= Solution
Use the three left-coset representatives $I,S,ST$ for $\Gamma_0(2)\backslash\mathrm{SL}_2(\mathbb Z)$. If a nonzero weight-four <cusp form> $h$ existed, its <coset norm of a modular form>
$$
H=h\,(h|_4S)\,(h|_4ST)
$$
would be a nonzero weight-twelve <modular form> for the full group: right multiplication permutes its factors. At infinity $h$ has order at least one in $q$. The other two factors correspond to the zero <modular cusp> of <cusp width> two and each have order at least $1/2$ in $q$. Thus $H$ has order at least two there.
The ratio $H/\Delta$ is weight zero, holomorphic on the half-plane because $\Delta$ has no zeros there, and holomorphic at the <modular cusp> with value zero. It descends to a <holomorphic function> on the compact full modular curve. Such a function is constant, hence zero, contradicting $H\ne0$. Therefore
$$
\boxed{S_4(\Gamma_0(2))=0.}
$$
Constant terms at the two <modular cusps> define a <linear map> $M_4(\Gamma_0(2))\to\mathbb C^2$ whose kernel is this <modular cusp> space. It is injective, so the dimension is at most two. The forms $E_4(\tau)$ and $E_4(2\tau)$ are holomorphic weight-four forms for this group. For the second, the same conjugation used in 1(c) proves transformation, and $E_4(2S\tau)|$ after the weight factor is $2^{-4}E_4(\tau/2)$, proving holomorphy at zero. Their constant terms are both one but their $q$ coefficients are respectively $240$ and zero, so they are independent. We obtain
$$
\boxed{M_4(\Gamma_0(2))=\mathbb CE_4(\tau)\oplus\mathbb CE_4(2\tau),\qquad\dim M_4(\Gamma_0(2))=2.}
$$
This is the <weight-four Eisenstein basis at level two>.
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