11.1 Torsion and the variation of mass7B453010B5B8

Proof. Since $\mathrm d^{\mathrm c}f$ is a one-form, $\mathrm{d}(\mathrm d^{\mathrm c}f\wedge T)=\mathrm{dd}^{\mathrm{c}}f\wedge T-\mathrm d^{\mathrm c}f\wedge \mathrm{d}T$, and Stokes' theorem gives the first equality in

For the second, recall that the integral of $\mathrm d^{\mathrm c}\sigma $ vanishes for every form $\sigma $ of total degree $2n-1$, by Stokes' theorem for $\partial $ and $\bar\partial $. We apply this to $\sigma =f\, \mathrm{d}T$ and use $\mathrm d^{\mathrm c}\mathrm{d}=-\mathrm{d}\mathrm d^{\mathrm c}$. $\square$

Proof. For $q=1$ there is nothing to prove. For $q\geq 2$, since $\theta $ has even degree, $\mathrm d^{\mathrm c}(\theta ^ q)=q\, \theta ^{q-1}\wedge \mathrm d^{\mathrm c}\theta $. Applying $\mathrm{d}$ and using again that $\theta ^{q-1}$ has even degree, we obtain

Proof. Expand $(\omega +\mathrm{dd}^{\mathrm{c}}u)^ n$ by the binomial theorem. The term with $k=n$ integrates to zero, since it is exact:

For $1\leq k\leq n-1$, apply lemma 11.1.2 with $f=u$ and $T=(\mathrm{dd}^{\mathrm{c}}u)^{k-1}\wedge \omega ^{n-k}$. Since both $\mathrm{d}$ and $\mathrm d^{\mathrm c}$ annihilate $\mathrm{dd}^{\mathrm{c}}u$,

which gives the $k$th term of (11.1.6). When $n=1$, the sum is empty and the mass is constant. $\square$

Proof. Step 1: the first condition implies the second. Let $u$ be a smooth real function. For $|t|$ small, $tu$ belongs to $\mathcal H_\omega $ for both signs of $t$, so the polynomial $t\mapsto \int _ X(\omega +t\mathrm{dd}^{\mathrm{c}}u)^ n$ is constant on an interval and its nonconstant coefficients vanish. For $1\leq j\leq n-1$ this gives

Replace $u$ by a linear combination of $n-j$ smooth real functions $u_1,\ldots ,u_{n-j}$ and compare the coefficients of the monomial in which every parameter appears once. We obtain

Integrating by parts in the first factor and using that $u_1$ is arbitrary, we obtain the pointwise identity

for every choice of the remaining functions. At a given point, the complex Hessians of these functions can be prescribed arbitrarily: multiply a quadratic polynomial in a coordinate chart by a cutoff function equal to one near that point. Since the identity is multilinear, we may complexify it and use the quadratic monomials $z_ a\bar z_ b$. Wedge products of their complex Hessians give all coordinate forms of the complementary bidegree, and pairing with these forms shows that every coefficient of $\mathrm{dd}^{\mathrm{c}}(\omega ^ j)$ vanishes. When $j=n-1$ there are no remaining factors, and the same conclusion follows directly from the arbitrariness of $u_1$. This proves the second condition.

Step 2: the remaining equivalences for smooth potentials. The second condition implies the first by theorem 11.1.5. For $n\geq 3$, the identities for $j=1,2$ and lemma 11.1.3 give $\mathrm{dd}^{\mathrm{c}}\omega =0$ and

so the second condition implies the third. For $n=2$, the second condition is $\mathrm{dd}^{\mathrm{c}}\omega =0$, while $\mathrm{d}\omega \wedge \mathrm d^{\mathrm c}\omega $ vanishes for degree reasons; the implication is again immediate. Conversely, lemma 11.1.3 derives all the identities of the second condition from the third.

Step 3: bounded potentials. Assume the first condition and let $u$ be a bounded $\omega $-plurisubharmonic function. By theorem 5.32.1, there are uniformly bounded functions $u_ j\in \mathcal H_\omega $ decreasing to $u$. Their masses all equal $\int _ X\omega ^ n$, and the continuity of the bounded products of section 5.28 along such sequences passes this common value to the limit. Conversely, every element of $\mathcal H_\omega $ is a bounded $\omega $-plurisubharmonic function, so constancy for bounded potentials implies the first condition. The zero potential shows that the common mass is $\int _ X\omega ^ n$. $\square$

Proof. Since $\mathrm M_\omega (u+tv)$ is a polynomial in $t$, its derivative at $t=0$ is

and lemma 11.1.2 turns this into the right-hand side of (11.1.9). For the second assertion, fix $u\in \mathcal H_\omega $. For every smooth real direction $v$, the segment $u+tv$ lies in $\mathcal H_\omega $ when $|t|$ is sufficiently small, so the right-hand side of (11.1.9) vanishes for all smooth real $v$. Hence the smooth top-degree form $\mathrm{dd}^{\mathrm{c}}(\omega _ u^{n-1})$ vanishes. $\square$

Proof. The formula is the case $n=2$ of (11.1.6), and it shows that $\mathrm{dd}^{\mathrm{c}}\omega =0$ implies constancy. Conversely, if the mass is constant, proposition 11.1.8 at $u=0$ gives $\mathrm{dd}^{\mathrm{c}}\omega =0$. $\square$

Proof. The potentials $u_ t$ of example 11.1.12 are smooth on the compact manifold $X$, so any two of them differ by a bounded function and have the same singularity type in the sense of section 5.33. For the same reason, their multiplier ideals are the structure sheaf for every positive multiple of the weight. By (11.1.13), their masses differ. $\square$

Proof. Fix $u,v\in \mathcal H_\omega $. By compactness there is a constant $C$ such that $u<v+C$ everywhere. Since $\omega _{v+C}=\omega _ v$, applying (11.1.17) to $u$ and $v+C$ gives $\mathrm M_\omega (v)\leq \mathrm M_\omega (u)$. Exchanging the roles of $u$ and $v$ gives the opposite inequality. $\square$

Proof. Let $X$, $\omega $, $u_ t$ and $B=\int _ X\beta _1\wedge \beta _2$ be as in example 11.1.12. For $-\pi ^{-2}<t<0$, set $u=0$ and $v=u_ t+C$ with $C>|t|$. Then $\{ u<v\} =X$, but

Thus this pair violates (11.1.17). $\square$

Proof. Let $u\in \mathcal H_\omega $. Since $C\eta +\mathrm{dd}^{\mathrm{c}}u\geq \omega +\mathrm{dd}^{\mathrm{c}}u>0$, the function $u/C$ belongs to $\mathcal H_\eta $. The top exterior power is monotone on positive $(1,1)$-forms: for positive Hermitian matrices $A\leq B$, all eigenvalues of $A^{-1/2}BA^{-1/2}$ are at least one, so $\det A\leq \det B$. Applying this pointwise gives

Taking the supremum over $u$ proves the inequality. Since any two Hermitian forms on the compact manifold $X$ are comparable in both directions, the last assertion follows. $\square$