10
$\begingroup$

I briefly recall the statement of the pentagon identity in quantum dilogarithm and cluster algebra.

For $b\in\mathbb{C}$ with $\operatorname{Re}(b)>0,\operatorname{Im}(b)\geq0$, Faddeev, Kashaev and Volkov et al. introduced the non-compact quantum dilogarithm $$\DeclareMathOperator{\Dmi}{d\!} e_b(z)=\exp\left(\frac{1}{4}\int_{C}\frac{e^{-2iz\zeta}}{\sinh(\zeta b)\sinh(\zeta b^{-1})\zeta}\Dmi\zeta\right),\quad|\operatorname{Im}(z)|<|\operatorname{Im}(c_b)|$$ where $C$ is a contour going along the real line from $−\infty$ to $+\infty$ surpassing the origin in a small semi-circle from above and $$c_b=\frac{i(b+b^{-1})}{2}.$$ One can analytically continue $e_b(z)$ to the whole complex plane by the following pair of functional equations $$e_b\left(z-\frac{ib^\pm}{2}\right)=\left(1+e^{2\pi zb^\pm}\right)e_b\left(z+\frac{ib^\pm}{2}\right).$$ We can show that $e_b$ is meromorphic with poles $$\{c_b+mib+nib^{-1}:m,n\in\mathbb{Z}_{\geq0}\}.$$

Now given a separable Hilbert space $\mathcal{H}$ (you may take $\mathcal{H}$ to be $L^2(\mathbb{R})$ or $\ell^2(\mathbb{Z})$), with a pair of densely defined self-adjoint operators $\hat{P},\hat{X}$ satisfying the commutation relation $[\hat{P},\hat{X}]=\frac{1}{2\pi i}$. While restricted to the real line, we can see that $e_b\in L^\infty(\mathbb{R})\cap C(\mathbb{R})$. So by Borel functional calculus, we have well-defined bounded operators $e_b(\hat{P}),e_b(\hat{X})$ and $e_b(\hat{P}+\hat{X})$.

The pentagon identity states that $$e_b(\hat{P})e_b(\hat{X})=e_b(\hat{X})e_b(\hat{P}+\hat{X})e_b(\hat{P}).$$

However, the above statement is not satisfactory in the sense of functional analysis. What's the meaning of $[\hat{P},\hat{X}]$? Generally, we are not allowed to composite unbounded operators.

Moreover, the proof given in the appendix of https://link.springer.com/article/10.1007/s002200100412 uses many singular integrations. For example, they use the "identity" $$\phi_+(x)\phi_+(y)e^{2\pi ixy}=\int_\mathbb{R}\phi_+(z)\phi_+(x-z)\phi_+(y-z)e^{i\pi z^2}\Dmi z,$$ where $\phi_+(z)$ is the inverse Fourier transform of $e_b$. However, this inverse Fourier transform can only be taken in the sense of tempered distributions, and in fact $\phi_+$ is not a function. Even if you use some functions to approximate $\phi_+$ and deform the integral contour around singularities, the integration in RHS is still by no means Lebesgue integrable.

I'm asking for a mathematically rigorous proof. This may be well known to experts, but I never find a reference.

Thank you,

Estwald

$\endgroup$

1 Answer 1

10
$\begingroup$

The pentagon relation for the quantum dilogarithm and quantized $M_{0,5}$ by A.B. Goncharov explicitly attempts to be more rigorous than the paper cited in the OP.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.