2
$\begingroup$

Trying to follow the computation in Song and Tian - The Kähler–Ricci flow on surfaces of positive Kodaira dimension, page 7, theorem 3.1 which proved a parabolic Schwarz lemma. Specifically, they computed $\Delta \operatorname{tr}_{g}h = g^{i \bar l} g^{k \bar j} R_{k \bar l} f^{\alpha}_{i} f^{\bar \beta}_{\bar j} h_{\alpha \beta} + g^{i \bar j} g^{k \bar l}f^{\alpha}_{i, k} f^{\bar \beta}_{\bar j, \bar l}h_{\alpha \bar \beta} - g^{i \bar j}g^{k \bar l} S_{\alpha \bar \beta \gamma \bar \delta }f^{\alpha}_{i}f^{\bar \beta}_{\bar j} f^{\gamma}_{k} f^{\bar \delta}_{\bar l}$.

The last term comes from the double derivative hitting on the metric $h$, but I am confused as to where does the extra $f^{\gamma}_{k} f^{\bar \delta}_{\bar l}$ come from.

$\endgroup$
2
  • 1
    $\begingroup$ In such a case, the quickest way to an answer might be an email to the authors. Mathematicians are usually happy to answer. $\endgroup$
    – user473423
    Commented Apr 6, 2022 at 6:26
  • 2
    $\begingroup$ Chain rule, since $h$ is the pullback of a metric on $Y$ via $f$... $\endgroup$
    – YangMills
    Commented Apr 6, 2022 at 12:18

1 Answer 1

2
$\begingroup$

You are computing $\Delta_g \operatorname{tr}_g(f^{\ast}h) = g^{i \bar{j}} \partial_i \partial_{\bar{j}} (g^{k\bar{\ell}} h_{\gamma \bar{\delta}} f_k^{\gamma} \bar{f_{\ell}^{\delta}})$ in normal coordinates (since the metrics are Kähler, but the computation can be done in much more general situations). This yields three terms, the particular term you are asking about is $g^{i\bar{j}} g^{k\bar{\ell}} \partial_i \partial_{\bar{j}} (h_{\gamma \bar{\delta}}) f_k^{\gamma} \bar{f_{\ell}^{\delta}}$. We have $\partial_{\bar{j}} h_{\gamma \bar{\delta}} = \partial_{\bar{\beta}} h_{\gamma \bar{\delta}} \overline{f_j^{\beta}}$ and $$\partial_i \partial_{\bar{j}}h_{\gamma \bar{\delta}} = (\partial_i\partial_{\bar{\beta}} h_{\gamma \bar{\delta}}) \bar{f_j^{\delta}} = \partial_{\alpha} \partial_{\bar{\beta}}h_{\gamma \bar{\delta}} f_i^{\alpha}\bar{f_j^{\beta}}.$$ This is just the chain rule, as pointed out by @YangMills.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .