Kähler potential in $\Bbb C^n$

Question

I wanted to check that $\varphi\colon \Bbb C^n \to \Bbb R$ given by $$\varphi(z^1,\ldots, z^n) = \log\left(\sum_{k=1}^n |z^k|^2 + 1 \right)$$is a Kähler potential. One can compute $$\frac{\partial \varphi}{\partial \overline{z}^k}(z^1,\ldots, z^n) = \frac{z^k}{\sum_{k=1}^n |z^k|^2 + 1}$$and so $$\frac{\partial^2\varphi}{\partial z^j \partial \overline{z}^k}(z^1,\ldots, z^n) = \frac{\delta^k_{\;j}\left(\sum_{k=1}^n |z^k|^2 + 1 \right) - \overline{z}^jz^k}{\left(\sum_{k=1}^n |z^k|^2 + 1 \right)^2}.$$It is easy to see that the matrix $(\partial^2\varphi/\partial z^j\partial \overline{z}^k)_{j,k=1}^n$ is Hermitian, but I don't have a clue of how to check that it is positive definite. Sylvester's criterion seems awful to apply here. Help?

Answer 1

Note that if you take $v = \sum a^k\frac{\partial}{\partial z^k}$, then let's denote by $\alpha=(a^1,\dots,a^n)$ and $z=(z^1,\dots,z^n)\in\Bbb C^n$ and by $\langle\cdot,\cdot\rangle$ the standard Hermitian inner product on $\Bbb C^n$. \begin{align*} \big(1+\sum_k |z^k|^2\big)^2\sum\frac{\partial^2\varphi}{\partial z^j\partial\bar z^k}(v,\bar v) &= \sum_j |a^j|^2\big(1+\sum_k|z^k|^2\big) - \sum_{j,k} a^j\bar a^k z^k\bar z^j \\ &= \langle\alpha,\alpha\rangle + \langle\alpha,\alpha\rangle\langle z,z\rangle - \langle\alpha,z\rangle\langle z,\alpha\rangle. \end{align*} By Cauchy-Schwarz, $\langle\alpha,z\rangle\langle z,\alpha\rangle = |\langle\alpha,z\rangle|^2\le \|\alpha\|^2\|z\|^2$, and so the sum is positive unless $\alpha=0$.