A survey on positive scalar curvature metrics

Abstract

In this note we survey a selection of classical results and recent advances concerning our understanding of spaces of positive scalar metrics on closed manifolds, and describe how the basic questions can be transplanted to compact manifolds with boundary, a setting that naturally connects to the study of data sets in general relativity. Special emphasis is devoted to highlighting links with nearby fields and discussing some promising future directions.

Appendices

Appendix A. Left-invariant metrics on $S^3$

Here we present the proof of the following simple, yet partly surprising result:

Proposition A.1

The three-dimensional sphere $S^3$, identified with the Lie group ${\text {SU}}(2)$, supports left-invariant metrics of negative scalar curvature.

Proof

Given the identification in the statement, the tangent space at the identity (namely: $\mathfrak {s}\mathfrak {u}(2)$) is spanned by the basis

$$\begin{aligned} e_1 = \begin{pmatrix} i &{} 0 \\ 0 &{} -i \end{pmatrix}, \quad e_2 = \begin{pmatrix} 0 &{} 1 \\ -1 &{} 0 \end{pmatrix}, \quad e_3 = \begin{pmatrix} 0 &{} i \\ i &{} 0 \end{pmatrix}. \end{aligned}$$

Hence, we define on $\mathfrak {su}(2)$:

$$\begin{aligned} g(e_i,e_j) = {\left\{ \begin{array}{ll} \mu _1 &{} \hbox {if }i=j=1\\ \mu _2 &{} \hbox {if }i=j=2\\ \mu _3 &{} \hbox {if }i=j=3\\ 0&{}\hbox {else} \end{array}\right. } \end{aligned}$$

Set $\epsilon _i = e_i/\sqrt{\mu _i}$ for $i=1,\ldots ,3$. Then, as it easily checked (cf. e.g. [36] for all details), the sectional curvatures of $(S^3,g)$ are given by

$$\begin{aligned} K_g(\epsilon _1,\epsilon _2) = -3\frac{\mu _3}{\mu _1\mu _2} + \frac{\mu _2}{\mu _1\mu _3} + \frac{\mu _1}{\mu _2\mu _3} + \frac{2}{\mu _1} + \frac{2}{\mu _2} - \frac{2}{\mu _3} \end{aligned}$$

and cyclic permutations. Hence, the scalar curvature of this manifold is given by

$$\begin{aligned} R_g&= 2 (K_g(\epsilon _1,\epsilon _2) + K_g(\epsilon _1,\epsilon _3) + K_g(\epsilon _2,\epsilon _3))\\&= -2\left( \frac{\mu _3}{\mu _1\mu _2} + \frac{\mu _2}{\mu _1\mu _3} + \frac{\mu _1}{\mu _2\mu _3} \right) + 4\left( \frac{1}{\mu _1} + \frac{1}{\mu _2} + \frac{1}{\mu _3} \right) \end{aligned}$$

constrained to the open octant $\{\mu _1>0,\, \mu _2>0,\, \mu _3>0\}$. Therefore, choosing $\mu _1=\sigma$, $\mu _2=\sigma ^2$, $\mu _3=\sigma ^3$ for a parameter $\sigma >0$, we obtain

$$\begin{aligned} R_g = -2(1 + \sigma ^{-2} + \sigma ^{-4}) + 4(\sigma ^{-1} + \sigma ^{-2} + \sigma ^{-3}) \end{aligned}$$

thus $R_g \simeq -2\sigma ^{-4}$ as $\sigma \rightarrow 0^+$, whence the conclusion is straightforward. $\square$

Appendix B. Trichotomy theorem

We shall state here, for the sake of completeness a basic but fundamental fact about the conformal geometry of closed manifolds of dimension at least three, that is sometimes cited in the literature as trichotomy theorem.

Theorem B.1

Let $(X^n,g_0)$ be a closed Riemannian manifold. Define the Yamabe invariant of the conformal class $[g_0] = \{ g = e^{2f}g_0 {\ :\ }f\in C^\infty (X) \}$ as

$$\begin{aligned} Y([g_0]) = \inf \{ E(g) {\ :\ }g\in [g_0], {{\,\mathrm{vol}\,}}_g = 1 \}, \end{aligned}$$

where $E(g) = \int _X R_g$. Then, there are exactly three mutually distinct possibilities:

• (1) $Y([g_0])>0$, if and only if there exists $g\in [g_0]$ with $R_g>0$, if and only if $\lambda _1(-L) > 0$.

• (2) $Y([g_0])=0$, if and only if there exists $g\in [g_0]$ with $R_g=0$, if and only if $\lambda _1(-L) = 0$.

• (3) $Y([g_0])<0$, if and only if there exists $g\in [g_0]$ with $R_g<0$, if and only if $\lambda _1(-L) < 0$.

Here L is the conformal Laplace operator, i.e.

$$\begin{aligned} Lu = \Delta _{g_0} u - c(n) R_{g_0} u, \qquad c(n) = \frac{n-2}{4(n-1)}. \end{aligned}$$

It is perhaps appropriate to note here how, in sketching the proof of Theorem 1.4, we did not rely on the full strength of the statement above but only the following straightforward lemma.

Lemma B.2

Given a closed Riemannian manifold $(X^n,g_0)$, if $E(g_0)<0$, then $\lambda _1(-L) < 0$. As a result, there exists $g\in [g_0]$ with $R_g<0$.

Proof

We recall the variational characterization of the first eigenvalue of an elliptic operator, which in this case reads

$$\begin{aligned} \lambda _1(-L) = \inf _{u\in H^1{\setminus }\left\{ 0\right\} } \frac{ \int _X |{\nabla u|}^2 +c(n) R_{g_0} u^2 }{\int _X u^2}. \end{aligned}$$

Now, an admissible competitor for the above minimization problem is $u=1$, so

$$\begin{aligned} \lambda _1(-L) \le c(n) \frac{\int _X R_{g_0} }{{{\,\mathrm{vol}\,}}_{g_0}(X)}=c(n)\frac{E(g_0)}{{{\,\mathrm{vol}\,}}_{g_0}(X)} < 0. \end{aligned}$$

If we set $g=u^{4/(n-2)}g_0$ and recall that $R(g) = -c(n)^{-1} u ^{-\frac{n+2}{n-2}} Lu$, we can just deform the Riemannian manifold $(X,g_0)$ using the first (positive) eigenfunction of the conformal Laplacian. This is an elliptic way of spreading the curvature, as we explained above.