L^2 space of holomorphic functions with given weight - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T11:35:37Z http://mathoverflow.net/feeds/question/27825 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/27825/l2-space-of-holomorphic-functions-with-given-weight L^2 space of holomorphic functions with given weight Daniel 2010-06-11T15:30:59Z 2010-06-11T20:42:15Z <p>Hi folks, what is known about the $L^2$ space of holomorphic functions of 1 complex variable with the scalar product</p> <p>$\langle f, g \rangle = \int dzd{\bar z} \frac{ {\bar f(z)} g(z) }{(1 + z{\bar z})^x}$</p> <p>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, $f(z) = \sum_{n = -\infty}^\infty f_n z^n$.</p> <p>Is this a well-known space? Is an orthogonal basis readily available?</p> <p>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.</p> <p>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</p> <p>$(gf)(z) = (bz + d)^{2j} f\left( \frac{az+c}{bz+d} \right)$</p> <p>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 $x=2j+2$.</p> <p>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?</p> http://mathoverflow.net/questions/27825/l2-space-of-holomorphic-functions-with-given-weight/27844#27844 Answer by Helge for L^2 space of holomorphic functions with given weight Helge 2010-06-11T18:00:14Z 2010-06-11T18:00:14Z <p>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 $\| f \| &lt; \infty$ 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.</p> <p>Helge</p> http://mathoverflow.net/questions/27825/l2-space-of-holomorphic-functions-with-given-weight/27868#27868 Answer by BR for L^2 space of holomorphic functions with given weight BR 2010-06-11T20:42:15Z 2010-06-11T20:42:15Z <p>To riff on the final part of your question:</p> <p>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. </p> <p>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 L<sup>2</sup>(G) by sending v in V to the function c<sub>v</sub>(g)=&lang;g&#8901;v,v'&rang; 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.)</p> <p>Thus V is finite dimensional: its image is not necessarily closed in L<sup>2</sup>(G), but, since V is irreducible, its image has to lie within a single irreducible component of L<sup>2</sup>(G), which are all finite dimensional.</p> <p>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<sup>p</sup> with p&lt;1. </p>