Hi,


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.


Thanks.


Asked that question already on Math.SE and got no answer: <https://math.stackexchange.com/q/112175/7110>



Full isometry groups of Stiefel and Grassmann manifolds
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](http://arxiv.org/abs/0906.1802).


**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 $\mathfrak{so}(8)$. 


