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Let $(X,\omega)$ be a hermitian $n$-dimensional manifold. Fix a point $x_0\in X$ an choose local holomorphic coordinates $(z_1,\dots,z_n)$ centered at $x_0$ and such that $(\partial/\partial z_1,\dots,\partial/\partial z_n)$ is unitary at $x_0$. Let $$ \Theta_{x_0}(T_X,\omega)=\sum_{j,k,l,m=1}^nc_{jklm}\hspace{0.3mm}dz_j\wedge d\bar z_k\otimes\left(\frac\partial{\partial z_l}\right)^*\otimes\frac\partial{\partial z_m} $$ be the Chern curvature at the point $x_0$. Consider the induced hermitian form on rank one tensors of $T_X\otimes T_X$ given by $$ \theta_{T_{X,x_o}}(v\otimes w)=\sum_{j,k,l,m}^nc_{jklm}\hspace{0.3mm}v_j\bar v_k w_l\bar w_m, $$ where $$ v,w\in T_{X,x_0},\quad v=\sum v_j\hspace{0.3mm}\frac\partial{\partial z_j},\quad w=\sum w_j\hspace{0.3mm}\frac\partial{\partial z_j}. $$ With this notation, the holomorphic sectional curvature in the direction of $v\in T_{X,x_0}\setminus\{0\}$ is given by $$ \frac{1}{||v||_\omega^4}\theta_{T_{X,x_o}}(v\otimes v). $$ The idea now is to take the average on the $\omega$-unit sphere $S^{2n-1}$ and try to deduce something on the scalar curvature at the point $x_0$ which is given by $$ s(x_0)=\sum_{j,k=1}^nc_{jjkk} $$ (or maybe twice this). So, let's compute the integral $$ \int_{S^{2n-1}}\sum_{j,k,l,m}^nc_{jklm}\hspace{0.3mm}\xi_j\bar \xi_k \xi_l\bar \xi_m\hspace{0.3mm}d\sigma(\xi), $$ where $d\sigma(\xi)$ is the probability Haar measure on $S^{2n-1}$. It is not hard to see that the integral $$ \int_{S^{2n-1}}\xi_j\bar \xi_k \xi_l\bar \xi_m\hspace{0.3mm}d\sigma(\xi) $$ vanishes unless $j=k$ and $l=m$. Thus, we have to compute $$ \int_{S^{2n-1}}|\xi_j|^2|\xi_k|^2\hspace{0.3mm}d\sigma(\xi),\quad j,k=1,\dots,n. $$ If we set $$ \lambda=\int_{S^{2n-1}}|\xi_j|^4\hspace{0.3mm}d\sigma(\xi),\quad j=1,\dots,n, $$ and $$ \mu=\int_{S^{2n-1}}|\xi_j|^2|\xi_k|^2\hspace{0.3mm}d\sigma(\xi),\quad 1\le j\ne k\le n, $$ then, from $1=|\xi|^4=\sum_{j,k=1}^n|\xi_j|^2|\xi_k|^2$, we get $$ \int_{S^{2n-1}}\sum_{j,k,l,m}^nc_{jklm}\hspace{0.3mm}\xi_j\bar \xi_k \xi_l\bar \xi_m\hspace{0.3mm}d\sigma(\xi)=\sum_{j,k=1}^nc_{jjkk}\bigl(\mu+\delta_{jk}(\lambda-\mu)\bigr), $$ where $n\lambda+(n^2-n)\mu=1$. At this point I block since I am not able to deduce that from $$ \sum_{j,k=1}^nc_{jjkk}\bigl(\mu+\delta_{jk}(\lambda-\mu)\bigr)<0 $$ follows $$ \sum_{j,k=1}^nc_{jjkk}<0. $$ I have the impression that on the book that Gunnar cites the author says that the average integral I compute here gives directly (a constant times) $s(x_0)$, which seems to me to be false.

On the other hand, I do know that what I claim in my first post is true: it is contained in

Berger, M.

Sur les variétés d'Einstein compactes. 1966 Comptes Rendus de la IIIe Réunion du Groupement des Mathématiciens d'Expression Latine (Namur, 1965) pp. 35–55 Librairie Universitaire, Louvain,

but I was not able to find this "Comptes Rendus".

Up to here, it does not seem to be that any kählerianity assumption is needed and that any kählerianity assumption would help. But, of course, maybe I am wrong.

Finally, observe that, on the other hand, if we perform the "double average" $$ \int_{S^{2n-1}}\int_{S^{2n-1}}\sum_{j,k,l,m}^nc_{jklm}\hspace{0.3mm}\xi_j\bar \xi_k \eta_l\bar \eta_m\hspace{0.3mm}d\sigma(\xi)d\sigma(\eta), $$ then we get $$ \frac 1{n^2}\sum_{j,k=1}^nc_{jjkk}, $$ which coincides with $\frac 1{n^2}s(x_0)$, since $\int_{S^{2n-1}}\xi_j\bar \xi_k\hspace{0.3mm}d\sigma(\xi)=\frac 1n\delta_{jk}$. Thus, the holomorphic bisectional curvature "dominates" the scalar curvature.

Here is the answer.

Let $(X,\omega)$ be a Kähler $n$-dimensional manifold. Fix a point $x_0\in X$ an choose local holomorphic coordinates $(z_1,\dots,z_n)$ centered at $x_0$ and such that $(\partial/\partial z_1,\dots,\partial/\partial z_n)$ is unitary at $x_0$. Let $$ \Theta_{x_0}(T_X,\omega)=\sum_{j,k,l,m=1}^nc_{jklm}\hspace{0.3mm}dz_j\wedge d\bar z_k\otimes\left(\frac\partial{\partial z_l}\right)^*\otimes\frac\partial{\partial z_m} $$ be the Chern curvature at the point $x_0$. Consider the induced hermitian form on rank one tensors of $T_X\otimes T_X$ given by $$ \theta_{T_{X,x_o}}(v\otimes w)=\sum_{j,k,l,m}^nc_{jklm}\hspace{0.3mm}v_j\bar v_k w_l\bar w_m, $$ where $$ v,w\in T_{X,x_0},\quad v=\sum v_j\hspace{0.3mm}\frac\partial{\partial z_j},\quad w=\sum w_j\hspace{0.3mm}\frac\partial{\partial z_j}. $$ With this notation, the holomorphic sectional curvature in the direction of $v\in T_{X,x_0}\setminus\{0\}$ is given by $$ \frac{1}{||v||_\omega^4}\theta_{T_{X,x_o}}(v\otimes v). $$ The idea now is to take the average on the $\omega$-unit sphere $S^{2n-1}$ and try to deduce something on the scalar curvature at the point $x_0$ which is given by $$ s(x_0)=2\sum_{j,k=1}^nc_{jjkk}. $$ So, let's compute the integral $$ \int_{S^{2n-1}}\sum_{j,k,l,m}^nc_{jklm}\hspace{0.3mm}\xi_j\bar \xi_k \xi_l\bar \xi_m\hspace{0.3mm}d\sigma(\xi), $$ where $d\sigma(\xi)$ is the probability Haar measure on $S^{2n-1}$. It is not hard to see that the integral $$ \int_{S^{2n-1}}\xi_j\bar \xi_k \xi_l\bar \xi_m\hspace{0.3mm}d\sigma(\xi) $$ vanishes unless $j=k$ and $l=m$ or $j=m$ and $k=l$. Thus, we have to compute $$ \int_{S^{2n-1}}|\xi_j|^2|\xi_k|^2\hspace{0.3mm}d\sigma(\xi),\quad j,k=1,\dots,n. $$ It is classically known that $$ \int_{S^{2n-1}}|\xi_j|^4\hspace{0.3mm}d\sigma(\xi)=\frac 2{n(n+1)},\quad j=1,\dots,n, $$ and $$ \int_{S^{2n-1}}|\xi_j|^2|\xi_k|^2\hspace{0.3mm}d\sigma(\xi)=\frac 1{n(n+1)},\quad 1\le j\ne k\le n. $$ Then, we get $$ \begin{aligned} \int_{S^{2n-1}}\sum_{j,k,l,m}^nc_{jklm}\hspace{0.3mm}\xi_j\bar \xi_k \xi_l\bar \xi_m\hspace{0.3mm}d\sigma(\xi) & =\sum_{j,k=1}^nc_{jjkk}\left(\delta_{jk}\frac 2{n(n+1)}+(1-\delta_{jk})\frac 2{n(n+1)}\right) \\ & = \frac 2{n(n+1)}\sum_{j,k=1}^nc_{jjkk}=\frac 1{n(n+1)}s(x_0), \end{aligned} $$ where we have used the Kähler identity $c_{jklm}=c_{jmlk}$.

Thus, if $\frac{1}{||v||_\omega^4}\theta_{T_{X,x_o}}(v\otimes v)$ is negative, so is its average and we are done.

Let $(X,\omega)$ be a hermitian $n$-dimensional manifold. Fix a point $x_0\in X$ an choose local holomorphic coordinates $(z_1,\dots,z_n)$ centered at $x_0$ and such that $(\partial/\partial z_1,\dots,\partial/\partial z_n)$ is unitary at $x_0$. Let $$ \Theta_{x_0}(T_X,\omega)=\sum_{j,k,l,m=1}^nc_{jklm}\hspace{0.3mm}dz_j\wedge d\bar z_k\otimes\left(\frac\partial{\partial z_l}\right)^*\otimes\frac\partial{\partial z_m} $$ be the Chern curvature at the point $x_0$. Consider the induced hermitian form on rank one tensors of $T_X\otimes T_X$ given by $$ \theta_{T_{X,x_o}}(v\otimes w)=\sum_{j,k,l,m}^nc_{jklm}\hspace{0.3mm}v_j\bar v_k w_l\bar w_m, $$ where $$ v,w\in T_{X,x_0},\quad v=\sum v_j\hspace{0.3mm}\frac\partial{\partial z_j},\quad w=\sum w_j\hspace{0.3mm}\frac\partial{\partial z_j}. $$ With this notation, the holomorphic sectional curvature in the direction of $v\in T_{X,x_0}\setminus\{0\}$ is given by $$ \frac{1}{||v||_\omega^4}\theta_{T_{X,x_o}}(v\otimes v). $$ The idea now is to take the average on the $\omega$-unit sphere $S^{2n-1}$ and try to deduce something on the scalar curvature at the point $x_0$ which is given by $$ s(x_0)=\sum_{j,k=1}^nc_{jjkk} $$ (or maybe twice this). So, let's compute the integral $$ \int_{S^{2n-1}}\sum_{j,k,l,m}^nc_{jklm}\hspace{0.3mm}\xi_j\bar \xi_k \xi_l\bar \xi_m\hspace{0.3mm}d\sigma(\xi), $$ where $d\sigma(\xi)$ is the probability Haar measure on $S^{2n-1}$. It is not hard to see that the integral $$ \int_{S^{2n-1}}\xi_j\bar \xi_k \xi_l\bar \xi_m\hspace{0.3mm}d\sigma(\xi) $$ vanishes unless $j=k$ and $l=m$. Thus, we have to compute $$ \int_{S^{2n-1}}|\xi_j|^2|\xi_k|^2\hspace{0.3mm}d\sigma(\xi),\quad j,k=1,\dots,n. $$ If we set $$ \lambda=\int_{S^{2n-1}}|\xi_j|^4\hspace{0.3mm}d\sigma(\xi),\quad j=1,\dots,n, $$ and $$ \mu=\int_{S^{2n-1}}|\xi_j|^2|\xi_k|^2\hspace{0.3mm}d\sigma(\xi),\quad 1\le j\ne k\le n, $$ then, from $1=|\xi|^4=\sum_{j,k=1}^n|\xi_j|^2|\xi_k|^2$, we get $$ \int_{S^{2n-1}}\sum_{j,k,l,m}^nc_{jklm}\hspace{0.3mm}\xi_j\bar \xi_k \xi_l\bar \xi_m\hspace{0.3mm}d\sigma(\xi)=\sum_{j,k=1}^nc_{jjkk}\bigl(\mu+\delta_{jk}(\lambda-\mu)\bigr), $$ where $n\lambda+(n^2-n)\mu=1$. At this point I block since I am not able to deduce that from $$ \sum_{j,k=1}^nc_{jjkk}\bigl(\mu+\delta_{jk}(\lambda-\mu)\bigr)<0 $$ follows $$ \sum_{j,k=1}^nc_{jjkk}<0. $$ I have the impression that on the book that Gunnar cites the author says that the average integral I compute here gives directly (a constant times) $s(x_0)$, which seems to me to be false.

On the other hand, I do know that what I claim in my first post is true: it is contained in

Berger, M.

Sur les variétés d'Einstein compactes. 1966 Comptes Rendus de la IIIe Réunion du Groupement des Mathématiciens d'Expression Latine (Namur, 1965) pp. 35–55 Librairie Universitaire, Louvain,

but I was not able to find this "Comptes Rendus".

Up to here, it does not seem to be that any kählerianity assumption is needed and that any kählerianity assumption would help. But, of course, maybe I am wrong.

Let $(X,\omega)$ be a hermitian $n$-dimensional manifold. Fix a point $x_0\in X$ an choose local holomorphic coordinates $(z_1,\dots,z_n)$ centered at $x_0$ and such that $(\partial/\partial z_1,\dots,\partial/\partial z_n)$ is unitary at $x_0$. Let $$ \Theta_{x_0}(T_X,\omega)=\sum_{j,k,l,m=1}^nc_{jklm}\hspace{0.3mm}dz_j\wedge d\bar z_k\otimes\left(\frac\partial{\partial z_l}\right)^*\otimes\frac\partial{\partial z_m} $$ be the Chern curvature at the point $x_0$. Consider the induced hermitian form on rank one tensors of $T_X\otimes T_X$ given by $$ \theta_{T_{X,x_o}}(v\otimes w)=\sum_{j,k,l,m}^nc_{jklm}\hspace{0.3mm}v_j\bar v_k w_l\bar w_m, $$ where $$ v,w\in T_{X,x_0},\quad v=\sum v_j\hspace{0.3mm}\frac\partial{\partial z_j},\quad w=\sum w_j\hspace{0.3mm}\frac\partial{\partial z_j}. $$ With this notation, the holomorphic sectional curvature in the direction of $v\in T_{X,x_0}\setminus\{0\}$ is given by $$ \frac{1}{||v||_\omega^4}\theta_{T_{X,x_o}}(v\otimes v). $$ The idea now is to take the average on the $\omega$-unit sphere $S^{2n-1}$ and try to deduce something on the scalar curvature at the point $x_0$ which is given by $$ s(x_0)=\sum_{j,k=1}^nc_{jjkk} $$ (or maybe twice this). So, let's compute the integral $$ \int_{S^{2n-1}}\sum_{j,k,l,m}^nc_{jklm}\hspace{0.3mm}\xi_j\bar \xi_k \xi_l\bar \xi_m\hspace{0.3mm}d\sigma(\xi), $$ where $d\sigma(\xi)$ is the probability Haar measure on $S^{2n-1}$. It is not hard to see that the integral $$ \int_{S^{2n-1}}\xi_j\bar \xi_k \xi_l\bar \xi_m\hspace{0.3mm}d\sigma(\xi) $$ vanishes unless $j=k$ and $l=m$. Thus, we have to compute $$ \int_{S^{2n-1}}|\xi_j|^2|\xi_k|^2\hspace{0.3mm}d\sigma(\xi),\quad j,k=1,\dots,n. $$ If we set $$ \lambda=\int_{S^{2n-1}}|\xi_j|^4\hspace{0.3mm}d\sigma(\xi),\quad j=1,\dots,n, $$ and $$ \mu=\int_{S^{2n-1}}|\xi_j|^2|\xi_k|^2\hspace{0.3mm}d\sigma(\xi),\quad 1\le j\ne k\le n, $$ then, from $1=|\xi|^4=\sum_{j,k=1}^n|\xi_j|^2|\xi_k|^2$, we get $$ \int_{S^{2n-1}}\sum_{j,k,l,m}^nc_{jklm}\hspace{0.3mm}\xi_j\bar \xi_k \xi_l\bar \xi_m\hspace{0.3mm}d\sigma(\xi)=\sum_{j,k=1}^nc_{jjkk}\bigl(\mu+\delta_{jk}(\lambda-\mu)\bigr), $$ where $n\lambda+(n^2-n)\mu=1$. At this point I block since I am not able to deduce that from $$ \sum_{j,k=1}^nc_{jjkk}\bigl(\mu+\delta_{jk}(\lambda-\mu)\bigr)<0 $$ follows $$ \sum_{j,k=1}^nc_{jjkk}<0. $$ I have the impression that on the book that Gunnar cites the author says that the average integral I compute here gives directly (a constant times) $s(x_0)$, which seems to me to be false.

On the other hand, I do know that what I claim in my first post is true: it is contained in

Berger, M.

Sur les variétés d'Einstein compactes. 1966 Comptes Rendus de la IIIe Réunion du Groupement des Mathématiciens d'Expression Latine (Namur, 1965) pp. 35–55 Librairie Universitaire, Louvain,

but I was not able to find this "Comptes Rendus".

Up to here, it does not seem to be that any kählerianity assumption is needed and that any kählerianity assumption would help. But, of course, maybe I am wrong.

Finally, observe that, on the other hand, if we perform the "double average" $$ \int_{S^{2n-1}}\int_{S^{2n-1}}\sum_{j,k,l,m}^nc_{jklm}\hspace{0.3mm}\xi_j\bar \xi_k \eta_l\bar \eta_m\hspace{0.3mm}d\sigma(\xi)d\sigma(\eta), $$ then we get $$ \frac 1{n^2}\sum_{j,k=1}^nc_{jjkk}, $$ which coincides with $\frac 1{n^2}s(x_0)$, since $\int_{S^{2n-1}}\xi_j\bar \xi_k\hspace{0.3mm}d\sigma(\xi)=\frac 1n\delta_{jk}$. Thus, the holomorphic bisectional curvature "dominates" the scalar curvature.

Let $(X,\omega)$ be a hermitian $n$-dimensional manifold. Fix a point $x_0\in X$ an choose local holomorphic coordinates $(z_1,\dots,z_n)$ centered at $x_0$ and such that $(\partial/\partial z_1,\dots,\partial/\partial z_n)$ is unitary at $x_0$. Let $$ \Theta_{x_0}(T_X,\omega)=\sum_{j,k,l,m=1}^nc_{jklm}\hspace{0.3mm}dz_j\wedge d\bar z_k\otimes\left(\frac\partial{\partial z_l}\right)^*\otimes\frac\partial{\partial z_m} $$ be the Chern curvature at the point $x_0$. Consider the induced hermitian form on rank one tensors of $T_X\otimes T_X$ given by $$ \theta_{T_{X,x_o}}(v\otimes w)=\sum_{j,k,l,m}^nc_{jklm}\hspace{0.3mm}v_j\bar v_k w_l\bar w_m, $$ where $$ v,w\in T_{X,x_0},\quad v=\sum v_j\hspace{0.3mm}\frac\partial{\partial z_j},\quad w=\sum w_j\hspace{0.3mm}\frac\partial{\partial z_j}. $$ With this notation, the holomorphic sectional curvature in the direction of $v\in T_{X,x_0}\setminus\{0\}$ is given by $$ \frac{1}{||v||_\omega^4}\theta_{T_{X,x_o}}(v\otimes v). $$ The idea now is to take the average on the $\omega$-unit sphere $S^{2n-1}$ and try to deduce something on the scalar curvature at the point $x_0$ which is given by $$ s(x_0)=\sum_{j,k=1}^nc_{jjkk} $$ (or maybe twice this). So, let's compute the integral $$ \int_{S^{2n-1}}\sum_{j,k,l,m}^nc_{jklm}\hspace{0.3mm}\xi_j\bar \xi_k \xi_l\bar \xi_m\hspace{0.3mm}d\sigma(\xi), $$ where $d\sigma(\xi)$ is the probability Haar measure on $S^{2n-1}$. It is not hard to see that the integral $$ \int_{S^{2n-1}}\xi_j\bar \xi_k \xi_l\bar \xi_m\hspace{0.3mm}d\sigma(\xi) $$ vanishes unless $j=k$ and $l=m$. Thus, we have to compute $$ \int_{S^{2n-1}}|\xi_j|^2|\xi_k|^2\hspace{0.3mm}d\sigma(\xi),\quad j,k=1,\dots,n. $$ If we set $$ \lambda=\int_{S^{2n-1}}|\xi_j|^4\hspace{0.3mm}d\sigma(\xi),\quad j=1,\dots,n, $$ and $$ \mu=\int_{S^{2n-1}}|\xi_j|^2|\xi_k|^2\hspace{0.3mm}d\sigma(\xi),\quad j,k=1,\dots,n, $$ then, from $1=|\xi|^4=\sum_{j,k=1}^n|\xi_j|^2|\xi_k|^2$, we get $$ \int_{S^{2n-1}}\sum_{j,k,l,m}^nc_{jklm}\hspace{0.3mm}\xi_j\bar \xi_k \xi_l\bar \xi_m\hspace{0.3mm}d\sigma(\xi)=\sum_{j,k=1}^nc_{jjkk}\bigl(\mu+\delta_{jk}(\lambda-\mu)\bigr), $$ where $n\lambda+(n^2-n)\mu=1$. At this point I block since I am not able to deduce that from $$ \sum_{j,k=1}^nc_{jjkk}\bigl(\mu+\delta_{jk}(\lambda-\mu)\bigr)<0 $$ follows $$ \sum_{j,k=1}^nc_{jjkk}<0. $$ I have the impression that on the book that Gunnar cites the author says that the average integral I compute here gives directly (a constant times) $s(x_0)$, which seems to me to be false.

On the other hand, I do know that what I claim in my first post is true: it is contained in

Berger, M.

Sur les variétés d'Einstein compactes. 1966 Comptes Rendus de la IIIe Réunion du Groupement des Mathématiciens d'Expression Latine (Namur, 1965) pp. 35–55 Librairie Universitaire, Louvain,

but I was not able to find this "Comptes Rendus".

Up to here, it does not seem to be that any kählerianity assumption is needed and that any kählerianity assumption would help. But, of course, maybe I am wrong.

Let $(X,\omega)$ be a hermitian $n$-dimensional manifold. Fix a point $x_0\in X$ an choose local holomorphic coordinates $(z_1,\dots,z_n)$ centered at $x_0$ and such that $(\partial/\partial z_1,\dots,\partial/\partial z_n)$ is unitary at $x_0$. Let $$ \Theta_{x_0}(T_X,\omega)=\sum_{j,k,l,m=1}^nc_{jklm}\hspace{0.3mm}dz_j\wedge d\bar z_k\otimes\left(\frac\partial{\partial z_l}\right)^*\otimes\frac\partial{\partial z_m} $$ be the Chern curvature at the point $x_0$. Consider the induced hermitian form on rank one tensors of $T_X\otimes T_X$ given by $$ \theta_{T_{X,x_o}}(v\otimes w)=\sum_{j,k,l,m}^nc_{jklm}\hspace{0.3mm}v_j\bar v_k w_l\bar w_m, $$ where $$ v,w\in T_{X,x_0},\quad v=\sum v_j\hspace{0.3mm}\frac\partial{\partial z_j},\quad w=\sum w_j\hspace{0.3mm}\frac\partial{\partial z_j}. $$ With this notation, the holomorphic sectional curvature in the direction of $v\in T_{X,x_0}\setminus\{0\}$ is given by $$ \frac{1}{||v||_\omega^4}\theta_{T_{X,x_o}}(v\otimes v). $$ The idea now is to take the average on the $\omega$-unit sphere $S^{2n-1}$ and try to deduce something on the scalar curvature at the point $x_0$ which is given by $$ s(x_0)=\sum_{j,k=1}^nc_{jjkk} $$ (or maybe twice this). So, let's compute the integral $$ \int_{S^{2n-1}}\sum_{j,k,l,m}^nc_{jklm}\hspace{0.3mm}\xi_j\bar \xi_k \xi_l\bar \xi_m\hspace{0.3mm}d\sigma(\xi), $$ where $d\sigma(\xi)$ is the probability Haar measure on $S^{2n-1}$. It is not hard to see that the integral $$ \int_{S^{2n-1}}\xi_j\bar \xi_k \xi_l\bar \xi_m\hspace{0.3mm}d\sigma(\xi) $$ vanishes unless $j=k$ and $l=m$. Thus, we have to compute $$ \int_{S^{2n-1}}|\xi_j|^2|\xi_k|^2\hspace{0.3mm}d\sigma(\xi),\quad j,k=1,\dots,n. $$ If we set $$ \lambda=\int_{S^{2n-1}}|\xi_j|^4\hspace{0.3mm}d\sigma(\xi),\quad j=1,\dots,n, $$ and $$ \mu=\int_{S^{2n-1}}|\xi_j|^2|\xi_k|^2\hspace{0.3mm}d\sigma(\xi),\quad 1\le j\ne k\le n, $$ then, from $1=|\xi|^4=\sum_{j,k=1}^n|\xi_j|^2|\xi_k|^2$, we get $$ \int_{S^{2n-1}}\sum_{j,k,l,m}^nc_{jklm}\hspace{0.3mm}\xi_j\bar \xi_k \xi_l\bar \xi_m\hspace{0.3mm}d\sigma(\xi)=\sum_{j,k=1}^nc_{jjkk}\bigl(\mu+\delta_{jk}(\lambda-\mu)\bigr), $$ where $n\lambda+(n^2-n)\mu=1$. At this point I block since I am not able to deduce that from $$ \sum_{j,k=1}^nc_{jjkk}\bigl(\mu+\delta_{jk}(\lambda-\mu)\bigr)<0 $$ follows $$ \sum_{j,k=1}^nc_{jjkk}<0. $$ I have the impression that on the book that Gunnar cites the author says that the average integral I compute here gives directly (a constant times) $s(x_0)$, which seems to me to be false.

On the other hand, I do know that what I claim in my first post is true: it is contained in

Berger, M.

Sur les variétés d'Einstein compactes. 1966 Comptes Rendus de la IIIe Réunion du Groupement des Mathématiciens d'Expression Latine (Namur, 1965) pp. 35–55 Librairie Universitaire, Louvain,

but I was not able to find this "Comptes Rendus".

Up to here, it does not seem to be that any kählerianity assumption is needed and that any kählerianity assumption would help. But, of course, maybe I am wrong.