What are the important geometric-topological consequences of 4-dimensional version of Gauss-Bonnet-Chern theorem?

The Gauss–Bonnet theorem in differential geometry is an important statement about surfaces which connects their geometry to their topology and has very important applications to Riemann surface theory. The generalized Gauss–Bonnet theorem (Gauss-Bonnet-Chern) in dimension $n=4$, for a compact oriented manifold states that $$\chi(M)=\frac{1}{32\pi^2}\int_M\left(|Rm|^2-4|Rc|^2+r^2\right)d\mu,$$ where $Rm$ is the full Riemann curvature tensor, $Rc$ is the Ricci curvature tensor, $r$ is the scalar curvature. My question is

What are the important local-global results of the 4-dimensional version of Gauss-Bonnet-Chern theorem similar to the 2-Dimensional case?

Update:(2,September,2017)

This update is just an additional information. If $W$ denote the Weyl curvature of $(M,g)$ then $${32\pi^2}\chi(M)=\int_M(|W|^2+ 8Q_g)d\mu,$$ where $$Q_g:=-\frac{1}{12}(\Delta_gr-r^2+3|Rc|^2),$$ is the Paneitz $Q$ curvature introduced by Branson.

• I don't understand which part of your highlighted statements you consider to be local. For example, in the first statement, both the Euler characteristic and total curvature are not local quantities. Aug 7, 2017 at 14:12
• The second statement also contains no local information. A two-dimensional torus cannot admit a metric which has positive sectional curvature at each point - this is a global statement. Aug 7, 2017 at 18:43

One can find some applications in Arthur Besse's book "Einstein manifolds", $\S6\ D$.

Proposition. If a closed manifold $M^4$ admits an Einstein metric, then $\chi(M)\ge 0$. Moreover $\chi(M)=0$ iff $M$ admits a flat metric.

Proof. Since $Rc(g)=\lambda g$, we have $|Rc|^2=4\lambda^2=r^2/4$. This implies $\chi(M)=1/32\pi^2 \int|Rm|^2dV\ge 0$ with equality iff $Rm=0$. $\square$

Combining with another formula for the signature of a 4-manifold $$\sigma(M)=\frac{1}{12\pi^2}\int_M(|W_+|^2-|W_-|^2)dV,$$ where $W_{\pm}$ are compontens of the Weyl tensor, one can prove a stronger inequality: for a closed Einstein manifold $M^4$ $$\chi(M)\ge 3/2|\sigma(M)|.$$

The equality case was studied by Nigel Hitchin, Compact four-dimensional Einstein manifolds, (1974). He proved that if $(M^4, g)$ is an Einstein manifold with $\chi(M)= 3/2|\sigma(M)|$, then $M$ is Ricci flat and either universal cover of $M$ is K3-surface, or $M$ is flat.

These are important results, since topological obstructions for the existence of Einstein metrics are very scarce. In particular in dimensions $\ge 5$ there are no known such obstructions.

• Are this only known application of Gauss-Bonnet-Chern theorem? What we can say about $\mathbb{S}^2\times\Bbb S^2$ and Hopf conjecture? Aug 9, 2017 at 6:26
• @C.F.G I don't think one can say anything smart relying solely on the Gauss-Bonnet-Chern formula in the context of Hopf conjecture. Aug 9, 2017 at 15:15
• @YuryUstinovskiy The equality case was studied by Nigel Hitchin <-- reference? Aug 27, 2017 at 12:55
• @DavidRoberts Reference added. Aug 27, 2017 at 19:12

A good paper in this direction is "Some implications of the generalized Gauss-Bonnet theorem" written by Bishop and Goldberg which proved the following two theorems

Theorem 1.1. A compact and oriented Riemannian manifold of dimension $4$ whose sectional curvatures are non-negative or nonpositive has non-negative Euler-Poincare characteristic. If the sectional curvatures are always positive or always negative, the Euler-Poincare characteristic is positive.

Theorem 1.2. In order that a $4$-dimensional compact and orientable manifold M carry an Einstein metric, i.e., a Riemannian metric of constant Ricci or mean curvature $R$, it is necessary that its Euler-Poincare characteristic be non-negative.

and a corollary

Corollary 1.2. If $V$ is the volume of $M$, $$\chi(M)\geq\frac{VR^2}{12\pi^2}$$ equality holding if and only if $M$ has constant curvature.