Let $G$ be a connected domain in $\mathbb{C}^{n}$, let $H$ be a Hilbert space and let $f,g:G\to H\backslash \{0\}$ be holomorphic (in my particular situation they are also injective, but I don't think it helps).

Is it true that if $\frac{\partial^2}{\partial z_i\partial \overline{z_j}}\log \|f(z)\|=\frac{\partial^2}{\partial z_i\partial \overline{z_j}}\log \|g(z)\|$, for all $i,j\le n$, then there is a holomorphic function $h:G\to\mathbb{C}$ and an isometry $U:H\to H$, such that $g(z)=h(z)Uf(z)$?

Motivation. Consider the canonical quotient $p:H\to PH$, where the latter is the projective space over $H$. Suppose $\varphi:G\to G$ is a biholomorphism, such that $f$ and $g=f\circ\varphi$ satisfy that condition. Then $pf\varphi(pf)^{-1}$ is an isometry of $pf(G)$ with respect to the Fubini-Study metric on $PH$. I wonder, if this isometry can be lifted to an isometry $H$, i.e. if there is an isometry $U:H\to H$, such that $pUf=pf\varphi$. Finally the last condition means that there is a holomorphic function $h:G\to\mathbb{C}$, such that $h(z)Uf(z)=f(\varphi(z))$.


1 Answer 1


Yes, this is true, and this is called the Calabi rigidity, though it was proved (for finite dimensional Hilbert space) long before Calabi, see, for example, Polya-Szego, revised edition, part IV, problem 207, where there is a reference on the original paper. Calabi's paper is: Isometric imbedding of complex manifolds, Ann. Math. 58 (1953), 1–23.

Another source is Theorem 2 in https://arxiv.org/pdf/math/0007030.pdf.

  • $\begingroup$ Thank you! Both Sodin and Calabi indeed prove the fact I was wondering about, but I don't understand why is it equivalent to the problem from Polya-Szego? (and Sodin seems to agree with you on that) $\endgroup$
    – erz
    May 6, 2017 at 7:03
  • $\begingroup$ I do not have Polya-Szego next to me at this moment to check, my answer was based on Sodin's statement. $\endgroup$ May 6, 2017 at 7:31
  • $\begingroup$ The statement there says that if $\|f(z)\|=\|g(z)\|$, then $g=Uf$, which is similar, and proven kind of similarly, but does not seem to be exactly analogous. $\endgroup$
    – erz
    May 6, 2017 at 8:20
  • $\begingroup$ You probably mean, $\log F$ is pluriharmonic and then $F=|h|$, right? The transition from harmonic to holomorphic usually requires that the domain is simply connected. So yes, we get the statement locally, and then using holomorphy we can "globalize" it, but this last step is not exactly trivial (at least the proof that I have), or am I missing something? $\endgroup$
    – erz
    May 8, 2017 at 15:22
  • $\begingroup$ @erz: I edited my coment: In fact your derivatives condition is equivalent to $\| f\|=|F|\| g\|$, where $\log|F|$ is pluriharmonic, so take $F=\log|h|$ and you obtain your statement from the Polya-Szego statement. $\endgroup$ May 8, 2017 at 18:18

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