:

  1. $\nabla^{LC}$ and $\nabla$ have the same geodesics (Theorem 2.10 and Theorem 3.3 in [1])
  2. Geodesic of $\nabla$ in the direction $X\in \mathfrak k^\perp\simeq T_p M$ is given by the curve $\exp (tX)p$ (Corollary 2.5 in [1]).
  3. Any invariant tensor field on $M$ is $\nabla$-parallel.

Finally, I would like to mention that naturally reductive metric is not the only invariant metric on $M$ and other invariant metrics might have different geodesics.

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  • $\begingroup$ in your last sentence, you probably mean left-invariant? the bi-invariant one is unique up to scalar multiplication. "invariant" means both left or bi-invariant according to authors. $\endgroup$
    – YCor
    Commented Jan 24, 2017 at 20:50
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    $\begingroup$ Note that, I said complex coadjoint orbit, .ie $G^{\mathbb C}/P$, which is projective variety $\endgroup$
    – user21574
    Commented Jan 25, 2017 at 4:12
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    $\begingroup$ there is a difference between real coadjoint orbit, and complex coadjoint orbit. real coadjoint orbit is affine, but complex coadjoint orbit is projective . see my old question mathoverflow.net/questions/156394/is-g-t-a-projective-variety $\endgroup$
    – user21574
    Commented Jan 25, 2017 at 4:28
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    $\begingroup$ A note about @YuryUstinovskiy's comment: the K\"ahler metric on a coadjoint orbit of a compact group is not the metric restricted from $\mathfrak g^*$. Indeed, for $SU(2)$ where the coadjoint orbits are concentric $2$-spheres, the K\"ahler area of an orbit is proportional to its radius, not the square of its radius. $\endgroup$
    – Allen Knutson
    Commented Jan 25, 2017 at 5:01
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    $\begingroup$ Allen was right, see corollary page 5, cambridge.org/core/services/aop-cambridge-core/content/view/… $\endgroup$
    – user21574
    Commented Jan 27, 2017 at 3:07

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