4 removed an unnecessary remark about Haar measure

Just to expand on a comment I made above: I'm not exactly sure what operators generate the Hecke algebra of $\Gamma_0(p^2)$, but the Hecke algebras of the principal congruence subgroups $\Gamma(p^r)$ are easier to handle.

Let's write $K_n$ for the principal congruence subgroup of level $p^n$ in $G = {\rm GL}_2(\mathbb{Z}_p)$.

CLAIM: For any $n > 0$, the Hecke algebra $H(G // K_n)$ is generated by the double cosets $K_n x K_n$ for $x$ in the set

$S = \left\{ \begin{pmatrix} 1 & 0 \\ 0 & p \end{pmatrix}, \begin{pmatrix} p & 0 \\ 0 & p \end{pmatrix}, \begin{pmatrix} p^{-1} & 0 \\ 0 & p^{-1} \end{pmatrix}, \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} a & 0 \\ 0 & 1 \end{pmatrix} \right\}$

where a is your favourite generator of $\mathbb{Z}_p^\times$.

(These correspond, classically, to $T_p$, $S_p$, $S_p^{-1}$, a "twisting operator at p", something close to the Atkin-Lehner $w$, and the diamond operator $\langle a \rangle$.)

Proof: It suffices to show that the subalgebra generated by these operators contains the double coset $[K_n g K_n]$ for any given $g \in G$. Let $X$ be the monoid of elements of the form $\begin{pmatrix} p^r & 0 \\ 0 & p^s\end{pmatrix}$ with $r \le s \in \mathbb Z$. The Cartan decomposition tells us that any $g \in G$ can be written as $g = k x k'$ for some $k, k' \in K_0$.

We now write $K_n\ g\ K_n = K_n\ k\ x\ k'\ K_n$

$= K_n\ k\ K_n\ x\ K_n\ k'\ K_n$ (using the normality of $K_n$ in $K_0$)

$= [K_n\ k\ K_n]\ [K_n\ x\ K_n]\ [K_n\ k'\ K_n]$

The first and last terms are obviously in the subring subalgebra $H(K_0 // K_n)$ of $H(G // K_n)$, which is isomorphic to the group algebra of the finite group $K_0 / K_n = {\rm GL}_2(\mathbb Z / p^n)$if we choose our Haar measure so $\mu(K_n) = 1$. In any case, this subalgebra This is clearly generated by the images of the last three elements of $S$, since these are topological generators of ${\rm GL}_2(\mathbb{Z}_p)$.

Meanwhile, the middle term is in the subring subalgebra of $H(G // K_n)$ generated by $X$, and it's easy to see that for $x, y \in X$ we have $K_n\ x\ K_n\ y\ K_n = K_n\ xy\ K_n$. Hence this subring subalgebra is just the monoid algebra of $X$, which is generated by the first three elements of $S$.

Now, as for your original question, the subgroup $U_0(p^2) \subseteq {\rm GL}_2(\mathbb{Z}_p)$ of matrices that are upper triangular modulo $p^2$ contains a conjugate of $K_1$, so its Hecke algebra is isomorphic to a subalgebra of the Hecke algebra of $K_1$. So although I can't give generators for your algebra, I can exhibit it as a subalgebra of something we know generators for.

3 added interpretation of these operators in classical terms

Just to expand on a comment I made above: I'm not exactly sure what operators generate the Hecke algebra of $\Gamma_0(p^2)$, but the Hecke algebras of the principal congruence subgroups $\Gamma(p^r)$ are easier to handle.

Let's write $K_n$ for the principal congruence subgroup of level $p^n$ in $G = {\rm GL}_2(\mathbb{Z}_p)$.

CLAIM: For any $n > 0$, the Hecke algebra $H(G // K_n)$ is generated by the double cosets $K_n x K_n$ for $x$ in the set

$S = \left\{ \begin{pmatrix} 1 & 0 \\ 0 & p \end{pmatrix}, \begin{pmatrix} p & 0 \\ 0 & p \end{pmatrix}, \begin{pmatrix} p^{-1} & 0 \\ 0 & p^{-1} \end{pmatrix}, \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} a & 0 \\ 0 & 1 \end{pmatrix} \right\}$

where a is your favourite generator of $\mathbb{Z}_p^\times$.

(These correspond, classically, to $T_p$, $S_p$, $S_p^{-1}$, a "twisting operator at p", something close to the Atkin-Lehner $w$, and the diamond operator $\langle a \rangle$.)

Proof: It suffices to show that the subalgebra generated by these operators contains the double coset $[K_n g K_n]$ for any given $g \in G$. Let $X$ be the monoid of elements of the form $\begin{pmatrix} p^r & 0 \\ 0 & p^s\end{pmatrix}$ with $r \le s \in \mathbb Z$. The Cartan decomposition tells us that any $g \in G$ can be written as $g = k x k'$ for some $k, k' \in K_0$.

We now write $K_n\ g\ K_n = K_n\ k\ x\ k'\ K_n$

$= K_n\ k\ K_n\ x\ K_n\ k'\ K_n$ (using the normality of $K_n$ in $K_0$)

$= [K_n\ k\ K_n]\ [K_n\ x\ K_n]\ [K_n\ k'\ K_n]$

The first and last terms are obviously in the subring $H(K_0 // K_n)$ of $H(G // K_n)$, which is isomorphic to the group algebra of the finite group $K_0 / K_n = {\rm GL}_2(\mathbb Z / p^n)$ if we choose our Haar measure so $\mu(K_n) = 1$. In any case, this subalgebra is clearly generated by the images of the last three elements of $S$, since these are topological generators of ${\rm GL}_2(\mathbb{Z}_p)$.

Meanwhile, the middle term is in the subring of $H(G // K_n)$ generated by $X$, and it's easy to see that for $x, y \in X$ we have $K_n\ x\ K_n\ y\ K_n = K_n\ xy\ K_n$(as $n > 0$), so . Hence this subring is just the monoid algebra of $X$, which is generated by the first three elements of $S$.

Now, as for your original question, the subgroup $U_0(p^2) \subseteq {\rm GL}_2(\mathbb{Z}_p)$ of matrices that are upper triangular modulo $p^2$ contains a conjugate of $K_1$, so its Hecke algebra is isomorphic to a subalgebra of the Hecke algebra of $K_1$. So although I can't give generators for your algebra, I can exhibit it as a subalgebra of something we know generators for.

2 fixed a minor typo

Just to expand on a comment I made above: I'm not exactly sure what operators generate the Hecke algebra of $\Gamma_0(p^2)$, but the Hecke algebras of the principal congruence subgroups $\Gamma(p^r)$ are easier to handle. You mention

Let's write $K_n$ for the principal congruence subgroup of level $p^n$ in $G = {\rm GL}_2(\mathbb{Z}_p)$.

CLAIM: For any $n > 0$, the Hecke algebra $H(G // K_n)$ is generated by the double cosets $K_n x K_n$ for $x$ in the set

$S = \left\{ \begin{pmatrix} 1 & 0 \\ 0 & p \end{pmatrix}, \begin{pmatrix} p & 0 \\ 0 & p \end{pmatrix}, \begin{pmatrix} p^{-1} & 0 \\ 0 & p^{-1} \end{pmatrix}, \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} a & 0 \\ 0 & 1 \end{pmatrix} \right\}$

where a is your favourite generator of $\mathbb{Z}_p^\times$.

Proof: It suffices to show that the subalgebra generated by these operators contains the double coset $[K_n g K_n]$ for any given $g \in G$. Let $X$ be the monoid of elements of the form $\begin{pmatrix} p^r & 0 \\ 0 & p^s\end{pmatrix}$ with $r \le s \in \mathbb Z$. The Cartan decomposition tells us that any $g \in G$ can be written as $g = k x k'$ for some $k, k' \in K_0$.

We now write $K_n\ g\ K_n = K_n\ k\ x\ k'\ K_n$

$= K_n\ k\ K_n\ x\ K_n\ k'\ K_n$ (using the normality of $K_n$ in $K_0$)

$= [K_n\ k\ K_n]\ [K_n\ x\ K_n]\ [K_n\ k'\ K_n]$

The first and last terms are obviously in the subring $H(K_0 // K_n)$ of $H(G // K_n)$, which is isomorphic to the group algebra of the finite group $K_0 / K_n = {\rm GL}_2(\mathbb Z / p^n)$ if we choose our Haar measure so $\mu(K_n) = 1$. In any case, this subalgebra is clearly generated by the images of the last three elements of $S$, since these are topological generators of ${\rm GL}_2(\mathbb{Z}_p)$.

Meanwhile, the middle term is in the subring of $H(G // K_n)$ generated by $X$, and it's easy to see that $K_n\ x\ K_n\ y\ K_n = K_n\ xy\ K_n$ (as $n > 0$), so this subring is just the monoid algebra of $X$, which is generated by the first three elements of $S$.

Now, as for your original question, the subgroup $U_0(p^2) \subseteq {\rm GL}_2(\mathbb{Z}_p)$ of matrices that are upper triangular modulo $p^2$ contains a conjugate of $K_1$, so its Hecke algebra is isomorphic to a subalgebra of the Hecke algebra of $K_1$. So although I can't give generators for your algebra, I can exhibit it as a subalgebra of something we know generators for.

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