This construction arises when constructing the Szego projector.

Let's consider the dual disc bundle $\overline{D}$ of a positive Hermitian line bundle ($L$,$h$) over a compact Kahler manifold $M$, i.e $\overline{D}$ = $\{$ $v$ $\in$$L^*$ : $\|L^*\|$ $\leq$$1$ $\}$.

Let $H^2$($\overline{D}$) = $Ker$ $\bar\partial$ $\bigcap$ $L^2$($\overline{D}$). Now $S^1$ acts on $H^2$($\overline{D}$): given a funtion $f\in$$H^2$($\overline{D}$) and $\lambda\in S^1$ the action gives the function $(\lambda,f)\in H^2$($\overline{D}$) such that $(\lambda,f)(p,v)=f(p,\lambda v) $ with $(p,v)\in \overline{D}, v\in \pi^{-1}(p)$

We can check easily that we have a unitary representation of $S^1$ on $H^2$($\overline{D}$): $\lambda$ $\longmapsto$ $T_{\lambda}$ $\in\mathbb{U} $($H^2$($\overline{D}$)) (unitary operators), such that $T_{\lambda}(f)=(\lambda,f)$

Using basic representation theory of Abelian groups we know that this representation decompose in irreducible representations indexed by $\mathbb{Z}$ (the dual group of $S^1$). This decompose $H^2$($\overline{D}$) in subspaces of dimension 1 (because $S^1$ is Abelian). We can easily check that the subspace corresponding to $k\geq0$ is given by those functions which are k-linear in the direction of the fibers $i.e$ $H^2_k$($\overline{D}$) = $\{$ $f\in$$H^2$($\overline{D}$): $f(p,\lambda v)=\lambda^k f(p, v)$ $\forall (p, v)\in \overline{D}$, $\lambda\in S^1 \}$

Thinking about the subspaces $H^2_k$ I wonder how it is possible that those have dimension 1. What is a basis for each subspace $H^2_k?$ I know the decomposition of the well-know space $L^2(S^1)$ in one dimesional subspaces but I am strugling to use this in the dual disc bundle case. Did I misunderstood something in the explaniation I wrote?

Thanks for your answers, if you could also give some references with more details about the Szego projector in the dual disc bundle case I would be very grateful.


2 Answers 2


It was shown by Zelditch that the spaces $H_k$ with the standard metric on $\mathbb{C}$ are isometric to the spaces $H^0(L^k)$ of holomorphic sections of the $k$-th power of the line bundle $L$, please see arXiv: math-ph/0002009v1 (propositions 6,7).

  • $\begingroup$ Many thanks for the reference, it was very useful. Let me remark one point that it is still not clear for me: In page 4 of the reference it says: "The S1 action commutes with $\bar\partial$, hence H2(X) =⊕$H^2_k$" I think that the action of S1 commutes with $\bar\partial$ implies that we have a representation of S1 and the decomposition in the direct sum is the decomposition in irreducible representations.However this seems to be a contradiction because S1 is abelian thus dim H^2_k=1 (all irreps have dim=1 when the group is Abelian). So let me ask why H^2 splits in the direct sum? Thanks David $\endgroup$
    – Coffee
    Oct 10, 2013 at 0:54
  • $\begingroup$ @josh the Abelian group action implies that the irreducible components are one dimensional, but they can have multiplicities. The space$H_k$ is a direct sum of all one dimensional subspaces of eigenvalue (weight) $k$. $\endgroup$ Oct 13, 2013 at 7:53

Josh: The finite-dimensionality of the spaces $H^2_k$ comes about because the underlying manifold $M$ is compact. Otherwise even these spaces would be infinite-dimensional.

The spaces $H^2_k$ and $H^2_{\ell}$ are orthogonal subspaces of $H^2$. This is basically because of Fubini's Theorem. Locally $\bar D$ looks like a product of a disk and a piece of $M$, and the integration respects that product in the sense of Fubini. It is then the usual Fourier series argument to see that these spaces are orthogonal to each other.

Here is another way to see this picture; a more pedestrian way. Take any holomorphic function $f$ on $\overline{D}$. You can restrict it to a disk $D_x$ in this disk bundle (lying in the fiber of $L^*$ over the point $x \in M$), and there you can expand it in a taylor series as follows. First, you need a complex variable on this disk. You choose a frame $\xi_x$ for $L^*$ and so any point of $\overline D_x$ is of the form $z\xi _x$. Then $z$ is your complex variable in $\overline{D}_x$. Now write $$ f(z\xi _x) = \sum _{j=1} ^{\infty} s_k(x) z^k. $$ You will find that $s_k(x) /\xi _x^{\otimes k} \in L_x^{\otimes k}$, and therefore in global terms you have $$ f(v) = \sum _{j=1} ^{\infty} \left < \sigma_k, v^{\otimes k}\right >, \qquad v \in \overline{D}. $$ The pairing in this formula is as follows. If you have a section $s$ of $L^{\otimes k}$ and a vector $v \in L^*_x$ then you can pair the dual objects $s(x)$ and $v^{\otimes k}$.

In words, you can write any holomorphic function as a homogeneous expansion with the $k$-homogeneous term a section of $L^{\otimes k}$. This already shows you that even non-$L^2$ holomorphic functions decompose in this "direct sum" way (which at the moment have nothing to do with $L^2$). You can now impose the $L^2$ structure and use the Fubini idea I mentioned above to see the pairwise orthogonality of these subspaces.

The finite dimensionality of $H^0(M,L^{\otimes k})$ when $M$ is compact follows from the Hodge Theorem, which is a much longer story. But it is easy to see in general that these spaces are not $1$-dimensional. From the Rep theory point of view, you don't have irreducible representations here, so there is no reason to think they are $1$-dimensional.


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