Background: Let $k$ be an even integer. The Eisenstein series are defined by $$E_{k} = 1 - \frac{2k}{B_{k}}\sum_{n=1}^{\infty} \sigma_{k-1}(n)q^{n}$$ where $$\sigma_{k-1}(n)= \sum\limits_{d \mid n,\;d \geq 1} d^{k-1}$$ and $B_{k}$ is the $k$-th Bernoulli number. If $k \gt 2$ then $E_{k}$ is a modular form of weight $k$. In the case when $k = 2$, $E_{k}$ is not a modular form. Let $P = E_{2}$. Thus $P = 1 - 24\sum_{n = 1}^{\infty} \sigma_{1}(n)q^{n}$.

Let $\theta$ be the differential operator defined by $\theta = q \frac{d}{dq}$. If $f$ is a modular form of weight $k$, define $\Delta f = 12 \theta f - kPf$. (Note that the symbol $\circ$ means composition.)

Let $$\alpha = \left( \begin{array}{cc} a & b \\ c & d \\ \end{array} \right) \in \textrm{SL}_{2}(\mathbf{R}).$$ Define $$\alpha \tau = \frac{a \tau + b}{c \tau + d}.$$ Define $f \circ [a]_{k}(\tau) = f(\alpha \tau)(c \tau + d)^{-k}$.

My question:

Lang claims (as a lemma on page 161 of his "Introduction to Modular Forms") that $\Delta(f \circ [\alpha]_{k}) = (\Delta f) \circ [\alpha]_{k + 2}$. My question is why is this true?

  • $\begingroup$ I think the formula holds in places other than Lang's book, so I changed the title slightly. (-; $\endgroup$ Nov 22, 2010 at 2:39
  • $\begingroup$ @user3798: when you first say that $\circ$ means composition, you have not yet used the symobl at all; it doesn't show up until the next paragraph. Are you missing it somewhere? $\endgroup$ Nov 22, 2010 at 3:20
  • $\begingroup$ @Arturo Magidin: i'm not missing it anywhere. i just defined it earlier than i needed it. $\endgroup$ Nov 22, 2010 at 3:35

1 Answer 1


Note to begin with that the symbol $\circ$ does not mean composition; rather, it means the operation given by the formula further on in your question, namely: $$f\circ[\alpha]_k(\tau) = (c\tau + d)^k f(\alpha \tau).$$

Now to say that $f$ is a modular form of weight $k$ is to say that $$f\circ[\alpha]_k(\tau) = f(\tau),$$ for all $\alpha \in SL_2(\mathbb Z),$ or equivalently, that $$f(\alpha \tau) = (c\tau + d)^k f(\tau).$$ Differentiating both sides with respect to $\tau,$ we find that $$f'(\alpha \tau) (c\tau + d)^{-2} = (c\tau + d)^k f'(\tau) + c k (c\tau + d)^{k-1}f(\tau).$$ Since $\theta = \dfrac{1}{2\pi i} \dfrac{d}{d\tau},$ we may rewrite this as $$(c\tau + d)^{-(k+2)}(\theta f)(\alpha\tau) = (\theta f)(\tau) + \dfrac{c k}{2 \pi i(c\tau + d)} f(\tau),$$ or equivalently, that $$(\theta f)\circ [\alpha]_{k+2}(\tau) = (\theta f)(\tau) + \dfrac{c k}{c \tau + d} f(\tau).$$

On the other hand, $$(k P f)\circ[\alpha]_{k+2}(\tau) = k (P\circ[\alpha]_2)(f\circ [\alpha]_k)(\tau),$$ and so the formula you ask about is equivalent to the formula $$P\circ [\alpha]_2(\tau) = P(\tau) + \dfrac{12 c}{2 \pi i(c\tau + d)}.$$

This is a standard formula for $P$ (which shows that although $P$ is not a modular form of weight 2, it is not so far off). One way to obtain it is to consider the weight 12 cuspform $\Delta = q\prod_{n = 1}^{\infty}(1 - q^n)^{24}.$ (Sorry for the conflict in notation, but $\Delta$ is the standard notation here. The operator that you have labelled $\Delta$ is often denoted $\delta$, perhaps to avoid precisely this notational conflict.)

A computation (using the product formula for $\Delta$) shows that $\theta(\log \Delta) = P.$ Now writing down the modularity equation $\Delta(\alpha \tau) = (c\tau + d)^{12} \Delta$ and applying $\theta$ to the log of each side, we find that indeed $P$ satisfies the required functional equation, and thus the formula you asked about holds.

As an aside, note that one can also reverse this last argument: suppose that we know that the formula in the question is true. Applying it to the modular form $\Delta$, we find that $\theta \Delta - P \Delta$ is a cuspform of weight $14$, but there are no non-zero such cuspforms. Thus we find that $\theta\Delta = P\Delta,$ or equivalently, that $\theta(\log \Delta) = P.$ If you look in Serre's Course in arithmetic, you will see that this is how he proves the product formula for $\Delta$: he verifies the functional equation for $P$ directly (in the special case when $\alpha = \begin{pmatrix}0 & 1 \\ - 1 & 0\end{pmatrix}$; but since the functional equation is obvious for $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix},$ and since these two matrices together generate $PSL_2({\mathbb Z})$, verifying it for this single choice of $\alpha$ is enough), and then essentially applies the argument I just gave to deduce the product formula for $\Delta$.

Incidentally, I highly recommend Serre's book; as an introduction to modular forms it is both shorter and clearer than Lang's. If you then want to go further with the theory, there are other, better sources. (For example, after reading Serre's book you could try directly reading Atkin and Lehner's paper on newforms.)

  • $\begingroup$ Thanks Matt! I really appreciate it. $\endgroup$ Nov 23, 2010 at 5:08
  • $\begingroup$ Other useful keywords to connect to more contemporary literature are "Rankin-Cohen brackets". $\endgroup$ Jun 25, 2011 at 18:22

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