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Theta function squared is a weight $1$ modular form

eisenstein part of theta functionModular forms on the theta groupOn the square coeffecients of a modular formModular forms on $Gamma_0(N)$ with character in SageReverse the twisting of modular formCusp condition of modular forms of half-integral weightWriting an integer as a sum of two squares: Why is this a modular form of weight 1?*WHY* are half-integral weight modular forms defined on congruence subgroups of level 4N?Hecke operator on half integral weight modular formGenerating function for sum of two squares is a modular form

2

$begingroup$

Let
$$vartheta(tau) = sum_{ninmathbb{Z}}e^{pi in^2tau}.$$
I know that $vartheta$ satisfies the transfromation properties
$$vartheta(tau + 2) = vartheta(tau), quad varthetaleft(-frac{1}{tau}right) = sqrt{frac{tau}{i}}vartheta(tau).$$
What I am interested in is the transformation properties of
$$f(tau) = vartheta(2tau)^2.$$
I have seen it mentioned in some notes by Kevin Buzzard and elsewhere that $f(tau)$ is a weight $1$ modular form on the congruence subgroup $Gamma_0(4)$ with Nebentypus $chi$, the Dirichlet character coming from the Legendre symbol mod $4$. In other words,
$$fleft(frac{atau + b}{ctau + d}right) = chi(d)(ctau + d)f(tau), quad begin{pmatrix} a & b \ c & d end{pmatrix} in Gamma_0(4).$$
I cannot find a reference for this fact nor does it seem obvious from the transformation properties of $vartheta(tau)$. I know that given a modular form $g(tau)$ of weight $k$ on $Gamma_0(N)$ for some $N$, the function $g(2tau)^2$ is modular of weight $2k$ on $Gamma_0(2N)$. But the problem is that the subgroup on which $vartheta(tau)$ is modular (of weight $1/2$) is not one of the congruence subgroups $Gamma_0(N)$. I would greatly appreciate any help tracking down a reference for this fact.

number-theory modular-forms theta-functions

share|cite|improve this question

asked Mar 7 at 4:58

Ethan AlwaiseEthan Alwaise

6,446717

$endgroup$

add a comment | 

2

$begingroup$

Let
$$vartheta(tau) = sum_{ninmathbb{Z}}e^{pi in^2tau}.$$
I know that $vartheta$ satisfies the transfromation properties
$$vartheta(tau + 2) = vartheta(tau), quad varthetaleft(-frac{1}{tau}right) = sqrt{frac{tau}{i}}vartheta(tau).$$
What I am interested in is the transformation properties of
$$f(tau) = vartheta(2tau)^2.$$
I have seen it mentioned in some notes by Kevin Buzzard and elsewhere that $f(tau)$ is a weight $1$ modular form on the congruence subgroup $Gamma_0(4)$ with Nebentypus $chi$, the Dirichlet character coming from the Legendre symbol mod $4$. In other words,
$$fleft(frac{atau + b}{ctau + d}right) = chi(d)(ctau + d)f(tau), quad begin{pmatrix} a & b \ c & d end{pmatrix} in Gamma_0(4).$$
I cannot find a reference for this fact nor does it seem obvious from the transformation properties of $vartheta(tau)$. I know that given a modular form $g(tau)$ of weight $k$ on $Gamma_0(N)$ for some $N$, the function $g(2tau)^2$ is modular of weight $2k$ on $Gamma_0(2N)$. But the problem is that the subgroup on which $vartheta(tau)$ is modular (of weight $1/2$) is not one of the congruence subgroups $Gamma_0(N)$. I would greatly appreciate any help tracking down a reference for this fact.

number-theory modular-forms theta-functions

share|cite|improve this question

asked Mar 7 at 4:58

Ethan AlwaiseEthan Alwaise

6,446717

$endgroup$

add a comment | 

$begingroup$

Let
$$vartheta(tau) = sum_{ninmathbb{Z}}e^{pi in^2tau}.$$
I know that $vartheta$ satisfies the transfromation properties
$$vartheta(tau + 2) = vartheta(tau), quad varthetaleft(-frac{1}{tau}right) = sqrt{frac{tau}{i}}vartheta(tau).$$
What I am interested in is the transformation properties of
$$f(tau) = vartheta(2tau)^2.$$
I have seen it mentioned in some notes by Kevin Buzzard and elsewhere that $f(tau)$ is a weight $1$ modular form on the congruence subgroup $Gamma_0(4)$ with Nebentypus $chi$, the Dirichlet character coming from the Legendre symbol mod $4$. In other words,
$$fleft(frac{atau + b}{ctau + d}right) = chi(d)(ctau + d)f(tau), quad begin{pmatrix} a & b \ c & d end{pmatrix} in Gamma_0(4).$$
I cannot find a reference for this fact nor does it seem obvious from the transformation properties of $vartheta(tau)$. I know that given a modular form $g(tau)$ of weight $k$ on $Gamma_0(N)$ for some $N$, the function $g(2tau)^2$ is modular of weight $2k$ on $Gamma_0(2N)$. But the problem is that the subgroup on which $vartheta(tau)$ is modular (of weight $1/2$) is not one of the congruence subgroups $Gamma_0(N)$. I would greatly appreciate any help tracking down a reference for this fact.

number-theory modular-forms theta-functions

share|cite|improve this question

asked Mar 7 at 4:58

Ethan AlwaiseEthan Alwaise

6,446717

$endgroup$

share|cite|improve this question

asked Mar 7 at 4:58

Ethan AlwaiseEthan Alwaise

6,446717

share|cite|improve this question

asked Mar 7 at 4:58

Ethan AlwaiseEthan Alwaise

6,446717

asked Mar 7 at 4:58

Ethan AlwaiseEthan Alwaise

6,446717

asked Mar 7 at 4:58

Ethan AlwaiseEthan Alwaise

6,446717

1 Answer
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1

$begingroup$

$g = vartheta^2(2tau)$ is invariant under $Gamma_1(4)$ is easily checked: since $Gamma_1(4)$ is generated by
$$begin{pmatrix}1 & 1 \ 0 & 1 end{pmatrix}qquad begin{pmatrix}1 & 0 \ 4 & 1 end{pmatrix}$$
Obviously $g$ is weight-$1$ invariant under the first one. For second one,
$$begin{aligned}gleft[begin{pmatrix}1 & 0 \ 4 & 1 end{pmatrix}right]_1 (tau)&= (4tau +1)^{-1}vartheta^2(frac{2tau}{4tau+1}) \&= (4tau +1)^{-1} frac{i(4tau+1)}{2tau}vartheta^2(-frac{4tau+1}{2tau})
\& = frac{i}{2tau}vartheta^2(frac{-1}{2tau}) = vartheta^2(2tau)
end{aligned}$$

To check $g$ has nebentypus $chi$, it suffices to show $$gleft[begin{pmatrix}-1 & 0 \ 4 & -1 end{pmatrix}right]_1(tau)=-g(tau)$$
I shall leave this checking to you.

share|cite|improve this answer

edited Mar 12 at 5:28

answered Mar 12 at 5:19

piscopisco

11.9k21743

$endgroup$

add a comment | 

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1

$begingroup$

$g = vartheta^2(2tau)$ is invariant under $Gamma_1(4)$ is easily checked: since $Gamma_1(4)$ is generated by
$$begin{pmatrix}1 & 1 \ 0 & 1 end{pmatrix}qquad begin{pmatrix}1 & 0 \ 4 & 1 end{pmatrix}$$
Obviously $g$ is weight-$1$ invariant under the first one. For second one,
$$begin{aligned}gleft[begin{pmatrix}1 & 0 \ 4 & 1 end{pmatrix}right]_1 (tau)&= (4tau +1)^{-1}vartheta^2(frac{2tau}{4tau+1}) \&= (4tau +1)^{-1} frac{i(4tau+1)}{2tau}vartheta^2(-frac{4tau+1}{2tau})
\& = frac{i}{2tau}vartheta^2(frac{-1}{2tau}) = vartheta^2(2tau)
end{aligned}$$

To check $g$ has nebentypus $chi$, it suffices to show $$gleft[begin{pmatrix}-1 & 0 \ 4 & -1 end{pmatrix}right]_1(tau)=-g(tau)$$
I shall leave this checking to you.

share|cite|improve this answer

edited Mar 12 at 5:28

answered Mar 12 at 5:19

piscopisco

11.9k21743

$endgroup$

add a comment | 

1

$begingroup$

$g = vartheta^2(2tau)$ is invariant under $Gamma_1(4)$ is easily checked: since $Gamma_1(4)$ is generated by
$$begin{pmatrix}1 & 1 \ 0 & 1 end{pmatrix}qquad begin{pmatrix}1 & 0 \ 4 & 1 end{pmatrix}$$
Obviously $g$ is weight-$1$ invariant under the first one. For second one,
$$begin{aligned}gleft[begin{pmatrix}1 & 0 \ 4 & 1 end{pmatrix}right]_1 (tau)&= (4tau +1)^{-1}vartheta^2(frac{2tau}{4tau+1}) \&= (4tau +1)^{-1} frac{i(4tau+1)}{2tau}vartheta^2(-frac{4tau+1}{2tau})
\& = frac{i}{2tau}vartheta^2(frac{-1}{2tau}) = vartheta^2(2tau)
end{aligned}$$

To check $g$ has nebentypus $chi$, it suffices to show $$gleft[begin{pmatrix}-1 & 0 \ 4 & -1 end{pmatrix}right]_1(tau)=-g(tau)$$
I shall leave this checking to you.

share|cite|improve this answer

edited Mar 12 at 5:28

answered Mar 12 at 5:19

piscopisco

11.9k21743

$endgroup$

add a comment | 

$begingroup$

$g = vartheta^2(2tau)$ is invariant under $Gamma_1(4)$ is easily checked: since $Gamma_1(4)$ is generated by
$$begin{pmatrix}1 & 1 \ 0 & 1 end{pmatrix}qquad begin{pmatrix}1 & 0 \ 4 & 1 end{pmatrix}$$
Obviously $g$ is weight-$1$ invariant under the first one. For second one,
$$begin{aligned}gleft[begin{pmatrix}1 & 0 \ 4 & 1 end{pmatrix}right]_1 (tau)&= (4tau +1)^{-1}vartheta^2(frac{2tau}{4tau+1}) \&= (4tau +1)^{-1} frac{i(4tau+1)}{2tau}vartheta^2(-frac{4tau+1}{2tau})
\& = frac{i}{2tau}vartheta^2(frac{-1}{2tau}) = vartheta^2(2tau)
end{aligned}$$

To check $g$ has nebentypus $chi$, it suffices to show $$gleft[begin{pmatrix}-1 & 0 \ 4 & -1 end{pmatrix}right]_1(tau)=-g(tau)$$
I shall leave this checking to you.

share|cite|improve this answer

edited Mar 12 at 5:28

answered Mar 12 at 5:19

piscopisco

11.9k21743

$endgroup$

share|cite|improve this answer

edited Mar 12 at 5:28

answered Mar 12 at 5:19

piscopisco

11.9k21743

answered Mar 12 at 5:19

piscopisco

11.9k21743

answered Mar 12 at 5:19

piscopisco

11.9k21743