2026-09-23

something just clicked, felt very happy.

1. We all know the classical story, $[x, p] = \hbar $ (sorry, I will drop $i$ as math people don't care about constant...). We can realize this algebra (or faithful representation) of it in two ways (view $\hbar$ as some complex number)

• They act on $\C[x]$ by $$x \cdot f(x) = x f(x), \quad p \cdot f(x) = -\hbar \d_x f(x) $$
• The act on $\C[p]$ by $$ x \cdot g(p) = \hbar \d_p g(p), \quad p \cdot g(p) = p g(p). $$

actually, can we have more? can we just rotate the polarization as we want? what formula do we have? Fourier transformation $e^{-xp/\hbar}$ is an extreme version of rotation generated by Hamiltonian function $x^2+p^2$. metaplectic representation?

These are quantization of the symplectic space $T_p^*\C_x$, and with different choice of Lagrangian polarization (polarization just means foliating by Lagrangians).

2. Now comes the 'cylinder'. $T^*_w \C^*_u = C^*_u \times \C_w$. We have commutator relation $$ [w, u] = \hbar u $$ Again, we have two ways to quantize, let $u = e^x$, we get

• Representation is function on $u$, we get $$ u \mapsto u \cdot, \quad w \mapsto \hbar \d_x = \hbar u \d_u $$
• Representation is function on $w$, we get $$ u \mapsto e^{-\hbar \d_w}, \quad w \mapsto w \cdot $$

3. Now comes the '2 torus'. $C^*_u \times C^*_w$, say we set $u = e^x, w = e^y$, if $[x,y] = \hbar$, and $q= e^\hbar$, then we can do 'Baker-Hausdorff-Campbell'? $e^x e^y = e^{x+y + 1/2\hbar}, e^y e^x = e^{x+y-1/2 \hbar}$, so we get $$e^x e^y = e^\hbar e^y e^x, \quad uw = q wu $$ The so called quantum torus relation. It can be quantized symmetrically in two ways, either

• acting on function of $u$, $w f(u) = f(u/q)$
• acting on function of $w$, $u g(w) = g(qw)$.

hmm, it seems the rational version and elliptic version are most natural ones, where as the trig version is kind of half-baked...?