On the variational properties of the prescribed Ricci curvature functional

We first recall some basics of the geometry of a homogeneous space. Let $H\subset G$ be two compact Lie groups with Lie algebras ${\mathfrak{h}}\subset{\mathfrak{g}}$ such that $G/H$ is an almost effective homogeneous space. We fix a bi-invariant metric $Q$ on ${\mathfrak{g}}$, which defines a $Q$-orthogonal $\operatorname{Ad}_{H}$-invariant splitting ${\mathfrak{g}}={\mathfrak{h}}\oplus{\mathfrak{m}}$, i.e., ${\mathfrak{m}}={\mathfrak{h}}^{\perp}$. The tangent space $T_{eH}(G/H)$ is identified with ${\mathfrak{m}}$, and $H$ acts on ${\mathfrak{m}}$ via the adjoint representation $\operatorname{Ad}_{H}$. A $G$-invariant metric on $G/H$ is determined by an $\operatorname{Ad}_{H}$-invariant inner product on ${\mathfrak{m}}$. We denote by $\mathcal{M}(G/H)$, or sometimes simply by $\mathcal{M}$, the space of $G$-invariant metrics on $G/H$. We will assume throughout the paper that $G/H$ is not a torus since in this case all $G$-invariant metrics are flat, in particular, $\operatorname{Ric}(g)=0$ for all $g\in\mathcal{M}$.

We describe metrics in $\mathcal{M}$ in terms of $\operatorname{Ad}_{H}$-invariant splittings. Under the action of $\operatorname{Ad}_{H}$ on ${\mathfrak{m}}$, we decompose

where $\operatorname{Ad}_{H}$ acts irreducibly on ${\mathfrak{m}}_{i}$. Some of these summands may need to be one-dimensional if there exists a subspace of ${\mathfrak{m}}$ on which $\operatorname{Ad}_{H}$ acts as the identity. We denote by $\mathcal{D}$ the space of all such decompositions and use the letter $D\in\mathcal{D}$ for a particular choice of decomposition. The space $\mathcal{D}$ has a natural topology induced from the embedding into the products of Grassmannians $G_{k}({\mathfrak{g}})$ of $k$-planes in ${\mathfrak{g}}$. Clearly, $\mathcal{D}$ is compact.

If $T$ is a $G$-invariant symmetric bi-linear form field on $G/H$, it is determined by its value on ${\mathfrak{m}}$. We are interested when such a bi-linear form field is (up to scaling) the Ricci curvature of a metric $g\in\mathcal{M}$, i.e., when

Throughout the paper we will assume that $T$ is positive-definite. We may also assume that $c>0$ in (1.1) since a compact homogeneous space does not admit any metrics with $\operatorname{Ric}\leq 0$ unless it is a torus (see [4, Theorem 1.84]), which we excluded above.

Define the hypersurface

where $\operatorname{tr}_{g}$ is the trace with respect to $g$. We denote it simply by $\mathcal{M}_{T}$ when the homogeneous space is clear from context. As shown in [25], a solution to (1.1) can we viewed as a critical point of the functional

where $S(g)$ is the scalar curvature of $g$. More precisely, the following result holds.

Here $c$ is the Lagrange multiplier of the variational problem. Our main interest in this paper is to describe the geometry of the functional $S$ and its implications for when (1.1) has a solution.

We now recall the formulas for the scalar curvature and the Ricci curvature of a homogeneous metric. Given $g\in\mathcal{M}$, we have $Q_{|{\mathfrak{m}}_{i}}=x_{i}\,Q_{|{\mathfrak{m}}_{i}}$ for some constant $x_{i}>0$. In general, ${\mathfrak{m}}_{i}$ and ${\mathfrak{m}}_{j}$ do not have to be orthogonal if some of these summands are equivalent. But we can diagonalize $g$ and $Q$ simultaneously, and hence there exists a decomposition $D\in\mathcal{D}$ such that the metric has the form

We call such metrics diagonal with respect to our choice of $D$ and denote their set by $\mathcal{M}^{D}(G/H)$, or simply $\mathcal{M}^{D}$. Thus, $\mathcal{M}=\cup_{D\in\mathcal{D}}\mathcal{M}^{D}$. We also denote $\mathcal{M}_{T}^{D}=\mathcal{M}_{T}\cap\mathcal{M}^{D}$. The scalars $x_{i}$ are simply the eigenvalues of $g$ with respect to $Q$. When these eigenvalues have multiplicity, and some of the modules in the corresponding eigenspace are equivalent under the action of $\operatorname{Ad}_{H}$, we note that $g\in\mathcal{M}^{D}$ for a compact infinite family of decompositions $D$. In order to describe all homogeneous metrics on $G/H$, we can thus restrict ourselves to diagonal metrics but allow the decomposition to change. The advantage is that, while the scalar curvature of a homogeneous metric for a fixed decomposition is quite complicated and hence somewhat intractable, for a diagonal metric it has a much simpler form. This idea was first used in [28] to study $G$-invariant Einstein metrics.

We define the structure constants

where $(e_{\alpha})$, $(e_{\beta})$ and $(e_{\gamma})$ are $Q$-orthonormal bases of ${\mathfrak{m}}_{i}$, ${\mathfrak{m}}_{j}$ and ${\mathfrak{m}}_{k}$. Clearly, $[ijk]\geq 0$, and $[ijk]=0$ if and only if $Q([{\mathfrak{m}}_{i},{\mathfrak{m}}_{j}],{\mathfrak{m}}_{k})=0$. We will denote by $B$ the Killing form of $G$. By the irreducibility of ${\mathfrak{m}}_{i}$, there exist constants $b_{i}\geq 0$ such that

with $b_{i}=0$ if and only if ${\mathfrak{m}}_{i}$ lies in the center of ${\mathfrak{g}}$. Furthermore, not all of $b_{i}$ vanish since otherwise ${\mathfrak{m}}$ is in the center ${\mathfrak{z}}({\mathfrak{g}})$ and $G/H$ is a torus. Using this notation, and $d_{i}=\dim{\mathfrak{m}}_{i}$, the scalar curvature is given by

(see [28]). For the tensor $T$, we introduce the constants $T_{i}$ such that

Varying over all decompositions, these constants determine $T$ uniquely.

We now recall some formulas for Riemannian submersions which will be useful for us. Let $K$ be a compact subgroup with Lie algebra ${\mathfrak{k}}$ lying between $H$ and $G$, i.e., $H\subset K\subset G$. We then have a homogeneous fibration

We will often consider metrics with respect to which the projection $G/H\to G/K$ is a Riemannian submersion. Assume that the decomposition ${\mathfrak{m}}=({\mathfrak{k}}\cap{\mathfrak{m}})\oplus{\mathfrak{k}}^{\perp}$ is orthogonal with respect to both $Q$ and $g$. Then $g\in\mathcal{M}$ is a Riemannian submersion metric if and only if $g_{|{\mathfrak{k}}^{\perp}}$ is $\operatorname{Ad}_{K}$-invariant. In this case, $g_{|{\mathfrak{k}}\cap{\mathfrak{m}}}$ can be thought of as a homogeneous metric on the fiber $F=K/H$, and $g_{|{\mathfrak{k}}^{\perp}}$ a homogeneous metric on the base $B=G/K$. We can introduce a new submersion metric $g_{s,t}$ on $G/H$ by scaling the fiber and the base, i.e.,

One has the following formula for the scalar curvature (see, e.g., [4, Proposition 9.70]):

where $A$ is the O’Neill tensor of the submersion, while $S_{F}$ and $S_{B}$ are the scalar curvatures of $g_{F}$ and $g_{B}$. The proof of this formula is a pointwise calculation and hence extends to the situation where $K$ is not compact (i.e., $G/K$ is not a manifold). In our case it can indeed happen that for some Lie algebra ${\mathfrak{k}}\subset{\mathfrak{g}}$ the connected subgroup $K\subset G$ with Lie algebra ${\mathfrak{k}}$ is not compact. However, this will not affect our discussions.

Let $N(H)$ denote the normalizer of $H$. Every element $n\in N(H)$ acts on $G/H$ as a diffeomorphism via right translation: $R_{n}(gH)=gn^{-1}H$. Combining this with the action by left translation, we obtain an action on the space of metrics via pullback: $\mathcal{M}\ni g\mapsto(\operatorname{Ad}_{n})^{*}(g)\in\mathcal{M}$. This induces an isometry between $g$ and $(\operatorname{Ad}_{n})^{*}(g)$, and hence $S((\operatorname{Ad}_{n})^{*}(g))=S(g)$ for the scalar curvature. This also holds for the possibly larger group $\operatorname{Aut}(G,H)$ of automorphisms of $G$ that preserve $H$. Note though that $S_{|\mathcal{M}_{T}}$ is invariant under $\operatorname{Aut}(G,H)$, or one of its subgroups, only if $T$ is as well, in which case the orbit of a critical point consists of further critical points.

The following remark, based on Palais’s principle of symmetric criticality, will be useful for us. If $T$ is invariant under a subgroup $L\subset\operatorname{Aut}(G,H)$, and if $\mathcal{M}_{T}^{L}$ is the set metrics in $\mathcal{M}_{T}$ invariant under $L$, then critical points of the restriction $S_{|\mathcal{M}_{T}^{L}}$ are also critical points of $S_{|\mathcal{M}_{T}}$. Indeed, given $g\in\mathcal{M}_{T}^{L}$, the Ricci curvature $\operatorname{Ric}(g)$ and hence the gradient $\operatorname{grad}S_{|\mathcal{M}_{T}}(g)$ is invariant under $L$. Consequently, $\operatorname{grad}S_{|\mathcal{M}_{T}}(g)$ must be tangent to $\mathcal{M}_{T}^{L}$.

In the remainder of the paper we will assume for simplicity that $G$, $H$ and all the intermediate subgroups are connected. Let us explain why this is in fact not necessary. One only needs to make the following modification. If $G$ or $H$ is not connected, we consider only intermediate subalgebras ${\mathfrak{k}}$ that are Lie algebras of intermediates subgroups $H\subset K\subset G$. This is easily seen to be equivalent to saying that ${\mathfrak{k}}$ must be invariant under $\operatorname{Ad}_{H}$. The proofs of all of our results apply without any changes, and the conclusions are also the same. This is useful when the space of invariants has large dimension or there are many intermediate subgroups. Adding components to $G$ and $H$ will reduce $\dim\mathcal{M}_{T}$ and may easily imply the existence of some critical points. This happens, for instance, when $G/H$ is isotropy irreducible; see [29] for many examples.

In Sections 2–3 we fix the decomposition $D\in\mathcal{D}$ and study the set of metrics $\mathcal{M}^{D}$ diagonal with respect to $D$.

It will be convenient for us to describe a homogeneous metric in terms of its inverse since this makes the space of metrics precompact. If $g\in\mathcal{M}_{T}^{D}$ is given by (1.3), we set $y_{i}=\frac{1}{x_{i}}$ and obtain the following formulas for the scalar curvature and its constraint:

We need to study the behavior of $S_{|\mathcal{M}_{T}^{D}}$ at infinity, which means that at least one of the variables $y_{i}$ goes to $0$. It is natural to introduce a simplicial complex and its stratification. Specifically, let

Notice that the numbers $T_{i}$, and thus the simplex $\Delta$, depend on the choice of $D$. This simplex is a natural parametrization of the set $\mathcal{M}_{T}^{D}$. We identify a metric $g\in\mathcal{M}_{T}^{D}$ with $y=(y_{1},\cdots,y_{r})\in\Delta$.

The boundary of $\Delta$ consists of lower-dimensional simplices. For every nonempty proper subset $J$ of the index set $I=\{1,\cdots,r\}$, let

Thus $\Delta_{J}$ is a $|J|$-dimensional simplex, which we call a stratum of $\partial\Delta$. The closure of $\Delta_{J}$ satisfies

and we call $\Delta_{J^{\prime}}$ a stratum adjacent to $\Delta_{J}$ if $J^{\prime}$ is a nonempty proper subset of $J$. It will also be useful for us to consider tubular $\epsilon$-neighborhoods of strata for $\epsilon>0$:

Finally, we associate to each stratum an $\operatorname{Ad}_{H}$-invariant subspace of ${\mathfrak{m}}$:

We can fill out the closure $\bar{\Delta}$ with geodesics starting at the center. To this end, consider the unit sphere

of dimension $r-2$. Define a geodesic $\gamma_{v}\colon[0,t_{v}]\to\bar{\Delta}$ by setting $\gamma_{v}(t)=v_{0}-tv$, where

The stratification of $\partial\Delta$ induces one of the sphere:

Our first observation is that we can mark the strata with subalgebras.

We also need to control how fast the scalar curvature goes to $-\infty$. For this purpose we prove the following result.

If the space $G/H$ has pairwise inequivalent isotropy summands, Proposition 2.3 implies the following result, originally proved in [25].

We can add a marking to the strata in $\partial\Delta$. If ${\mathfrak{h}}\oplus{\mathfrak{m}}_{J}={\mathfrak{k}}$ is a subalgebra, we denote the stratum $\Delta_{J}$ by $(\Delta_{J},{\mathfrak{k}})$ or simply $\Delta_{\mathfrak{k}}$; if it is not, we denote the stratum by $(\Delta_{J},\infty)$ or $\Delta_{\infty}$.

Next, we need an estimate for $S$ near $(\Delta_{J},{\mathfrak{k}})$. Let $K$ be the connected subgroup of $G$ with Lie algebra ${\mathfrak{k}}$. Define

where the letter $D$ is preserved for the decomposition of the tangent space to $K/H$ induced by $D$. Since a normal homogeneous metric has non-negative scalar curvature, $\alpha_{\mathfrak{k}}^{D}\geq 0$. Also, $\alpha_{\mathfrak{k}}^{D}=0$ if and only if $K/H$ is flat. It is important for us to note that, since $\Delta_{\mathfrak{k}}$ is not closed in general, the supremum in the definition of $\alpha_{\mathfrak{k}}^{D}$ may not be achieved in $\Delta_{\mathfrak{k}}$, which will complicate our discussion.

Consider a metric in $\mathcal{M}_{T}^{D}$ identified with $y=(y_{1},\ldots,y_{r})\in\Delta$. If ${\mathfrak{k}}={\mathfrak{h}}\oplus{\mathfrak{m}}_{J}$ for some $J\subset I$, then

We may regard $y_{|{\mathfrak{k}}\cap{\mathfrak{m}}}$ as a metric on $K/H$. Its scalar curvature is given by

where the Killing form of $K$ restricted to ${\mathfrak{m}}_{i}$ equals $-\bar{b}_{i}Q_{|{\mathfrak{m}}_{i}}$. One easily shows that

We can reformulate Proposition 2.5 as follows.

Combining Propositions 2.3 and 2.5, we arrive at the following conclusion.

Proposition 2.5 shows that $\alpha_{\mathfrak{k}}^{D}$ is an upper bound for the possible values of the scalar curvature as we approach points in $\Delta_{\mathfrak{k}}$. However, it is important to keep in mind that $S$ does not, in general, extend continuously to the closure of $\Delta$.

We end this section with the following observation. Recall that even if $G$ and $H$ are connected and compact, and if ${\mathfrak{k}}$ is an intermediate subalgebra, then the connected (intermediate) subgroup with Lie algebra ${\mathfrak{k}}$ is not necessarily compact.