## A new, worse, proof that the holomorphic sectional curvature dominates the scalar curvature

If $$R$$ is the curvature tensor of a Kähler metric, we define its holomorphic sectional curvature to be $H(x) = R(x, \bar x, x, \bar x) / |x|^4$ for nonzero $$x$$. If $$(e_1, \ldots, e_n)$$ is a frame of the tangent bundle that is orthonormal at the center of a coordinate system, we also define the scalar curvature of $$R$$ at the center to be $s = \sum_{j,k=1}^n R(e_j, \bar e_j, e_k, \bar e_k).$ A classical result of Berger is that the sign of the holomorphic sectional curvature determines the sign of the scalar curvature. More precisely he proved that $\int_{S^{2n-1}} H(x) \, d\sigma(x) = \frac{1}{\binom{n+1}{2}} s,$ where $$d\sigma$$ is the Lebesgue measure on the sphere normalized to have volume one. Berger’s proof is by a straightforward calculation that proceeds by expanding $$H(x)$$ in terms of a local frame and then integrating each polynomial factor over the sphere. There is also a slightly less painful proof that uses representation theory to compute the integral.

We are going to explain a new, worse, proof of the fact that the holomorphic sectional curvature determines the sign of the scalar curvature. The classical proof of Berger’s result is not very illuminating, as it seems to hinge on the fact that the integrals of $$|z_j|^4$$ and $$|z_j|^2 |z_k|^2$$ over the unit sphere agree up to a factor of two. The representation-theoretic calculation of the integral explains why this happens; the integral is invariant under such a large group action that the whole thing has to be the trace, which evaluates to the scalar curvature. Our new proof offers no explanations, only brute force calculations that work out, somehow. If we conspire to write $$R(x) := R(x, \bar x, x, \bar x)$$ then we will show that $\sum_{j=1}^n \hat R(e_j) + \hat R\biggl( \sum_{j=1}^n e_j \biggr) = 2 s$ for a different curvature tensor $$\hat R$$ whose holomorphic sectional curvature has the same sign as that of $$R$$. Since $$\hat R(x)$$ has the same sign as $$H(x)$$, we conclude that the sign of $$H$$ determines that of $$s$$.

We’re going to work at the center of a normal coordinate system, so we will pick an orthonormal basis $$(e_1, \ldots, e_n)$$ of the tangent space at that point. If $$(x_1, \ldots, x_n)$$ are local coordinates in $$\mathbf C^n$$ the torus $$T^n := S^1 \times \cdots \times S^1$$ acts on $$\mathbf C^n$$ by $\psi(x) = ( e^{it_1} x_1, \ldots, e^{it_n} x_n ).$ Given a curvature tensor $$R$$ of a Kähler metric, we then define $\hat R(x, \bar y, z, \bar w) = \frac{1}{(2\pi)^n} \int_{T^n} \psi^* R(x, \bar y, z, \bar w) \, dt_1 \cdots dt_n.$ We note some properties of $$\hat R$$:

1. $$\hat R_{jklm} = R_{jklm}$$ if $$j=k$$ and $$l=m$$ or $$j=m$$ and $$k=l$$ and is zero otherwise. This is clear by evaluating the integral.

2. $$\hat R$$ has the symmetries of a Kähler curvature tensor, by 1.

3. $$\hat{\hat R} = \hat R$$, by 1.

4. If $$R(x) > 0$$ for all $$x$$ then $$\hat R(x) > 0$$ for all $$x$$, because then the integrand is positive for all $$x$$ and $$(t_1, \ldots, t_n)$$.

5. The scalar curvatures of $$R$$ and $$\hat R$$ are equal, because $$s = \sum_{j,k=1}^n R_{jjkk} = \sum_{j,k=1}^n \hat R_{jjkk}$$.

We can explain what $$\hat R$$ is: Recall that a Hermitian curvature tensor on $$\mathbf C^n$$ can be viewed as a Hermitian form on $$\mathbf C^n \otimes \mathbf C^n$$, and that such a tensor has the symmetries of a Kähler curvature tensor if and only if it is the pullback of a Hermitian form on $$\operatorname{Sym}^2 \mathbf C^n$$. Then $$\hat R$$ is just the projection of $$R$$ onto the subspace of diagonal Hermitian forms on $$\operatorname{Sym}^2 \mathbf C^n$$. This way however it’s not so clear that $$\hat R$$ also has positive holomorphic sectional curvature if $$R$$ does.

We now note that $\hat R\biggl( \sum_{j=1}^n e_j \biggr) = \sum_{j=1}^n R_{jjjj} + 4 \sum_{j\not=k} R_{jjkk}$ and so $\hat R\biggl( \sum_{j=1}^n e_j \biggr) + \sum_{j=1}^n \hat R(e_j) = 2s.$ If the holomorphic sectional curvature of $$R$$ is positive, then so is the one of $$\hat R$$, and thus $$s > 0$$.