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Theta Functions Tutorial


Post Fri Jul 26, 2013 8:03 am
Shobhit Site Admin
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This tutorial is in progress. You can send me your suggestions if any. The solutions to exercises will be posted in the end.

1. The Theta Functions

The four theta functions are defined as

\(\displaystyle \begin{align*}
\vartheta_1(z,q) & = -i \sum_{n=-\infty}^\infty (-1)^n q^{n(n+1/2)^2}e^{i (2n+1)z} = 2\sum_{n=0}^\infty (-1)^n q^{(n+1/2)^2}\sin(2n+1)z \tag{1}\\
\vartheta_2(z,q) & = \sum_{n=-\infty}^\infty q^{n(n+1/2)^2}e^{i (2n+1)z} = 2\sum_{n=0}^\infty q^{(n+1/2)^2}\cos(2n+1)z \tag{2}\\
\vartheta_3(z,q) & = \sum_{n=-\infty}^\infty q^{n^2}e^{2in z} = 1+2\sum_{n=1}^\infty q^{n^2} \cos(2nz) \tag{3}\\
\vartheta_4(z,q) & = \sum_{n=-\infty}^\infty (-1)^n q^{n^2}e^{2in z} =1+ 2\sum_{n=0}^\infty (-1)^n q^{n^2}\cos(2nz)\tag{4}
\end{align*}\)

where \(z\) and \(q\) are complex numbers. To ensure convergence, we let \(|q|<1\). An alternative notation is to write
\[q=e^{i \pi \tau}\]
\(q\) is called the nome of the Theta Function and \(\tau\) is it's parameter.

1.1 Identities involving products of Theta Functions

A lot of interesting identities of the theta functions can be obtained by the multiplication of two of their series, followed by rearrangement of terms in the product series.

Problem
Show that
\[\vartheta_1(x,q) \vartheta_1(y,q) = \vartheta_3(x+y,q^2)\vartheta_2(x-y,q^2)-\vartheta_2 (x+y,q^2)\vartheta_3(x-y,q^2)\]

Solution

We have
\[\vartheta_1(x,q) \vartheta_1(y,q) =- \sum_{m=-\infty}^\infty \sum_{n=-\infty}^\infty (-1)^{m+n} q^{(m+1/2)^2+(n+1/2)^2}e^{i(2m+1)x+i(2n+1)y}\]

We now change the integer variables of summation from \((m,n)\) to \((r,s)\) by the following transformation equations
\[m+n=r, \quad m-n=s\]
If \((m,n)\) are both even or odd, \((r,s)\) will both be even. If, however, \((m,n)\) have opposite parity, \((r,s)\) will both be odd. Hence, by permitting \((r,s)\) to range over pairs of even integers and over all pairs of odd integers, every pair of integers \((m,n)\) will be obtained exactly once.

Thus, we can write
\[\begin{align*}
\left( m+1/2\right)^2+\left(n+1/2 \right)^2 &= \frac{1}{2}(r+1)^2+\frac{s^2}{2} \\ \\
(2m+1)x+(2n+1)y &= (r+1)(x+y)+s(x-y)
\end{align*}\]
So
\[\vartheta_1(x,q) \vartheta_1(y,q) =-\sum (-1)^r q^{(1/2)(r+1)^2+(1/2)s^2}e^{i(r+1)(x+y)+is(x-y)}\]
The summation is taken over all even and odd pairs of \((r,s)\), as just explained. It follows that
\[\vartheta_1(x,q) \vartheta_1(y,q) = -\sum_r \sum_s q^{2(r+1/2)^2+2s^2}e^{i(2r+1)(x+y)+2is(x-y)}+\sum_r\sum_s q^{2r^2+2(s+1/2)^2}e^{2ir(x+y)+i(2s+1)(x-y)}\]
where the summation are now to be taken over all pairs of integers. Now, each of the double series can be written as products of two single series, as follows

\(\displaystyle \begin{align*}
\vartheta_1(x,q) \vartheta_1(y,q) &= -\sum_r q^{2(r+1/2)^2+2s^2}e^{i(2r+1)(x+y)}\sum_s q^{2s^2}e^{2is(x-y)} \\
&\quad +\sum_r q^{2r^2}e^{2ir (x+y)}\sum_s q^{2(s+1/2)^2}e^{i(2s+1)(x-y)} \\
&= \vartheta_3(x+y,q^2)\vartheta_2(x-y,q^2)-\vartheta_2 (x+y,q^2)\vartheta_3(x-y,q^2) \tag{5}
\end{align*}\)

Similarly, the following identities can be established.

\(\displaystyle \begin{align*}
\vartheta_1(x,q) \vartheta_2(y,q) &= \vartheta_1(x+y,q^2)\vartheta_4(x-y,q^2)+\vartheta_4 (x+y,q^2)\vartheta_1(x-y,q^2) \tag{6} \\
\vartheta_2(x,q) \vartheta_2(y,q) &= \vartheta_2(x+y,q^2)\vartheta_3(x-y,q^2)+\vartheta_3 (x+y,q^2)\vartheta_4(x-y,q^2) \tag{7} \\
\vartheta_3(x,q) \vartheta_3(y,q) &= \vartheta_3(x+y,q^2)\vartheta_3(x-y,q^2)+\vartheta_2 (x+y,q^2)\vartheta_2(x-y,q^2) \tag{8} \\
\vartheta_3(x,q) \vartheta_4(y,q) &= \vartheta_4(x+y,q^2)\vartheta_4(x-y,q^2)-\vartheta_1 (x+y,q^2)\vartheta_1(x-y,q^2) \tag{9} \\
\vartheta_4(x,q) \vartheta_4(y,q) &= \vartheta_3(x+y,q^2)\vartheta_3(x-y,q^2)-\vartheta_2 (x+y,q^2)\vartheta_2(x-y,q^2) \tag{10}
\end{align*}\)

These are left as an exercise for you.

Shobhit Site Admin
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Posts: 852
Location: Jaipur, India

1.2 Identities of the Basic Theta Functions

Define
\[\vartheta_i(q) \equiv \vartheta_i(z=0,q)\]
Now, by equations (2), (3) and (4) we see that
\[\begin{align*}
\vartheta_2(q) &= \sum_{n=-\infty}^\infty q^{(n+1/2)^2} \tag{11}\\
\vartheta_3(q) &= \sum_{n=-\infty}^\infty q^{n^2} \tag{12}\\
\vartheta_4(q) &= \sum_{n=-\infty}^\infty (-1)^n q^{n^2} \tag{13}
\end{align*}\]
Adding equations (12) and (13) we obtain
\[\vartheta_3(q)+\vartheta_4(q) = 2 \sum_{n \text{ even}} q^{n^2} =2\vartheta_3(q^4) \tag{15}\]
Also, by squaring equations (12) and (13) we see that
\[\begin{align*} \vartheta_3^2(q) &= \sum_{n=0}^\infty r_2(n) q^n \tag{16} \\ \vartheta_4^2(q) &=\sum_{n=0}^\infty (-1)^n r_2(n) q^n\tag{17} \end{align*}\]
where \(r_2(n)\) counts the number of ways of writing \(n=j^2+k^2\). Here we distinguish sign and permutation so that for example, \(r_2(5)=8\) since \((\pm 2)^2+(\pm 1)^2 = (\pm 1)^2+(\pm 2)^2\) and set \(r_2(0)=1\).

Note that \(r_2(2n)=r_2(n)\). This may be proved elementarily by knowing that \(2(a^2+b^2)=(a+b)^2+(a-b)^2\). Thus by adding equations (16) and (17) we obtain the beautiful identity
\[\begin{align*}
\vartheta_3^2(q)+\vartheta_4^2(q) &= 2 \sum_{n=0}^\infty r_2(2n) q^{2n} \\
&= 2\sum_{n=0}^\infty r_2(n) q^{2n} \\
&= 2\vartheta_3^2(q^2) \tag{18}
\end{align*}\]
Three more important relations are obtained by setting \(x,y=0\) in equations (8),(9) and (10):
\[\begin{align*}
\vartheta_3^2(q^2)+\vartheta_2^2(q^2) &=\vartheta_3^2(q) \tag{19}\\
\vartheta_3(q) \vartheta_4(q) &=\vartheta_4^2(q^2) \tag{20} \\
\vartheta_3^2(q^2)-\vartheta_2^2(q^2) &= \vartheta_4^2(q) \tag{21}
\end{align*}\]

Problem
Prove the Jacobi's Identity:
\[\vartheta_3^4(q)=\vartheta_4^4(q)+\vartheta_2^4(q) \tag{22}\]
Solution
Multiply equations (19) and (21) and then use identity (20).

Shobhit Site Admin
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Posts: 852
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1.3 Periodicity Properties

Sometimes, the alternate notation
\[q=e^{i\pi \tau}\]
is used to express the theta function. Thus, by (1) we can write
\[i\vartheta_1(z|\tau) = \sum_{n=-\infty}^\infty (-1)^n e^{(n+1/2)^2i\pi \tau +(2n+1)iz}\]
Now, increment the argument \(z\) by \(\pi \tau\).
\[\begin{align*} \vartheta_1(z+\pi \tau |\tau) &= -i \sum_{n=-\infty}^\infty (-1)^n e^{(n+1/2)^2 i\pi \tau +i(2n+1)(z+\pi \tau)} \\
&= i e^{-i\pi \tau-2iz} \sum_{n=-\infty}^\infty (-1)^{n+1}e^{(n+3/2)^2 i\pi \tau +i(2n+3)z} \\
&= -(q e^{2iz})^{-1} \vartheta_1(z|\tau)
\end{align*}\]
\(\pi \tau\) is termed as a quasi-period of \(\vartheta_1(z|\tau)\) with periodicity factor \(-(q e^{biz})^{-1}\). Writing \(\lambda = qe^{2iz}\), the following identities may be obtained similarly:

\(\displaystyle \begin{align*}
\vartheta_1(z|\tau) &= -\vartheta_1(z+\pi|\tau) = -\lambda \vartheta_1(z+\pi \tau|\tau) = \lambda \vartheta_1(z+\pi+\pi \tau|\tau) \tag{23}\\
\vartheta_2(z|\tau) &= -\vartheta_2(z+\pi|\tau) = \lambda \vartheta_2(z+\pi \tau|\tau) = -\lambda \vartheta_2(z+\pi+\pi \tau|\tau) \tag{24}\\
\vartheta_3(z|\tau) &= \vartheta_3(z+\pi|\tau) = \lambda \vartheta_3(z+\pi \tau|\tau) = \lambda \vartheta_3(z+\pi+\pi \tau|\tau) \tag{25} \\
\vartheta_4(z|\tau) &= \vartheta_4(z+\pi|\tau) = -\lambda \vartheta_4(z+\pi \tau|\tau) = -\lambda \vartheta_4(z+\pi+\pi \tau|\tau) \tag{26}
\end{align*}\)

Exercise

By incrementing \(z\) by the half periods \(\frac{\pi}{2},\frac{1}{2}\pi \tau\) and \(\frac{1}{2}(\pi +\pi \tau)\), verify the following identities:

\[\begin{align*}
\vartheta_1(z|\tau) &= - \vartheta_2(z+\pi/2|\tau) =-i \mu \vartheta_4(z+\pi \tau/2|\tau)=-i\mu \vartheta_3(z+(\pi+\pi\tau)/2|\tau) \tag{27} \\
\vartheta_2(z|\tau) &= \vartheta_1(z+\pi/2|\tau) = \mu \vartheta_3(z+\pi \tau/2|\tau)=\mu \vartheta_4(z+(\pi+\pi\tau)/2|\tau) \tag{28} \\
\vartheta_3(z|\tau) &= \vartheta_4(z+\pi/2|\tau) = \mu \vartheta_2(z+\pi \tau/2|\tau)=\mu \vartheta_1(z+(\pi+\pi\tau)/2 |\tau) \tag{29} \\
\vartheta_4(z|\tau) &= \vartheta_3(z+\pi/2|\tau) =-i \mu \vartheta_1(z+\pi \tau/2|\tau)=i\mu \vartheta_2(z+(\pi+\pi\tau)/2|\tau) \tag{27}
\end{align*}\]

where \(\mu =q^{1/4}e^{iz}\)

Shobhit Site Admin
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Posts: 852
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1.4 Further Identities of Theta Functions

Now, we are going to deduce a large number of identities involving products of four theta functions, all having the same nome.

For example, squaring and subtracting (5) and (10) we get

\(\displaystyle \vartheta_4^2(x,q)\vartheta_4^2(y,q) = [\vartheta_3^2(x+y,q^2)-\vartheta_2^2(x+y,q^2)]\times [\vartheta_3^2(x-y,q^2)-\vartheta_2^2(x+y,q^2)] \tag{28}\)

Putting \(y=0\), gives

\(\displaystyle \vartheta_3^2(x,q^2)-\vartheta_2^2(x,q^2)=\vartheta_4(x,q) \vartheta_4(0,q) \tag{29}\)

We now deduce from (28) that

\(\displaystyle \vartheta_4(x+y,q)\vartheta_4(x-y,q) \vartheta_4^2(0,q) = \vartheta_4^2(x,q)\vartheta_4^2(y,q)-\vartheta_1^2(x,q)\vartheta_1^2(y,q) \tag{29}\)

In a similar manner, we obtain the following set of identities:

\(\displaystyle \begin{align*}
\vartheta_1(x+y,q)\vartheta_1(x-y,q) \vartheta_3^2(0,q) &= \vartheta_1^2(x,q)\vartheta_3^2(y,q)-\vartheta_3^2(x,q)\vartheta_3^2(y,q) \tag{30} \\
\vartheta_2(x+y,q)\vartheta_2(x-y,q) \vartheta_3^2(0,q) &= \vartheta_2^2(x,q)\vartheta_3^2(y,q)-\vartheta_4^2(x,q)\vartheta_1^2(y,q) \tag{31} \\
\vartheta_3(x+y,q)\vartheta_3(x-y,q) \vartheta_3^2(0,q) &= \vartheta_1^2(x,q)\vartheta_1^2(y,q)+\vartheta_3^2(x,q)\vartheta_3^2(y,q) \tag{32} \\
\vartheta_4(x+y,q)\vartheta_4(x-y,q) \vartheta_3^2(0,q) &= \vartheta_1^2(x,q)\vartheta_2^2(y,q)+\vartheta_3^2(x,q)\vartheta_4^2(y,q) \tag{33} \\
\end{align*}\)

Shobhit Site Admin
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Posts: 852
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Exercise

Show that

\(\displaystyle \begin{align*}
\vartheta_3^2(x,q)\vartheta_3^2(0,q) &= \vartheta_4^2(x,q)\vartheta_4^2(0,q)+\vartheta_2^2(x,q)\vartheta_2^2(0,q) \tag{34}\\
\vartheta_4^2(x,q)\vartheta_3^2(0,q) &= \vartheta_1^2(x,q)\vartheta_2^2(0,q)+\vartheta_3^2(x,q)\vartheta_4^2(0,q) \tag{35}\\
\vartheta_4^2(x,q)\vartheta_2^2(0,q) &= \vartheta_1^2(x,q)\vartheta_3^2(0,q)+\vartheta_2^2(x,q)\vartheta_4^2(0,q) \tag{36}\\
\vartheta_3^2(x,q)\vartheta_2^2(0,q) &= \vartheta_1^2(x,q)\vartheta_4^2(0,q)+\vartheta_2^2(x,q)\vartheta_3^2(0,q) \tag{37}\\
\end{align*}\)

1.5 The identity \(\vartheta_1'(0,q)=\vartheta_2(0,q)\vartheta_3(0,q)\vartheta_4(0,q)\)

Differentiating equation (6) with respect to \(x\) and then putting \(x=y=0\), we have that
\[\vartheta_1'(0,q)\vartheta_2(0,q)=2\vartheta_1'(0,q^2)\vartheta_4(0,q^2)\tag{*}\]
Next, putting \(x=y=0\) in (7) and (9) we show that
\[
\begin{align*}
\vartheta_2^2(0,q) &= 2\vartheta_2(0,q^2)\vartheta_4(0,q^2) \\
\vartheta_3(0,q)\vartheta_4(0,q) &= \vartheta_4^2(0,q^2)
\end{align*}
\]
Dividing (*) by these two equations
\[\frac{\vartheta_1'(0,q)}{\vartheta_2(0,q)\vartheta_3(0,q)\vartheta_4(0,q)}=\frac{\vartheta_1'(0,q^2)}{\vartheta_2(0,q^2)\vartheta_3(0,q^2)\vartheta_4(0,q^2)}\]
By repeated application
\[\frac{\vartheta_1'(0,q)}{\vartheta_2(0,q)\vartheta_3(0,q)\vartheta_4(0,q)}=\frac{\vartheta_1'(0,q^{2^n})}{\vartheta_2(0,q^{2^n})\vartheta_3(0,q^{2^n})\vartheta_4(0,q^{2^n})}\]
Letting \(n\to \infty\) in this identity, \(q^{2^n} \to 0\). Hence
\[\frac{\vartheta_1'(0,q)}{\vartheta_2(0,q)\vartheta_3(0,q)\vartheta_4(0,q)}=\lim_{q\to 0}\frac{\vartheta_1'(0,q)}{\vartheta_2(0,q)\vartheta_3(0,q)\vartheta_4(0,q)}\tag{**}\]
Now, from equations (1)-(4), we find that
\[
\begin{align*}
\vartheta_1'(0,q) &=2q^{1/4}+\mathcal{O}(q^{9/4}) \\
\vartheta_2(0,q) &=2q^{1/4}+\mathcal{O}(q^{9/4}) \\
\vartheta_3(0,q) &=1+\mathcal{O}(q) \\
\vartheta_4(0,q) &=1+\mathcal{O}(q)
\end{align*}
\]
Substituting these in (**) and simplifying the limit we get
\[\vartheta_1'(0,q)=\vartheta_2(0,q)\vartheta_3(0,q)\vartheta_4(0,q) \tag{38}\]

Shobhit Site Admin
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Posts: 852
Location: Jaipur, India

1.6 Infinite Product Representation of Theta Functions

We have

\(\displaystyle \begin{align*}
\vartheta_1(z,q) &= 2q^{\frac{1}{4}}\sin z \prod_{n=1}^\infty (1-q^{2n})(1-2q^{2n}\cos(2z)+q^{4n}) \tag{39}\\
\vartheta_2(z,q) &= 2q^{\frac{1}{4}}\cos z \prod_{n=1}^\infty (1-q^{2n})(1+2q^{2n}\cos(2z)+q^{4n}) \tag{40}\\
\vartheta_3(z,q) &= \prod_{n=1}^\infty (1-q^{2n})(1+2q^{2n-1}\cos(2z)+q^{4n-2}) \tag{41}\\
\vartheta_4(z,q) &= \prod_{n=1}^\infty (1-q^{2n})(1-2q^{2n-1}\cos(2z)+q^{4n-2}) \tag{42}\\
\end{align*}\)

These can be proved using the Jacobi Triple Product identity:
\[\sum_{n=-\infty}^\infty x^{n^2}y^{n} = \prod_{m=1}^\infty \left(1-x^{2m} \right)\left(1-x^{2m-1} y\right)\left(1-x^{2m-1} y^{-1}\right) \tag{43}\]

There are many methods to prove Jacobi Triple Product but we shall not discuss it's proof in this tutorial.


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