Rewriting $\tau(p)\Delta(\tau)$ when $p$ is prime

Question

$p$ is a prime, and $\tau$ is Ramanujan's tau function: $$p^{11}\Delta(p\tau)+\frac{1}{p}\sum_{k=0}^{p-1}\Delta\bigg(\frac{\tau + k}{p}\bigg)=p^{11}\Delta(\tau)+\frac{p^{12}}{p}\sum_{k=0}^{p-1}\Delta(\tau) =p^{11}\Delta(\tau)+p^{12}\Delta(\tau)=p^{11}\Delta(\tau)(1+p)$$ I am unclear how the latest expression is equivalent to $\tau(p)\Delta(\tau)$, where $\Delta $ is the modular discriminant.

Answer 1

This follows from the fact that $\Delta$ is an eigenvector for the Hecke operators acting on $M_{12}$. A classical calculation of the effect of Hecke operators at the level of $q$-expansions shows that the Hecke eigenvalues of an eigenform are precisely its Fourier coefficients (more precisely, the $n$-th Hecke eigenvalue is the $n$-th Fourier coefficient). The expression you have written down is $T_p\Delta$, and therefore this equals $\tau(p)\Delta$.