# A question about Quantized closed Kaehler manifolds

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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 :)

This post imported from StackExchange MathOverflow at 2014-10-17 11:02 (UTC), posted by SE-user Hassan Jolany
retagged Nov 9, 2014
It would probably help if you explained exactly what you mean by «homogeneous space».

This post imported from StackExchange MathOverflow at 2014-10-17 11:03 (UTC), posted by SE-user Mariano Suárez-Alvarez
By the way, if this is a known conjecture/open problem, then it is a good idea to give a reference to its origin.

This post imported from StackExchange MathOverflow at 2014-10-17 11:03 (UTC), posted by SE-user Mariano Suárez-Alvarez
I edited it again, :)

This post imported from StackExchange MathOverflow at 2014-10-17 11:03 (UTC), posted by SE-user Hassan Jolany
homogeneous space, here is of the form $G/H$ which $G,H$ are Lie groups

This post imported from StackExchange MathOverflow at 2014-10-17 11:03 (UTC), posted by SE-user Hassan Jolany
A strong necessary condition is that the automorphism group of $M$ act transitively on $M$.

This post imported from StackExchange MathOverflow at 2014-10-17 11:03 (UTC), posted by SE-user Peter Crooks
You mean $Aut(M)\cap Isom(M,\omega)$ act transitively on $M$?

This post imported from StackExchange MathOverflow at 2014-10-17 11:03 (UTC), posted by SE-user Hassan Jolany

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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.

This post imported from StackExchange MathOverflow at 2014-10-17 11:03 (UTC), posted by SE-user Peter Crooks
answered May 24, 2014 by (70 points)
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

This post imported from StackExchange MathOverflow at 2014-10-17 11:03 (UTC), posted by SE-user Hassan Jolany
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.

This post imported from StackExchange MathOverflow at 2014-10-17 11:03 (UTC), posted by SE-user Peter Crooks
So, let remore simply connected part. Then it is still correct,?

This post imported from StackExchange MathOverflow at 2014-10-17 11:03 (UTC), posted by SE-user Hassan Jolany
If $X$ is not simply-connected, then it cannot be written as a homogeneous space of a semisimple simply-connected $G$. A projective homogeneous $G$-variety is equivariantly isomorphic to a partial flag variety $G/P$. These are always simply-connected. So, the elliptic curve I mentioned above is not a homogeneous $G$-variety.

This post imported from StackExchange MathOverflow at 2014-10-17 11:03 (UTC), posted by SE-user Peter Crooks
In assumption, I assumed closed manifolds , i.e compact manifolds without boundary, so here you have considered compact elliptic curves without boundary ?

This post imported from StackExchange MathOverflow at 2014-10-17 11:03 (UTC), posted by SE-user Hassan Jolany
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.

This post imported from StackExchange MathOverflow at 2014-10-17 11:03 (UTC), posted by SE-user Peter Crooks
But the quotient of $\mathbb C$ by an integral lattice of rank 2 is simply connected really? You said "A projective homogeneous G-variety is equivariantly isomorphic to a partial flag variety $G/P$" can you give a reference for it?

This post imported from StackExchange MathOverflow at 2014-10-17 11:03 (UTC), posted by SE-user Hassan Jolany
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.

This post imported from StackExchange MathOverflow at 2014-10-17 11:03 (UTC), posted by SE-user Peter Crooks
Yes, I know$G/P$ is simply connected , but I am looking for a reason for your statement "A projective homogeneous $G$-variety is equivariantly isomorphic to a partial flag variety $G/P$""

This post imported from StackExchange MathOverflow at 2014-10-17 11:03 (UTC), posted by SE-user Hassan Jolany
You can find this in several books on algebraic groups. A parabolic subgroup is one for which $G/P$ is a complete variety. Also, a projective variety is always complete. So, I would suggest looking for material on parabolics and complete varieties.

This post imported from StackExchange MathOverflow at 2014-10-17 11:03 (UTC), posted by SE-user Peter Crooks
Thanks a lot, this conjecture was from Andrea loi

This post imported from StackExchange MathOverflow at 2014-10-17 11:03 (UTC), posted by SE-user Hassan Jolany
Homogeneous space for what sort of groups? The elliptic curve iis homogeneous over itself! :-)

This post imported from StackExchange MathOverflow at 2014-10-17 11:03 (UTC), posted by SE-user Mariano Suárez-Alvarez
Fair enough. In my comments, I assumed $G$ was simply-connected and semisimple. This seemed reasonable in the context of the question.

This post imported from StackExchange MathOverflow at 2014-10-17 11:03 (UTC), posted by SE-user Peter Crooks

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