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For integer $k<0$, the operator $D=(2\pi i{)}^{-1}d/dz=qd/dq$ sends a weight-$k$ level-one weak modular form to a weight-$2-k$ one after $1-k$ iterations. In the derivative transformation of a weak modular form, every lower derivative coefficient then contains a zero factor. This also holds on the subspace $M_k^{!}$ of forms meromorphic at the cusp.

For modular forms $f,g$ of weights $k,\ell$, respectively, the displayed combination cancels the anomaly in derivative transformation of a weak modular form and is a weight-$k+\ell +2$ cusp form. Equivalently, for weights $k,\ell$ in the order $f,g$, a conventional first bracket is $kfDg-\ell gDf$; its negative is equally modular. Differentiation kills each constant Fourier coefficient, proving cusp vanishing.

The required anomaly cancellation depends on the exact Eisenstein series of weight two lawHere is a way to establish it without assuming that $E_2$ is a modular form. The lattice definition of the completed nonholomorphic Eisenstein series in Question 6 is modular invariant. Its Poisson summation calculation, proved there, gives the finite Laurent coefficient at $s=1$ asThe residue is constant, so $C$ is also invariant. This calculation uses only convergent Gaussian integrals and the analytic infinite product, not modularity of that product. Using the Wirtinger derivative ${\partial }_{\tau }$, with ${\partial }_{\tau }y=1/(2i)$, differentiation of the locally uniformly convergent series givesDifferentiate $C(\gamma \tau )=C(\tau )$ to see that $E_2^{\ast }=E_2-3/(\pi y)$ transforms with weight two. Substituting $\Im (\gamma \tau )=y/|j{|}^2$ yieldssince $\overline{j}-j=-2icy$. This is the weight-two transformation from an invariant Eisenstein limit.

Combine this law with the derivative transformation of a weak modular form proved in part (i). The anomalous terms in $Df-(k/12)E_{2}f$ cancel exactly, leaving the weight-$k+2$ transformation law. Both functions are holomorphic on the complex upper half-plane, and their Fourier expansions contain no negative powers. Hence the Serre derivative satisfiesIts constant coefficient is $-k{a}_0(f)/12$; it need not be a cusp form when $f$ is not cuspidal.

Let $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in S{L}_2(\mathbb{Z})$ and $j=c\tau +d$. Differentiating the transformation law of a weight-$k$ modular form gives the derivative transformation of a weak modular formThe corresponding anomaly for $Dg$ has coefficient $\ell c/(2\pi i)$. Consequently these anomalies cancel in $\ell gDf-kfDg$, which transforms with weight $k+\ell +2$. The derivatives are holomorphic on the complex upper half-plane. In a Fourier expansion, $D({\sum }_{n\ge 0}{a}_n{q}^n)={\sum }_{n\ge 1}n{a}_n{q}^n$, so each term in the combination has zero constant coefficient and no negative powers. There is just one modular cusp up to the full modular group. ThereforeThis is, up to the chosen sign, the first Rankin-Cohen bracket.

Use the convention that a weak modular form is a holomorphic function $f$ on the complex upper half-plane satisfyingwith no condition at infinity. In particular $f(z+1)=f(z)$, so it has a Laurent series $f(z)={\sum }_{n\in \mathbb{Z}}{a}_n{q}^n$ on $0<|q|<1$, where $q=e^{2\pi iz}$. It is a modular form when it is holomorphic at a cusp, here the unique level-one cusp at infinity, equivalently $a_n=0$ for $n<0$. It is a cusp form when additionally $a_0=0$. If one uses the narrower weakly holomorphic convention, one also requires finitely many negative terms; the derivative argument below holds in that subspace as well. The matrix $-I$ forces odd-weight forms to be zero.

Put $(a{)}_r=a(a+1)\cdots (a+r-1)$, with $(a{)}_0=1$, andWe prove the derivative transformation of a weak modular form in the uniform formFor $\ell =0$ this is the weight-$k$ transformation under $S$. Differentiating the formula and multiplying by $z^2$, since $d(-1/z)/dz=z^{-2}$, gives the recurrencewhere out-of-range coefficients are zero. The displayed explicit coefficients satisfy it. Indeed, for interior $j$, extract the common polynomial $(k+j{)}_{\ell -j}$; the remaining identity isThis follows from Pascal's identity and $j\left(\genfrac{}{}{0ex}{}{\ell }{j}\right)=(\ell -j+1)\left(\genfrac{}{}{0ex}{}{\ell }{j-1}\right)$. The endpoint coefficients satisfy the recurrence directly. This polynomial proof also applies when a factor is zero, avoiding division by that factor. As $C_{\ell ,\ell }=1$, the result isThe letter $l$ in the printed exponent is the same derivative order $\ell$.

For integer $k<0$, take $\ell =1-k$. If $j<\ell$, the product runs from $k+j\le 0$ through $k+\ell -1=0$, and therefore vanishes. Consequently $f^{(1-k)}$ has weight $k+2(1-k)=2-k$ under $S$. It is also invariant under $T$, by differentiating $f(z+1)=f(z)$, and $S,T$ generate the modular group. Its holomorphy is preserved by differentiation. With $D=(2\pi i{)}^{-1}d/dz=qd/dq$, termwise differentiation of the locally convergent Laurent series gives the Bol identity for modular forms:For negative indices the exponent $1-k$ is a positive integer, and the constant term is killed, so every term is unambiguous.