Set up: Let $K$ be a number field. Let $M_K$ be the places of $K$, and define the standard height on $\mathbb{P}^n(K)$ as $$H([x_0, \cdots, x_n]) = \prod_{v \in M_K} \max\{|x_0|_v, \cdots, |x_n|_v\}$$ Schanuel's theorem gives an asymptotic of number of points up to height $B$, as $B \to \infty$ via geometry of numbers.
Note that there is built-in non-smoothness of this height at archimedean place. The presence of $\max$ and absolute value means this is not a smooth function. Of course, one can pick other metrics so that the resulting height is smooth (perhaps up to some power). For example, using the $L^2$ norm at archimedean place we can get the height
$$H'([x_0, \cdots, x_n]) = \prod_{v | \infty} (|x_0|_v^2 + \cdots + |x_n|_v^2)^{1/2} \prod_{v < \infty} \max\{|x_0|_v, \cdots, |x_n|_v\}$$
which removes the non-smoothness from $\max$ and $|\cdot|$ (via squaring). The 1/2 power at the end makes this not smooth, but at least $H'([x_0, \cdots, x_n])^2$ is a smooth function in $x_i$'s.
Alternative proof of Schanuel's theorem In Section 3 of the paper of Chambert-Loir and Tschinkel, "POINTS OF BOUNDED HEIGHT ON EQUIVARIANT COMPACTIFICATIONS OF VECTOR GROUPS", they give another proof of Schanuel's theorem on $\mathbb{P}^n$ via the Poisson summation formula.
It seems like the non-smoothness of standard height would forbid them to use Poisson summation, due to the Fourier transform not having good decay (so that sum of $\widehat{H}$ over additive characters do not converge). Maybe for this reason, they use smoothed version of height instead (such as the one from $L^2$ norm above), instead of the standard height.
Question Is the non-smoothness of standard height a hard blocker of the Fourier method? In other words,
- is there an abstraction where Poisson summation formula can still make sense and their argument can go through even for standard height?
- Or is there a way to deduce Schanuel's theorem of standard height via the version of smoothed height?
Thanks!
