The logarithmic Monge–Ampère operator generates a parabolic flow whose linearization is the operator $L_ u$ of (11.23.1). Along this flow, the ratio of the Monge–Ampère measure to a fixed target measure satisfies a heat-type equation. Integrating a convex function of this ratio by parts produces, besides the usual nonpositive gradient term, a term involving $\mathrm{dd}^{\mathrm{c}}(\omega _{u_ t}^{n-1})$. If the forms along the flow are Gauduchon, this additional term vanishes and all convex entropies decrease. Requiring decrease for every smooth positive target at the initial time characterizes the Gauduchon condition.
Recall that $M=\int _ X\omega ^ n$, and fix a smooth strictly positive probability volume form $\nu $ on $X$. Consider a smooth solution on an interval $[0,T)$ of the logarithmic Monge–Ampère flow
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11.23.7
\begin{equation} \dot u_ t=\log \rho _ t,\qquad \rho _ t=\frac{\omega _{u_ t}^ n}{M\nu },\qquad u_0=0,\qquad \omega _{u_ t}>0. \end{equation}
We keep the initial normalizing constant $M$ fixed; the evolving mass $\int _ X\omega _{u_ t}^ n$ need not equal $M$. Define the linearized operator $\Delta _ t$ by
\[ \Delta _ t v\, \omega _{u_ t}^ n =n\mathrm{dd}^{\mathrm{c}}v\wedge \omega _{u_ t}^{n-1}; \]
it is the operator $L_{u_ t}$ of (11.23.1). Logarithmic differentiation gives $\partial _ t\log \rho _ t=\Delta _ t\dot u_ t=\Delta _ t\log \rho _ t$. Integration by parts against $\omega _{u_ t}^{n-1}$ then produces the same derivative of the background power that appears in the mass variation of proposition 11.1.8.
Proposition 11.23.8.AE4D0608184F Let $\Phi \in C^2((0,+\infty ),\mathbb R)$, and choose a function $G$ with $G'(r)=\Phi '(r)/r$. Along (11.23.7),
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11.23.9
\begin{equation} \begin{split} \frac{\mathrm{d}}{\mathrm{d}t}\int _ X\Phi (\rho _ t)\nu ={}& \frac nM\int _ XG(\rho _ t)\mathrm{dd}^{\mathrm{c}}(\omega _{u_ t}^{n-1})\\ & -\frac nM\int _ X\frac{\Phi ''(\rho _ t)}{\rho _ t} \mathrm{d}\rho _ t\wedge \mathrm d^{\mathrm c}\rho _ t\wedge \omega _{u_ t}^{n-1}. \end{split} \end{equation}
In particular, for $p>1$,
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11.23.10
\begin{equation} \begin{split} \frac{\mathrm{d}}{\mathrm{d}t}\int _ X\rho _ t^ p\nu ={}& \frac{np}{(p-1)M}\int _ X\rho _ t^{p-1}\mathrm{dd}^{\mathrm{c}}(\omega _{u_ t}^{n-1})\\ & -\frac{np(p-1)}M\int _ X\rho _ t^{p-3} \mathrm{d}\rho _ t\wedge \mathrm d^{\mathrm c}\rho _ t\wedge \omega _{u_ t}^{n-1}. \end{split} \end{equation}
Proof.
Using $\partial _ t\rho _ t=\rho _ t\Delta _ t\log \rho _ t$ and $\rho _ t\nu =M^{-1}\omega _{u_ t}^ n$, we find that the derivative of the entropy is
\[ \frac nM\int _ X\Phi '(\rho _ t)\mathrm{dd}^{\mathrm{c}}\log \rho _ t \wedge \omega _{u_ t}^{n-1}. \]
The chain rule and the definition of $G$ give
\[ \Phi '(r)\mathrm{dd}^{\mathrm{c}}\log r =\mathrm{dd}^{\mathrm{c}}G(r)-\frac{\Phi ''(r)}r\mathrm{d}r\wedge \mathrm d^{\mathrm c}r, \]
and lemma 11.1.2 applies to the first term. Adding a constant to $G$ does not change the result, since $\int _ X\mathrm{dd}^{\mathrm{c}}(\omega _{u_ t}^{n-1})=0$. Taking $\Phi (r)=r^ p$ and $G(r)=p r^{p-1}/(p-1)$ proves the second formula.
$\square$
Corollary 11.23.11.1C7A6B39E520 Suppose that every form $\omega _{u_ t}$ in (11.23.7) is Gauduchon. Then $\int _ X\rho _ t\nu =1$, and $t\mapsto \int _ X\Phi (\rho _ t)\nu $ is nonincreasing for every convex function $\Phi \in C^2((0,+\infty ),\mathbb R)$. In particular,
\[ \| \rho _ t\| _{L^ p(\nu )}\leq \| \rho _ s\| _{L^ p(\nu )} \quad (0\leq s\leq t<T,\ p>1). \]
If $\Phi ''>0$, the derivative of the entropy vanishes at a time $t$ precisely when $\rho _ t=1$, in which case $\dot u_ t=0$.
The hypothesis holds on a Gauduchon surface, and in any dimension it holds if $\mathrm{dd}^{\mathrm{c}}(\omega ^ j)=0$ for $1\leq j\leq n-1$.
Proof.
The derivative of the mass is
\[ \frac{\mathrm{d}}{\mathrm{d}t}\int _ X\rho _ t\nu =\frac nM\int _ X\dot u_ t\mathrm{dd}^{\mathrm{c}}(\omega _{u_ t}^{n-1})=0, \]
and its initial value is one. In (11.23.9) the first term vanishes, while the second is nonpositive, because $\mathrm{d}\rho _ t\wedge \mathrm d^{\mathrm c}\rho _ t$ is a semipositive $(1,1)$-form and $\Phi ''\geq 0$. If $\Phi ''>0$, the vanishing of this term forces $\mathrm{d}\rho _ t=0$, and connectedness together with the mass normalization gives $\rho _ t=1$. The converse and the assertion about $\dot u_ t$ follow from the equation.
On a surface, $\mathrm{dd}^{\mathrm{c}}\omega _{u_ t}=\mathrm{dd}^{\mathrm{c}}\omega =0$. Under the power identities, expand $\omega _{u_ t}^{n-1}$ by the binomial theorem; since the factors $\mathrm{dd}^{\mathrm{c}}u_ t$ are annihilated by $\mathrm d$ and $\mathrm d^{\mathrm c}$, each term has vanishing complex Hessian. These are also the persistence conditions of proposition 11.23.4, theorem 11.1.7.
$\square$
For $\Phi (r)=r\log r$, the same formula gives the dissipation of the relative entropy. In higher dimension, the Gauduchon condition on the initial form alone need not persist along potential perturbations, as example 11.23.5 shows. The next criterion determines the role of this condition at the initial time, without assuming that the flow exists.
Proposition 11.23.12.F45FF03B558B Fix $p>1$. For each smooth strictly positive probability volume form $\nu $, put $\rho _\nu =\omega ^ n/(M\nu )$ and define
\[ \mathcal D_ p(\omega ,\nu ) :=\left.\frac{\mathrm{d}}{\mathrm{d}t}\right|_{t=0} \int _ X\left(\frac{(\omega +t\mathrm{dd}^{\mathrm{c}}\log \rho _\nu )^ n}{M\nu }\right)^ p\nu . \]
Then $\mathcal D_ p(\omega ,\nu )\leq 0$ for every such $\nu $ if and only if $\omega $ is Gauduchon. If $\omega $ is not Gauduchon, there is such a $\nu $ with $\mathcal D_ p(\omega ,\nu )>0$.
Proof.
The form in the definition is positive for sufficiently small $|t|$. The path $u_ t=t\log \rho _\nu $ starts at $u_0=0$ with the velocity $\log \rho _\nu $ prescribed by (11.23.7) at time zero. The proof of proposition 11.23.8 at a given time uses only the potential and its velocity at that time, so (11.23.10) holds for the derivative at $t=0$. If $\omega $ is Gauduchon, its first term vanishes and the derivative is nonpositive.
Conversely, suppose that $\lambda :=\mathrm{dd}^{\mathrm{c}}(\omega ^{n-1})\ne 0$. Write $\lambda =f\omega ^ n/M$, where $f$ is smooth and real. Stokes' theorem gives $\int _ Xf\omega ^ n=0$, whereas
\[ \int _ Xf\lambda =\frac1M\int _ Xf^2\omega ^ n>0. \]
For small $\varepsilon >0$, set
\[ c_\varepsilon =\frac1M\int _ X\mathrm{e}^{-\varepsilon f}\omega ^ n, \qquad r_\varepsilon =c_\varepsilon \mathrm{e}^{\varepsilon f}, \qquad \nu _\varepsilon =\frac{\omega ^ n}{M r_\varepsilon }. \]
Then $\nu _\varepsilon $ is a smooth strictly positive probability volume form, and $\rho _{\nu _\varepsilon }=r_\varepsilon $. Since $c_\varepsilon =1+O(\varepsilon ^2)$, $\int _ X\lambda =0$ and $\mathrm{d}r_\varepsilon =\varepsilon r_\varepsilon \mathrm{d}f$, the two terms in (11.23.10) give
\[ \mathcal D_ p(\omega ,\nu _\varepsilon ) =\frac{np}M\varepsilon \int _ Xf\lambda +O(\varepsilon ^2)>0 \]
for sufficiently small $\varepsilon >0$. This proves necessity and the last assertion.
$\square$
The criterion involves only the initial velocity: every smooth solution of (11.23.7) with target $\nu $ has the same initial entropy derivative, independently of its existence for later times or its convergence. Quantification over all targets is essential. For the target $\nu =\omega ^ n/M$, the flow $u_ t=0$ is stationary for every Hermitian $\omega $, and all entropy derivatives vanish.
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