The conjugation operator on newforms

Question

Let $f\in S_k^{\text{new}}(N,\chi)$ be a newform, and let $K$ the conjugation operator defined by $$ (Kf)(z)= \bar{f}(-\bar{z}) $$ Is it true that $Kf\in S_k^{\text{new}}(N,\bar{\chi})$ ?

Answer 1

Yes, it is true.

It is not difficult to prove that $K(f)$ is modular of weight $k$ and level $N$, that it is cuspidal, and that its character is $\bar\chi$; I'm guessing you already know this, since you asked specifically about newforms.

So let me explain why $K(f)$ is new whenever $f$ is. Recall that the space of newforms is defined as the orthogonal complement (with respect to the Petersson product) of the space of oldforms, which is the sum of the images of the degeneracy maps $i_{1, p}$ and $i_{2, p}$ from $S_k(N/p, \chi)$ for each prime divisor $p$ of $N$.

So the argument goes as follows:

• The degeneracy maps $i_{1, p}$ and $i_{2, p}$ commute with the map $K$ (clear from the definitions).
• Hence the space of oldforms is a sum of $K$-invariant subspaces, so it is $K$-invariant.
• The Petersson product is compatible with $K$: an easy computation (substituting $z \mapsto -\bar{z}$ in the integral) shows that $\langle K(f), K(g) \rangle = \overline{\langle f, g \rangle}$.
• Putting these together, we deduce that the space of newforms is preserved by the action of $K$.

This is an instance of a much more general theorem, due to Shimura: if $\sigma$ is any field automorphism of $\mathbf{C}$, and $f = \sum a_n q^n \in S_k(N, \chi)$, then the form $f^\sigma = \sum \sigma(a_n) q^n$ is in $S_k(N, \chi^\sigma)$, and $f^\sigma$ is new if $f$ is. The above argument shows this when $\sigma$ is complex conjugation. However, if $\sigma$ isn't complex conjugation (or the identity) then this is much more difficult to prove.