14.22 Polarized cofactors and propagation9769545E981A

Uniform ellipticity is sufficient for the propagation of an extremum, but it is not necessary: the directions in which the cofactor is positive may already connect enough points to force constancy. For $C^2$ potentials, the quadratic form $Q$ in (14.21.7) is continuous, and its polarization defines $Q^\sharp :T^*X\to TX$ by $\eta (Q^\sharp \xi )=Q(\xi ,\eta )$. The reachable set of a point consists of the endpoints of absolutely continuous curves starting there whose velocity lies in $\operatorname {im}Q^\sharp $ almost everywhere. The control orbit of a point consists of the endpoints of finite concatenations of flow lines, starting there, of locally Lipschitz vector fields with values in $\operatorname {im}Q^\sharp $.

A positive weight $h$ satisfying $\mathrm{dd}^{\mathrm{c}}(hT)=0$ gives an energy identity that controls $\mathrm{d}w$ everywhere. Without such a weight, a degenerate maximum principle gives propagation only from the extremal sets. This distinction accounts for the different density assumptions on reachable sets and on orbits in the following theorem.

14.22.3 Weighted energy1961884C0812

Testing the weighted cofactor against $w^2$ produces a nonnegative gradient integral. Only distributional $\mathrm{dd}^{\mathrm{c}}$-closedness of $hT$ is needed: for a continuous $h>0$, the condition is

The argument uses only the test function $w^2$. For general test functions, integration by parts introduces derivatives of $hT$; the condition $\mathrm{dd}^{\mathrm{c}}(hT)=0$ alone does not make $L\varphi =(\mathrm{dd}^{\mathrm{c}}\varphi \wedge T)/\mathrm{d}V$ a symmetric operator in divergence form.

14.22.8 The real symbol and exterior-ball propagation1FF2DD5C252C

To propagate an extremum without a weight, we write the linearized operator in real coordinates. Its symbol is continuous and semipositive. At a point where a ball outside the maximum set touches it, the normal direction must lie in the kernel of the symbol; this forces the maximum set to be invariant under the flows of vector fields with values in the range of the symbol.

In a holomorphic chart $z_ j=x_ j+\mathrm{i} y_ j$, let $C_{\rm cf}$ be the Hermitian matrix characterized by

for every Hermitian matrix $H=(H_{j\bar k})$. Positivity of mixed products makes $C_{\rm cf}$ positive semidefinite. Write $C_{\rm cf}=P_{\rm cf}+\mathrm{i} Q_{\rm cf}$, where $P_{\rm cf}$ is real symmetric and $Q_{\rm cf}$ is real skew-symmetric. In the real coordinate order $(x_1,\ldots ,x_ n,y_1,\ldots ,y_ n)$,

Indeed $\varphi _{j\bar k}=(\varphi _{x_ jx_ k}+\varphi _{y_ jy_ k} +\mathrm{i}\varphi _{x_ jy_ k}-\mathrm{i}\varphi _{y_ jx_ k})/4$; inserting this expression in (14.22.9) proves (14.22.10). For real coordinate vectors $\xi ,\eta \in \mathbb R^ n$, moreover,

Thus $a$ is a continuous positive semidefinite principal symbol, the operator $L$ has no first-order term in holomorphic coordinates, and the range of $a$ is that of $Q^\sharp $.

An exterior-ball argument propagates the maximum for this operator. Let $W\in C^2$ satisfy $LW\geq 0$ in a chart, suppose $W\leq M$, and put $K=\{ W=M\} $. If $z\in K$ is a point of $K$ nearest to a point $y\notin K$ of the chart, then

To prove this, suppose the left side is positive, set $r_0=|y-z|$, and shrink the tangent ball toward $z$: $y'=z+\theta (y-z)$, $R=\theta r_0$, with $0<\theta <1$ small. The closed ball $\overline{B(y',R)}$ lies in the chart and meets $K$ only at $z$. This follows from the triangle inequality; equality in $|k-y|\leq |k-y'|+|y'-y|\leq r_0$, for $k\in K$, forces $k=z$. For large $\alpha >0$, the function

satisfies

in a neighborhood of $z$, because $z-y'$ is a nonzero multiple of $y-z$, so that $(z-y')^ Ta(z)(z-y')>0$ by assumption. Choose a sufficiently small $r<R$, and consider the lens $D=B(z,r)\cap B(y',R)$. On the part of its boundary lying on the sphere $\partial B(y',R)$, we have $\psi =0$ and $W-M\leq 0$. On the remaining compact boundary portion, $W-M<0$, since the tangent ball meets $K$ only at $z$. Hence, for sufficiently small $\varepsilon >0$, $W-M\leq \varepsilon \psi $ on $\partial D$. At a positive interior maximum of $W-M-\varepsilon \psi $, the Hessian would be nonpositive, whereas $L(W-M-\varepsilon \psi )>0$; this is impossible since $a\geq 0$. Hence the inequality $W-M\leq \varepsilon \psi $ holds in $D$. Outside the ball $B(y',R)$ we have $\psi \geq 0$, so this inequality holds in a full neighborhood of $z$. Since equality holds at $z$, the function $W-M-\varepsilon \psi $ has a local maximum at $z$, and therefore

proving (14.22.11).

Since $a\geq 0$, its range is the orthogonal complement of its kernel, and (14.22.11) implies that $y-z$ lies in this kernel. Consequently $(y-z)\cdot Z(z)=0$ for every vector field $Z$ with values in the range of $a$. If $Z$ is locally Lipschitz and $\gamma $ is one of its flow curves starting in $K$, let $d(t)$ denote the squared distance from $\gamma (t)$ to $K$ in a sufficiently small compact chart neighborhood. Shortening the time interval ensures that nearest points $z_ t$ lie in the chart interior. Write $D^+d(t)=\limsup _{h\downarrow 0}(d(t+h)-d(t))/h$ for the upper right derivative. Holding such a nearest point fixed when differentiating the distance from above gives

As $d(0)=0$, Gronwall's inequality gives $d(t)=0$. The same argument for $-Z$, followed by continuation, proves invariance under both signs of time and under finite concatenations of such flows.

For $C^{2,1}$ potentials, the columns of $a$ are locally Lipschitz, so the orbit criterion still applies to them. The bracket criterion requires smooth vector fields.

14.22.12 The unweighted energy defectB7138A0B3C69

Integration against $w^2$ also measures the obstruction to using the unweighted cofactor, that is, the case $h=1$. For smooth potentials, the error is exactly the pairing of $w^2$ with $\mathrm{dd}^{\mathrm{c}}T$. On a surface, the Gauduchon condition makes this error vanish for every pair of potentials; in dimension three, $\mathrm{dd}^{\mathrm{c}}T$ contains an additional term depending on the potentials.

In dimension three, direct expansion gives

The Gauduchon condition cancels the first term but need not cancel the second. For instance, take both potentials equal to the perturbation $sv$ in example 11.23.5. The common form $A=B$ is positive and satisfies $\mathrm{dd}^{\mathrm{c}}(A^2)\ne 0$, so $\mathrm{dd}^{\mathrm{c}}T=3\mathrm{dd}^{\mathrm{c}}(A^2)\ne 0$. Thus even for two equal positive solution forms, the cofactor need not be $\mathrm{dd}^{\mathrm{c}}$-closed.