PomeProposition 11.23.2.0B0003A3A873 For every smooth real $v$,
Moreover, for smooth real $h$, the equation $L_ uv=h$ has a smooth solution if and only if $\int _ Xh\, \mathrm{d}\mu _ u=0$. When it exists, the solution is unique modulo constants.
11.23 The adjoint measure of the linearizationC01896004229
C01896004229
The derivative of the logarithmic Monge–Ampère operator $u\mapsto \log (\omega _ u^ n/\mathrm dV)$ annihilates the constants, so its linearized equation can be solved only under a compatibility condition. The Gauduchon multiplier of theorem 11.4.2 identifies this condition. Fix $u\in \mathcal H_\omega $ and write $\eta =\omega _ u$. For a smooth real direction $v$, logarithmic differentiation gives the linearized operator
BEA90D7B5428
11.23.1
\begin{equation} L_ u v=n\, \frac{\mathrm{dd}^{\mathrm{c}}v\wedge \eta ^{n-1}}{\eta ^ n}. \end{equation}
Choose the Gauduchon multiplier $\rho _ u$ of $\eta $ given by theorem 11.4.2, normalized by
\[ \int _ X\rho _ u\eta ^ n=1, \qquad \text{and put}\qquad \mathrm{d}\mu _ u:=\rho _ u\eta ^ n. \]
The adjoint measure $\mu _ u$ is a smooth positive probability measure.
Proof. The integral in (11.23.3) is
\[ n\int _ X\mathrm{dd}^{\mathrm{c}}v\wedge \rho _ u\eta ^{n-1} =n\int _ Xv\, \mathrm{dd}^{\mathrm{c}}(\rho _ u\eta ^{n-1})=0. \]
This proves that the compatibility condition is necessary.
By the Fredholm argument in the proof of theorem 11.4.2, the range of $L_ u:C^{k+2,\alpha }(X)\to C^{k,\alpha }(X)$ has codimension one for every integer $k\geq 0$ and $0<\alpha <1$. The nonzero functional $h\mapsto \int _ Xh\, \mathrm{d}\mu _ u$ annihilates that range, so its kernel is the range. Elliptic regularity gives a smooth solution, and since the kernel of $L_ u$ consists of the constants, the solution is unique modulo constants. $\square$
By its defining equation and the uniqueness in theorem 11.4.2, the multiplier $\rho _ u$ is constant precisely when $\eta $ is Gauduchon. Thus
\[ \mu _ u=\frac{\eta ^ n}{\int _ X\eta ^ n} \quad \text{if }\mathrm{dd}^{\mathrm{c}}(\eta ^{n-1})=0. \]
On a surface, $\mathrm{dd}^{\mathrm{c}}\omega _ u=\mathrm{dd}^{\mathrm{c}}\omega $, so every positive potential perturbation of a Gauduchon form is still Gauduchon. In higher dimension this persistence requires more than the Gauduchon condition at $u=0$. Write $M=\int _ X\omega ^ n$.
Proposition 11.23.4.A914D4D9C6CB Every form $\omega _ u$ with $u\in \mathcal H_\omega $ is Gauduchon if and only if
\[ \int _ X\omega _ u^ n=M \quad \text{for every }u\in \mathcal H_\omega . \]
In that case the normalized multiplier is $\rho _ u=M^{-1}$ and $\mu _ u=M^{-1}\omega _ u^ n$ for every $u\in \mathcal H_\omega $.
Proof. If all the forms $\omega _ u$ are Gauduchon, the mass derivative (11.1.9) vanishes at every point of $\mathcal H_\omega $ and in every direction. Since $\mathcal H_\omega $ is convex, integration along the segment from $0$ to $u$ gives the mass identity. Conversely, proposition 11.1.8 shows that constant mass forces $\mathrm{dd}^{\mathrm{c}}(\omega _ u^{n-1})=0$ for every $u\in \mathcal H_\omega $. In that case the normalized multiplier is the constant $M^{-1}$, by theorem 11.4.2. $\square$
By theorem 11.1.7, this persistence is equivalent to $\mathrm{dd}^{\mathrm{c}}(\omega ^ j)=0$ for all integers $1\leq j\leq n-1$. The following example distinguishes it from the single Gauduchon identity in dimension three.
Example 11.23.5.28F39F2B7C2C Let $E=\mathbb C/(\mathbb Z+\mathrm i\mathbb Z)$ and $X=E^3$. For $1\leq j\leq 3$, write $z_ j=x_ j+\mathrm i y_ j$ for the coordinates on the covering space, and let $\eta _ j$ be the form on $X$ induced by $(\mathrm i/2)\mathrm dz_ j\wedge \mathrm d\bar z_ j$. The periodic functions
\[ b=\exp (\cos (2\pi x_1)),\qquad c=b^{-1} \]
define a Hermitian form
\[ \omega =\eta _1+b\eta _2+c\eta _3. \]
Since $bc=1$,
\[ \omega ^2=2b\eta _1\wedge \eta _2 +2c\eta _1\wedge \eta _3+2\eta _2\wedge \eta _3. \]
The functions $b$ and $c$ depend only on $x_1$, so $\mathrm{dd}^{\mathrm{c}}b$ and $\mathrm{dd}^{\mathrm{c}}c$ are multiples of $\eta _1$, and $\mathrm{dd}^{\mathrm{c}}(\omega ^2)=0$. Thus $\omega $ is Gauduchon.
Put $v=\cos (2\pi x_2)$ and $h=-\pi \cos (2\pi x_2)$, so that $\mathrm{dd}^{\mathrm{c}}v=h\eta _2$. For a real $s$ with $0<|s|<1/(\pi \mathrm e)$, the form
\[ \omega _{sv}=\eta _1+(b+sh)\eta _2+c\eta _3 \]
is positive, since $b\geq \mathrm e^{-1}$ and $|h|\leq \pi $. However,
\[ \mathrm{dd}^{\mathrm{c}}(\omega _{sv}^2) =2s\mathrm{dd}^{\mathrm{c}}\omega \wedge \mathrm{dd}^{\mathrm{c}}v =2sh\mathrm{dd}^{\mathrm{c}}c\wedge \eta _2\wedge \eta _3\ne 0. \]
Indeed, $\mathrm{dd}^{\mathrm{c}}c=(c''/(4\pi ))\eta _1$, where the prime denotes differentiation in $x_1$, and $c''(0)=4\pi ^2\mathrm e^{-1}$, while $h=-\pi $ at $x_2=0$. Thus a Gauduchon form can have arbitrarily small positive potential perturbations that are not Gauduchon.
When persistence holds, the one-form $v\mapsto \int _ Xv\, \mathrm d\mu _ u$ on $\mathcal H_\omega $ is the Monge–Ampère one-form of section 11.2 divided by $M$. By theorem 11.2.4, it therefore has a primitive exactly when $\omega $ is closed.
11.23.6 Convex entropies along a Monge–Ampère flow5C3C01BC6D13
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
DD9886FF9427
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),
313276789DFD
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 $p>1$,
0BC0A4E1D2A4
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.
11.23.13 Total variation between two evolving probabilitiesF9C293B132B8
The entropies of 11.23.6 compare one evolving probability with the fixed target. Two solutions of the flow with different initial potentials can nevertheless move apart. On a flat torus, even with a closed background, their total-variation distance can increase at the initial time, while no convex entropy relative to the target increases. We show this by computing the initial derivatives explicitly. By a logarithmic Monge–Ampère flow with target $\nu $ we mean a smooth solution of the equation in (11.23.7), with an arbitrary smooth initial potential.
For this example, let $E=\mathbb C/(\mathbb Z+\mathrm i\mathbb Z)$ and $X=E^2$, with covering coordinates $z_ j=x_ j+\mathrm i y_ j$. Write
\[ \eta _ j=\frac{\mathrm i}2\mathrm dz_ j\wedge \mathrm d\bar z_ j, \qquad \kappa =\eta _1+\eta _2, \qquad \mu =\eta _1\wedge \eta _2=\frac{\kappa ^2}2. \]
Thus $\mu $ is a probability volume form and $\int _ X\kappa ^2=2$. Put
\[ c_ j=\cos (2\pi x_ j),\qquad S=c_1+c_2,\qquad P=c_1c_2. \]
Lemma 11.23.14.6C85B622ADCF The averages of $|S|$ and $P|S|$ are
\[ \int _ X|S|\mu =\frac8{\pi ^2}, \qquad \int _ XP|S|\mu =\frac8{3\pi ^2}. \]
Proof. The integrands do not depend on $y_1,y_2$. The degree-two map of real tori $(a,b)\mapsto (a+b,a-b)$ preserves normalized Haar averages. With the angles $A=2\pi a$ and $B=2\pi b$, it gives
\[ S=2\cos A\cos B,\qquad P=\frac{\cos (2A)+\cos (2B)}2. \]
The one-variable averages of $|\cos A|$ and $|\cos A|^3$ are $2/\pi $ and $4/(3\pi )$, respectively, so the average of $\cos (2A)|\cos A|$ is $2/(3\pi )$. Multiplying the averages over the two independent angles gives both formulas. $\square$
For a fixed real number $K>3$, define the smooth target probability $\nu _ K$ by
\[ Z_ K=\int _ X\mathrm e^{KP}\mu , \qquad f_ K=Z_ K^{-1}\mathrm e^{KP}, \qquad \nu _ K=f_ K\mu . \]
For $0<\varepsilon <1/\pi $, the initial potentials $\varepsilon S$ and zero give positive forms. Set
\[ A_\varepsilon =1-\pi \varepsilon c_1, \qquad B_\varepsilon =1-\pi \varepsilon c_2, \qquad r_\varepsilon =A_\varepsilon B_\varepsilon , \]
so that $\kappa +\mathrm{dd}^{\mathrm{c}}(\varepsilon S)=A_\varepsilon \eta _1+B_\varepsilon \eta _2$ and its normalized Monge–Ampère probability is $r_\varepsilon \mu $.
Consider the short smooth paths
7732A51ABE6E
11.23.15
\begin{equation} U_\varepsilon (t)=\varepsilon S+t\log \frac{r_\varepsilon }{f_ K}, \qquad V(t)=-t\log f_ K. \end{equation}
They remain positive for sufficiently small $|t|$, and at $t=0$ their velocities are exactly those prescribed by the logarithmic Monge–Ampère flow with target $\nu _ K$. Their normalized probabilities are
\[ \mu _\varepsilon (t)=\frac{(\kappa +\mathrm{dd}^{\mathrm{c}}U_\varepsilon (t))^2}2, \qquad \mu _0(t)=\frac{(\kappa +\mathrm{dd}^{\mathrm{c}}V(t))^2}2, \]
which have mass one because the background $\kappa $ is closed. Define their total-variation distance, with the convention for probabilities, by
\[ D_\varepsilon (t) =\frac12\int _ X|\mu _\varepsilon (t)-\mu _0(t)|. \]
Proposition 11.23.16.621E75C5A9F4 For the paths (11.23.15),
\[ D'_\varepsilon (0) =4\varepsilon \left(\frac K3-1\right)+o(\varepsilon ) \quad \text{as }\varepsilon \downarrow 0. \]
In particular, this derivative is positive for sufficiently small $\varepsilon >0$. Any two logarithmic Monge–Ampère flows with target $\nu _ K$ and these initial potentials have the same positive initial derivative of their total-variation distance.
Proof. For a real function $h$ depending only on $x_1,x_2$, the coefficient matrix of $\mathrm{dd}^{\mathrm{c}}h$ with respect to the forms $(\mathrm i/2)\mathrm dz_ j\wedge \mathrm d\bar z_ k$ is $(4\pi )^{-1}(h_{x_ jx_ k})$. Put $R_\varepsilon (t)=\mathrm d\mu _\varepsilon (t)/\mathrm d\mu $ and $R_0(t)=\mathrm d\mu _0(t)/\mathrm d\mu $. Differentiating the determinant at the diagonal initial matrices gives
\[ \begin{split} \left.\partial _ t(R_\varepsilon -R_0)\right|_{t=0} =\frac1{4\pi }\bigl (& B_\varepsilon \partial _{x_1}^2\log A_\varepsilon +A_\varepsilon \partial _{x_2}^2\log B_\varepsilon \\ & -K(B_\varepsilon -1)\partial _{x_1}^2P -K(A_\varepsilon -1)\partial _{x_2}^2P\bigr ). \end{split} \]
Since $\partial _{x_ j}^2P=-4\pi ^2P$, Taylor expansion in $\varepsilon $ yields the uniform estimate
\[ \left.\partial _ t(R_\varepsilon -R_0)\right|_{t=0} =\pi ^2\varepsilon S(1-KP)+O(\varepsilon ^2). \]
At $t=0$, the difference of the densities is
\[ q_\varepsilon =r_\varepsilon -1 =-\pi \varepsilon S+\pi ^2\varepsilon ^2P. \]
For each fixed $x_2$, its coefficient of $c_1$ is $-\pi \varepsilon (1-\pi \varepsilon c_2)\ne 0$, so as a function of $x_1$ it has at most two zeros in a period, and its zero set has measure zero. Differentiation under the integral sign, justified by the uniform bound on the derivatives of the smooth densities, gives
\[ D'_\varepsilon (0)=\frac12\int _ X\operatorname {sgn}(q_\varepsilon ) \left.\partial _ t(R_\varepsilon -R_0)\right|_{t=0}\mu . \]
The set $\{ S=0\} $ also has measure zero, and $\operatorname {sgn}(q_\varepsilon )\to -\operatorname {sgn}(S)$ off this set. Dominated convergence and lemma 11.23.14 give
\[ \lim _{\varepsilon \downarrow 0}\frac{D'_\varepsilon (0)}\varepsilon =\frac{\pi ^2}2\int _ X|S|(KP-1)\mu =4\left(\frac K3-1\right). \]
Every flow with the stated initial data has the velocities (11.23.15) at $t=0$, so its initial density derivatives, and hence the derivative of the distance, coincide with this computation. $\square$
Corollary 11.23.17.21368BD09EBB For these paths, the initial derivative of every $C^2$ convex entropy of each probability relative to $\nu _ K$ is nonpositive, while the total-variation distance between the two probabilities has positive initial derivative for sufficiently small $\varepsilon >0$.
Proof. The initial velocities are those of the logarithmic flow, and both initial forms are closed. The chain-rule calculation in the proof of proposition 11.23.8 therefore applies at $t=0$: its background term vanishes and its gradient term is nonpositive. The preceding proposition gives the assertion about the pair. $\square$
Since only initial velocities enter, the example shows, without any existence or convergence theorem for the flow, that the logarithmic Monge–Ampère flow need not contract the total-variation distance infinitesimally, even though no convex entropy relative to the target increases at the initial time.
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