# Examples of manifolds that do not admit scalar flat metrics

The Kazdan-Warner trichotomy states that for $n\ge 3$, a compact $n$-manifold falls into one of three categories:

(A) Every (smooth) function is a scalar curvature.

(B) The manifold is strongly scalar flat.

(C) The manifold only admits scalar curvatures which are negative somewhere.

Of course class (A) is nonempty in all dimensions because it contains $S^n$. Gromov and Lawson showed that (B) contains all tori $T^n$. However, it's not clear to me that (C) is nonempty in all dimensions. Kazdan and Warner (Prescribing Curvatures, Proc. Symp. Pure Math. 27) showed:

Let $M$ be a spin manifold with $\hat A(M)\ne 0$ and $b_1(M)=\dim M$. Then $M$ does not admit a metric of zero scalar curvature.

Consequently, any such manifold must be type (C). They only give the example $T^4\#K3$. Are there examples in dimensions $3$ and $\ge 5$ of type (C) manifolds? Presumably one could use the Kazdan-Warner result above and then apply some knowledge of manifolds with nonzero A-roof genus. They mention Hitchin told them one can strengthen the hypothesis to $b_1(M)\ne 0$.

• Note that $(T^4\# K3)^m$ is a $4m$-dimensional manifold in class (C) using the same criterion. So the missing dimensions are $n \equiv 2, 3, 5, 6, 7 \bmod 8$. By the classification of manifolds in class (A) mentioned in Igor Belegradek's answer, if $n \equiv 3, 5, 6, 7 \bmod 8$, any $n$-manifold in class (C) has non-trivial fundamental group. – Michael Albanese Mar 13 '18 at 3:47
• For $n \equiv 2 \bmod 8$, there are examples similar to the ones mentioned by Igor Belegradek (see my comment on his answer). So the only dimensions that have not been covered so far are $n \equiv 3, 5, 6, 7 \bmod 8$. – Michael Albanese Mar 18 '18 at 3:44

Bourguignon showed that if a compact manifold does not admit positive scalar curvature metrics, then any scalar flat metric (actually, any non-negative scalar curvature metric) is Ricci-flat; I suppose this is what you mean when you write "strongly scalar flat". But in three dimensions, the Ricci curvature determines the full curvature tensor, in particular, a Ricci-flat metric is flat. So any non-flat three-manifold which does not admit positive scalar curvature metrics will provide an example.

If I'm not mistaken, Gromov and Lawson proved that a compact three-manifold admits positive scalar curvature if and only if its prime decomposition contains no aspherical factors; note, this was before the Poincaré conjecture had been verified, so there would have been a caveat at the time of publication.

So $T^3\#S^1\times S^2$ is an example of a compact three-manifold of type (C). Note, $T^3\#S^1\times S^2$ is not flat as a non-trivial connected sum of compact manifolds of dimension at least three is never aspherical, but flat $n$-dimensional manifolds have universal cover $\mathbb{R}^n$.

In dimension four, you can sometimes use Seiberg-Witten invariants to rule out the existence of positive scalar curvature metrics, and then use the Hitchin-Thorpe inequality to rule out the existence of a Ricci-flat metric. For example, a compact Kähler surface with $b^+ \geq 2$ does not admit positive scalar curvature metrics; blowing up doesn't change this, but it eventually violates the Hitchin-Thorpe inequality.

A proof of Bourguignon's result can be found in Kazdan and Warner's paper Prescribing Curvatures, namely Lemma 5.2. As for the result of Gromov and Lawson, see Chapter IV, Theorem 6.18 of Lawson and Michelsohn's Spin Geometry and the discussion which follows.

• The example for $n=3$ is nice. Why doesn't it admit a scalar flat metric though? Do you know about $n\ge 5$? – Ryan Unger Mar 3 '18 at 21:54
• If it were scalar flat it would be Ricci-flat (Bourguignon's result), and because it is three-dimensional, it would also be flat. – Michael Albanese Mar 3 '18 at 23:31
• Ah, right, of course, that's what you wrote up there. – Ryan Unger Mar 4 '18 at 0:40

There is a substantial literature on the subject. Let me focus on closed simply-connected manifolds of dimension at least $5$. With these assumptions

1. A manifold in the class (A) if and only if it is either non-spin, or it is spin and the Lichnerowicz-Hitchin obstruction vanishes.

2. A manifold is in class (B) if and only if its Lichnerowicz-Hitchin obstruction is nonzero, and it is the product of Ricci-flat Kaehler or Spin(7) manifolds. In this case the manifold is spin and even-dimensional. This is proved by A.Futaki in "Scalar-flat closed manifolds not admitting positive scalar curvature metrics" [Inventiones Math., 1993].

3. In any dimension $9+8k$ where $k$ is nonnegative integer there is a homotopy sphere with nonzero Lichnerowicz-Hitchin obstruction. Thus it lies in class (C). It cannot lie in (B) because it is odd-dimensional. I am sure there are many more examples.

• There is also an exotic sphere $\Sigma$ in dimensions $10 + 8k$ with $\alpha(\Sigma) \neq 0$. I think this also lies in class (C). It can't be written as a product involving a Kähler manifold, otherwise it would have non-trivial second cohomology, and it can't be a product of Spin(7) manifolds as it has the wrong dimension. – Michael Albanese Mar 13 '18 at 14:53
• In fact, a homotopy sphere (in any dimension) can't be a product. – Michael Albanese Mar 17 '18 at 4:40

Christos Mantoulidis showed me how to construct examples in (C) in all dimensions. Namely, if $$\Sigma_g^2$$ denotes a genus $$g$$ surface with $$g\ge 2$$, then $$\Sigma_g^2\times T^{n-2}$$ does is in class (C).

It does not carry a PSC metric because it is enlargeable (because it carries a metric of nonpositive sectional curvature or because it is a product of enlargeable manifolds). It is not in class (B) because it does not carry a Ricci flat metric. This follows from the general fact that $$\mathrm{Ric}\ge 0\implies b_1\le n$$. But by the Kuenneth formula, $$b_1(\Sigma_g^2\times T^{n-2})=b_1(\Sigma_g^2)+b_1(T^{n-2})=2g+n-2,$$ which is $$>n$$ when $$g\ge 2$$. Therefore these manifolds lie in class (C).