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?



L^2 space of holomorphic functions with given weight
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


