Is it "well-known" which odd powers of the theta function are eigenforms for the half-integral weight Hecke operators? If so, what is a good reference? Is there a slick algorithm for proving computationally that a specific weight k/2 non-cusp form is an eigenform?

The case of even powers was (asked and) solved by P. T. Bateman in "Problem E 2051", Amer. Math. Monthly 76 (1969), the solution being that the powers 2, 4, 8 are the only even powers of $\Theta$ that are eigenforms. A similar idea shows that indeed the only odd powers of theta that are eigenforms are 1, 3, 5 and 7:

Proposition.If $k$ is an odd integer greater than $7$, then $\Theta^k$ is not an eigenform for $T_{3^2}$.

**Proof.** The idea is to compute explicit expressions for the first two fourier coefficients of $\Theta^k$ and $T_{3^2} \Theta^k$, depending only on $k$. For a fixed $n$, in order to compute the $n$-th coefficient of $\Theta^k$ we consider "square-partitions" of $n$ - all the ways to write $n$ as a sum of squares of positive numbers, regardless of order.

Each such partition contributes to all large enough powers of $\Theta$, specifically, as soon as the power is greater than the number of summands in the partition. We can calculate exactly how much a square-partition contributes to a coefficient. Namely, the partition: $$n = \alpha_1 \cdot n_1^2 + \alpha_2 \cdot n_2^2 + \ldots + \alpha_r \cdot n_r^2$$

where the $n_i$ are distinct, constributes to the $n$-th coefficient of $\Theta^k$: $$2^{\alpha_1+\ldots+\alpha_r} \binom{k}{\alpha_1} \binom{k-\alpha_1}{\alpha_2}\cdot \ldots\cdot \binom{k-\alpha_1-\ldots-\alpha_{r-1}}{\alpha_r}$$

The power of 2 accounts for different choices of sign, and each binomial coefficient accounts for the number of ways left to choose the positions of $n_i$'s.

If $n$ is fixed, each such expression is a polynomial in $k$. So, we can sum over all square-partitions of $n$ to get the $n$-th coefficient of $\Theta^k$.

Now we need to get the coefficients of $T_{3^2}\Theta^k$. This is well known. We use theorem 1.7 of "On Modular Forms of Half Integral Weight", G. Shimura (1973), which can also be found as proposition 13 of chapter 4 in "Introduction to Elliptic Curves and Modular Forms", N. Koblitz (1984) (I am slightly weakening the theorem to exclude $p=2$):

Theorem (Shimura).Let $p$ be an odd prime number, and $f \in G_k(N, \chi)$ (an integral modular form of half-integer weight $k/2$, level $N$, and character $\chi$).

Put: $$f=\sum_{n=0}^{\infty} a_n q^n$$ $$T_{p^2}f=\sum_{n=0}^{\infty} b_n q^n$$ Then: $$b_n = a_{p^2n} + \chi (p) \bigg(\frac{(-1)^\lambda n}{p}\bigg) p^{\lambda -1}a_n + \chi(p^2) p^{k-2}a_{n/p^2}$$ where $\lambda = \frac{k-1}{2}$, and $a_{n/p^2}=0$ if $p^2 \not| n$.

For $\Theta^k$ we have $N = 4$ and $\chi = 1$. Using the above with $p=3$: $$b_1 = a_9 - (-3)^{\lambda-1}a_1$$ $$b_2 = a_{18} + (-3)^{\lambda-1}a_2$$

And $a_1$, $a_2$, $a_9$, $a_{18}$ are polynomials in $k$, or $\lambda$ (which is more convenient). If $\Theta^k$ is an eigenform for $T_{3^2}$, we must have $\frac{b1}{a1}=\frac{b2}{a2}$. We will show this is doesn't happen for $k\gt 7$.

The idea is simple: writing $b_1a_2-b_2a_1$ out we see a polynomial of $\lambda$ of degree 19, and $(-3)^{\lambda-1}$ times a quadratic polynomial in $\lambda$. As $\lambda \rightarrow \infty$ (so $k\rightarrow \infty$ as well), the exponential factor dominates, and the expression cannot be zero. We compute a lower bound for this, and check remaining cases manually.

The last part isn't very enlightening, but since you expressed keenness for an algorithm (and someone might want to check that I'm not rambling nonsense), I will include the sage computation below.

```
def square_partitions(n, min=1):
""" Returns a list of partitions of n into squares, the least of which can be min^2. """
if n == 0:
return [[(-1,1)]]
else:
l = []
for i in [min..sqrt(n)]:
for p in square_partitions(n-i^2, i):
if p[0][0] == i:
p[0] = (i, p[0][1]+1)
else:
p = [(i,1)] + p
l += [p]
return l
def theta_coeff(k, n):
""" Computes the nth coefficient of $\theta^k$. """
f = 0
for p in square_partitions(n):
g = 1
j = k
for a in p:
if a[0] != -1:
g *= 2^a[1] * binomial(j, a[1])
j -= a[1]
if (not k in ZZ) or j >= 0:
f += g
return f
R.<l,s> = QQ['l','s']
# s = (-3)^(l-1)
k = 2*l+1
a1, a2, a9, a18 = theta_coeff(k,1), theta_coeff(k,2), theta_coeff(k,9), theta_coeff(k,18)
b1, b2 = a9 - s*a1, a18 + s*a2
g = b1*a2-b2*a1
lc = LCM([denominator(c) for c in g.coefficients()])
g *= lc
# Check the quadratic factor increases in absolute value when $x\ge7$
x = QQ['x'].0
print factor(g.coefficient(s)(l=x)) # (-6252318072000) * x * (x + 1/2)^2
# Lower bound where exponential factor dominates. Can do better since we're not using the quadratic factor.
m1 = sum(abs(c) for c in g(s=0).coefficients())
print m1*116^19 < 3^115 # True
# Check remaining cases manually.
print [2*i+1 for i in [0..120] if g(i, (-3)^(i-1)) == 0] # [1, 3, 5, 7]
```

guessthen here we go: the moment the space of modular forms of level 4 (or 1 or whatever your convention is) and weight n+1/2 has dimension bigger than one, why would you expect a random product of eigenforms to be an eigenform? So I would guess that it would only happen for very small powers. As for an algorithm---given a modular form, hit it with a Hecke operator and compare q-expansions. You'll quickly guess the eigenvalue from the first few terms, if it's an eigenform. The Sturm bound tells you when you have to stop checking q-expansion coeffts. Did I misunderstand? $\endgroup$ – Kevin Buzzard Jul 30 '10 at 10:15theoremabout powers of theta. Here might be one way to go about it: the power of theta will necessarily be an Eisenstein series, and specific formulae for these will surely be known. Check that for a sufficiently big power, the coefficient of $q^2$ isn't what it is supposed to be. This might be ugly and might even be so ugly as to be unworkable, but it might work. As for being an eigenform for all Hecke operators---you'll certainly only have to check for $p$ up to a certain point because the space is finite-dimensional, right? But I don't know what that point is :-/ $\endgroup$ – Kevin Buzzard Jul 30 '10 at 12:25