Hecke Operators and Eigenfunctions, Fourier coefficients

[binbagsss]

Homework Statement

Consider the action of $T_2$ acting on $M_k(\Gamma_{0}(N))$, and show that $\theta^4(n)+16F$ and $F(t)$ are both eigenfunctions.

Functions are given by:

Homework Equations

For the Hecke Operators $T_p$ acting on $M_k(\Gamma_{0}(N))$, the Hecke conguence subgroup of $M_k (SL_2(Z))$ , the coefficients are given as the same as acting on (SL_2(Z)) if $p$ does not divide the level $N$ $b_n=a_{pn}+p^{k-1}a_{n/p}$ but $b_n=a_{pn}$ if $p$ does divide $N$.

Where $T_p f(t) = \sum\limits^{\infty}_{n=0}a_nq^n$ and $f(t) = \sum\limits^{\infty}_{n=0} a_n q^n$
For this question $2$ does divide $4$ and so we are using: $b_n=a_{2n}$

The Attempt at a Solution

So we have $b_0 = a_0$
$b_1= a_2$
$b_2=a_4$
$b_4=a_6$

solution here:

I agree that $T_2 \theta^4(t)= \theta^4(t)+16F(t)$ from $b_0$ and $b_1$ since $b_0=a_0$ is given by $a_0$ of $\theta^4(t)$ since $f(t)$ only has odd coefficients. similarly $b_1=a_2$= the first coeffieicnt of $\theta^4(t)$ + 16 * the first coefficient of $F(t)$.

However can not see how this can be true for any even $n >2$. (Since $F(t)$ has only odd $n$ contributions.

Let me use the result that the $r_4(n)$ are given by $8 \Omega_1(n)$ for $n$ odd and $24 \sigma_1(n_0)$ for $n_0$ the odd divisors of $n$. So I get $r_4(4)=24(3+1)=96=a_4$ and so since $T_2 \theta^4(t)= \theta^4(t)+16F(t) => b_2 = a_4$. But now since $F(t)$ only has odd coefficients, this implies that $a_2 = b_2=a_4$ which of course it isn't.

Am i doing something really stupid?
Many thanks

 

[binbagsss]

Homework Statement

Consider the action of $T_2$ acting on $M_k(\Gamma_{0}(N))$, and show that $\theta^4(n)+16F$ and $F(t)$ are both eigenfunctions.

Functions are given by:View attachment 224651

Homework Equations

For the Hecke Operators $T_p$ acting on $M_k(\Gamma_{0}(N))$, the Hecke conguence subgroup of $M_k (SL_2(Z))$ , the coefficients are given as the same as acting on (SL_2(Z)) if $p$ does not divide the level $N$ $b_n=a_{pn}+p^{k-1}a_{n/p}$ but $b_n=a_{pn}$ if $p$ does divide $N$.

Where $T_p f(t) = \sum\limits^{\infty}_{n=0}a_nq^n$ and $f(t) = \sum\limits^{\infty}_{n=0} a_n q^n$
For this question $2$ does divide $4$ and so we are using: $b_n=a_{2n}$

The Attempt at a Solution

So we have $b_0 = a_0$
$b_1= a_2$
$b_2=a_4$
$b_4=a_6$

solution here: View attachment 224652View attachment 224652

I agree that $T_2 \theta^4(t)= \theta^4(t)+16F(t)$ from $b_0$ and $b_1$ since $b_0=a_0$ is given by $a_0$ of $\theta^4(t)$ since $f(t)$ only has odd coefficients. similarly $b_1=a_2$= the first coeffieicnt of $\theta^4(t)$ + 16 * the first coefficient of $F(t)$.

However can not see how this can be true for any even $n >2$. (Since $F(t)$ has only odd $n$ contributions.

Let me use the result that the $r_4(n)$ are given by $8 \Omega_1(n)$ for $n$ odd and $24 \sigma_1(n_0)$ for $n_0$ the odd divisors of $n$. So I get $r_4(4)=24(3+1)=96=a_4$ and so since $T_2 \theta^4(t)= \theta^4(t)+16F(t) => b_2 = a_4$. But now since $F(t)$ only has odd coefficients, this implies that $a_2 = b_2=a_4$ which of course it isn't.

Am i doing something really stupid?
Many thanks

Apolgies I have made a typo in the OP where it reads $T_p f(t) = \sum\limits^{\infty}_{n=0}a_nq^n$ , it should read $T_p f(t) = \sum\limits^{\infty}_{n=0}b_nq^n$ ,

I am still stuck on this so any help much appreciated, thanks