In this proof here in the paper by Tunnell on Congruent Numbers, how does one conclude that
the span of the weight $\frac{3}{2}$ forms $g\theta_2$ and $g\theta_8$ for $\Gamma_0(128)$ are $T(p^2)$ stable? i think there are some facts I need about half weight modular forms and how the $T(p^2)$ act on them to see this, for instance, it is mentioned in Koblitz's book on modular forms that the space of cusp forms of weight half-integer weight $k/2$ for $\Gamma_0(N)$ have a basis of eigenforms for Hecke operators $T(p^2)$ for all primes $p$ except maybe the primes dividing $N$. it is probably also relevant that it is mentioned earlier that the the three forms $g\theta_2, g\theta_8$ and $g\theta_{32}$ form a basis for the weight 3/2 forms for $\Gamma_0(128)$.
$g(\theta_2 - \theta_8)$ and $g\theta_8$ are eigenforms for all the $T(p^2)$? I follow the work on exponents modulo 8 before this claim, but I don't see how this is conclude from that. here, $g$ is the modular form corresponding to the elliptic curve $E_1: y^2 = x^3 - x$, with a $q$ expansion of the form $g(z) = \sum (-1)^m q^{(4n+1)^2+8m^2}$. Also, $\theta_2 = \sum q^{2n^2}, \theta_8 = \sum q^{8n^2}$.
First, the Hecke operators $T(p^2)$ act on the space of weight $3/2$ cusp forms of fixed level $N$ and trivial character, and this space is generated, for $N=128$, by $g\theta_2$, $g\theta_8$ and $g\theta_{32}$ (allegedly).
Furthermore, it is known that the above space contains a basis of cusp forms that are eigenfunctions for every $T(p^2)$ where $p\nmid N$ (we’ll call them the good Hecke operators).
What Tunnell does (if I understood well) is show (I would guess he’s using the direct $q$-expansion of Hecke operators, but you should check this actually exists in half-integer weight) that $f_1=g\theta_2, f_2=g\theta_8$ and $f_3=g(2\theta_{32}-\theta_8)$ are eigenfunctions for $T(3^2)$ and $T(5^2)$.
Now, $f_1$ and $f_2$ have (allegedly) the same eigenvalues for these specific operators, which are distinct from the eigenvalues of $f_3$. Because the good Hecke operators are self-adjoint and pairwise commute, this means that any linear combination of good Hecke operators applied to $f_1$ and $f_2$ also has the same eigenvalues for $T(3^2),T(5^2)$, and is therefore orthogonal to $f_3$.
So, if $S$ is the subspace generated by $f_1,f_2$ and $T$ the subspace generated by the images under the good Hecke operators of $f_1,f_2$, then $S \subset T \subset f_3^{\perp}$. Since $\dim{S}=2=\dim{f_3^{\perp}}$, $S=T=f_3^{\perp}$ and $S$ is stable under the good Hecke operators.
Since the good Hecke operators are self-adjoint, they also preserve $\mathbb{C}f_3$, and thus $f_3$ is an eigenform.
Then Tunnell discusses the basis $(f_1-f_2,f_2)$ of the space $S$. Let $A$ be a good Hecke operator: $A(f_1-f_2)=a(f_1-f_2)+bf_2$ for scalars $a,b$, by the above.
Let $n$ be any integer such that the coefficient $\lambda$ of degree $n$ of the $q$-expansion of $f_2$ is nonzero. Then $n \equiv 1\pmod{8}$, so the coefficient of degree $n$ of the $q$-expansion of $f_1-f_2$ vanishes. Therefore, the coefficient of degree $n$ of the $q$-expansion of $A(f_1-f_2)$ is $b\lambda$. But, by the explicit formulas, since $n \not\equiv 3\pmod{8}$, the coefficient of degree $n$ of the $q$-expansion of $A(f_1-f_2)$ must vanish.
Therefore, $b=0$, and $A(f_1-f_2)$ is proportional to $f_1-f_2$; hence $f_1-f_2$ is an eigenform. One proves in the same way that $f_2$ is an eigenform.