L^2 space of holomorphic functions with given weight - MathOverflow

L^2 space of holomorphic functions with given weight

2010-06-11T15:30:59Z

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?

Answer by Helge for L^2 space of holomorphic functions with given weight

2010-06-11T18:00:14Z

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

Answer by BR for L^2 space of holomorphic functions with given weight

2010-06-11T20:42:15Z

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 c_v(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 L^p with p<1.