Congruences between Modular forms of different weight

Question

I'm trying to fill in the details of a computation from example 2.4 here on page 12.

To set up, consider the spaces of modular forms $M_k(1)$ for weights $k=4,6$. These are both one dimensional and generated by the Eisenstein series $E_k$ whose $n$-th Fourier coefficient for positive $n$ is $$a_{n}=\sum_{d|n}d^{k-1}.$$

Now consider the Hecke algebras $\mathbb T_4$ and $\mathbb T_6$, the algebras generated by the operators $T_l$ and $lS_l$. Since $M_4(1),M_6(1)$ are both one dimensional so that $T_l$ an $lS_l$ are determined by their simultaneous eigenvalues $l^{k-1}+1$ and $l^{k-1}$ on $M_k(1)$. Since their difference is $1$, the map taking the operators to their eigenvalues yields an isomorphism of both $M_4(1)$ and $M_6(1)$ with $\mathbb Z$.

Issue 1: Now I've checked that $$1+l^3\equiv 1+l^5 \pmod {24} $$

for all primes $l$. But the author stops short of this by just checking this modulo $12$ and says that no analogous congruence holds for a higher modulus. What am I doing wrong here?

Issue 2: If we go one step ahead and consider the operators $T_l$ and $S_l$ as acting on the direct sum $M_4(1)\oplus M_6(1)$ just acting componentwise, the $\mathbb Z$-algebra these generate is $\mathbb T_{\leq 6}$. Then by restriction, we can consider an injection $$\mathbb T_{\leq 6}\hookrightarrow\mathbb T_4\times\mathbb T_6$$The subalgebra generated by omitting the operators for a prime $p$ is denoted $\mathbb T^{(p)}_{\leq 6}$. Now the author claims that $$\mathbb T^{(2)}_{\leq 6}\simeq \{(u,v)\in \mathbb Z\times \mathbb Z|u\equiv v \pmod {12}\}$$ and $$\mathbb T^{(3)}_{\leq 6}\simeq \{(u,v)\in \mathbb Z\times \mathbb Z|u\equiv v \pmod {6}\}$$

I have no idea where this is coming from. Any help would be appreciated!

Answer 1

Regarding your issue 1, the author most likely made a computational blunder; it looks like $12$ should be replaced by $24$.

Regarding issue 2, the is a two dimensional $\mathbb Q$-vector space spanned by $\mathbb Q$-linear combinations of $E_4$ and $E_6$, and the Hecke algebras (with $T_2$ omitted) generate a commutative algebra of endomorphisms. Since $E_4$ and $E_6$ are eigenvectors for these endomorphisms, each Hecke algebra just acts by a diagonal matrix, and we get an embedding

$$\mathbb T_{\mathbb Q} \hookrightarrow \mathbb Q \times \mathbb Q$$

(here I've omitted the various decorations on $\mathbb T$, but included a subscript $\mathbb Q$ to indicate that everything is over $\mathbb Q$ for the moment) which, concretely, maps $T_{\ell}$ to the pair $(1+\ell^3, 1+ \ell^5$). Since the entries in this ordered pair are distinct, we see that the above embedding is actually an isomorphism.

Now lets look at the Hecke algebra over $\mathbb Z$, which preserves the $\mathbb Z$-lattice spanned by $E_4$ and $E_6$. Each Hecke algebra now acts by a diagonal matrix with integral entries, and so the above embedding restricts to an embedding $$\mathbb T \to \mathbb Z\times\mathbb Z,$$ of course given by the same formula $T_{\ell} \mapsto (1+\ell^3, 1+\ell^5)$.

Since this map becomes an isomorphism after tensoring with $\mathbb Q$, it is an embedding whose image has finite index in $\mathbb Z\times\mathbb Z$, but it does not necessarily have to be an isomorphism, and in fact it is not.

Indeed, you can check that the $\mathbb Z$-subalgebra of $\mathbb Z\times\mathbb Z$ generated by the pairs $(1+\ell^3, 1+\ell^5)$ is precisely the algebra $\{(u,v) \, | \, u \equiv v \bmod 24\}.$ (This is basically a restatement of the fact that $1 + \ell^3 \equiv 1+ \ell^5 \bmod 24,$ but not modulo any higher modulus (if $\ell$ runs over all primes, or even all but finitely many primes).)