关于Luo和Sarnak论文中涉及Hecke特征值的两个求和等式的推导疑问

Great question—this equality relies on a key identity for normalized Hecke eigenvalues plus a standard divisor-based change of variables. Let's walk through it step by step to clear up your confusion!

First, let's recap the key facts we need:

• $\lambda_f$ is the normalized Hecke eigenvalue of a weight-$k$ modular form $f$, which means it's completely multiplicative (i.e., $\lambda_f(ab) = \lambda_f(a)\lambda_f(b)$ for all positive integers $a,b$) and satisfies the critical Hecke product identity:

  For any positive integers $n,m$, $\lambda_f(n)\lambda_f(m) = \sum_{d \mid \gcd(n,m)} \lambda_f\left(\frac{nm}{d^2}\right)$

• $m$ is a fixed positive integer, and $h \in C_0^\infty(0,\infty)$ is compactly supported (so our sums are absolutely convergent, justifying any order changes).

Step 1: Rewrite the product of eigenvalues using the Hecke identity #

Start with the left-hand side (LHS) sum:
$$
\sum_{r\geq 1}\lambda_f(r)\lambda_f(r+m)h\left(\frac{k-1}{4\pi\left(r+\frac{m}{2}\right)}\right)
$$

Apply the Hecke product identity to $\lambda_f(r)\lambda_f(r+m)$ (here $n=r$, $m=r+m$). Note that $\gcd(r, r+m) = \gcd(r, m)$, so we can rewrite the product as:
$$
\lambda_f(r)\lambda_f(r+m) = \sum_{d \mid \gcd(r,m)} \lambda_f\left(\frac{r(r+m)}{d^2}\right)
$$

Substitute this back into the LHS sum:
$$
\sum_{r\geq1} \left( \sum_{d \mid \gcd(r,m)} \lambda_f\left(\frac{r(r+m)}{d^2}\right) \right) h\left(\frac{k-1}{4\pi\left(r+\frac{m}{2}\right)}\right)
$$

Step 2: Swap the order of summation #

Since our sum is absolutely convergent (thanks to $h$'s compact support and the slow growth of $\lambda_f$), we can swap the order of summation: instead of summing over $r$ first, we sum over all divisors $d$ of $m$ (since $d \mid \gcd(r,m)$ implies $d \mid m$), then sum over all $r$ where $d \mid \gcd(r,m)$ (i.e., $d$ divides $r$):
$$
\sum_{d \mid m} \sum_{\substack{r\geq1 \ d \mid r}} \lambda_f\left(\frac{r(r+m)}{d^2}\right) h\left(\frac{k-1}{4\pi\left(r+\frac{m}{2}\right)}\right)
$$

Step 3: Change variables to simplify the sum #

For each fixed $d \mid m$, let $r = d r'$ (so $r' \geq 1$ is a new dummy variable). Substitute this into the sum:

1. First, compute the argument of $\lambda_f$:
   $$
   \frac{r(r+m)}{d^2} = \frac{d r' \cdot (d r' + m)}{d^2} = r'\left(r' + \frac{m}{d}\right)
   $$
   (Since $d \mid m$, $\frac{m}{d}$ is an integer, so this is a valid input for $\lambda_f$.)
2. Next, compute the argument of $h$:
   $$
   r + \frac{m}{2} = d r' + \frac{m}{2} = d\left(r' + \frac{m}{2d}\right)
   $$
   So the $h$ term becomes:
   $$
   h\left(\frac{k-1}{4\pi d\left(r' + \frac{m}{2d}\right)}\right)
   $$

Step 4: Clean up the dummy variable #

Finally, rename $r'$ back to $r$ (it's just a dummy variable, so this doesn't change the sum). We end up with:
$$
\sum_{d\mid m}\sum_{r\geq 1}\lambda_f\left(r\left(r+\frac{m}{d}\right)\right)h\left(\frac{k-1}{4\pi d\left(r+\frac{m}{2d}\right)}\right)
$$

Which is exactly the right-hand side (RHS) of the equality from the paper!

Addressing your confusion #

You were worried about the $d=1$ term matching the LHS, but that's not how the equality works—we're not canceling terms, we're reorganizing the original sum using the Hecke identity. The LHS sums over all $r$, while each term in the RHS sums over $r$ that are multiples of $d$ (after variable change), with the Hecke identity splitting the product of eigenvalues into a sum over divisors of $\gcd(r,m)$. There's no cancellation involved—just a standard number-theoretic trick to rewrite the sum using divisor relations and Hecke eigenvalue properties.

备注：内容来源于stack exchange，提问作者Steven Creech