MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$\def\C{\mathbb{C}}\def\CP{\mathbb{CP}}$Every complex projective space $\CP^n$ has a natural Riemannian metric, the Fubini–Study metric, which is defined via the quotient definition of $\CP^n = \C^n/\C^* \cong S^{2n - 1}/S^1$ to be the metric descending from the ($S^1$-invariant) round metric on $S^{2n - 1}$. This is not the only metric one could define, but it is a nice one; for me, it is nice because I don't have to choose coordinates to define it.

My question is about the grassmannian varieties or, more generally, the flag varieties. Of course, one could define metrics on them by restriction along their Plücker embeddings in projective space, but this is a little indirect: one step removed from the end result for grassmannians, and two steps for finer flag varieties (which it seems to me would have to be embedded first into products of grassmannians).

If you agree upon the standard basis on $\C^n$, then you can fix a complex inner product $(,)$ on it and then define a metric on, say, the grassmannian $G(k, n)$ by specifying the real inner product on each tangent space $T_W$ to $W \subset \C^n$ via: $$(\phi, \psi) = \Re\operatorname{tr}(\psi^* \phi), \qquad \text{for } \phi, \psi \in T_W \cong \operatorname{Hom}(W, \C^n/W) \text{ with } \C^n/W \cong W^\perp \subset \C^n.$$ This has a nice natural feel other than the choice of basis on $\C^n$. In the special case where $k = 1$, I suppose one could check directly that this agrees with the Fubini–Study metric up to a scalar multiple; I suppose a nice abstract reason is that it is homogeneous (I am uneducated: does this uniquely-up-to-scalars determine it?)

Is there a good "intrinsic" way of defining a metric on $G(k,n)$ or more generally, any flag variety of (possibly partial) flags in $\C^n$, that generalizes the Fubini–Study metric?

Edit: I have streamlined the definition of the metric from the original version of the question. I'm satisfied with Will's answer now that it occurs to me that in fact, even the Fubini–Study metric requires a little non-naturality, in that to specify the isomorphism $\C^n/\C^* \cong S^{2n - 1}/S^1$, you need to be able to pick out the sphere in $\C^n$, which requires identifying the standard inner product (metric). So this really is no worse.

share|cite|improve this question
The answer is "yes". Just view the Grassmannian or flag manifold as a quotient of $SU(n)$ by the appropriate subgroup. The bi-invariant metric on $SU(n)$ induces a natural Riemannian metric on the flag manifold. For me it's all easiest to work out using moving frames, as presented, say, in Griffiths's paper (Duke Math. J. 41 (1974), 775–814). – Deane Yang Jan 22 '12 at 21:51
Is it obvious that this restricts to the correct metrick on $G(1,n)$? – Will Sawin Jan 23 '12 at 3:11
It's obvious after you work out the details. In particular, $SU(n)$ acts transitively on each of these spaces, and the natural Riemannian metric is the unique one (up to a constant scale factor) that is invariant under this action. So you just have to check that the Fubini-Study metric is indeed invariant under the group action. – Deane Yang Jan 23 '12 at 9:53
up vote 3 down vote accepted

A truly coordinate-free metric would be preserved by the action of $GL_n(\mathbb C)$, which I think is impossible, for instance because it would produce an invariant measure, which should be impossible.

Consider the manifold of (ordered or unordered) sets of $k$ orthonormal vectors. This is the appropriate analogue to $S^{2n-1}$. The analogue to $S^1$ is $U^k$. We thus need to find a nice $U^k$-invariant metric on this space.

If $v_1,...,v_k$ is a set of orthonormal vectors, the tangent space consists of derivatives $dv_1,...,dv_k$ satisfying $v_i\cdot dv_i=0$, $v_i \cdot dv_j+v_j\cdot dv_i=0$. The obvious metric is just $\sum_{i=1}^k dv_i^a \cdot dv_i^b$, which is $U(n)$-invariant.

This is clearly equivalent to Fubini-Study when $k=1$. If I understand your metric correctly, it's equivalent to yours, since the sum should just be calculating the trace using a basis.

share|cite|improve this answer

The Fubini-Study metric on $\mathbb P^n$ arises as the curvature of a line bundle. More precisely, once you fix a hermitian inner product in the complex vector space you are projectivizing, you get a natural metric on the tautological line bundle $\mathcal O(-1)$ on the projective space. If you take the dual metric on $\mathcal O(1)$ and you compute its Chern curvature you find the Fubini-Study metric.

Once you look at things in this way, on the Grassmannian you have a completely analogous construction (which, after all, gives you the induced metric of the Plücker embedding in a somewhat more direct way).

Call $S\to G(k,n)$ the tautological holomorphic vector bundle of rank $k$ (the fiber over each point is the $k$-dimensional subspace of the vector space $V$ you are considering). Then, once you are given a hermitian inner product on $V$, you can endow naturally $S$ with a smooth hermitian metric. In particular, you can induce a smooth hermitian metric on the holomorphic line bundle $\det S^*\to G(k,n)$. Now, compute its Chern curvature to discover that it is a (strictly) positive $(1,1)$-form, so that it gives you a hermitian metric on $G(k,n)$.

Observe that the map given by the complete linear system associated to $\det S^*$ is nothing more than the Plücker embedding, so that the metric we constructed here is what you were looking for, but given in a more intrinsic way.

Notice finally that this is really a generalization of the Fubini-Study metric, since when you look to $G(1,n)$, then $S=\mathcal O(-1)$ and the construction is as above.

There are of course more general constructions for complete or incomplete flag varieties, but the guiding philosophy is again the same: to put a natural hermitian metric on some (positive) line bundle and then to take the metric on your space to be its Chern curvature. For more details and computations you can look for instance at this paper by J.-P. Demailly.

share|cite|improve this answer
Ooh, I was hoping something like this would work. – Ryan Reich Jan 24 '12 at 17:48
Great! Happy for you! :) – diverietti Jan 24 '12 at 18:01

Here's another rather easy way to define a natural Riemannian metric on a Grassmannian $G(k,n)$. Let $V$ be an $n$-dimensional complex vector space with a hermitian inner product. Given any $k$-dimensional subspace $L \subset V$, there is a natural isomorphism $$ T_LG(k,n) = \operatorname{Hom}(L, V/L) = L^* \otimes V/L $$ The hermitian inner product on $V$ induces a natural hermitian inner on $L^*\otimes V/L$. This defines a hermitian metric on $G(k,n)$ that is easily checked to be invariant under the action of $SU(n)$ on $G(k,n)$. What's less obvious from this construction is that the metric is K\"ahler.

This approach also works for flag manifolds.

I haven't looked, but I'm pretty sure this is all explained in Griffiths and Harris.

share|cite|improve this answer
I think this is the same as my construction, actually. For flag manifolds, the tangent space is just a direct sum of expressions like that for each step of the flag. (Now would be a great time for me not to have put G&H in a box and left it at my mother's.) – Ryan Reich Jan 24 '12 at 17:34
Yes, you're right. I didn't read everything you wrote carefully enough. The only new observation is that the metric does not depend on a choice of the basis at all, only on the hermitian inner product. – Deane Yang Jan 24 '12 at 18:05

Let $G$ be a (semi)simple compact connected Lie group. Any flag variety $M=G^{\mathbb{C}}/P=G/K$, where $K=C(S)$ is the centralizer of a torus $S$ of $G$, admits a reductive decomposition $\frak{g}=\frak{k}\oplus\frak{m}$ with respect to the inner product $( \ , \ ) =-B( \ , \ )$ induced by the negative of the Killing form of $G$, that is

$\left[\frak {k}, \frak{ m}\right] \subset \frak{ m}$

Then, we identify ${\frak{m}} = T_{o} M$ with the tangent space $T_{o} M$ of $M$ at $o=eK$. Moreover, since the isotropy representation $\chi : K\to SO(\frak{m})$ is equivalent with the restriction of the adjoint representation $Ad\big|_{K}$ on the subaspace $\frak{m}\subset\frak{g}$, any $G$-invariant Riemannian metric on $M$, say $\rho=\rho( \ , \ )$ corresponds to an $Ad(K)$-invariant inner product on $\frak{m}$, say $\langle \ , \ \rangle$.

If we assume that $ {\frak m} $ admits a decomposition into mutually non-equivalent irreducible $\mbox{Ad}(K)$-modules as follows (such a decomposition always exists and is given in terms of $\frak{t}$-roots, see Alekseevsky's papers):

$ {\frak m} = {\frak m}_1 \oplus \cdots \oplus {\frak m}_q $

then a $G$-invariant metric on $G/K$ can be written as

$ \rho= \langle \ , \ \rangle = x_1 ( \ , \ )\big|_{\frak{m}_1} + \cdots + x_q ( \ , \ )\big|_{\frak{m}_q} $

for positive real numbers

$(x_1, \cdots, x_q)\in\mathbb{R}^{q}_{+}$

(such metrics are usually called diagonal). Thus, the space of $G$-invariant metrics on flag manifolds can be parametrizided by real positive nymbers. Note that since the $Ad(K)$-submodules $\frak{m}_{i}$ are always pairwise inequivalent, any $G$-invariant metric has the above form. In particular, any $G$-invariant symmetric covariant 2-tensors on $G/K$ are of the same form as the Riemannian metrics (although they are not necessarilly positive definite).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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