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I am reading through a proof in W. Ding and G. Tian's 1992 paper on the generalised Futaki invariant. To provide context, we are looking for obstructions to the existence of Kähler--Einstein metrics with positive scalar curvature. The Futaki invariant provides an example of such an obstruction. My confusion is not in the Kähler geometry, but in some standard Riemannian Geometry/Geometric Analysis.

Suppose that $X$ is a compact $\mathbb{Q}$-Fano variety, i.e., there is an ample line bundle $L \longrightarrow X$ which restricts to the pluri-anticanonical line bundle $K_X^{-m}$ over the regular part of $X$ for some $m$. Let $\omega$ be an admissible Kähler metric on $X$ which represents $\frac{1}{m} c_1(L)$, where $c_1(L)$ denotes the first Chern class of $L$. Note that by $\omega$ being admissible, we mean that it is given by the pullback of the Fubini-Study metric on $\mathbb{CP}^n$, i.e., $$\omega = \frac{\alpha}{m} \phi_m^{\ast} \left( \omega_{\text{FS}} + \frac{\sqrt{-1}}{2\pi} \partial \overline{\partial} \psi \right),$$ $\psi \in \mathscr{C}^{\infty}(\mathbb{CP}^n, \mathbb{R})$ and $\phi_m$ is the Kodaira embedding furnished from the global sections of $L \to X$.

Now let $\pi : \widetilde{X} \longrightarrow X$ be a smooth resolution of $X$, with $\pi$ given simply by projection. It is clear that the support of the cohomology class $\text{Ric}(\widetilde{\omega}) - \omega$ is contained the exceptional divisors of $\widetilde{X}$. Here $\widetilde{\omega}$ denotes a Kähler metric on $\widetilde{X}$ such that $\pi^{\ast} \omega \leq \omega$. Let $E_1, ..., E_{\ell}$ be the exceptional divisors, then $$\text{Ric}(\tilde{\omega}) - \pi^{\ast} \omega = \sum_{k=1}^{\ell} \alpha_k C_1([E_k]),$$ where $C_1([E_k])$ denotes the Poincaré duals to $E_k$ in $\widetilde{X}$.

For each $k$, let $\| \cdot \|_k$ denote a Hermitian metric on the line bundle $[E_k]$ and $S_k$ a section of $[E_k]$ whose zero locus is exactly $E_k$. Then in the sense of distributions, $$\text{Ric}(\tilde{\omega}) - \pi^{\ast} \omega = \frac{\sqrt{-1}}{2\pi} \partial \overline{\partial} \left( - \sum_{k=1}^{\ell} \alpha_k \log \| S_k \|_k^2 + \varphi \right),$$ where $\varphi \in \mathscr{C}^{\infty}(\widetilde{X}, \mathbb{R})$.

Continuing the proof, we obtain a function $f$ of the form $$f = - \sum_{k=1}^{\ell} \alpha_k \log \| S_k \|_k^2 + \varphi + \log \left( \frac{\widetilde{\omega}^n}{\pi^{\ast} \omega^n} \right) + \text{constant}$$ on the regular part of $X$.

Question: How does one show that $f \in L^p(X, \omega^n)$, i.e., $$\int_X \left| f \right|^p \omega^n < \infty,$$ where $p > 1$?

Of course, $\varphi$ is smooth, and $X$ is compact, so this is no concern. I cannot control the logarithm terms however, is there something I am missing that is blindingly obvious?

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In polar coordinates

$ \int_{|z|<1} |\log(|z|)|^p = 2 \pi \int_0^1 (-\log r)^p r dr < +\infty .$

Ok..

In a compact space it is enough to check the intergrability condition locally, only the terms $\log \|S\|$ might give trouble at points on $E = \{z=0\}$ (in local coordinates), so $ \log \|S\| = H + \log|z| $ with $H$ a smooth function (the log of the norm of a trivilalizing section). Fubini together with the above computation imply that $ \log \|S\|$ is in $L^p$; therefore also is $f$..unless I missunderstood the question.

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  • $\begingroup$ I've flagged this as "not an answer", but may flag got disputed. Here I go again, this time flagging it as "very low quality". Hopefully, reviewers will pay more attention this time. $\endgroup$ – Alex M. Aug 7 '18 at 18:47
  • $\begingroup$ "Very low quality" is certainly safer than "not an answer", because deciding whether a response qualifies as an answer (even if a wrong one) often involves gray areas. (As a moderator, I struggle virtually every day deciding which way to process such "not an answer" flags.) This might explain why the flag was disputed. Of course, you could also ask the respondent to explain in what way this answers the question. $\endgroup$ – Todd Trimble Aug 7 '18 at 19:52

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