# L^2 space of holomorphic functions with given weight

Hi folks, what is known about the $L^2$ space of holomorphic functions of 1 complex variable with the scalar product

$\langle f, g \rangle = \int dzd{\bar z} \frac{ {\bar f(z)} g(z) }{(1 + z{\bar z})^x}$

where $x > 2$ is a real number? The domain of integration is the entire complex plane. Poles are allowed in the functions so all possible powers in the Laurent expansion are allowed, $f(z) = \sum_{n = -\infty}^\infty f_n z^n$.

Is this a well-known space? Is an orthogonal basis readily available?

If $f(z)$ is a polynomial with sufficiently low degree then certainly it is in the above defined $L^2$ space. But there are much more functions that are okay, it seems, for instance $f(z) = \exp( -z )$. Or anything that falls off sufficiently fast.

The background is this: if $x=2j+2$ where $j$ is a half-integer and the holomorphic functions can only be at most $2j$ order polynomials, then the above defined space is the $2j+1$ dimensional irreducible unitary representation of $SU(2)$. The action of $g = [ [ a, b ], [ c, d ] ] \in SU(2)$ is

$(gf)(z) = (bz + d)^{2j} f\left( \frac{az+c}{bz+d} \right)$

Clearly, if $f(z)$ is a polynomial at most of order $2j$ then $(gf)(z)$ is also one. And the scalar product is the one I gave above, with $x=2j+2$.

Okay, this was the case for half-integer $j$. What is the deal with arbitrary $j$? Then I can still define the above scalar product with arbitrary $x$. The action above still preserves the scalar product. It is still a group action by $SU(2)$. Do I get an infinite dimensional representation of $SU(2)$? Is it reducible/irreducible?

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What domain is the integral over? IF it's all C, I would suspects, it's just polynomial of degree < x - 2. –  Helge Jun 11 '10 at 15:49
It seems to me that if the domain of integration contains the origin, you might as well forget about having singularities there. We really need some information on the domain. Also, how about a little background? Do you see a need for these spaces somewhere, or is it just idle speculation? (Not that there is necessarily something wrong with that.) –  Harald Hanche-Olsen Jun 11 '10 at 16:06
Edited the question to answer these questions. –  Daniel Jun 11 '10 at 17:02
You seem to infer that $\exp(-z)$ "falls off sufficiently fast". Not as $z$ goes to $-\infty$ along the real axis, it doesn't. –  Robin Chapman Jun 11 '10 at 17:58
Oh, that's true! Sorry. –  Daniel Jun 11 '10 at 18:06

Hi Daniel. As already said in my comment the space consists just of order polynomials of degree $\lfloor x - 1 \rfloor$. First, one can check that any function in the space must be holomorphic, since the weight doesn't help to integrate over poles. Then one gets from $\\| f \\| < \infty$ that $|f(x)| \leq |z|^{x-1 }$, so one has that $f$ is a degree $\lfloor x - 1 \rfloor$ polynomial by a consequence of Liouville's Theorem.

Helge

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Okay, great, thanks! –  Daniel Jun 11 '10 at 18:09

To riff on the final part of your question:

By the Peter-Weyl Theorem, all irreducible Hilbert space representations of a compact group (e.g. SU(2)) are finite dimensional. Thus, any infinite dimensional Hilbert space representation will be reducible.

What can be said for non-Hilbert space representations? Given a compact group G acting irreducibly (and continuously) on a locally convex topological vector space V, you can inject V into L2(G) by sending v in V to the function cv(g)=⟨g⋅v,v'⟩ where v' is basically any nonzero element of the continuous dual of V (this is where local convexity is used). (Note that the irreducibility of G implies the injectivity of the map.)

Thus V is finite dimensional: its image is not necessarily closed in L2(G), but, since V is irreducible, its image has to lie within a single irreducible component of L2(G), which are all finite dimensional.

So to find a representation that is infinite dimensional and irreducible, you'd have to look at non-locally convex vector spaces (actually, you need the dual space to not separate points). Like Lp with p<1.

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