Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.


I'm looking for a reference for the full isometry groups of the

(i) complex Stiefel manifolds $U(m)/U(m-l)$, either for the Euclidean metric (i.e. identifying it with orthonormal $m \times l$-matrices and using the inner product $\operatorname{tr}(X^\ast Y)$) or for the induced quotient metric (where we use on $U(m)$ the Euclidean metric), and of the

(ii) real and / or complex (doesn't matter for me now) Grassmann manifolds with the induced quotient metric from the Euclidean metric on O(n), resp. U(n).

It is easy to see that left, resp. right multiplication with unitary / orthogonal matrices are isometries, so I'm interested whether there are more isometries and especially how to show that there are no more.


Asked that question already on Math.SE and got no answer:

share|cite|improve this question

2 Answers 2

up vote 4 down vote accepted

As long as connected groups of isometries are concerned, Grassmann manifolds are symmetric spaces, so the identity component of its isometry group is $G$ in its symmetric presentation $G/H$ ($G$ connected) as a homogeneous space, namely, $SO(n)$ for $n$ odd and $SO(n)/\mathbf Z_2$ for $n$ even in the real case, and $PU(n)=SU(n)/\mathbf Z_n$ in the complex case. (Note that $U(n)$ acts on the left on the Grassmannian with a $U(1)$-kernel (its center), so the effectivized group is the projectivization $PU(n)$. Moreover the center of $U(n)$ meets $SU(n)\subset U(n)$ along its center, which consists of $\omega I$ where $\omega$ is an $n$-th root of unit.)

Further, Cartan described the full isometry groups of symmetric spaces, and an explicit result is easy to figure out in the case of Grassmann manifolds. I do not remember now, but you can find Cartan's description in the book of O. Loos on symmetric spaces, the second volume. I tend to agree with Ryan when he writes that in the case of Grassmann manifolds, the full isometry group should be $G\times N_G(H)$.

About Stiefel manifolds: with the metric you describe, they are normal homogeneous spaces $G/H$, i. e. have the metric induced from a bi-invariant Riemannian metric on $G$. There is a recent paper by S. Reggiani with a very effective way of computing the identity component of isometry groups of normal homogeneous spaces in here.

Added: I looked up Loos, "Symmetric spaces, II", Theorem 4.4 and the ensuing Table 10 on page 156 for the full isometry group of the real and complex Grassmannians. If I understand correctly, indeed in the case of complex Grassmannians
$SU(n)/S(U(p)\times U(n-p))$, every isometry comes from left multiplication by elements from $SU(n)$ except for two cases: an isometry induced by complex conjugation; and mapping a $p$-plane to its orthogonal complement in case $n=2p\geq4$. In the case of real unoriented Grassmannians $SO(n)/S(O(p)\times O(n-p))$, every isometry comes from left multiplication by an element of $O(n)$ except for: mapping a $p$-plane to its orthogonal complement in case $n=2p\geq4$; the symmetric group $S_3$ in case $n=2p=8$, coming from outer automorphisms of $\mathfrak{so}(8)$.

share|cite|improve this answer
Thanks a lot for the reference and the link! So do I see it correctly that in both cases (the complex Grassmannians and the complex Stiefel manifolds) every isometry of them can be liftet (to an isometry of U(m))? –  AlexE Mar 3 '12 at 13:52

There's a lot of basic facts about Grassmannians and Stiefel manifolds like yours, for which I imagine there are plenty of references out there, yet they also don't appear to show-up in commonly-used textbooks. Or if they do, they're disguised in too much verbiage for me to notice.

For Grassmannians $G_{n,l}$ the isometry group can be larger than just $O(n)$ or $U(n)$ respectively.

For example, there is an isometry $G_{2n,n} \to G_{2n,n}$ with no fixed points. This is the involution that exchanges a subspace with its orthogonal complement. It isn't induced by multiplication by an element of $O(2n)$ or $U(2n)$ (whichever case) provided $n>1$.

Let's just consider the real case to reduce notational baggage, $G_{n,k} = O_n / (O_{n-k} \times O_k)$.

$Isom(G_{n,k})$ has to contain at least the two groups

  • $O(n)$ acting on the left.


  • The quotient of normalizer of $O_{n-k} \times O_k$ in $O_n$ by $O_{n-k} \times O_k$, i.e. $N_{O_n}(O_{n-k}\times O_k) / O_{n-k} \times O_k$, acting on the right.

This is how the involution of $G_{2n,n}$ shows-up since $O_{2n}$ has the matrix that acts (on the right) by switching $O_n \times O_n$ factors. A Lie algebra computation tells you this normalizer is $O_{n-k} \times O_k$ unless $n$ is even and $k$ is $n/2$.

The isometry group of a generic homogeneous space for a group $G$ and point stabilizer $H$ has to always contain $G$ (acting on the left) and $N_G(H)/H$ (acting on the right). I suspect in the Grassmannian case that's all there is but a proof isn't jumping to mind.

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.