# Rigorous proof of the pentagon identity

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

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.