Theorem 10.26.19A07C571BD3F

Proof. Rigidity when $H=X$. Suppose first that $H=X$. By lemma 20.4.10, a positive current in $\alpha $ can be written $T=\theta +\mathrm{dd}^{\mathrm{c}}u$ with $u$ quasi-plurisubharmonic, in particular $u\in L^1(X)$. For a constant real vector $v\in K$, the coefficient $T(v,Jv)$ is a positive Radon measure. Its total integral against Haar measure is zero, because $\theta (v,Jv)=0$ and the remaining part of the coefficient is a combination of constant-coefficient derivatives of $u$. Hence this coefficient vanishes. Writing $D_ v$ for differentiation along $v$ and using that $K$ is complex, we obtain

For a real linear functional $\ell $ with $\ell (\Lambda )\subset \mathbb Z$, let $\widehat u(\ell )$ be the Fourier coefficient of $u$ at the character $\exp (2\pi \mathrm i\ell )$. The identity above gives

Since the image of $K$ is dense, a nonzero character cannot vanish identically on $K$. Thus every nonconstant Fourier coefficient of $u$ is zero, and convolution with a smooth approximate identity shows that $u$ is constant. Hence $T=\theta $.

Nonrigidity when $H\ne X$. If $H\ne X$, a nontrivial character of the real torus $X/H$ gives a nonzero $\ell $ that vanishes on $K$. Put $\phi =\cos (2\pi \ell )$. Since $K$ is complex, the form $\mathrm{dd}^{\mathrm{c}}\phi $ vanishes whenever one of its arguments lies in $K$. On a fixed complement of $K$, $\theta $ is positive definite and $\mathrm{dd}^{\mathrm{c}}\phi $ is uniformly bounded. Consequently, the forms $\theta \pm \varepsilon \mathrm{dd}^{\mathrm{c}}\phi $ are semipositive for sufficiently small $\varepsilon >0$. They represent the same class, and they are distinct because a nonconstant smooth function on a compact torus cannot have zero $\mathrm{dd}^{\mathrm{c}}$. Hence $\alpha $ is not rigid.

Collapse when $H=X$. Suppose again that $H=X$, and fix $\delta >0$. By density and compactness, there are finitely many vectors $k_1,\ldots ,k_ N\in K$ whose images form a $\delta $-net in the flat metric of $\kappa $. Put $R=\max _ j|k_ j|_\kappa $. Since $\theta (k_ j,Jk_ j)=0$, the straight path from $0$ to the image of $k_ j$ has $(\theta +t\kappa )$-length $\sqrt t\, |k_ j|_\kappa $. For $0<t\leq 1$ these metrics are bounded above by a fixed multiple of $\kappa $. Joining this path to a point within $\delta $ therefore gives

since by translation invariance the diameter equals the supremum of the distances from $0$. Letting first $t\downarrow 0$ for this fixed net and then $\delta \downarrow 0$ proves the third condition.

No collapse when $H\ne X$. If $H\ne X$, choose $\ell $ as above. Its differential vanishes on $K$, so $|\mathrm d\ell (v)|\leq L|v|_\theta $ for some $L>0$, where $|v|_\theta $ is the seminorm defined by $\theta $. Since $\theta +t\kappa \geq \theta $, the surjective map $\ell \bmod \mathbb Z$ onto the circle of length one is $L$-Lipschitz for every $t>0$. Two points mapping to opposite points of the circle therefore have distance at least $1/(2L)$, and the diameters cannot tend to zero. This completes the proof of the equivalences. Since the second condition does not involve $\kappa $, neither does the third. $\square$