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

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.

• scalar curvature scales like metric inverse. Volume form scales like metric to the power $n/2$. So if $n/2 - 2 \neq 0$ and if your manifold admits any non-flat metric, then you can get what you want by rescaling. May 17, 2022 at 4:23

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 $$\displaystyle\int_{\Sigma_2\times\Sigma_2}s_g^2 d\mu_g = c$$.