Consider the real vector space of smooth real-valued functions on the circle modulo constants (i.e. mean-zero functions on the ), equipped with the (real) inner product Let be the completion of with respect to the above inner product. The Hilbert transform forms a complex structure on . Now, let us apply the procedure described at the beginning of this section to construct a symplectic Hilbert space from . Then the underlying set of is , the orthogonal subspaces and consists of functions that can extend holomorphically to and respectively. The symplectic form is given by Note that is real for real functions and .
Proposition
Suppose , then the operator .
Proof is symplectic, because To show that is in , it suffices to show that can be written as a Hilbert-Schmidt perturbation of the identity. Once we can write , we can apply the polar decomposition and functional calculus to write for some unitary .
We first show that is Hilbert-Schmidt. Note that Therefore, the kernel of is Utilize the Laurent expansion of at , we can write , where is a real-analytic function vanishes at . Let Hence, Observe that the only potentially non-smooth term is . Let . Then , , we can write for some smooth such that . Therefore, the potentially non-smooth term can be written as which is smooth. Hence, is smooth and thus is Hilbert-Schmidt.
Since is real, let us write according to the decomposition . Computing the commutator gives so and have to be Hilbert-Schmidt. As is symplectic, we have Then hence can be written as .
Embed into
The embedding is natural in the sense that the change of variable is the only thing we can do given a diffeomorphism. Similarly, if we want to treat an annulus as an operator on , the natural way is to holomorphically “extend” some function from the one boundary to the other boundary, and then restrict it back to the circle. The following proposition shows this idea precisely.
Proposition
There is a well-defined injective semigroup homomorphism extending and sending an annulus to the closure of an (unbounded) operator on satisfying for some holomorphic .
Proof
We shall now check that always lands in . Let us pick the standard representative of an annulus , and consider the trivial round annulus that is of the same modulus. Denote . For , means there exists such that These are exactly which are precisely conditions for . Therefore, This further means Note that we have already showed that embeds diffeomorphisms to , so it remains to show that lands in . Suppose , we have the following equations hold: Since is holomorphic on , we shall write in Laurent series: . We can also write in Fourier series , hence the first equation implies , and the second equation implies , so Therefore, satisfies where is a contraction. Now it is clear .
Next we show that is multiplicative. Let and be two annuli, and be the welded annulus. We want to show that . Let , then means there exists such that Similarly, means there exists such that Since is identified with in the welding, we can glue and together to get a holomorphic function on , which shows that .