Let $(M,\omega)$ be a Quantized closed Kaehler manifold then by Koderia embedding theorem , $M$ must be algebraicly projective i.e, we have the embedding

$$\phi: (M,\omega)\to  (\mathbb CP^N, \omega_{FS})$$ So $$\phi^*\omega_{FS}=\omega+\frac{i}{2\pi}\partial\bar \partial \epsilon $$

where $\epsilon$ is a smooth function and is defined as follows:

Definition of $\epsilon$ function: Let $\pi:(L,h)\to (M,\omega)$ be a prequantum line bundle and let $x\in M$ and $q\in L^+$ such that $\pi(q)=x$ and $H$ is the Hilbert space of global holomorphic sections ($h$ is hermitian metric). Then we can write $s(x)=\delta_q(s)q$ where $\delta_q:H\to \mathbb C$ is a linear continous functional of $s$ and by Riesz theorem $\delta_q(s)=\langle s,e_q \rangle_h$ where $e_q\in H$ and thus $s(x)= \langle s,e_q\rangle_hq$ and we can define the real valued function on $M$ by the formula

$$\epsilon(x)=h(q,q)\left \| e_q \right \|_h^2$$

Now the conjecture is that, if $\epsilon$ be constant then $M$ is homogeneous space? Is there any counterexample or proof for it?

This question is known as Andrea Loi's conjecture in his doctoral thesis

Peter Crooks gave a counterexample and I removed the part simply connected, I want to see this conjecture still is conjecture :)

  • $\begingroup$ It would probably help if you explained exactly what you mean by «homogeneous space». $\endgroup$ May 25, 2014 at 19:33
  • $\begingroup$ By the way, if this is a known conjecture/open problem, then it is a good idea to give a reference to its origin. $\endgroup$ May 25, 2014 at 19:34
  • $\begingroup$ I edited it again, :) $\endgroup$
    – user21574
    May 25, 2014 at 19:36
  • $\begingroup$ homogeneous space, here is of the form $G/H$ which $G,H$ are Lie groups $\endgroup$
    – user21574
    May 25, 2014 at 19:38
  • $\begingroup$ A strong necessary condition is that the automorphism group of $M$ act transitively on $M$. $\endgroup$ May 25, 2014 at 20:47

1 Answer 1


I do not believe this is the case. If you have any smooth complex submanifold $X$ of $\mathbb{CP}^n$, then the Kahler form on $\mathbb{CP}^n$ pulls-back to a Kahler form $\omega$ on $X$ (so $\phi^*\omega_{FS}=\omega$). You can take $X$ to be any smooth projective variety. These are not all simply-connected. For instance, let $X$ be an elliptic curve.

  • $\begingroup$ Quantized Kaehler manifolds leds to $[\omega]\in H^2(M,\mathbb Z)$, then the curvature form is belong to the image of $H^2(M,\mathbb Z)\to H^2(M,\mathbb C)$ so we will have Koderia embedding theorem. I edited my question and defined $\epsilon$ function $\endgroup$
    – user21574
    May 24, 2014 at 17:19
  • $\begingroup$ If $\epsilon$ is constant, then your condition seems to be $\phi^*(\omega_{FS})=\omega$. This occurs in my setup by construction. Also, a Kahler form $\omega$ on a projective variety defines a class in integral cohomology, so I think this is fine. $\endgroup$ May 24, 2014 at 17:33
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    $\begingroup$ Yes, that's right. By elliptic curve, I mean the quotient of $\mathbb{C}$ by an integral lattice of rank $2$. It has no boundary. $\endgroup$ May 24, 2014 at 19:06
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    $\begingroup$ The point is that this quotient is NOT simply-connected. Therefore, it is not a homogeneous $G$-variety. This gives you the counter-example you requested. That the spaces $G/P$ are simply-connected follows from the Schubert cell decomposition of $G/P$. I would suggest searching for this. $\endgroup$ May 24, 2014 at 19:52
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    $\begingroup$ Homogeneous space for what sort of groups? The elliptic curve iis homogeneous over itself! :-) $\endgroup$ May 25, 2014 at 19:33

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