# Homogeneous metrics of given volume element, on the n-sphere.

Let $\omega$ be an $n$-form on $S^n$, nowhere vanishing.

Is there a Riemannian metric $g$ on $S^n$, so that its volume form is $\omega$, and $(S^n,g)$ is homogeneous? Is it unique, and if not, what transformations of $S^n$ relates such metrics?

For example, for $n=1$, the unique metric with these properties is $\text{d} u^2$, where $u=\frac{1}{2\pi}\int\omega$.

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There is. By Moser's Theorem, there exists a diffeomorphism $f:S^n\to S^n$ such that $f^*\omega = c\omega_{std}$ where $c>0$ and $\omega_{std}$ is the standard volume form homogeneous under $\mathrm{SO}(n{+}1)$. Now use $f$ to transfer the right multiple of the standard metric (under which $S^n$ is homogeneous) back to one for which $\omega$ is the volume form.

About uniqueness: (I forgot to answer this question on the first pass; also I am fixing a few errors in the cases below in this edit.) The answer about uniqueness of homogeneous metrics with a given volume form depends on the dimension $n$. Obviously, when $n=1$, it is unique. It's also unique (up to volume-preserving diffeomorphism) when $n=2$ (and, in fact, for all even $n$). However, it is not unique (up to volume-preserving diffeomorphism) when $n$ is odd.

The reason has to do with this: Suppose $g$ is a metric on $S^n$ that is homogeneous, and let $G$ be the identity component of the group of $g$-isometries. Then $G$ is compact, and $S^n = G/H$ where $H\subset G$ is a closed, connected subgroup of $G$ acting effectively on $S^n$. It is known, from the work of Borel, which pairs such $(G,H)$ have the property that $G/H$ is diffeomorphic to $S^n$ for some $n$, so we know all of the possibilities and can work out the isometry groups in each case.

When $n=2k$, the only possibility is $G = \mathrm{SO}(2k{+}1)$ with $H = \mathrm{SO}(2k)$. The $G$-invariant metric $g$ is then uniquely determined by specifying its total volume. (While $S^6 = \mathrm{G}_2/\mathrm{SU}(3)$ would appear to be a counterexample, in this case $\mathrm{G}_2$ is not the full isometry group of the invariant metric, $\mathrm{SO}(7)$ is.)

When $n = 4k{+}1>1$, we can have $G = \mathrm{SU}(2k{+}1)$ and $H = \mathrm{SU}(2k)$, and there is a $1$-parameter family of $G$-invariant metrics with a given total volume. The full isometry group of such a metric is $\mathrm{U}(2k{+}1)$, except that one element of the family (the metric of constant sectional curvature) has (connected) isometry group $\mathrm{SO}(4k{+}2)$.

When $n=4k{+}3$, we can have $G = \mathrm{Sp}(k{+}1)$ and $H=\mathrm{Sp}(k)$, and there is a $5$- (when $k=0$) or $6$- (when $k>0$) parameter family of $G$-invariant metrics on $S^n$ with a given total volume. Some of these are isometric though; one can get rid of $3$ of the parameters that way. Some of these metrics have slightly larger isometry groups than $\mathrm{Sp}(k{+}1)$, such as $\mathrm{Sp}(k{+}1){\cdot}S^1$ or even $\mathrm{Sp}(k{+}1){\cdot}\mathrm{Sp}(1)$, and one, the metric of constant sectional curvature, has (connected) isometry group $\mathrm{SO}(4k{+}4)$.

Finally, there is the exceptional case of $S^{15} = \mathrm{Spin}(9)/\mathrm{Spin}(7)$, which has a $1$-parameter family of inequivalent $\mathrm{Spin}(9)$-invariant metrics with the same total volume. (Even though we also have $S^{15}=\mathrm{Sp}(4)/\mathrm{Sp}(3)$, these homogeneous metrics are not comparable, except at the constant curvature metric, which belongs to both families.)

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Thanks, that's exactly what I needed! – Cristi Stoica Jul 23 '12 at 16:27
Here's a link to Moser's paper ams.org/journals/tran/1965-120-02/S0002-9947-1965-0182927-5/… – Cristi Stoica Jul 23 '12 at 16:28