4 Hecke Theory

4.1 Double coset operators

Let \(\Gamma_1\) and \(\Gamma_2\) be two congruence subgroups, and let \(\alpha\in\operatorname{GL}_2^+({\mathbb{Q}})\).

Definition 4.1 The \(\Gamma_1\alpha\Gamma_2\) is the set \[\Gamma_1\alpha\Gamma_2=\{\gamma_1\alpha\gamma_2 ~|~ \gamma_1\in\Gamma_1,\gamma_2\in\Gamma_2\}.\]

Multiplication gives a left action of \(\Gamma_1\) on \(\Gamma_1\alpha\Gamma_2\) and another right action of \(\Gamma_2\). Consider a decomposition of this double coset into (disjoint) orbits: \[\Gamma_1\alpha\Gamma_2=\cup \Gamma_1 \beta_j.\]

Lemma 4.1

If \(\Gamma\) is a congruence subgroup and \(\alpha\in\operatorname{GL}_2^+({\mathbb{Q}})\), then \(\alpha^{-1}\Gamma\alpha\cap {\operatorname{SL}}_2({\mathbb{Z}})\) is also congruence subgroup.
Any two congruence subgroups \(\Gamma_1\), \(\Gamma_2\) are . That is, \[[\Gamma_1\colon \Gamma_1\cap\Gamma_2]<\infty\quad\text{ and }\quad [\Gamma_2\colon \Gamma_1\cap\Gamma_2]<\infty.\]

Proof. Let \(N\) be a positive integer such that \(\Gamma(N)\subseteq \Gamma\), \(N\alpha\in M_2({\mathbb{Z}})\) and \(N\alpha^{-1}\in M_2({\mathbb{Z}})\). Set \(M=N^3\). Then one can check that \(\alpha\Gamma(M)\alpha^{-1} \subseteq \Gamma(N)\), which implies that \(\Gamma(M)\subseteq \alpha^{-1}\Gamma\alpha\). Since \(\Gamma(M)\) is also contained in \({\operatorname{SL}}_2({\mathbb{Z}})\), we are done with the first statement.

For the second assertion, just note that there is some \(M\) such that \(\Gamma(M)\subseteq \Gamma_1\cap\Gamma_2\). Therefore the indices to compute are bounded above by \([{\operatorname{SL}}_2({\mathbb{Z}})\colon\Gamma(M)]\), which is finite.

Proposition 4.1 Let \(\Gamma_1\) and \(\Gamma_2\) be two congruence subgroups, and let \(\alpha\in\operatorname{GL}_2^+({\mathbb{Q}})\). Set \(\Gamma_3\) to be the congruence subgroup: \[\Gamma_3 = (\alpha^{-1}\Gamma_1\alpha)\cap \Gamma_2.\] The map \(\gamma_2\mapsto \Gamma_1\alpha\gamma_2\) induces a bijection \[\Gamma_3\backslash \Gamma_2 \cong \Gamma_1\backslash \Gamma_1\alpha\Gamma_2.\]

Proof. Consider the map \[\Gamma_2\longrightarrow\Gamma_1\backslash (\Gamma_1\alpha\Gamma_2),\quad \gamma_2\mapsto \Gamma_1\alpha\gamma_2.\] It is clearly surjective. Moreover, two elements \(\gamma_2\) and \(\gamma_2'\) get mapped to the same orbit if and only if: \[\Gamma_1\alpha\gamma_2=\Gamma_1\alpha\gamma_2'\iff \gamma_2'\gamma_2^{-1}\in\alpha^{-1}\Gamma_1\alpha,\] and the latter happens if and only if \(\gamma_2\) and \(\gamma_2'\) are in the same coset for \((\alpha^{-1}\Gamma_1\alpha) \cap\Gamma_2=\Gamma_3\).

Corollary 4.1 Let \(\Gamma_2=\cup \Gamma_3\gamma_j\) be a coset decomposition of \(\Gamma_3\backslash\Gamma_2\). Then \[\Gamma_1\alpha\Gamma_2=\cup \Gamma_1\alpha\gamma_j\] is an orbit decomposition. In particular, the number of orbits of \(\Gamma_1\alpha\Gamma_2\) under the action of \(\Gamma_1\) is finite.

Let \(f\in M_k(\Gamma_1)\) be a modular form of weight \(k\) for a congruence subgroup \(\Gamma_1\). Let \(\Gamma_1\alpha\Gamma_2\) be a double coset, where \(\Gamma_2\) is a congruence subgroup and \(\alpha\in\operatorname{GL}_2^+({\mathbb{Q}})\). The on \(f\) is defined as: \[f|_k (\Gamma_1\alpha\Gamma_2) = \sum f|_k \beta_j,\] if \(\Gamma_1\alpha\Gamma_2=\cup \Gamma_1\beta_j\) is any orbit decomposition.

The action is well defined, independent of the choice of the \(\beta_j\). This is so because \(f\) is \(k\)-invariant under \(\Gamma_1\).

The next goal is to show that the double coset operator maps \(M_k(\Gamma_1)\) to \(M_k(\Gamma_2)\) and preserves cusps forms. We will need a technical lemma to treat the cusp conditions.

Lemma 4.2 Suppose that for all \(\gamma\in{\operatorname{SL}}_2({\mathbb{Z}})\) the function \(f|_k\gamma\) has an expansion of the form \[\sum_{n\geq n_0} a(n)q_N^n,\] with \(n_0\) and \(a(n)\) depending on \(\gamma\). Let \(\alpha\in\operatorname{GL}_2^+({\mathbb{Q}})\). Then for all \(\gamma\in{\operatorname{SL}}_2({\mathbb{Z}})\) the function \(f|_k(\alpha\gamma)\) has the expansion \[\sum_{n\geq an_0} b(n)q_{Nd}^n,\] where \(a\) and \(d\) are positive integers depending only on \(\alpha\).

Proof. First, note that, for \(a>0\), \[f|_k\left(\begin{matrix}a&0\\0&a\end{matrix}\right) = a^{2(k-1)}a^{-k}f = a^{k-2}f.\] So without loss of generality we may assume that \(\alpha\in M_2({\mathbb{Z}})\). Let \(\gamma_0\in{\operatorname{SL}}_2({\mathbb{Z}})\) be such that \(\gamma_0^{-1}\alpha=\left(\begin{smallmatrix}a&b\\0&d\end{smallmatrix}\right)\) (upper-triangular), with \(a\) and \(d\) being positive integers. Then: \[\begin{align*} f|_k\alpha &= (f|_k\gamma_0)|_k\left(\begin{matrix}a&b\\0&d\end{matrix}\right) = \left(\sum_{n\geq n_0}a(n)e^{\frac{2\pi i n z}{N}}\right)|_k\left(\begin{matrix}a&b\\0&d\end{matrix}\right)\\ &=(\cdots)\sum_{n\geq n_0} a(n)e^{\frac{2\pi i n(az+b)}{dN}} = (\cdots) q_{Nd}^{an_0}+\cdots \end{align*}\] This concludes the proof.

Proposition 4.2 Let \(\Gamma_1\) and \(\Gamma_2\) be two congruence subgroups, and let \(\alpha\in\operatorname{GL}_2^+({\mathbb{Q}})\). The rule \(f\mapsto f|_k\Gamma_1\alpha\Gamma_2\) induces a map \(M_k(\Gamma_1)\longrightarrow M_k(\Gamma_2)\).

Proof. Write \(\Gamma_3=(\alpha^{-1}\Gamma_1\alpha)\cap \Gamma_2\), and consider a coset decomposition \(\Gamma_2= \cup \Gamma_3\gamma_j\). One can take as set of representatives \(\beta_j=\alpha\gamma_j\). If \(\gamma_2\in\Gamma_2\), then \(\{\gamma_j\gamma_2\}_j\) is a complete set of representatives for \(\Gamma_3\backslash \Gamma_2\), and hence \(\{\alpha\gamma_j\gamma_2\}_j\) is a complete set of representatives for \(\Gamma_1\backslash\Gamma_1\alpha\Gamma_2\). This implies that \(f|_k\Gamma_1\alpha\Gamma_2\) is \(k\)-invariant for \(\Gamma_2\).

If \(f\) is holomorphic on \({\mathbb{H}}\), then \(f|_k\beta_j\) is holomorphic on \({\mathbb{H}}\) for any \(\beta_j\in\operatorname{GL}_2^+({\mathbb{Q}})\), so it only remains to check the cusp conditions. But Lemma 4.2 precisely ensures that these are preserved.

4.1.1 First examples

Consider the case \(\Gamma_2\subseteq \Gamma_1\) and \(\alpha=1\). Then \(\Gamma_1\alpha\Gamma_2=\Gamma_1\), and \(\Gamma_1=\Gamma_11\) is an orbit decomposition. Therefore \(f|_k\Gamma_1\alpha\Gamma_2 = f|_k1 = f\). This just says that \(M_k(\Gamma_1)\) is a subspace of \(M_k(\Gamma_2)\).

As a more interesting example, given \(\alpha\in\operatorname{GL}_2^+({\mathbb{Q}})\) consider the conjugate \(\Gamma'=\alpha^{-1}\Gamma\alpha\). Then \(\Gamma\alpha\Gamma'=\Gamma\alpha\) is an orbit decomposition. This implies that acting by \(\alpha\) induces a map \[M_k(\Gamma)\longrightarrow M_k(\alpha^{-1}\Gamma\alpha).\] Since the inverse of this map is given by the action of \(\alpha^{-1}\), we conclude that \(M_k(\Gamma)\) and \(M_k(\alpha^{-1}\Gamma\alpha)\) are naturally isomorphic.

Finally, consider the case \(\Gamma_1\subseteq \Gamma_2\) and \(\alpha=1\). Then \(\Gamma_1\alpha\Gamma_2 = \cup \Gamma_1 \beta_j\), where \(\beta_j\) is a set of coset representatives for \(\Gamma_1\backslash \Gamma_2\). The map \[f\mapsto \sum_{j} f|_k \beta_j\] is to be seen as a trace operator from \(M_k(\Gamma_1)\longrightarrow M_k(\Gamma_2)\). In particular, it maps \(f\in M_k(\Gamma_2)\) to \([\Gamma_2\colon \Gamma_1]f\) and thus it is surjective.

4.2 Hecke operators for \(\Gamma_1(N)\)

Fix now \(\Gamma=\Gamma_1(N)\). We will describe the Hecke operators for the group \(\Gamma\).

4.2.1 The \(T_p\) operators

Let \(p\) be a prime. The at \(p\) is defined as: \[T_p f = f|_k\Gamma_1(N)\left(\begin{matrix}1&0\\0&p\end{matrix}\right)\Gamma_1(N).\] By Proposition 4.2, the operator \(T_p\) acts on \(M_k(\Gamma_1(N))\).

In order to describe the action of \(T_p\) more precisely, we need to understand the double coset \(\Gamma_1(N)\left(\begin{smallmatrix}1&0\\0&p\end{smallmatrix}\right)\Gamma_1(N)\). Note first that if \(\gamma\in \Gamma_1(N)\left(\begin{smallmatrix}1&0\\0&p\end{smallmatrix}\right)\Gamma_1(N)\) then:

\(\det \gamma = p\), and
\(\gamma\equiv\left(\begin{smallmatrix}1&*\\0&p\end{smallmatrix}\right)\pmod N\).

In fact, the converse is true:

Lemma 4.3 We have that \[\Gamma_1(N)\left(\begin{matrix}1&0\\0&p\end{matrix}\right)\Gamma_1(N)=\left\{\gamma\in M_2({\mathbb{Z}})~|~\det\gamma = p,\gamma\equiv \left(\begin{smallmatrix}1&*\\0&p\end{smallmatrix}\right)\pmod{N}\right\}.\]

Proof. To prove the remaining inclusion, let \(\gamma\in M_2({\mathbb{Z}})\) have determinant \(p\), and satisfy \(\gamma\equiv \left(\begin{smallmatrix}1&*\\0&p\end{smallmatrix}\right)\pmod N\). Consider \(L={\mathbb{Z}}^2\) and \[L_0=L_0(N)=\{\left(\begin{smallmatrix}x\\y\end{smallmatrix}\right)\in L\colon y\equiv 0\pmod N\}.\] Note that \(\gamma L_0\subseteq L_0\). Since \(\det \gamma=p>0\), we have: \[[L\colon\gamma L_0]=[L\colon L_0][L_0\colon\gamma L_0] = Np.\] Choose a basis of \(L\) adapted to \(\gamma L_0\). That is, a basis \(u\), \(v\) such that \(\det(u|v)=1\) and such that \[\gamma L_0=mu{\mathbb{Z}}\oplus nv{\mathbb{Z}}, \text{ with } 0<m\mid n, mn=Np.\] We will show:

\(\gamma L_0=u{\mathbb{Z}}\oplus Np v{\mathbb{Z}}\).
\(L_0 = u{\mathbb{Z}}\oplus Nv{\mathbb{Z}}\).
\(\gamma L = u{\mathbb{Z}}+pv{\mathbb{Z}}\).

In fact, since \(\gamma\left(\begin{smallmatrix}1\\0\end{smallmatrix}\right)\in \gamma L_0\), we have that \(\left(\begin{smallmatrix}a\\b\end{smallmatrix}\right)\equiv\left(\begin{smallmatrix}0\\0\end{smallmatrix}\right)\pmod m\). Since \(\gcd(a,N)=1\), this implies that \(\gcd(m,N)=1\). Now, if \(p\mid m\), then \(p\mid n\), and so \(p^2\mid mn=Np\). Therefore \(p\mid N\), which is a contradiction with \(\gcd(m,N)=1\). Therefore \(p\nmid m\) and hence \(m=1\) and \(n=Np\).

The two other facts follow because \(L_0\subset L\) is a subgroup of index \(N\), and \(\gamma L\subset L\) is of index \(p\) in \(L\). This proves the above three statements.

Next, set \(\gamma_1=(u|v)\), which belongs to \(\Gamma_0(N)\) because \(u\) belongs to \(L_0\). Set also \(\gamma_2=(\gamma_1\left(\begin{smallmatrix}1&0\\0&p\end{smallmatrix}\right))^{-1}\gamma\), so that \(\gamma=\gamma_1\left(\begin{smallmatrix}1&0\\0&p\end{smallmatrix}\right)\gamma_2\). Note that \(\gamma_2\) belongs to \({\operatorname{SL}}_2({\mathbb{Q}})\). It remains to show that \(\gamma_1\) and \(\gamma_2\) belong to \(\Gamma_1(N)\). This will follow if we can prove:

\(\gamma_2\in \Gamma_0(N)\).
\(\Gamma_0(N)\left(\begin{smallmatrix}1&0\\0&p\end{smallmatrix}\right) \Gamma_0(N)=\Gamma_1(N)\left(\begin{smallmatrix}1&0\\0&p\end{smallmatrix}\right)\Gamma_0(N)\).
If \(\gamma=\gamma_1\left(\begin{smallmatrix}1&0\\0&p\end{smallmatrix}\right)\gamma_2\) with \(\gamma_1\in \Gamma_1(N)\) and \(\gamma_2\in \Gamma_0(N)\), then \(\gamma_2\) belongs to \(\Gamma_1(N)\).

Each of these statements can be easily proved, and we ommit these proofs.

Proposition 4.3 Let \(f\in M_k(\Gamma_1(N))\). Then \(T_p f\) is given by: \[T_p f = \begin{cases} \sum_{j=0}^{p-1} f|_k \left(\begin{matrix}1&j\\0&p\end{matrix}\right)&p\mid N,\\ \sum_{j=0}^{p-1} f|_k\left(\begin{matrix}1&j\\0&p\end{matrix}\right) +f|_k\left(\begin{matrix}mp&n\\Np&p\end{matrix}\right)&p\nmid N. \end{cases}\] Here the matrix \(\left(\begin{smallmatrix}mp&n\\Np&p\end{smallmatrix}\right)\) is chosen such that \(\gamma_\infty=\left(\begin{smallmatrix}mp&n\\N&1\end{smallmatrix}\right)\) belongs to \(\Gamma_1(N)\).

Proof. We just need to trace the definition of the double coset operator. That is, we need to find an explicit coset decomposition of \(\Gamma_3\backslash \Gamma_1(N)\), where \[\Gamma_3={\left(\begin{matrix}1&0\\0&p\end{matrix}\right)}^{-1} \Gamma_1(N) \left(\begin{matrix}1&0\\0&p\end{matrix}\right) \cap \Gamma_1(N).\] Define \(\Gamma^0(p)\) to be the group of matrices which are lower triangular modulo \(p\). It is easy to see that \[\Gamma_3 = \Gamma_1(N)\cap \Gamma^0(p).\] Consider the matrices \(\gamma_j=\left(\begin{smallmatrix}1&j\\0&1\end{smallmatrix}\right)\), with \(j\) ranging from \(0\) to \(p-1\) inclusive. These are all distinct modulo \(\Gamma_1(N)\cap \Gamma^0(p)\) (check it). Given any matrix \(\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right) \in\Gamma_1(N)\), note that \[\left(\begin{matrix}a&b\\c&d\end{matrix}\right) \left(\begin{matrix}1&-j\\0&1\end{matrix}\right) = \left(\begin{matrix}a&-aj+b\\c&-cj+d\end{matrix}\right).\] Therefore if \(p\nmid a\) we can make the right-hand side to belong to \(\Gamma^0(p)\) for some \(j\). This means that if \(p\) divides \(N\) then \(p\) will not divide \(a\) (because of the determinant condition), and thus the set \(\{\gamma_j\}\) is a complete set of representatives. If \(p\nmid N\), we need to consider matrices \(\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)\) with \(p\mid a\). Choose some \(\gamma_\infty=\left(\begin{smallmatrix}mp&n\\N&1\end{smallmatrix}\right)\in\Gamma_1(N)\). Then: \[\left(\begin{matrix}a&b\\c&d\end{matrix}\right) \gamma_\infty^{-1} = \left(\begin{matrix}*&-na+bmp\\0&*\end{matrix}\right).\] Since \(p\) divides \(-na+bmp\), the matrix \(\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)\) is in the coset of \(\gamma_\infty\) modulo \(\Gamma^0(p)\). Hence \(\{\gamma_j\}\cup \{\gamma_\infty\}\) forms a complete set of representatives. In order to get the orbit representatives for the double coset, we just need to multiply the \(\gamma_j\) by the fixed element \(\alpha = \left(\begin{smallmatrix}1&0\\0&p\end{smallmatrix}\right)\).

4.2.2 The diamond \(\langle d \rangle\) operators

We define another (finite) set of operators on \(M_k(\Gamma_1(N))\), called the diamond operators. First we need some preliminaries on characters.

Definition 4.2 A modulo \(N\) is a group homomorphism \[\chi\colon ({\mathbb{Z}}/N{\mathbb{Z}})^\times\longrightarrow{\mathbb{C}}^\times.\]

It can be extended to a map \(\chi\colon {\mathbb{Z}}\longrightarrow{\mathbb{C}}\) by the recipe \[\chi(d) = \begin{cases} \chi(d\bmod N)&(d,N)=1\\ 0&(d,N)\neq 1. \end{cases}\] The resulting function is totally multiplicative: it satisfies \[\chi(d_1d_2)=\chi(d_1)\chi(d_2)\quad \forall d_1,d_2\in{\mathbb{Z}}.\] Consider the map \(\Gamma_0(N)\longrightarrow{\mathbb{Z}}/N{\mathbb{Z}}^\times\) sending a matrix \(\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)\) to \(d\bmod N\). Its kernel is precisely \(\Gamma_1(N)\), and therefore we obtain an isomorphism \[\Gamma_0(N)/\Gamma_1(N)\cong ({\mathbb{Z}}/NZ)^\times,\quad \left(\begin{matrix}a&b\\c&d\end{matrix}\right) \Gamma_1(N)\mapsto d\bmod N.\]

Definition 4.3 Given \(d\in {\mathbb{Z}}\) coprime to \(N\), the \(\langle d\rangle\) is the operator on \(M_k(\Gamma_1(N))\) defined as \[\langle d \rangle f = f|_k \left(\begin{matrix}a&b\\c&d'\end{matrix}\right),\] where \(a\), \(b\), \(c\), \(d'\) are chosen so that \(\left(\begin{smallmatrix}a&b\\c&d'\end{smallmatrix}\right)\) belongs to \(\Gamma_0(N)\) and \(d'\equiv d\pmod N\).

Note that the above is well defined, and only depends on the class of \(d\) modulo \(N\). This is precisely because \(\Gamma_0(N)/\Gamma_1(N)\cong ({\mathbb{Z}}/NZ)^\times\). The operator \(\langle d\rangle\) is a linear invertible map, and thus it makes sense to look at its eigenspaces.

Definition 4.4 The space of modular forms with character \(\chi\) is \[M_k(\Gamma_0(N),\chi)=\{f\in M_k(\Gamma_1(N))~|~ f|_k\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right) = \chi(d) f,\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)\in \Gamma_0(N)\}.\] The \(\chi\) is defined similarly and written \(S_k(\Gamma_0(N),\chi)\).

Note that \(M_k(\Gamma_0(N),\chi)\) can also be defined as \[M_k(\Gamma_0(N),\chi)=\{f\in M_k(\Gamma_1(N))~|~ \langle d\rangle f = \chi(d) f,\quad d\in ({\mathbb{Z}}/N{\mathbb{Z}})^\times\}.\]

Theorem 4.1 There is a decomposition of \({\mathbb{C}}\)-vector spaces \[M_k(\Gamma_1(N))=\bigoplus_{\chi\bmod N} M_k(\Gamma_0(N),\chi),\] where the sum runs over the \(\phi(N)=\#({\mathbb{Z}}/N{\mathbb{Z}})^\times\) Dirichlet characters modulo \(N\).

Proof. Picking a basis of \(M_k(\Gamma_1(N))\), we get a representation \[\rho\colon ({\mathbb{Z}}/N{\mathbb{Z}})^\times\longrightarrow\operatorname{GL}_n({\mathbb{C}}),\quad \rho(d)=\langle d\rangle,\] where \(n\) is the dimension of \(M_k(\Gamma_1(N))\). Since \(({\mathbb{Z}}/N{\mathbb{Z}})^\times\) is abelian, the representation \(\rho\) decomposes as a sum of irreducible representations, which are necessarily one-dimensional. This means that we can pick a basis for \(M_k(\Gamma_1(N))\) such that \[\rho(d) = \operatorname{diag}(\chi_1(d),\ldots,\chi_n(d)).\] This means that \(\langle d\rangle\) acts as \(\chi_i(d)\) on the \(i\)th component. One just needs to collect then the repeated \(\chi\) to form \(M_k(\Gamma_0(N),\chi)\).

4.2.3 Hecke operators on \(q\)-expansions

In order to study the action of Hecke operators on \(q\)-expansion, we introduce two simple operators: if \(f=\sum a_n q^n\), define: \[U_p f = \sum a_{np} q^n=\sum a_n q^{n/p}.\] The second equality is an abuse of notation: we define \(q^{n/p}=0\) if \(p\nmid n\). We define also: \[V_p f = f(pz) = \sum a_n q^{np} = \sum a_{n/p} q^n.\]

Lemma 4.4 If \(f=\sum a_n q^n\), then

\[U_p f = \frac 1p \sum_{j=0}^{p-1} f\left(\frac{z+j}{p}\right) = \sum_{j=0}^{p-1} f|_k \left(\begin{matrix}1&j\\0&p\end{matrix}\right).\]
\[V_p f= p^{1-k} f|_k\left(\begin{matrix}p&0\\0&1\end{matrix}\right).\]

Proof. Note that if \(\zeta_p=e^{\frac{2\pi i}{p}}\) is a primitive \(p\)th root of unity, then \[\sum_{j=0}^{p-1} \zeta_p^{nj} = \begin{cases} p&p\mid n\\ 0&p\nmid n. \end{cases}\] Now compute: \[\sum_{j=0}^{p-1} f|_k \left(\begin{matrix}1&j\\0&p\end{matrix}\right) = p^{k-1} p^{-k}\sum_j f\left(\frac{z+j}{p}\right).\] Since \(f\) is \(1\)-periodic, this is the same as: \[\frac 1p \sum_j \sum_n a_n e^{2\pi i \frac{z+j}{p}} = \sum_n a_n e^{\frac{2\pi i n z}{p}} \frac 1p \sum_j \zeta_p^{nj}.\] This proves the first statement. The second statement is clear.

Putting together what we have seen so far, we get a description of \(T_p\) in terms of \(U_p\), \(V_p\) and the diamond operators.

Theorem 4.2 We have: \[T_p f = \begin{cases} U_p f&p\mid N,\\ U_p f + p^{k-1} V_p\langle p\rangle f&p\nmid N. \end{cases}\]

Corollary 4.2 If \(f\in M_k(\Gamma_0(N),\chi)\) then for all \(p\) we have: \[T_p f= U_p f +\chi(p)p^{k-1}V_p f.\] In particular, if \(f\in M_k(\Gamma_0(N))\) then: \[T_p f = \begin{cases} U_p f&p\mid N,\\ U_p f +p^{k-1}V_pf& p\nmid N. \end{cases}\]

Moreover, the relation between \(U_p\) and \(T_P\) allows us to think of \(U_p\) as an operator on modular forms, which possibly raises the level.

Corollary 4.3

If \(p\mid N\) then \(U_p\) maps \(M_k(\Gamma_1(N))\) to itself.
If \(p\nmid N\) then \(U_p\) maps \(M_k(\Gamma_1(N))\) to \(M_k(\Gamma_1(Np))\).

Example 4.1 Consider the Eisenstein series \[E_k(z)=1-\frac{2k}{B_k}\sum_{n=1}^\infty \sigma_{k-1}(n)q^n\in M_k(\Gamma_1(1)).\]

Proposition 4.4 We have: \[T_p E_k = \sigma_{k-1}(p) E_k = (1+p^{k-1})E_k.\] That is, \(E_k\) is an eigenform for all \(T_p\), with eigenvalue \(\sigma_{k-1}(p)\).

Proof. In general we have seen that, since \(E_k\in M_k(\Gamma_0(1))\)), \[a_n(T_pf)= a_n(U_p f) + p^{k-1} a_n(V_p f) = a_{np}(f)+p^{k-1} a_{n/p}(f).\] So \[a_0(T_pE_k) = a_0(E_k) + p^{k-1} a_0(E_k) = \sigma_{k-1}(p) a_0(E_k).\] For \(n\geq 1\), we get \[a_n(T_pE_k) = \frac{-2k}{B_k}\left(\sigma_{k-1}(np)+p^{k-1}\sigma_{k-1}(n/p)\right),\] where we understand that \(\sigma_{k-1}(n/p)=0\) if \(p\nmid n\). We claim that: \[\sigma_{k-1}(pn) + p^{k-1}\sigma_{k-1}(n/p) = \sigma_{k-1}(p)\sigma_{k-1}(n),\quad \forall n\geq 1.\] When \(p\nmid n\), this is just the multiplicativity of \(\sigma_{k-1}\). If \(p\mid n\), write \(n=p^em\) with \(p\nmid m\). Then we need to show that for all \(e\geq 1\) \[\sigma_{k-1}(p^{e+1}m) + p^{k-1}\sigma_{k-1}(p^{e-1}m) = \sigma_{k-1}(p)\sigma_{k-1}(p^em).\] This follows easily by dividing both sides by \(\sigma_{k-1}(m)\), which is a common factor of both sides of the equation again by multiplicativity of \(\sigma_{k-1}\).

If \(f=1+\sum_{n\geq 1} a_n q^n\) is a modular form for \({\operatorname{SL}}_2({\mathbb{Z}})\) of weight \(k\) and it is an eigenform for \(T_p\), then the eigenvalue must be \(\sigma_{k-1}(p)\), by the first calculation of the above proof. The real content of the proposition is thus that \(E_k\) is actually an eigenform.

4.3 The Hecke algebra

Definition 4.5 Let \(N\geq 1\) and \(k\in {\mathbb{Z}}\). The acting on \(M_k(\Gamma_1(N))\) is the \({\mathbb{C}}\)-subalgebra of \({\operatorname{End}}_{\mathbb{C}}M_k(\Gamma_1(N))\) generated by \[\Big\langle T_p\colon p\text{ prime};\text{ and } \langle d\rangle \colon d\in ({\mathbb{Z}}/N{\mathbb{Z}})^\times\Big\rangle.\] The Hecke algebra is denoted by \({\mathbb{T}}(M_k(\Gamma_1(N)))\). Similarly we define \({\mathbb{T}}(S_k(\Gamma_1(N)))\) as a subalgebra of \({\operatorname{End}}_{\mathbb{C}}S_k(\Gamma_1(N))\).

Theorem 4.3 For every \(N\geq 1\) the Hecke algebra \({\mathbb{T}}(M_k(\Gamma_1(N)))\) is commutative.

Proof. We must show that for all primes \(p\), \(q\) and all elements \(e\) and \(d\) of \(({\mathbb{Z}}/N{\mathbb{Z}})^\times\) we have:

\(\langle d\rangle T_p = T_p \langle d\rangle\),
\(\langle d \rangle \langle e\rangle = \langle e\rangle\langle d\rangle\), and
\(T_p T_q = T_q T_p\).

First we show (2) and (3) assuming (1). Note that (1) means that \(T_p\) preserves the spaces \(M_k(\Gamma_0(N),\chi)\) and so it’s enough to check (2) and (3) for forms \(f\in M_k(\Gamma_0(N),\chi)\). This makes (2) obvious. As for (3), we can use the \(q\)-expansions: if \(f=\sum a_n q^n\), then \[a_n(T_p f)= a_{pn}(f) + \chi(p)p^{k-1} a_{n/p}(f).\] Then: \[\begin{align*} a_n(T_pT_qf) &= a_{pn}(T_qf) + \chi(p)p^{k-1}a_{n/p}(T_qf)\\ &=a_{pqn}(f) + \chi(q) q^{k-1} a_{pn/q}(f)+\chi(p)p^{k-1}(a_{nq/p}(f)+\chi(q) q^{k-1} a_{n/(pq)}(f)). \end{align*}\] This formula is symmetric in \(p\) and \(q\) so we are done.

Finally, to prove (1) we must write \(\langle d\rangle\) as a double coset. Let \(\gamma\equiv \left(\begin{smallmatrix}*&*\\0&d\end{smallmatrix}\right)\pmod N\). Write \(\Gamma=\Gamma_1(N)\). Then, since \(\Gamma\) is normal in \(\Gamma_0(N)\), we have \[\Gamma\gamma\Gamma=\Gamma\gamma,\] and thus \(\langle d\rangle f = f|_k \gamma\). We want to show that \(\langle d\rangle^{-1}T_p\langle d\rangle = T_p\). Write \(\Gamma\alpha\Gamma=\bigcup_j \Gamma \beta_j\) for the orbit decomposition of the double coset corresponding to \(T_p\). We thus need to show that \[\Gamma\alpha\Gamma=\bigcup_j \Gamma(\gamma\beta_j\gamma^{-1}).\] We note that \[\bigcup_j \Gamma (\gamma\beta_j\gamma^{-1}) = \gamma\left(\bigcup_j \Gamma \beta_j\right)\gamma^{-1} = \gamma(\Gamma\alpha\Gamma)\gamma^{-1} = \Gamma(\gamma\alpha\gamma^{-1})\Gamma,\] and one just checks then that \[\Gamma\alpha\Gamma = \Gamma(\gamma\alpha\gamma^{-1})\Gamma.\]

Next, we define operators \(T_n\) and \(\langle n\rangle\) for all \(n\geq 1\). First, define \(\langle p\rangle =0\) whenever \(p\mid N\). One can implicitly define \(T_n\) by the following formula: \[\sum_{n=1}^\infty T_n n^{-s} = \prod_p \frac{1}{1-T_pp^{-s} + \langle p\rangle p^{k-1-2s}}.\] This in turn is equivalent to the following conditions:

\(T_{nm} = T_n T_m\) if \((n,m)=1\),
\(T_1 = \operatorname{id}\), and
for all primes \(p\) and for all \(r\geq 2\), \[T_{p^r} = T_p T_{p^{r-1}} - p^{k-1}\langle p\rangle T_{p^{r-2}}.\]

From the definition we can see that each \(T_n\) is an explicit polynomial on the \(T_p\), and therefore all \(T_n\) commute with each other.

Theorem 4.4 Suppose \(f\in M_k(\Gamma_1(N))\) has an expansion of the form \(\sum a_m(f)q^m\). Then \(T_n(f)=\sum a_m(T_nf) q^m\), where \[a_m(T_nf) = \sum_{d\mid (m,n)} d^{k-1}a_{\frac{mn}{d^2}}(\langle d\rangle f).\] In particular, if \(f\in M_k(\Gamma_0(N),\chi)\) then \[a_m(T_nf)=\sum_{d\mid (m,n)} \chi(d) d^{k-1} a_{\frac{mn}{d^2}}(f).\]

Proof. A long computation.

We end this section with the notion of Hecke eigenforms.

Definition 4.6 A (or just eigenform) is a non-zero modular form \(f\in M_k(\Gamma_1(N))\) which is an eigenvector for all the Hecke algebra \({\mathbb{T}}(M_k(\Gamma_1(N))\). A (or normalized eigenform) is an eigenform satisfying \(a_1(f)=1\).

Let \(f\in M_k(\Gamma_1(N))\) be an eigenform, say \(T_nf=\lambda_n f\) for all \(n\). Then we obtain \[a_n(f) = a_1(T_n f) = \lambda_n a_1(f),\quad n\geq 1.\] So if \(a_1(f)=0\) then all \(a_n(f)=0\) and thus \(f=0\). Therefore a non-constant non-zero eigenform must have \(a_1(f)\neq 0\) and it may be scaled to a normalized eigenform. In particular, we have the following.

Theorem 4.5 Let \(f\in M_k(\Gamma_1(N))\) be a normalized eigenform. Then the eigenvalues of the Hecke operators on \(f\) are precisely the coefficients of the \(q\)-expansion of \(f\) at the cusp \(\infty\): \[ T_n f = a_n(f) f,\quad n\geq 1. \tag{4.1}\]

Proof. Write \(\lambda_n\) for the eigenvalue of the Hecke operator \(T_n\). By Equation 4.1 we have \(a_n(f) = a_1(T_n f) = \lambda_n a_1(f)\). Since \(f\) is normalized, \(a_1(f)=1\) and hence \(a_n(f) = \lambda_n\).

In fact, the Fourier coefficients of a modular form readily tell whether it is a normalized eigenform:

Proposition 4.5 Let \(f\in M_k(\Gamma_0(N),\chi)\) be a modular form with \(q\)-expansion \(\sum_{n=0}^\infty a_n(f) q^n\). Then \(f\) is a normalized eigenform if and only if:

\(a_1(f) = 1\),
\(a_{mn}(f) = a_m(f) a_n(f)\) whenever \((m,n)=1\), and
\(a_{p^r}(f) = a_p(f)a_{p^{r-1}}(f) - p^{k-1}\chi(p) a_{p^{r-2}}(f),\quad r\geq 2\).

Proof. The implication \(\implies\) follows directly from the previous proposition and the definition of the Hecke operators \(T_n\). For the converse, if \(f\in M_k(\Gamma_0(N),\chi)\) satisfies \((1)\), \((2)\) and \((3)\) then \(f\) is already normalized, so to be an eigenform we must show that it satisfies \[a_m(T_pf)=a_p(f)a_m(f),\quad \forall p\text{ prime}, \forall m\geq 1.\] If \(p\nmid n\) then it follows from the formula that we have for \(T_m\) on \(q\)-expansions that \(a_m(T_pf)=a_{pm}(f)\), which by \((2)\) is \(a_p(f)a_m(f)\). If \(p\mid m\) then writing \(m=p^rm'\) with \(r\geq 1\) and \(p\nmid m'\) we have by the same formula \[a_m(T_pf) = a_{p^{r+1}m'}(f) + \chi(p)p^{k-1}a_{p^{r-1}m'}(f).\] Using now conditions \((2)\) and \((3)\) this can be rewritten as \(a_p(f)a_m(f)\), as wanted.

4.4 Petersson inner product

4.4.1 Surface integrals

Let \(V\subseteq {\mathbb{C}}\). A \(2\)-form on \(V\) is an expression of the form \(\omega = f(z,\bar z)dz\wedge d\bar z\). Note that \[dz\wedge d\bar z = (dx + i dy)\wedge (dx-idy) = -2idx\wedge dy.\] The integral of \(\omega\) on \(V\) is: \[\int_V \omega = \int_V f(z,\bar z dz\wedge d\bar z=\int\int -2i f(x+iy,x-iy)dxdy.\] Consider now, for \(\alpha\in\operatorname{GL}_2^+({\mathbb{R}})\), the change \(z\mapsto \alpha z\). Then: \[\Im(\alpha z)=\frac{\det \alpha}{|cz+d|^2}\Im(z),\] and also \[d(\alpha z) = \frac{\det \alpha}{(cz+d)^2} dz,\quad \overline{d(\alpha z)} = \frac{\det\alpha}{\overline{(cz+d)^2}} d\bar z.\] This gives that: \[d(\alpha z)\wedge d\overline{\alpha z} = \frac{(\det \alpha)^2}{|cz + d|^4} dz\wedge d\bar z.\] Therefore the \(2\)-form \(\frac{dz\wedge d\bar z}{\Im(z)^2}\) is invariant under changes of the form \(z\mapsto \alpha z\). We will work instead with a certain multiple of this \(2\)-form. Define \[d\mu(z) = \frac{dx\wedge dy}{y^2} = \frac{-1}{2i} \frac{dz\wedge d\bar z}{\Im(z)^2}.\] We can define the of \({\operatorname{SL}}_2({\mathbb{Z}})\) as \[{\operatorname{covol}}({\operatorname{SL}}_2({\mathbb{Z}})) = \int_{D^*} d\mu(z).\] where \(D^*\) is a fundamental domain for \({\operatorname{SL}}_2({\mathbb{Z}})\).

Lemma 4.5 \[{\operatorname{covol}}({\operatorname{SL}}_2({\mathbb{Z}})) = \frac{\pi}{3}.\]

Proof. Exercise.

Corollary 4.4 If \(\varphi\) is a bounded function on \(D^*\), then \(\int_{D^*} \varphi(z)d\mu(z)\) is a well-defined complex number.

4.4.2 Integral over \(X(\Gamma)\)

Let \(\mathcal{D}\) be a fundamental domain for a congruence subgroup \(\Gamma\). Such a fundamental domain is the union (almost disjoint) of translates of \(D^*\): \[\mathcal{D}=\cup_j \alpha_j D^*,\] where \(\{\alpha_j\}\) is a set of coset representatives for \((\pm 1\cdot \Gamma)\backslash {\operatorname{SL}}_2({\mathbb{Z}})\). If \(\varphi\) is \(\Gamma\)-invariant, then we may define the integral of \(\varphi\) on \(X(\Gamma)=\Gamma\backslash {\mathbb{H}}\) as: \[\int_{X(\Gamma)} \varphi(\tau) d\mu(\tau) = \sum_j \int_{\alpha_j D^*} \varphi(\tau)d\mu(\tau) = \sum_j\int_{D^*} \varphi(\alpha_j\tau)d\mu(\alpha_j\tau)=\sum_j\int_{D^*} \varphi(\alpha_j\tau)d\mu(\tau).\] The last term in the above equality shows that the definition is independent of the choice of coset representatives. We may calculate the covolume of \(\Gamma\) as:

Lemma 4.6 Let \(\Gamma\subset{\operatorname{SL}}_2({\mathbb{Z}})\) be a congruence subgroup. Then \[{\operatorname{covol}}(\Gamma)= \int_{X(\Gamma)} d\mu(\tau) = [\operatorname{PSL}_2({\mathbb{Z}})\colon \overline\Gamma]{\operatorname{covol}}({\operatorname{SL}}_2({\mathbb{Z}})) = \frac{\pi}{3} [\operatorname{PSL}_2({\mathbb{Z}})\colon \overline\Gamma].\]

Let \(f\) and \(g\) be two cusp forms for \(\Gamma\) of weight \(k\), and set \(\varphi(\tau)=f(\tau)\overline{g(\tau)}\Im(\tau)^k\).

Lemma 4.7 The function \(\varphi\) is \(\Gamma\)-invariant and, for all \(\alpha\in {\operatorname{SL}}_2({\mathbb{Z}})\), the translate \(\varphi(\alpha\tau)\) is bounded on \(D^*\).

Proof. If \(\gamma\) belongs to \(\Gamma\), then we may compute: \[\varphi(\gamma\tau)=f|_k\gamma j(\gamma,\tau)^{-k}\overline{g|_k\gamma j(\gamma,\tau)^{-k}} j(\gamma,\tau)^{2k}\Im(z)^k = \varphi(\tau).\] If \(\alpha\) belongs to \({\operatorname{SL}}_2({\mathbb{Z}})\), then: \[\varphi(\alpha\tau)=f|_k\alpha \overline{g|_k\alpha}\Im(\tau)^k = O(q_h)\overline{O(q_h)} y^k = O(|q_h|^2 y^k).\] This approaches \(0\) as \(y\) approaches infinity, because \(q_h = e^{\frac{2\pi i (x+iy)}{z}}\). This gives boundedness.

The previous lemma allows us to define an inner product on the spaces of cusp forms:

Definition 4.7 The of \(f\) and \(g\) is: \[\langle f,g\rangle_\Gamma = \frac{1}{{\operatorname{covol}}(\Gamma)}\int_{X(\Gamma)} f(\tau)\overline{g(\tau)} \Im(\tau)^kd\mu(\tau).\]

For the above to converge it is enough that one of the forms \(f\) and \(g\) is in \(S_k\). Therefore the product of a modular form with a cusp form is well defined.

The reason to divide by \({\operatorname{covol}}(\Gamma)\) is that, in this way, if \(\Gamma\subseteq \Gamma'\) then \[\langle f,g\rangle_\Gamma=\langle f,g\rangle_{\Gamma'}.\]

Proposition 4.6 The Petersson inner product is a positive-definite hermitian product on the \({\mathbb{C}}\)-vector space \(S_k(\Gamma)\). That is:

\(\langle a_1 f_1 + a_2 f_2,g\rangle_\Gamma = a_1\langle f_1,g\rangle_\Gamma + a_2\langle f_2,g\rangle_\Gamma\).
\(\langle g,f\rangle_\Gamma = \overline{\langle f,g\rangle_\Gamma}\).
\(\langle f,f\rangle\geq 0\), with equality if and only if \(f=0\).

Although the Petersson inner product does not extend to all of \(M_k(\Gamma)\), it still allows us to define an “orthogonal complement to \(S_k(\Gamma)\):

Definition 4.8 The of \(M_k(\Gamma)\) is the space \[{\mathcal{E}}_k(\Gamma) = \{f\in M_k(\Gamma)~|~ \langle f,g\rangle_\Gamma =0 \quad \forall g\in S_k(\Gamma)\}.\]

4.4.3 Adjoint operators

If \(\langle\cdot,\cdot\rangle\) is an hermitian product on a \({\mathbb{C}}\)-vector space \(V\) and \(T\colon V\longrightarrow V\) is a linear operator, the of \(T\) is defined as the operator \(T^*\) which satisfies: \[\langle Tf,g\rangle = \langle f,T^*g\rangle.\] The goal of this subsection is to calculate the adjoint operators to the Hecke operators. We will need the following technical result.

Lemma 4.8 Let \(\Gamma\subset{\operatorname{SL}}_2({\mathbb{Z}})\) be a congruence subgroup and let \(\alpha\in \operatorname{GL}_2^+({\mathbb{Q}})\).

If \(\varphi\colon{\mathbb{H}}\longrightarrow{\mathbb{C}}\) is continuous, bounded and \(\Gamma\)-invariant then: \[\int_{\alpha^{-1}\Gamma\alpha\backslash{\mathbb{H}}} \varphi(\alpha\tau)d\mu(\tau) = \int_{\Gamma\backslash{\mathbb{H}}} \varphi(\tau)d\mu(\tau).\]
If \(\alpha^{-1}\Gamma\alpha\) is contained in \({\operatorname{SL}}_2({\mathbb{Z}})\) then \(\Gamma\) and \(\alpha^{-1}\Gamma\alpha\) have equal covolumes and indices in \({\operatorname{SL}}_2({\mathbb{Z}})\).
Let \(n=[\Gamma\colon\alpha^{-1}\Gamma\alpha\cap\Gamma] = [\Gamma\colon \alpha\Gamma\alpha^{-1}\cap\Gamma]\). There are matrices \(\beta_1,\ldots,\beta_n\in\operatorname{GL}_2^+({\mathbb{Q}})\) inducing disjoint unions \[\Gamma\alpha\Gamma = \bigcup \Gamma \beta_j = \bigcup \beta_j \Gamma.\]

Proof. The first two statements are easy and follow from the change of variables formula and Lemma 4.6. The equality of indices in (3) follows by applying (2) to \(\alpha\Gamma\alpha^{-1}\cap\Gamma\) instead of \(\Gamma\) and using multiplicativity of indices. Therefore there exist \(\gamma_1,\ldots\gamma_n\) and \(\tilde\gamma_1,\ldots,\tilde\gamma_n\) in \(\Gamma\) such that \[\Gamma = \bigcup (\alpha^{-1}\Gamma\alpha \cap \Gamma)\gamma_j = \bigcup (\alpha\Gamma\alpha^{-1}\cap \Gamma)\tilde\gamma_j^{-1}.\] By how coset representatives are linked to orbit representatives in a double coset, we get: \[\Gamma\alpha\Gamma = \bigcup \Gamma \alpha\gamma_j,\quad \Gamma\alpha^{-1}\Gamma=\bigcup \Gamma\alpha^{-1}\tilde\gamma_j^{-1}.\] By taking inverses in the second decomposition we get \[\Gamma\alpha\Gamma = \bigcup \tilde\gamma_j\alpha\Gamma.\] Suppose that \(\Gamma\alpha\gamma_j\cap \tilde\gamma_j\alpha\Gamma = \emptyset\). Then \[\Gamma\alpha\gamma_j\subset\bigcup_{i\neq j} \tilde\gamma_i\alpha\Gamma.\] Multiply from the right by \(\Gamma\) to get \(\Gamma\alpha\Gamma\subset \bigcup_{i\neq j} \tilde\gamma_i\alpha\Gamma\), a contradiction with the decomposition of \(\Gamma\alpha\Gamma\) into \(n\) orbits for \(\Gamma\). Therefore we deduce that \(\Gamma\alpha\gamma_j\) intersects \(\tilde\gamma_j\alpha\Gamma\), for each \(j\). Let \(\beta_j\) be any element in this intersection. This gives \[\Gamma\alpha\Gamma=\bigcup\Gamma\beta_j = \bigcup\beta_j\Gamma.\]

This allows us to compute adjoints of double coset operators.

Proposition 4.7 Let \(\Gamma\subseteq {\operatorname{SL}}_2({\mathbb{Z}})\) be a congruence subgroup and let \(\alpha\in\operatorname{GL}_2^+({\mathbb{Q}})\). Let \(\alpha^*=\det(\alpha)\alpha^{-1}\) be the classical adjoint to \(\alpha\). Then

If \(\alpha^{-1}\Gamma\alpha\subseteq {\operatorname{SL}}_2({\mathbb{Z}})\), and \(f\in S_k(\Gamma)\) and \(g\in S_k(\alpha^{-1}\Gamma\alpha)\), \[\langle f|_k\alpha,g\rangle_{\alpha^{-1}\Gamma\alpha}=\langle f,g|_k\alpha^*\rangle_\Gamma.\]
For all \(f,g\in S_k(\Gamma)\), \[\langle f|_k[\Gamma\alpha\Gamma],g\rangle = \langle f,g|_k[\Gamma\alpha^*\Gamma]\rangle.\]

Proof. We prove only (1). The second statement follows easily. We will use the equalities that we have already seen: \[j(\alpha,\alpha^*z) = j(\alpha\alpha^*,z)j(\alpha^*,z)^{-1} = \det \alpha j(\alpha^*,z)^{-1},\quad \Im(\alpha^*z) = \frac{\det\alpha^*}{|j(\alpha^*,z)|^2} \Im(z).\] Let \(M={\operatorname{covol}}(\Gamma)={\operatorname{covol}}(\alpha^{-1}\Gamma\alpha)\). Then we compute: \[\begin{align*} M\langle f|_k\alpha,g\rangle_{\alpha^{-1}\Gamma\alpha} &= \int_{\alpha^{-1}\Gamma\alpha\backslash{\mathbb{H}}} (\det\alpha)^{k-1}j(\alpha,z)^{-k}f(\alpha z) \overline{g(z)} \Im(z)^k d\mu(z)\\ &= \int_{\Gamma\backslash{\mathbb{H}}} (\det\alpha)^{k-1}j(\alpha,\alpha^*z)^{-k} f(z)\overline{g(\alpha^*z)}\Im(\alpha^*z)^kd\mu(z)\\ &=\int_{\Gamma\backslash {\mathbb{H}}} (\det\alpha)^{k-1}(\det\alpha)^{-k}f(z)j(\alpha^*,z)^k\overline{g(\alpha^*z)} \frac{(\det\alpha^*)^k}{|j(\alpha^*,z)|^{2k}}\Im(z)^kd\mu(z)\\ &=\int_{\Gamma\backslash{\mathbb{H}}} f(z)(\det \alpha)^{k-1} \overline{j(\alpha^*,z)}^{-k} \overline{g(\alpha^*z)} \Im(z)^kd\mu(z)\\ &=\int_{\Gamma\backslash{\mathbb{H}}} f(z)\overline{g|_k\alpha^*(z)} \Im(z)^kd\mu(z) = M\langle f,g|_k\alpha^*\rangle_\Gamma. \end{align*}\]

Definition 4.9 A linear operator \(T\) is if it commutes with its adjoint: \[T T^* = T^* T.\]

Theorem 4.6 Consider the \({\mathbb{C}}\)-vector space \(S_k(\Gamma_1(N))\). If \(p\nmid N\) then: \[\langle p\rangle^* = \langle p\rangle^{-1} = \langle p^{-1}\rangle,\text{ and } T_p^* = \langle p\rangle^{-1} T_p.\]

Proof. Write \(\langle p\rangle = [\Gamma\alpha\Gamma]\), where \(\alpha\in\Gamma_0(N)\) is such that modulo \(N\) is congruent to \(\left(\begin{smallmatrix}a&b\\0&p\end{smallmatrix}\right)\). By Proposition 4.7, we have that \(\langle p\rangle^*\) consists on acting with \(\alpha^*=\det\alpha \alpha^{-1}\). Since \(\det\alpha=1\), then \(\alpha^*=\alpha^{-1}\) and thus \(\langle p\rangle^* = \langle p^{-1}\rangle = \langle p\rangle^{-1}\).

As for the second part, we set \(\alpha=\left(\begin{smallmatrix}1&0\\0&p\end{smallmatrix}\right)\) and we need to compute \(\Gamma\alpha^*\Gamma\). Note that \[\alpha^*=\left(\begin{matrix}p&0\\0&1\end{matrix}\right)=\left(\begin{matrix}1&n\\N&mp\end{matrix}\right) ^{-1} \left(\begin{matrix}1&0\\0&p\end{matrix}\right) \left(\begin{matrix}p&n\\N&m\end{matrix}\right),\quad mp-nN=1.\] In the right-hand side, the first matrix is in \(\Gamma_1(N)\) and the last is in \(\Gamma_0(N)\). Since \(\Gamma_0(N)\) is normal in \(\Gamma_1(N)\), we get \[\Gamma_1(N)\left(\begin{matrix}p&0\\0&1\end{matrix}\right)\Gamma_1(N)=\Gamma_1(N)\left(\begin{matrix}1&0\\0&p\end{matrix}\right)\Gamma_1(N)\left(\begin{matrix}p&n\\N&m\end{matrix}\right).\] Since \(m\equiv p^{-1}\pmod N\), the matrix \(\left(\begin{matrix}p&n\\N&m\end{matrix}\right)\) acts as \(\langle p^{-1}\rangle\). Therefore: \[T_p^*f = \sum_j f|_k\beta_j \left(\begin{matrix}p&n\\N&m\end{matrix}\right) = (T_pf)|_k \left(\begin{matrix}p&n\\N&m\end{matrix}\right) = \langle p^{-1}\rangle T_p f.\]

Corollary 4.5 If \(n\) is coprime to \(N\), the Hecke operators \(T_n\) and \(\langle n\rangle\) are normal.

Theorem 4.7 Let \(T\) be a normal operator on a finite dimensional \({\mathbb{C}}\)-vector space. Then \(T\) has an orthogonal basis of eigenvectors.

Applying this theorem multiple times we deduce that if a \({\mathbb{C}}\)-vector space has a family of normal, pairwise commuting operators then it has a basis of simultaneous eigenvectors. Particularizing to our situation, we get the following result.

Corollary 4.6 The space \(S_k(\Gamma_1(N))\) has an orthogonal basis of simultaneous eigenforms for all the \(T_n\) and \(\langle n\rangle\) with \((n,N)=1\).

Proof. Apply the spectral theorem for the first of the \(T_n\), to get an orthogonal basis of eigenforms. To each of the subspaces one can apply the second of the \(T_n\) to refine the basis, thanks to the fact that the Hecke operators commute with each other and hence preserve eigenspaces. The process terminates after a finite number of steps because \(S_k(\Gamma_1(N))\) is finite-dimensional.

Consider \(S_k({\operatorname{SL}}_2({\mathbb{Z}})) = S_k(\Gamma_1(1))\). It has a basis of eigenforms for all the Hecke operators \(T_n\) (and \(\langle n\rangle\)). We may normalize the eigenforms \(f\) so that \(a_1(f)=1\). Then we will obtain: \[T_n f = a_n(f) f,\quad \forall n.\] Therefore each system of eigenvalues \(\{a_n(f)\}_{n\geq 1}\) corresponds to a unique eigenform \(f\). We say that \(S_k({\operatorname{SL}}_2({\mathbb{Z}}))\) satisfies . In other words, \(S_k({\operatorname{SL}}_2({\mathbb{Z}}))\) decomposes into a direct sum of one-dimensional eigenspaces. In the next section we investigate when this fails to be true, and what can be done to remedy it.

4.5 Atkin-Lehner-Li theory

Let us consider \(S_k(\Gamma_1(N))\) for an arbitrary \(N\). We have already seen that there is a basis of simultaneous eigenforms for the \(T_n\) and \(\langle n\rangle\) operators, as long as \(n\) is coprime to \(N\). We want to investigate if the components of this basis are also eigenforms for the remaining Hecke operators and if multiplicity one is satisfied.

Recall the operator \(V_d\colon M_k(\Gamma_1(M))\longrightarrow M_k(\Gamma_1(Md))\) which was introduced before for \(d\) a prime: \[(V_d f)(\tau) = f(d\tau) = d^{1-k} f|_k\left(\begin{matrix}d&0\\0&1\end{matrix}\right).\] If \((t,d)=1\) then it is easy to check that \(V_d U_t = U_t V_d\), and hence \(V_d T_n = T_n V_d\) whenever \((n,d)=1\).

4.5.1 Examples

Both \(\Delta(z)\) and \(\Delta(2z)\) are cusp forms in \(S_{12}(\Gamma_1(2))\). Write \[\Delta = \sum_{n\geq 1} \tau(n) q^n,\] so that \(T_p \Delta=\tau(p)\Delta\) for all \(p\). Here, by \(T_2\) we mean the Hecke operator as acting on \(S_{12}({\operatorname{SL}}_2({\mathbb{Z}}))\). By what we have seen above, we have: \[T_p(\Delta(2z)) = \tau(p)\Delta(2z),\quad p\neq 2.\] Therefore \(\Delta(z)\) and \(\Delta(2z)\) have, when considered in \(S_{12}(\Gamma_1(2))\), the same “system of eigenvalues” \(\{\tau(n)\}_{(n,2)=1}\). Therefore \(S_{12}(\Gamma_1(2))\) does not satisfy multiplicity one.

However, the Hecke operator \(T_2=U_2\) as acting on \(S_{12}(\Gamma_1(2))\) satisfies: \[U_2(\Delta(2z)) = \Delta(z),\text{ and } U_2(\Delta(z)) = T_2\Delta - 2^{11}V_2(\Delta) = -24\Delta(z) - 2^{11} \Delta(2z).\] Therefore \(U_2\) acts on \(S_{12}(\Gamma_1(2))\) with matrix \[[U_2] = \left(\begin{matrix}-24&1\\-2^{11}&0\end{matrix}\right),\] which is diagonalizable. The eigenvectors \[f_{\pm} = \Delta(z) + (12 \pm 4\sqrt{-119})\Delta(2z)\] can be completed to give a basis of eigenforms for all the \(T_n\).

The following example shows that sometimes one may not get a basis of eigenforms for all \(T_p\). Let \(f\in S_2(\Gamma_1(N))\) be an eigenform for \(\{T_q\}_{q\nmid N}\cup\{U_q\}_{q\mid N}\), and let \(p\nmid N\). Let \(S\) be the following \(4\)-dimensional \({\mathbb{C}}\)-vector subspace of \(S_2(\Gamma_1(Np^3))\): \[S = \operatorname{span}_{\mathbb{C}}\{f(\tau),f(p\tau),f(p^2\tau),f(p^3\tau)\}.\] Since \(T_q\) commutes with \(V_p\), the subspace \(S\) is stable under \(\{T_q\}_{q\nmid Np^3}\). Moreover, \(S\) is also stable under \(\{T_q=U_q\}_{q\mid N}\). The following result shows that \(S\) does not satisfy multiplicity one.

Proposition 4.8

\(S\) is stable under \(\{T_q\}_{q\nmid Np^3} \cup \{T_q=U_q\}_{q \mid N}\cup \{T_p=U_p\}\).
The matrix of \(U_p\) is not diagonalizable.

Proof. Exercise.

4.5.2 New and old forms

Suppose that \(M\mid N\) are two positive integers. There are many ways to embed \(S_k(\Gamma_1(M))\) into \(S_k(\Gamma_1(N))\). For example, for any \(d\) such that \(dM\mid N\), we can map \(f\) to \(V_df\).

Definition 4.10 The , denoted by \(S_k(\Gamma_1(N))^\text{old}\) is: \[S_k(\Gamma_1(N))^\text{old} = \operatorname{span}_{\mathbb{C}}\left\{ V_d(S_k(\Gamma_1(M))) ~\colon~ dM\mid N, M\neq N\right\}.\] The , denoted by \(S_k(\Gamma_1(N))^\text{new}\) is the orthogonal complement (with respect to the Petersson inner product) of \(S_k(\Gamma_1(N))^\text{old}\) in \(S_k(\Gamma_1(N))\).

Theorem 4.8 The spaces \(S_k(\Gamma_1(N))^\text{old}\) and \(S_k(\Gamma_1(N))^\text{new}\) are stable under all Hecke operators.

Proof. Let \(\ell\) be a prime dividing \(N\). We may define \[S_k(\Gamma_1(N))^{\ell-\text{old}} = \iota S_k(\Gamma_1(N/\ell)) + V_\ell S_k(\Gamma_1(N/\ell)),\] where \(\iota\) is embedding induced by \(f\mapsto f\). In this way, \[S_k(\Gamma_1(N))^\text{old} = \sum_{\ell \mid N} S_k(\Gamma_1(N))^{\ell-\text{old}},\] where the sum runs over prime divisors \(\ell\) of \(N\). What we will prove is that each of the spaces \(S_k(\Gamma_1(N))^{\ell-\text{old}}\) is stable under the diamond operators, the Hecke operators \(T_p\), and their adjoints. Note also that if \(V\subset S_k(\Gamma_1(N))\) is a subspace which stable under an operator \(T\), then the orthogonal complement to \(V\) is stable under the adjoint \(T^*\).

Let \(f\in S_k(\Gamma_1(N/\ell))\), and let \(T\) be one of the Hecke operators above. We must prove that \(T(\iota f)\) and \(T(V_\ell f)\) are in \(S_k(\Gamma_1(N))^{\ell-\text{old}}\). Consider the matrix \(\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right) \in \Gamma_0(N)\), which defines the operator \(\langle d\rangle\) on \(S_k(\Gamma_1(N))\) and on \(S_k(\Gamma_1(N/\ell))\). This shows that \(\langle d\rangle\) preserves \(\iota S_k(\Gamma_1(N/\ell))\). Next, note that that \[\left(\begin{matrix}\ell &0\\0&1\end{matrix}\right)\left(\begin{matrix}a&b\\c&d\end{matrix}\right) = \left(\begin{matrix}a&b\ell\\c/\ell&d\end{matrix}\right) \left(\begin{matrix}\ell &0\\0&1\end{matrix}\right).\] Since \(c/\ell\) is an integer which is divisible by \(N/\ell\), the matrix \(\left(\begin{smallmatrix}a&b\ell\\c/\ell&d\end{smallmatrix}\right)\) defines the operator \(\langle d\rangle\) on \(S_k(\Gamma_1(N/\ell))\). Therefore the matrix equality above gives \(\langle d\rangle(V_\ell f) = V_\ell(\langle d\rangle \iota f)\).

Next we prove that the operators \(T_p\) also preserve \(S_k(\Gamma_1(N))^{\ell-\text{old}}\). If \(p\) does not divide \(N\) this is easy to show that \(T_p\) preserves both \(V_\ell S_k(\Gamma_1(N/\ell))\) and \(\iota S_k(\Gamma_1(N/\ell))\). When \(p\) does divide \(N\) but \(p\neq \ell\), the same argument works. We now consider \(T_\ell\). Suppose that \(\ell\) divides \(N\) exactly once. Then \[T_\ell(\iota f) = \iota U_\ell f,\text{ and }\quad T_\ell(V_\ell f) = \iota f.\] However, in \(S_k(\Gamma_1(N/\ell))\) we have \[\iota T_\ell f = T_\ell (\iota f ) + \ell^{k-1}V_\ell(\langle \ell\rangle f),\text{ so } T_\ell(\iota f) = \iota T_\ell f - \ell^{k-1} V_\ell(\langle\ell\rangle f).\] In particular we see that \(T_\ell(\iota f)\) and \(T_\ell(V_\ell f)\) are in \(S_k(\Gamma_1(N))^{\ell-\text{old}}\).

Finally if \(\ell^2\) divides \(N\) then \(T_\ell\) acts as \(U_\ell\) in both \(S_k(\Gamma_1(N/\ell))\) and \(S_k(\Gamma_1(N))\), and hence \[T_\ell \iota f = \iota T_\ell f,\quad T_\ell V_d f = \iota f.\]

It only remains to show that the adjoint of \(T_p\) preserves \(S_k(\Gamma_1(N))^{\ell-\text{old}}\) when \(p\) divides \(N\) (when \(p\) does not divide \(N\), the adjoints of Hecke operators are in the Hecke algebra and hence preserves the old subspace. In this case, consider the Fricke operator \(w_N\) acting on \(S_k(\Gamma_1(N))\) by \[f\mapsto f|_k W_N,\quad W_N=\left(\begin{matrix}0&-1\\N&0\end{matrix}\right).\] (note that \(W_N\) normalizes \(\Gamma_1(N)\)). We can check: \[(w_N f)(z) = z^{-k}f(-1/(Nz)).\] Also, note that \(W_N\left(\begin{smallmatrix}1&0\\0&p\end{smallmatrix}\right) W_N^{-1} = \left(\begin{smallmatrix}p&0\\0&1\end{smallmatrix}\right)\), and thus \(T_p^* = w_N^{-1} T_p w_N\). One can then compute: \[w_N\iota f = \ell^k V_\ell w_{N/\ell} f,\text{ and } w_N V_\ell f = \iota w_{N/\ell} f.\] Therefore \(w_N\) (and hence \(T_p^*\)) preserves the old subspace.

We say that \(f\in S_k(\Gamma_1(N))^\text{new}\) is a if it is an eigenform for all Hecke operators, which is normalized so that the leading coefficient is \(1\).

Theorem 4.9 Consider the space \(S_k(\Gamma_1(N))^\text{new}\) for \(N\geq 1\).

The space \(S_k(\Gamma_1(N))^\text{new}\) has a basis of newforms.
If \(f\in S_k(\Gamma_1(N))^\text{new}\) is an eigenvector for \(\{T_q\}_{q\nmid N}\) then \(f\) is a scalar multiple of a newform, and hence an eigenvector for all the Hecke operators.
If \(f\in S_k(\Gamma_1(N))^\text{new}\) and \(g\in S_k(\Gamma_1(M))^\text{new}\) are both newforms satisfying \(a_q(f)=a_q(g)\) for all but finitely many primes \(q\), then \(N=M\) and \(f=g\).

Proof. This was proven by Atkin–Lehner in 1970 and a partial proof can be found in (Diamond and Shurman 2005).

Corollary 4.7

If \(f\) is a newform, then there is a Dirichlet character \(\chi\) such that \(f\in S_k(\Gamma_0(N),\chi)\).
If \(\{\lambda_n\}_{(n,N)=1}\) is a system of eigenvalues for the \(T_n\) such that \((n,N)=1\), then \(\exists !\) newform \(f\in S_k(\Gamma_1(M))^\text{new}\) for some \(M\mid N\), such that \(T_nf = \lambda_n f\) for all \(n\) satisfying \((n,N)=1\).

Finally, we see that the new subspaces give a complete description of \(S_k(\Gamma_1(N))\) and \(S_k(\Gamma_0(N))\).

Theorem 4.10 There are direct sum decompositions \[S_k(\Gamma_1(N)) = \bigoplus_{M\mid N} \bigoplus_{dM \mid N} V_d\left(S_k(\Gamma_1(M))^\text{new}\right),\] and \[S_k(\Gamma_0(N)) = \bigoplus_{M\mid N} \bigoplus_{dM \mid N} V_d\left(S_k(\Gamma_0(M))^\text{new}\right).\]

Proof. Write \(S_k(\Gamma_1(N))=W_1\oplus\cdots\oplus W_t\), where each of the \(W_i\) is a simultaneous eigenspace for \(\{T_n\}_{(n,N)=1}\cup \{\langle n\rangle\}\). Each form \(f\in W_i\) has the same “package” of eigenvalues \(\{\lambda_n\}_{(n,N)=1}\). Therefore by Corollary 4.7 this \(f\) comes from a unique newform \(f_i\in S_k(\Gamma_1(M_i))^\text{new}\) for some \(M_i\mid N\). Therefore \[W_i=\bigoplus_{dM_i\mid N} {\mathbb{C}}V_d(f_i)\] as wanted. Since each of these spaces is stable under the diamond operators, we get the second decomposition by further taking the subspaces on which they act trivially.