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The new subspace is the orthogonal complement of the old subspace of cusp forms for the Petersson inner product. A form belongs to it precisely when the adjoints of all lower-level oldform by argument dilation maps annihilate it.

Take a nonzero eigenform of positive weight, and put $h(\tau )=f(p\tau )$ and $C=\chi (p)p^{k-1}$. The original form belongs to level $Np$ by inclusion of groups. For $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in {\Gamma }_0(Np)$, the conjugate $\left(\begin{array}{cc}a& pb\\ c/p& d\end{array}\right)$ lies in ${\Gamma }_0(N)$, soIts Dirichlet character is therefore the specified reduced Dirichlet character; modular cusp holomorphy is permitted in the question. This is the Dirichlet character version of an oldform by argument dilation.

A nonzero positive-weight modular form cannot be constant, because the matrix $\left(\begin{array}{cc}1& 0\\ N& 1\end{array}\right)$ would force a nonzero constant to equal $(N\tau +1{)}^k$ times itself. Let $n_0>0$ be its first nonzero positive Fourier index. In a relation $Af+Bh=0$, the $q^{n_0}$ coefficient forces $A=0$, then $B=0$. Hence their span is two-dimensional, even when the original constant term is nonzero.

At the new level $p$ is a bad prime, so its operator is $U_p$. The good-prime eigenrelation at the old level and part (c) giveThus the matrix in the ordered basis $(f,h)$ iswith characteristic polynomial $X^2-\lambda X+C$. For its distinct roots,This is prime stabilization of an oldform.

The nonzero positive-weight qualification is necessary for the two-dimensional assertion. The zero form gives no such span, and if weight zero is allowed, $f=1$ at trivial Dirichlet character has distinct good-prime roots $1$ and $p^{-1}$, while $f(\tau )=f(p\tau )$ spans only one dimension. The asserted result uses the usual nonzero positive-weight eigenform setting.

A modular form of integer weight $k$ for a finite-index subgroup $\Gamma \le S{L}_2(\mathbb{Z})$ is a holomorphic function on the complex upper half-plane satisfying $f{|}_k\gamma =f$ for $\gamma \in \Gamma$ and holomorphic at every cusp of a modular group. Explicitly, for each $\sigma \in S{L}_2(\mathbb{Z})$ choose a positive integer $h_{\sigma }$ with $\sigma T^{h_{\sigma }}{\sigma }^{-1}\in \Gamma$. Then $f{|}_k\sigma$ is periodic with period $h_{\sigma }$, and its expansion in $q_{h_{\sigma }}=e^{2\pi iz/h_{\sigma }}$ must have no negative powers. Equivalently it must be bounded as $\mathrm{Im}z\to \infty$, uniformly in one period. This definition applies to noncongruence subgroups too. If $-I\in \Gamma$, nonzero forms require even weight.

Put $d=[S{L}_2(\mathbb{Z}):\Gamma ]$ and let $f\ne 0$. Form the product over left cosetsChanging a representative does not affect a factor. Right multiplication permutes the cosets, so the product is a nonzero level-one modular form of weight $kd$. It is holomorphic on the upper half-plane and at infinity, since every factor is bounded there by cusp holomorphy.

Let $h$ be the least positive integer with $T^h\in \Gamma$, and write $f={\sum }_{n\ge 0}{b}_n{q}_h^n$. If $r$ is its first nonzero index, the $h$ distinct cosets $\Gamma T^j$, $0\le j<h$, give the factors $f(z+j)$. Their product has leading term a nonzero constant times $e^{2\pi irz}=q^r$. All remaining factors are bounded at infinity. Hence $F(z)=O(e^{-2\pi r\mathrm{Im}z})$, so its level-one order satisfies $v_{\infty }(F)\ge r$. Notice that using just the single factor $f$ would only give $r/h$; the entire translation orbit is needed.

The permitted valence formula for the modular group bounds $v_{\infty }(F)\le kd/12$, because all its other weighted zero orders are nonnegative. Thus $r\le kd/12$. With $R=\lfloor kd/12\rfloor$, the map $f\mapsto (b_0,\dots ,b_R)$ is injective: a nonzero form in its kernel would have $r\ge R+1$. This proves the dimension bound for modular forms on a finite-index subgroup:

For the dilation assertion, the original PDF states $N_{1}D\mid N$, a product-divisibility condition, rather than the TeX's chain $N_1\mid D\mid N$. Use the PDF condition. Put $A=\mathrm{diag}(D,1)$. If $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in {\Gamma }_0(N)$, thensince $N_{1}D\mid c$. Therefore, for $g(z)=f(Dz)$, either direct substitution or the slash operator for modular forms gives $g{|}_k\gamma =g$. It is holomorphic on the upper half-plane.

To verify every cusp, take $\sigma \in S{L}_2(\mathbb{Z})$ and choose $\tau \in S{L}_2(\mathbb{Z})$ taking infinity to the rational cusp $A\sigma \infty$. Then $B={\tau }^{-1}A\sigma$ is rational upper triangular, with positive ratio of its two diagonal entries. Consequentlyis a nonzero constant factor times $f{|}_k\tau$ evaluated at an affine map $z\mapsto uz+v$ with $u>0$. The imaginary part tends to infinity, where $f{|}_k\tau$ is bounded. Thus $g{|}_k\sigma$ is bounded too. Together with its periodicity this proves cusp holomorphy by the Riemann removable singularity theorem. Hence the oldform by argument dilation satisfies

Finally ${\Gamma }_0(3)$ has index four: reduction modulo three gives a transitive $S{L}_2({\mathbb{F}}_3)$ action on the four points of ${\mathbb{P}}^1({\mathbb{F}}_3)$, and ${\Gamma }_0(3)$ is the inverse image of the stabilizer of infinity. Reduction is surjective, since elementary determinant-one matrices generate $S{L}_2({\mathbb{F}}_3)$ and lift integrally. The dimension bound gives $\dim M_4({\Gamma }_0(3))\le 2$.

The normalized Eisenstein series $E_4(z)$ and the oldform $E_4(3z)$ both belong to this space. Their Fourier expansion of a normalized Eisenstein series starts asA linear relation between them has zero constant term only if its two scalars sum to zero; its $q$ coefficient then forces both scalars to vanish. They are independent, proving the weight-four modular forms on Gamma 0 3 description

The group ${\Gamma }_0(3)$ has index four. The dimension bound for modular forms on a finite-index subgroup gives dimension at most two in weight four. The Eisenstein series $E_4(z)$ and its oldform by argument dilation $E_4(3z)$ are independent, since their constant coefficients agree while their $q$ coefficients are $240$ and zero.