To show the derivation of the explicit formula from the trace formula, here's an example using a sharp cutoff test function, which recovers Riemann's original formula for the prime counting function \(\pi(x) = \sum_{p \leq x} 1\). This test function is not smooth (as for strict convergence in Weil's formula), but it does serve as a distributional limit and demonstrates the process. This example will show how the bridge works between the dynamical trace formula to classical analytic number theory, where the stability condition (RH) defines the sum over zeros.
Musing on the sharp cutoff test function in the multiplicative variable; define the function \(f(y)\) as: \[ f(y) = \begin{cases} 1 & \text{if } 1 \leq y \leq x, \\ 1/2 & \text{if } y = x, \\ 0 & \text{otherwise}. \end{cases} \] This corresponds to a step function for the sum \(\sum_{n \leq x} \Lambda(n)/n^{1/2}\), but to align with the trace, the associated \(g(u) = e^{u/2} f(e^u)\), is used which is 1 for \(0 \leq u \leq \log x\) (shifted by the critical line factor).
The trace \[ \operatorname{Tr}(U(g)) \] will approximate the smoothed sum over primes and in the limit of sharp cutoff, it looks like it produces Riemann's formula for: \[ \sum_{n \leq x} \Lambda(n) = x - \sum_\rho \frac{x^\rho}{\rho} + \text{archimedean terms}. \]
Geometric Side from Trace Formula: Applying the sharp cutoff \(g(u)\) to the prime power sum: \[ \sum_p \sum_k \frac{\log p}{p^{k/2}} \left[ g(k \log p) + g(-k \log p) \right] \approx \sum_{p^k \leq x} \frac{\log p}{\sqrt{p^k}} + \sum_{p^k \leq x} \frac{\log p}{p^k \sqrt{p^k}}, \] where the second term (from negative logs) is small for large x (converges quickly). Main term is \(\sum_{n \leq x} \frac{\Lambda(n)}{\sqrt{n}}\), with \(\Lambda(n) = \log p\) for n = p^k.
Archimedean Term \(W_r(g)\): For the real place, the integral with sharp cutoff gives the logarithmic term: \(\log(2\pi) + (\log x)/2\), from the Gamma factor in the functional equation.
Spectral Side: The Fourier transform of the cutoff is \(\hat{g}(\gamma) = \frac{\sin(\gamma \log x)}{\gamma}\), so the sum over zeros is: \[ \sum_\rho \frac{x^\rho}{\rho} = \sum_\gamma \frac{x^{1/2 + i\gamma}}{1/2 + i\gamma}. \] (assuming RH for real \(\gamma\)).
Equating and Integrating: The explicit formula for the von Mangoldt sum is: \[ \sum_{n \leq x} \Lambda(n) = x - \sum_\rho \frac{x^\rho}{\rho} - \log(2\pi) - \frac{1}{2} \log(1 - x^{-2}) + \epsilon, \] where \(\epsilon\) is the error term (under RH, O(\sqrt{x} \log x)). To get \(\pi(x)\), integrate with weight 1/\log n: \[ \pi(x) = \operatorname{li}(x)\ - \sum_\rho \operatorname{li}(x^\rho) + \text{terms}, \] where \(\operatorname{li}(y)\ = \int_2^y \frac{dt}{\log t}\).
Its dynamic: Cutoff enforces balance; off the critical line sum over \(\rho\) would diverge, but stability (RH) keeps it oscillatory.
For x = 10, \(\pi(10) = 4\) (primes: 2,3,5,7). Riemann's formula gives the approximate match with known zeros (exact computation needs many terms). This shows how the trace leads to prime counting.