invariant inner product on eigenspace

Question

I have several questions about the following corollary:

"Let G/H be a riemannian homogeneous space where G is a compact Lie group. Let $E_{\lambda}=\lbrace f\in C^{\infty}(G/H) : -\Delta f= \lambda f\rbrace$ be a non-trivial eigenspace of Laplace-Beltrami operator, and introduce an inner product invariant under the natural action $f\mapsto f\circ g$ of G on $E_{\lambda}$. Choose an orthonormal basis $f_1,...,f_N$ for $E_{\lambda}$ in this inner product. Then the image of $f=(f_1,...,f_N)$ is contained in some sphere of radius r."

Usually a manifold M is called homogeneous, if some Lie group acts transitively on it. Then the manifold is diffeomorphic to $G/H$, for the isotropy group H. In this situation we have a Riemannian manifold $(M,g)$. Is G the group of isometries here? On the other hand, what kind of inner product is meant here? Is there a explicit way to write down the inner product? If G is the set of isometries, then the action of G on $E_\lambda$ is well defined. But how can I construct an inner product?

Answer 1

First take a Lie group $G$ and a closed subgroup $H$. Denote by $M$ the manifold $G/H$. The tangent space of $M$ at $eH$ can be canonically identified with $\mathfrak{g}/\mathfrak{h}$, where the Gothic letters are the Lie algebras of the corresponding Roman letters.

Since $\mathfrak{h}$ is a Lie subalgebra of $\mathfrak{g}$, the adjoint action of $H$ on $\mathfrak{g}$, leaves $\mathfrak{h}$ stable, hence there is an action of $H$ on $\mathfrak{g}/\mathfrak{h}$.

It is a standard fact in the theory of homogeneous spaces that the set of $G$-invariant Riemannian metric on $G/H$ is canonically in bijection with the set of inner products on $\mathfrak{g}/\mathfrak{h}$, invariant by the above action of $H$.

Such a metric always exists when $H$ is compact, by a general averaging argument: you start with any inner product on $\mathfrak{g}/\mathfrak{h}$ and you average it, using the Haar mesure on $H$ (think of the case of finite groups, it is the same argument).

By definition of a $G$-invariant metric, $G$ respects the Riemannian metric on $G/H$, so $G$ is a subgroup of the group of isometries of $G/H$ (see the edit). I don't think that generally $G$ will be the whole group of isometries, but I'm quite confused here.

Finally, you have a Riemannian metric on $M = G/H$ and you can use it to define an inner product on the Hilbert space $L^2(M)$. When, $M$ is compact (which is the case if and only if $G$ is compact, since $H$ is compact), you have $C^\infty(M) \subset L^2(M)$, and you have an inner product on $C^\infty(M)$. This inner product is invariant for the action of $G$ on $C^\infty(M)$.

Edit: I just add a remark, inspired by the other answer. It is indeed incorrect to say that $G$ is a subgroup of the group of isometries of $G/H$. Since $G$ acts by isometries on $G/H$, there is just one natural morphism $G \rightarrow Isom(G/H)$, but this has no reason to be injective.

Answer 2

I'm going to try to answer "is $G$ the isometry group of $G/H$?" at least when $G$ is compact (or more generally, when $H$ is.)

First, we restrict to the class of metrics on $G/H$ obtained by beginning with a left invariant, right $H$-invariant metric on $G$ and taking the induced submersion metric on $G/H$ (these are the same class of metrics considered in the other answer.) As in the previous answer, these definitely exist when $H$ is compact, but may not otherwise.

Fact 1. $G$ may not be a subgroup of the isometry group of $G/H$, but a quotient of $G$ always is. If $K$ is the largest normal subgroup of $G$ which is also a subgroup of $H$, then $G/K\subseteq Iso(G/H)$. For some examples, $Iso(SU(n)/U(n-1))$ doesn't contain $SU(n)$, but doesn't contain $SU(n)/Z_n$, where $Z_n$ denotes the $n$th roots of one times the identity matrix. More dramatically, $Iso(G/G) = \{e\}$ for any $G$ for stupid reasons.

Fact 2. Sometimes $G = Iso(G/H)$. For example, there is a unique embedding, up to conjugacy, of $SO(3)$ into $SO(5)$ which is not conjugate to the block embedding. For a metric induced from the bi-invariant metric on $SO(5)$, we have $Iso(G/H) = G$. See

Shankar, Ravi. Isometry groups of homogeneous, positively curved manifolds, Differential Geometry & Appl., 14(1) (2001), 57--78.

for details. (A free version may be found on his website.)

Fact 3. Often, $G$ (or $G/K$) is a proper subgroup of $Iso(G/H)$. This occurs, for example, for $S^n = SO(n+1)/SO(n)$ where $SO(n+1)\subsetneq O(n+1) = Iso(S^n)$. Moreover, $G$ (or $G/K$) doesn't have to be the identity component of $Iso(G/H)$. For example, we have $S^6 = G_2/SU(3)$, but $Iso(S^6) = O(7)$, and $G_2$ is a smaller dimensional subgroup of $O(7)$, so the $G$ isn't even the identity component in this case.