Can we prescribe the $L^2$ norm of the scalar curvature on a four-manifold?

Question

As mentioned by Willie Wong, I modified to the following version:

Let $M$ be a closed smooth $4$ manifold.

Q Suppose that $c>0$ is any positive number, can we find a Riemannian metric $g$ on $M$, such that the $\int_MScal^2_gdv_g=c$, where $Scal_g$ denotes the scalar curvature of $g$? If not, for any small $\epsilon>0$, can we find a metric $g_\epsilon$ such that $|\int_MScal^2_{g_\epsilon}dv_{g_\epsilon}-c|<\epsilon$?

PS I do not know whether the question is trivial or not. Any reference is welcome.

Answer 1

This is not always possible.

Let $M$ be a compact smooth manifold of dimension $n$. Consider the Einstein-Hilbert functional $\mathcal{E}$ given by

$$\mathcal{E}(g) = \dfrac{\displaystyle\int_Ms_g d\mu_g}{\operatorname{Vol}(M, g)^{\frac{n-2}{n}}}.$$

If $\mathcal{C}$ is a conformal class, then by using the conformal Laplacian and Hölder's inequality, one can show that $\mathcal{E}|_{\mathcal{C}}$ is bounded below. So we can define the Yamabe constant of $\mathcal{C}$ to be the finite quantity $Y(M, \mathcal{C}) = \inf_{g\in\mathcal{C}}\mathcal{E}(g)$. A result of Aubin shows that $Y(M, \mathcal{C}) \leq Y(S^n, [g_{\text{round}}])$, so we can define the Yamabe invariant of $M$ to be the finite quantity

$$Y(M) = \sup_{\mathcal{C}}\mathcal{E}(g) = \sup_{\mathcal{C}}\inf_{g\in\mathcal{C}}\mathcal{E}(g).$$

The following result is Proposition 1 from Kodaira Dimension and the Yamabe Problem by LeBrun.

Let $M$ be a smooth compact $n$-manifold, $n \geq 3$. Then $$\inf_{g}\int_M|s_g|^{n/2} d\mu_g= \begin{cases}0 & \text{if}\ Y(M) \geq 0\\ |Y(M)|^{n/2} & \text{if}\ Y(M) < 0.\end{cases}$$ Here the infimum on the left hand side is taken over all smooth Riemannian metrics $g$ on $M$.

In particular, if $n = 4$ and $Y(M) < 0$, then for all $c$ with $c < Y(M)^2$, there is no metric $g$ with $\displaystyle\int_Ms_g^2 d\mu_g = c$.

The question now becomes: is there an example of a compact smooth four-dimensional manifold with $Y(M) < 0$? The answer is yes as is shown in Theorem 2 of the same paper:

Let $M$ be the underlying $4$-manifold of a complex surface with Kodaira dimension $2$. Then $Y(M) < 0$. Moreover, if $X$ is the minimal model of $M$, then $$Y(M) = Y(X) = -4\pi\sqrt{2c_1(X)^2}.$$

For complex surfaces, $c_1(X)^2 = 2\chi(X) + 3\sigma(X)$, so we can actually compute the Yamabe invariant for the manifolds in the above theorem precisely.

Example: Let $X$ be the product of two complex curves of genus $2$. Then $X$ has Kodaira dimension $2$ and is minimal. Moreover, it has Euler characteristic $4$ and signature $0$, so $c_1^2(X) = 8$. Therefore

$$Y(X) = -4\pi\sqrt{2c_1(X)^2} = -4\pi\sqrt{16} = -16\pi.$$

The underlying smooth $4$-manifold is $\Sigma_2\times \Sigma_2$. So for any $c < (-16\pi)^2 = 256\pi^2$, we see there is no metric $g$ on $\Sigma_2\times\Sigma_2$ with $\displaystyle\int_{\Sigma_2\times\Sigma_2}s_g^2 d\mu_g = c$.