Recall that the scalar curvature of a Riemannian manifold is given by the trace of the Ricci curvature tensor. I will now summarize everything that I know about scalar curvature in three sentences:

  • The scalar curvature at a point relates the volume of an infinitesimal ball centered at that point to the volume of the ball with the same radius in Euclidean space.
  • There are no topological obstructions to negative scalar curvature.
  • On a compact spin manifold of positive scalar curvature, the index of the Dirac operator vanishes (equivalently, the $\hat{A}$ genus vanishes).

The third item is of course part of a larger story - one can use higher index theory to produce more subtle positive scalar curvature obstructions (e.g. on non-compact manifolds) - but all of these variations on the compact case.

I am also aware that the scalar curvature is an important invariant in general relativity, but that is not what I want to ask about here. This is what I would like to know:

  1. Are there any interesting theorems about metrics with constant scalar curvature? For example, are there topological obstructions to the existence of constant scalar curvature metrics, or are there interesting geometric consequences of constant scalar curvature?

  2. Can anything be said about manifolds with scalar curvature bounds (other than the result I quoted above about spin manifolds with positive scalar curvature), analogous to the plentiful theorems about manifolds with sectional curvature bounds? (Thus additional hypotheses like simple connectedness are allowed)

  3. Is anything particular known about positive scalar curvature for non-spin manifolds?

  • 9
    $\begingroup$ 1. is essentially the Yamabe problem. $\endgroup$ Commented Jun 30, 2010 at 11:45
  • 1
    $\begingroup$ I assume the Dirac operator theorem you are talking about is the one due to Gromov and Lawson? There is a generalization to almost-spin by Mathai springerlink.com/content/j2rm681616477777 and I seem to remember there being some work on positive scalar curvature for compact manifolds done by Schoen and Yau using non-spin techniques, but my memory maybe faulty. $\endgroup$ Commented Jun 30, 2010 at 11:59
  • $\begingroup$ Probably someone here on MO will know the literature better than I, but just in case, you may want to ask Simon Brendle or Szu-Yu (Sophie) Chen about question 2. I seem to remember that they are working (independently) on problems related to what you asked. $\endgroup$ Commented Jun 30, 2010 at 12:05
  • $\begingroup$ Thanks everyone! Between the four answers, I think I have a pretty coherent answer for all three questions. I upvoted everyone. I'll come back and try to pick a best answer soon... $\endgroup$ Commented Jul 1, 2010 at 9:46
  • $\begingroup$ One thing to realize about the scalar curvature is that it's not the only curvature invariant. In many cases, knowing all the curvature invariants tells you everything there is to know about the metric. But for semi-Riemannian signature in 3 or more dimensions, there are nontrivial spaces where all the curvature invariants vanish. See Schmidt, "Why do all the curvature invariants of a gravitational wave vanish?" arxiv.org/abs/gr-qc/9404037 Can anything be said about manifolds with scalar curvature bounds In the semi-Riemannian case, maybe not much, for the reasons described above. $\endgroup$
    – user21349
    Commented May 14, 2017 at 23:21

10 Answers 10


The Kazdan-Warner theorem goes a long way toward answering the first and second questions.

(For notes typed up by Kazdan, see http://www.math.upenn.edu/~kazdan/japan/japan.pdf.)

Here's what is says (taken almost verbatim from the notes, page 93): Divide the class of all closed manifolds (edit: of dimension > 2. See comments) into 3 types:

I. Those which admit a metric of nonnegative scalar curvature which is positive somewhere.

II. Those which don't but admit a metric of 0 scalar curvature.

III. All other closed manifolds.

The theorem is that if $M$ is in class I, then any $f:M\rightarrow\mathbb{R}$ is the scalar curvature of some metric.

If $M$ is in class II, then $f:M\rightarrow\mathbb{R}$ is the scalar curvature of some metric iff it's identically 0 or negative somewhere.

If M is in class III, then $f:M\rightarrow\mathbb{R}$ is the scalar curvature of some metric iff it's negative somewhere.

In particular, every closed manifold has a metric of constant negative scalar curvature. Those in class I or II have a metric of 0 scalar curvature, and those in class I have a metric of constant positive scalar curvature.

  • $\begingroup$ This version of the KW theorem only works for dimension > 2. Obviously $S^2$ is in class I with the standard metric, but by Gauss-Bonnet cannot admit a metric of negative scalar curvature. The more precise version of the KW theorem is Theorem 6.1 in the notes. $\endgroup$ Commented Jun 30, 2010 at 18:41
  • 4
    $\begingroup$ Also, if you don't care too much about reading the proofs, you should take a look at Jerry Kazdan's CBMS notes "Prescribing the Curvature of a Riemannian Manifold". It has a pretty extensive list of references to questions of the type you are asking. $\endgroup$ Commented Jun 30, 2010 at 19:05

Dear Paul Siegel,

Concerning your question 1, it is related to Yamabe problem (as it was pointed out by W. Wong). More precisely, this problems asks whether one can find a conformal deformation $g=u.g_0$ (where $u>0$ is a smooth function) of a given metric $g_0$ on a compact boundaryless manifold of dimension $n\geq 3$ such that the scalar curvature of $g$ is constant. In some sense, Yamabe problem is a higher-dimensional counterpart of Poincare-Koebe uniformization theorem (for Riemann surfaces).

Roughly speaking, the history of Yamabe problem is:

-Yamabe claimed in 1960 that he solved this problem (as a preliminary step towards Poincare conjecture), but, as it was pointed out by Trudinger, Yamabe's solution had a gap: basically, he converted the constraint "scalar curvature of $g$ is constant" into an non-linear elliptic PDE involving the conformal factor $u$, and Yamabe's assertion was that this PDE was solvable by "usual" elliptic theory, but this is wrong because the corresponding PDE has a so-called "critical" non-linearity (in the sense that this PDE is exactly a borderline case of the "standard" theory)

-The case of the round sphere was dealt with by Obata, who characterized all solutions to Yamabe PDE.

-In view of the stereographic projection, the case of the round sphere is called globally flat case.

-Aubin showed that the case of manifolds of dimension $n\geq 6$ which are not locally flat (i.e., there is some point having some neighborhood which is not conformal to the flat Euclidean space), Yamabe PDE is solvable. To do so, he construct some local test functions to prove that a quantity (related to Yamabe's PDE) called Yamabe quotient is strictly smaller than the same quotient for the round sphere. Once this is proven, the standard theory allows to conclude the solvability of Yamabe's problem.

-In the remaining cases (i.e., a compact manifold of low dimension $n=3,4,5$ or a locally (but not globally) flat manifold), Schoen replaced the use of local test functions by the use of global test function (obtained by suitable gluing of local test functions with appropriate Green functions of Yamabe PDE). Again, using these global test function, Schoen's goal was to show that the Yamabe quotient of our manifold was strictly smaller than the corresponding Yamabe quotient of the round sphere. To do so, Schoen applies a brilliant idea of expanding the Yamabe quotient in terms of the Yamabe quotient of the local function (which turns out to be almost the same of the round sphere) and then checking that the contribution of the Green function is robustly negative because of the so-called Positive Mass Theorem (of Schoen and Yau) (whose connections with General Relativity are well-known).

-After these answers to Yamabe's problem, Schoen asked about the compactness of the set of solutions of Yamabe PDE (in the non-globally flat case, of course). It turns out that this problem was completely solved by the works of Khuri, Marques, Schoen, Brendle and Brendle, Marques: assuming the Positive Mass Theorem, Schoen's question (or perhaps conjecture) has a positive answer in dimensions $3\leq n\leq 24$ but there are non-globally flat counterexamples in dimensions $n\geq 25$ (and, furthermore, this strange "dimensional" behavior is explicitly explained by the lack of definiteness of a certain bilinear forms.

For a nice introduction to Yamabe problem, please see this excellent classical paper of Lee and Parker.

  • 2
    $\begingroup$ This is a pretty nice summary. $\endgroup$ Commented Jun 30, 2010 at 18:26
  • $\begingroup$ Matheus writes a nice summary of the Yamabe conjecture. Unless I missed it, he does not mention this paper: MR1032773 (91a:53065) Jin, Zhi Ren A counterexample to the Yamabe problem for complete noncompact manifolds. Partial differential equations (Tianjin, 1986), 93--101, Lecture Notes in Math., 1306, Springer, Berlin, 1988. (Reviewer: Dennis M. DeTurck) 53C20 (35J60 58G30) $\endgroup$
    – user9087
    Commented Sep 8, 2010 at 19:01
  • 1
    $\begingroup$ Yes, the article by Lee and Parker is a very good reference. However it contains a gap at the end of Proposition 4.6, and this gap is more serious that one might think initially. I do not know an easy way to repair this. I am currently giving a lecture in wich the gap is repaired. if you can read German you can have a look at my lecture notes (with Christian Bär), see mathematik.uni-regensburg.de/ammann/lehre/2018s_yamabe. If not, then you have to wait until it is translated or you have to study the original literature by Schoen. $\endgroup$ Commented Dec 17, 2018 at 17:01

A big classification result that I'm aware of is due to Gromov and Lawson.

Theorem. Let $M$ be a compact simply connected manifold of dimension $\geq 5$, which is not a spin. Then $M$ admits a metric of positive scalar curvature. A spin manifold of dimension $\geq 5$ carries a metric of positive scalar curvature iff its $\alpha$-genus is zero.

The Seiberg-Witten invariants provide special obstructions to existence of a metric of positive scalar curvature in dimension 4.

There are two good survey articles on the subject by J. Rosenberg (link 1) and S. Stolz (link 2).

  • 3
    $\begingroup$ Here's the paper: "The Classification of Simply Connected Manifolds of Positive Scalar Curvature," Mikhael Gromov and H. Blaine Lawson, Jr., The Annals of Mathematics, Second Series, Vol. 111, No. 3 (May, 1980), pp. 423-434. jstor.org/stable/1971103?cookieSet=1 $\endgroup$ Commented Jun 30, 2010 at 13:36
  • 4
    $\begingroup$ It should be mentioned that it was Stephan Stolz who finished the proof of the second half of the theorem (simply connected spin manifolds of large dimension with trivial alpha-genus have positive scalar curvature). It has been published in the Annals. $\endgroup$ Commented Feb 28, 2011 at 9:45

For three manifolds (to "complete the set" started by Andrey's answer) the Schoen and Yau result I mentioned in the comments can be found in Existence of incompressible minimal surfaces and the topology of three - dimensional manifolds with nonnegative scalar curvature, Annals of Maths 110 (1979). A description is also in their book Seminar on Differential Geometry.

They use minimal surface techniques to obtain some topological restrictions for 3 manifolds to admit metrics of nonnegative scalar curvature. I don't actually remember the precise statement of the theorems off the top of my head, but it requires the existence of a specific homotopy type of minimal surfaces. In particular it rules out $T^3$.

  • $\begingroup$ Good point! +1. $\endgroup$ Commented Jun 30, 2010 at 19:23
  • 1
    $\begingroup$ A metric of nonnegative scalar curvature on a torus has to be flat. This is another result due to Gromov and Lawson. $\endgroup$ Commented Jun 30, 2010 at 19:30
  • $\begingroup$ In fact, G&L proved it for all tori. Their proof is based on Dirac operators and is developed somewhat parallel and independent to S&Y. Anyway, I mentioned $T^3$ because it is, I am ashamed to say, the only manifold I remember where the incompressible minimal surface is guaranteed to exist (it'd be a $T^2$ section). $\endgroup$ Commented Jun 30, 2010 at 19:35

Perhaps it's worth adding something about the obstructions to positive scalar curvature in four dimensions. As Andrey Rekalo remarks, these come from the Seiberg-Witten equations. As with their applications to diffeomorphism problems, these equations show that dimension 4 is anomalous, and tantalise us with questions of whether the obstructions that they provide are complete.

The $\hat{A}$-genus obstruction to positive scalar curvature comes about because, in even dimensions, this genus is the index of the Dirac operator $D: \Gamma(S^+)\to \Gamma(S^-)$ of a (Riemannian) spin manifold. $D$ satisfies a Weitzenboeck formula involving scalar curvature, $$ - D^2 = \nabla^\ast \nabla + \frac{s}{4}\mathrm{id},$$ and integration by parts shows that when $s>0$, a spinor $\psi$ satsifying the Dirac equation $D\psi =0$ must vanish.

One can try to generalise this to $spin^c$-manifolds and the Dirac operator $D_A$ coupled to a spin connection $A$, but the argument breaks because the Weitzenboeck formula has an extra term involving the curvature $F_A$. In 4 dimensions, the Seiberg-Witten equations are the Dirac equation $D_A\psi =0 $ together with a non-linear equation $$F_A^+ = q(\psi)$$ invoking a natural quadratic form $q$ on $S^+$. Witten used the Weitzenboeck formula to show that if $s > 0$, a solution to these equations must have $\psi=0$, so that $A$ is an anti-self-dual connection: $F_A^+=0$. But if the (oriented) manifold $X$ has $b^+(X)>0$ then for generic metrics there are no such ASD connections. So, if we can guarantee that $X$ has solutions to the SW equations in a chosen $spin^c$-structure for all metrics, it cannot admit a metric of positive scalar curvature.

We have such a guarantee, for instance, when $X$ is symplectic with $b^+>1$ (Taubes). Why there should be any relationship between scalar curvature and symplectic geometry is a curious puzzle. Existence results seem to be generally lacking - e.g., if $X$ is not spin, and has $b^+>0$, does the connected sum of $X$ with $S^2\times S^2$ have a metric of positive scalar curvature?


There is an important relation between variation of scalar curvature Kahler metric and Lichnerowicz operator. Lets explain it here

Let $(X,J,\omega)$ be Fano Kahler manifold with $\omega_0\in [\omega_0]=\kappa$, then by $\partial\bar\partial$-lemma any other Kahler 2-form can be written as $\omega_0+\sqrt{-1}\partial\bar\partial \varphi$. Now the set of Kahler metrics in $[\omega_0]$ can be identified with $$\mathcal H=\{\varphi \in C^\infty(X,\mathbb R)|\omega_0+\sqrt{-1}\partial\bar\partial \varphi>0\}/\mathbb R$$. Note that $T_\varphi\mathcal H=C^\infty(X,\mathbb R)/\mathbb R$

The Scalar curvature can de define as

$$Sc:\mathcal H\to C^{\infty}(X,\mathbb R)$$

$$Sc(\varphi)=\Lambda_\varphi\sqrt{-1}\partial\bar\partial\log \omega^n$$

and for $\psi\in T_{\varphi}\mathcal H$ we have

$$d \text{Sc}(\varphi).\psi=\Delta_{\omega_\varphi}^2\psi-Sc(\varphi)\Delta_{\omega_{\varphi}}\psi$$ and moreover by using following known relation of Lichnerowicz

$$\mathcal D_\varphi^*\mathcal D_\varphi \psi=\Delta_{\omega_\varphi}^2\psi-Sc(\varphi)\Delta_{\omega_\varphi}\psi+\frac{1}{2}g(dSc(\varphi),d\psi)$$ so we can rewrite the derivative of scalar curvature metric as follows

$$d \text{Sc}(\varphi).\psi=\mathcal D_\varphi^*\mathcal D_\varphi \psi-\frac{1}{2}g(dSc(\varphi),d\psi)$$

See also Oliver Biquard expository paper


Regarding question 2, Lohkamp proved here that for a given compact Riemannian manifold $(M,g_0)$ with scalar curvature $\kappa_0$, one can prescribe a function $f\le\kappa_0$ and find a metric $g$ on $M$ whose scalar curvature is $C^0$ close to $f$, such that $g$ is $C^0$ close to $g_0$. Thus, it does not make sense to ask for pointwise lower bounds for scalar curvature, even if one poses additional restrictions on the geometry.

On the other hand, inspired by a remark by Gromov, Llarull proved here that one cannot find a metric on $S^n$ that is pointwise at least as large as the round metric (where you actually compare sizes of 2-vectors $v\wedge w$), and whose scalar curvature is pointwise at least as large as on the round sphere. Following Gromov, one may call such Riemannian manifolds area-extremal and ask for more examples. There some results by Semmelmann and me, see here for a list.

Finally, you can study manifolds with boundary where you fix the geometry only close to the boundary, and ask for rigidity results. In this article, Brendle, Marques and Neves show that there exist metrics on a hemisphere that have a round sphere as a totally geodesic boundary, with scalar curvature larger than that of a round hemisphere with the same boundary.


Also: when one is talking about positive scalar curvature, the Yamabe invariant is important. The Yamabe invariant is the supremum over all conformal classes of the Yamabe constants of a manifold. The Yamabe invariant is positive if and only if the manifold supports a metric with positive scalar curvature. For a comprehensive guide to the somewhat recent state of affairs on the classification of such manifolds look for Jonathan Rosenberg's preprint on his website. Also, Bray and Neves have computed the Yamabe invariant for some three manifolds. This can be found on the wikipedia website for the Yamabe invariant.


In the setting of K\"ahler geometry, a necessary condition for the existence of a k\"ahlerian metric of constant scalar curvature is vanishing of the so-called Futaki invariant. (cf. Futaki On compact Kähler manifolds of constant scalar curvatures).


Another thing needs to mention is cscK on a polarized K\"{a}hler manifolds whose existence is conjectured to be equivalent to a notion of stability. See the work of Donaldson, Tian, et al.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.