On Abrikosov Lattice Solutions of the Ginzburg-Landau Equations

Abstract

We prove existence of Abrikosov vortex lattice solutions of the Ginzburg-Landau equations of superconductivity, with multiple magnetic flux quanta per fundamental cell. We also revisit the existence proof for the Abrikosov vortex lattices, streamlining some arguments and providing some essential details missing in earlier proofs for a single magnetic flux quantum per a fundamental cell.

Appendices

Appendix

Recall that x and z are related as in (5.11) and identify x = (x₁,x₂) with x₁ + ix₂.

A On Solutions of the Linearised Problem

Lemma A.1

There is no linear solution ψ,as in Section 5, such that

$$ \psi(\bar{x}) = e^{ig_{r}(z)} \psi(x) $$

(A.1)

for some real valued g.

Proof

Assume for the sake of contradiction that suchψ exists. Then

$$ \psi(x) = e^{\frac{n}{4}(z^{2}-|z|^{2})} \theta(z) $$

(A.2)

for some holomorphic 𝜃. Equation (A.1) becomes

$$ \theta(\bar{z}) = e^{\frac{n}{4}(z^{2}-\bar{z}^{2})} e^{ig_{r}(z)} \theta(z) $$

(A.3)

Taking ∂ _z on both sides, we see that

$$ 0 = e^{\frac{n}{4}(z^{2}-\bar{z}^{2})} e^{ig_{r}(z)} \left( -\frac{n}{2}\bar{z} \theta+i\theta \partial_{z} g_{r} + \theta^{\prime}\right) $$

(A.4)

This shows that the term in the bracket vanishes identically. In particular,

$$ \left( -\frac{n}{2}{z} +i \partial_{z} g_{r}\right)\theta = -\theta^{\prime} $$

(A.5)

Taking $\partial _{\bar {z}}$again, we see that

$$ \left( + i \partial_{\bar{z}}\partial_{z} g_{r}\right) \theta = 0 $$

(A.6)

Since 𝜃 has at most finitely many zeros, we conclude

$$ -{\Delta} g_{r} = 0 $$

(A.7)

This shows that g _r is harmonic. Since it is periodic, it is a constant. Equation (A.3) then shows that such solution isimpossible. □

B Theta Functions

In this appendix we review basic properties of theta functions, which are likely to be known but which we could not find in the literature. From now on, we fix a lattice shape τ and a lattice

$$ \mathcal{L}_{\tau} = \mathbb{Z} + \tau \mathbb{Z} $$

(B.1)

throughout this appendix (unless otherwise stated).

3.1 B.1 Basic Properties

In this section, we prove some basic properties of the theta functions. Let n be fixed. Define for 0 ≤ m ≤ n − 1,

$$\begin{array}{@{}rcl@{}} \theta_{n,m}(z) = \sum\limits_{l \in [m]_{n}} \gamma^{l^{2}} e^{2\pi i l z} \end{array} $$

(B.2)

where γ := e^πiτ/n and $[m]_{n} = \{ a \in \mathbb {Z} \ : \ a = m \bmod n \}$.

Theorem B.1

The 𝜃 _{n, m} ’s form a basis for V _n that satisfy

• 1. $\theta _{n,m}(z+\frac {1}{n}) = e^{2\pi i m/n} \theta _{n,m}(z)$

• 2. 𝜃_n,m(−z) = 𝜃_n,n−m(z)

• 3. 𝜃_n,m(z + τ/n) = γ^− 1e^2πiz𝜃_{n, m+ 1}(z)

Theorem B.2

Any n-theta function has exactly n zeros modulo translation by lattice elements. Moreover, any two theta functions that share the same zeros (counting multiplicity) are linearly dependent.

Theorem B.3

𝜃 _1,0 has a simple zero at $\frac {1}{2}(1+\tau )$

Proof

See Proposition 11.1.□

Theorem B.4

Suppose that 𝜃 ∈ V _n and σ ∈ V _m , then 𝜃σ ∈ V_{n + m}.

Proof

Inspection.□

The proof of the theorems consists of the following lemmas:

Proof

of Theorem A.2 Expanding in e^2πikz for $k \in \mathbb {Z}$, the coefficients of any element of V _n satisfies the recurssion c_{m + n} = c _m e^{i(2m + n)πτ}. This recursion implies that for 0 ≤ m ≤ n − 1, we have that

$$ c_{m+ln} = c_{m} e^{i\pi\tau(l^{2}n + 2lm)} $$

(B.3)

where l is an integer. So the functions

$$ \sum\limits_{k \in \mathbb{Z}} e^{i\pi\tau(k^{2}n + 2km)} e^{2\pi i(nk +m)z} ,\ m = 0,...,n-1 $$

(B.4)

form a basis for the eigenspace. If we let l = kn + m, then we can rewrite the above as

$$ \sum\limits_{l \in [m]_{n}} e^{i\pi\tau \frac{l^{2}-m^{2}}{n}} e^{2\pi ilz} = e^{-i\pi m^{2}/n} \theta_{m} $$

(B.5)

Now we prove the three bullet points. We note that

$$ \theta_{m}\left( z+\frac{1}{n}\right) =\sum\limits_{l \in [m]_{n}} \gamma^{l^{2}} e^{2\pi i l z} e^{2\pi i l/n} $$

(B.6)

Since l ∈ [m] _n , we have that $l/n - m/n \in \mathbb {Z}$. Hence

$$ \theta_{m}\left( z+\frac{1}{n}\right) = e^{2\pi i m/n} \theta_{m}(z) $$

(B.7)

Now for the second item, we note

$$\begin{array}{@{}rcl@{}} \theta_{m}(-z) &=& \sum\limits_{l \in [m]_{n}} \gamma^{l^{2}} e^{-2\pi i l z} \end{array} $$

(B.8)

$$\begin{array}{@{}rcl@{}} &=& \sum\limits_{l \in [m]_{n}} \gamma^{(-l)^{2}} e^{2\pi i (-l) z} \end{array} $$

(B.9)

$$\begin{array}{@{}rcl@{}} &=& \sum\limits_{l \in [n-m]_{n}} \gamma^{l^{2}} e^{2\pi i l z} \end{array} $$

(B.10)

$$\begin{array}{@{}rcl@{}} &=& \theta_{n-m \bmod n}(z) \end{array} $$

(B.11)

Finally, recalling that γ = e^πiτ/n, we note that

$$\begin{array}{@{}rcl@{}} \theta_{m}(z+\tau/n) &=& \sum\limits_{k \in \mathbb{Z}} \gamma^{(kn+m)^{2}} e^{2\pi i(kn+m)z + 2\pi i(kn+m)\tau/n} \end{array} $$

(B.12)

$$\begin{array}{@{}rcl@{}} &=& \gamma^{-1} \sum\limits_{k \in \mathbb{Z}} \gamma^{k^{2}n^{2}+ 2knm+m^{2}+ 2kn+ 2m + 1} e^{2\pi i(kn+m)z} \end{array} $$

(B.13)

$$\begin{array}{@{}rcl@{}} &=& \gamma^{-1} \sum\limits_{k \in \mathbb{Z}} \gamma^{(kn+m + 1)^{2}} e^{2\pi i(kn+m)z} \end{array} $$

(B.14)

$$\begin{array}{@{}rcl@{}} &=& \gamma^{-1} e^{-2\pi i z} \theta_{m + 1 \bmod n}(z) \end{array} $$

(B.15)

□

Proof

of Theorem A.3 First we prove that elements of V _n has exactly n zeros modulo translation by lattice elements. We compute the winding number of𝜃. First, since 𝜃 is holomorphic, its zeros are discrete. Hence we may assume WLOG that all the zeros are in the interior of the fundamental domain. Let Ω denote the fundamental domain. Then the total number of zeros of 𝜃 is given by

$$ \frac{1}{2\pi i} {\int}_{\partial {\Omega}} \frac{\theta^{\prime}}{\theta} dz $$

(B.16)

Since 𝜃(z) = 𝜃(z + 1), the integral along the tτ and t τ + 1 for t ∈ [0, 1] is zero. Let y(z) = e^−iπτ e^αz where α = − 2πi. Since 𝜃(z + τ) = yⁿ𝜃(z) and y^′ = αy, we see that 𝜃^′(z + τ) = yⁿ(z)𝜃^′(z) + nαyⁿ(z)𝜃(z). Hence, only the horizonal segment of the line integral contribute:

$$\begin{array}{@{}rcl@{}} \frac{1}{2\pi i} {\int}_{\partial {\Omega}} \frac{\theta^{\prime}}{\theta} dz &\,=\,& \frac{1}{2\pi i}{{\int}_{0}^{1}} \frac{\theta^{\prime}(t)}{\theta(t)} - \frac{\theta^{\prime}(1\,+\,\tau \,-\, t)}{\theta(1\,+\,\tau\,-\,t)} dt \end{array} $$

(B.17)

$$\begin{array}{@{}rcl@{}} &\,=\,& \frac{1}{2\pi i} {{\int}_{0}^{1}} \frac{\theta^{\prime}(t)}{\theta(t)} - \frac{y^{n}(1\,-\,t) \theta^{\prime}(1\,-\, t) \,+\, n \alpha y^{n}(1\,-\,t) \theta(1\,-\,t)}{y^{n}(1\,-\,t) \theta(1\,-\,t)} dt \end{array} $$

(B.18)

$$\begin{array}{@{}rcl@{}} &\,=\,& \frac{1}{2\pi i} {{\int}_{0}^{1}} \frac{\theta^{\prime}(t)}{\theta(t)} - \frac{\theta^{\prime}(1\,-\, t) \,+\, n \alpha \theta(1\,-\,t)}{\theta(1\,-\,t)} dt \end{array} $$

(B.19)

$$\begin{array}{@{}rcl@{}} &\,=\,& \frac{1}{2\pi i} {{\int}_{0}^{1}} \frac{\theta^{\prime}(t)}{\theta(t)} - \frac{\theta^{\prime}(1\,-\, t)}{\theta(1\,-\,t)} dt \,+\, n \end{array} $$

(B.20)

$$\begin{array}{@{}rcl@{}} &\,=\,& n \end{array} $$

(B.21)

Next, we show that any two theta functions that share the same zeros (counting multiplicity) are linearly dependent. Let 𝜃 and φ be the two nonzero zeta functions that shares the same zeros. Set f(z) = 𝜃(a)/φ(z). Wes how that

• 1. f(z) can be extended analytically to all of $\mathbb {C}$ and

• 2. f(z) is doubly periodic.

Certainly f is holomorphic away from zeros of φ. We only need to show that f can be extended analytically to zeros of φ. But this is precisely the requirement that 𝜃 and φ share the same zeros (counting multiplicity).

For the second item, we note that

$$ f(z + 1) =\theta(z + 1)/\varphi(z + 1) = \theta(z)/\varphi(z) = f(z) $$

(B.22)

and

$$ f(z+\tau) = \frac{\theta(z+\tau)}{\varphi(z+\tau)} = \frac{e^{-2\pi i nz} e^{-\pi i n \tau} \theta(z)}{e^{-2\pi i nz} e^{-\pi i n \tau} \varphi(z)} = \frac{\theta(z)}{\varphi(z)} = f(z) $$

(B.23)

This shows that f is doubly periodic.

Now, Liouville’s theorem shows that f must be constant. It follows that 𝜃 and φ are collinear. □

3.2 B.2 Classification of Singular n-theta Functions

Theorem B.5

Let X _n be the set of singular n-theta functions mod scaling. Then

$$ X_{n} = \left\{ {\theta_{0}^{n}}\left( z + \frac{1}{n}(a+b\tau) \right) e^{2\pi i b z} : a,b \in \mathbb{Z} \right\} $$

(B.24)

where 𝜃₀ is a basis for V₁. Moreover, |X _n | = n². The location of zeros of elements in X _n form the set

$$ \frac{1}{2}(1+\tau) + \frac{1}{n}(\mathbb{Z} +\tau\mathbb{Z}) $$

(B.25)

As before, we establish the theorem through various lemmas. The idea of the proof is as follows: by Theorem A.3, we may identify elements of X _n with the location of their zeros. We attempt to locate the zeros of singular n-theta function first and show that there are only n² possible locations in a fundamental cell. So |X _n | = n². Then we explicitly construct n² singular n-theta functions to complete the proof.

To locate the zeros of singular n-theta functions, we study the Wronskian of a particular set of nice basis element: ${\Theta }(z) := \det (\theta ^{(i)}_{j} )$ for i,j ∈{0,...,n − 1}, where $\theta ^{(i)}_{j}$ means the i-th derivative of 𝜃 _j (see (B.2) for definition 𝜃 _j ).

Proposition B.6

The function Θ is holomorphic and

• 1. The locations of the zeros of Θ are exactly the locations where a singular n-theta function can have zero.

• 2. Θ(−z) = (− 1)^n+ 1Θ(z),

• 3. Θ(z + 1/n) = (− 1)^n+ 1Θ(z),

• 4. Θ(z + τ/n) = (− 1)^n+ 1γ^n(n− 1)yⁿΘ(z) where y = e^−iπτe^αz and α = − 2πi.

Proof

We recall that the 𝜃 _m ’s form a basis for V _n . If $\theta (z) = {\sum }_{m} a^{m} \theta _{m}(z)$ has n zeros at z₀, then

$$ 0 = \theta^{(i)}(z_{0}) = \sum\limits_{m} a^{m} \theta^{(i)}_{m}(z) $$

(B.26)

for i = 0,...,n − 1. So the matrix $(\theta ^{(i)}_{j}(z_{0}))$ has a nonzero vector (a⁰,...,a^n− 1) in its kernel. Hence Θ(z₀) = 0. Conversely, if Θ(z₀) = 0, then we can find a nonzero vector (a⁰,....,a^n− 1) in the kernel of the matrix $(\theta ^{(i)}_{j}(z_{0}))$. Then 𝜃 = a^m𝜃 _m has n-zeros at z₀.

Recall from Theorem A.2 that 𝜃 _m (−z) = 𝜃_{n−m mod n}(z). It follows that $\theta ^{(k)}_{m}(-z) = (-1)^{k}\theta ^{(k)}_{n-m \bmod n}(z)$. If n is even, then after z↦ − z, every even row in the matrix $(\theta ^{(i)}_{j})$picks up a minus sign, and moreover, we need to interchange the m-th collumn with the(n − m mod n)-th collumn for0 < m < n/2. Together wepick up n/2 + n/2 − 1 minus signs for Θ. So Θ(−z) = −Θ(z). If n is odd, we pick up(n − 1)/2 minus signs from the even rows and need to interchange (n − 1)/2 columns. So Θ(−z) = Θ(z).

Recall from Theorem A.2 that 𝜃 _m (z + 1/n) = ζ^m𝜃 _m (z) where ζ = e^2πi/n. It follows after z↦z + 1/n, the m-th column of $(\theta ^{(i)}_{j})$picks up a factor of ζ^m− 1. Hence ${\Theta }(z + 1/n) = \zeta ^{{\sum }_{k = 0}^{n-1} k}{\Theta }(z) = (-1)^{n + 1}{\Theta }(z)$.

Finally, we recall from Theorem A.2 and the definition y = e^−iπτ e^2πiz = γ⁻ⁿ e^2πiz that

$$\begin{array}{@{}rcl@{}} \theta_{m}(z+\tau/n) &=& \gamma^{-1} e^{2\pi i z} \theta_{m + 1 \bmod n}(z) \end{array} $$

(B.27)

$$\begin{array}{@{}rcl@{}} &=& \gamma^{-1} \gamma^{n} y\theta_{m + 1 \bmod n}(z) \end{array} $$

(B.28)

$$\begin{array}{@{}rcl@{}} &=& \gamma^{n-1} \theta_{m + 1 \bmod n}(z) \end{array} $$

(B.29)

Repeated differentiation shows that

$$ \theta_{m}^{(k)}(z+\tau) = \gamma^{n-1} \sum\limits_{i = 0}^{k} {k \choose i} (y)^{(i)} \theta_{m + 1}^{(k-i)}(z) $$

(B.30)

Hence

$$ \left( \theta^{(i)}_{j}(z+\tau/n)\right) = \gamma^{n-1} E\left( \begin{array}{cccccc} y \\ (y)^{\prime} & y \\ (y)^{\prime\prime} & 2(y)\prime & y \\ {\vdots} &&& {\ddots} \\ (y)^{n} && {\cdots} && y \end{array} \right) \left( \theta^{(i)}_{j}(z)\right) $$

(B.31)

where E is the matrix that corresponds to a permutation of columns(1, 2,...,n)↦(2, 3,...,n, 1). It follows that

$$\begin{array}{@{}rcl@{}} &&\det \left( \theta^{(i)}_{j}(z+\tau/n)\right) \\ &=& (-1)^{n + 1}\det \left[ \gamma^{n-1} \left( \begin{array}{cccccc} y \\ (y)^{\prime} & y \\ (y)^{\prime\prime} & 2(y)^{\prime} & y \\ &&& {\ddots} \\ (y)^{n} &&&& y \end{array} \right) \left( \theta^{(i)}_{j}(z)\right) \right] \end{array} $$

(B.32)

(where (− 1)^n+ 1 = det E). Hence Θ(z + τ) = (− 1)^n+ 1γ^n(n− 1)yⁿΘ(z).□

Corollary B.7

${\Theta } \in V_{n^{2}}$

Proof

The lemma above shows that

$$ {\Theta}(z + 1) = {\Theta}\left( z+ \sum\limits_{i = 1}^{n} 1/n \right) = (-1)^{(n + 1)n} {\Theta}(z) = {\Theta}(z) $$

(B.33)

We repeat the above proof with τ/n replaced by τ. Note first that 𝜃(z + τ) = e^{− 2πinz−πinτ}𝜃(z) for all 𝜃 ∈ V _n . Set Y = e^{−2πinz−πinτ}, then we see that

$$ \left( \theta^{(i)}_{j}(z+\tau)\right) = \left( \begin{array}{cccccc} Y \\ (Y)^{\prime} & Y \\ (Y)^{\prime\prime} & 2(Y)^{\prime} & Y \\ {\vdots} &&& {\ddots} \\ (Y)^{n} && {\cdots} && Y \end{array} \right) \left( \theta^{(i)}_{j}(z)\right) $$

(B.34)

Taking det of both sides, we see that ${\Theta }(z+\tau ) = Y^{n} {\Theta }(z) = e^{-2\pi i n^{2} z-\pi i n^{2}} {\Theta }(z)$, which is precisely the defining conditions of elements of $V_{n^{2}}$.□

Corollary B.8

|X _n | = n².

Proof

The uniqueness Theorem A.3 shows us that |X _n | is equal to the number of possible locations of zeros of singular n-theta functions. Proposition A.7 shows that that this is equal to the size of the zero set of Θ mod L _τ . Since ${\Theta } \in V_{n^{2}}$. We conclude by Theorem A.3, again, that |X _n | = n².□

Now, we obtain explicit formulae for elements of X _n . To do this, we need the following lemma

Lemma B.9

If 𝜃 ∈ V _n , so is

$$ \gamma(z) = \theta\left( z + \frac{1}{n}(a+b\tau) \right) e^{2\pi i b z} $$

(B.35)

for $a,b \in \mathbb {Z}$.

Proof

We check that

$$\begin{array}{@{}rcl@{}} \gamma(z + 1) &=& \theta\left( z + \frac{1}{n}(a+b\tau) + 1 \right) e^{2\pi i b z + 2\pi ib} \end{array} $$

(B.36)

$$\begin{array}{@{}rcl@{}} &=& \gamma(z) \end{array} $$

(B.37)

since $b \in \mathbb {Z}$. Similarly,

$$\begin{array}{@{}rcl@{}} \gamma(z+\tau) &=& \theta\left( z + \frac{1}{n}(a+b\tau) +\tau \right) e^{2\pi i b z + 2\pi i b \tau} \end{array} $$

(B.38)

$$\begin{array}{@{}rcl@{}} &=& e^{-\pi i n \tau - 2\pi i n z - 2\pi i (a+b\tau)} \theta\left( z + \frac{1}{n}(a+b\tau) \right) e^{2\pi i b z + 2\pi i b \tau} \end{array} $$

(B.39)

$$\begin{array}{@{}rcl@{}} &=& e^{-\pi i n \tau - 2\pi i n z} \gamma(z) \end{array} $$

(B.40)

since $a,b \in \mathbb {Z}$.□

Now, let 𝜃₀ be a basis for V₁. From Theorems A.4 and A.5, we see that that ${\theta _{0}^{n}} \in X_{n}$, it follows by Lemma A.10 that

$$ \theta_{a,b}(z):={\theta_{0}^{n}}\left( z + \frac{1}{n}(a+b\tau) \right) e^{2\pi i b z} $$

(B.41)

are all in X _n for $a,b \in \mathbb {Z}$. But there are exactly n² = |X _n | number of distinct such functions (mod scaling). So X _n is contains exactly these elements. Moreover, by Proposition 11.1, the zero of 𝜃₀ is at $\frac {1}{2}(1+\tau )$. So the zeros of 𝜃_a,b are located at $\frac {1}{2}(1+\tau )-\frac {1}{n}(a+b\tau )$.

C Choice of χ _g

The action of point groups is given by

$$ \psi(g x) = e^{i\chi_{g}} \psi(x). $$

(C.1)

for some χ _g , which we determine below.

Proposition C.1

Let $g \in SH(\mathcal {L})$ and ψ is a linear solution satisfying (C.1), then χ _g are constant.

Proof

We identify $SH(\mathcal {L})$ as a subset of $\mathbb {C}$ so that gx is the multiplication of the two complex numbers g and x. Assume that χ _g satisfies (C.1). Since ψ is a linear solution, by (5.11), we can find a holomorphic theta function 𝜃 such that 𝜃(x) = h(x)ψ(x) for some smooth, nonvanishing, h with the property $(\bar \partial h)(x) = \frac {b}{2} xh(x)$ where $\bar {\partial }:=\frac {1}{2}(\partial _{x_1}+i \partial _{x_2})$. Then (C.1) is equivalent to the fact that

$$ H_{g}(x) := h(g x) e^{i\chi_{g}} h(x)^{-1} $$

(C.2)

is holomorphic. Taking $\overline {\partial }$, this requirement is equivalent to

$$\begin{array}{@{}rcl@{}} 0&=& \bar{\partial}(h(g x) e^{i\chi_{g}} h(x )^{-1}) \end{array} $$

(C.3)

$$\begin{array}{@{}rcl@{}} &=& \;\left( i\bar{\partial} \chi_{g} + \frac{b}{2}\bar{g}gx - \frac{b}{2}x\right)h(g x) e^{i\chi_{g}} h(x)^{-1} . \end{array} $$

(C.4)

Since |g| = 1 and $h(g x) e^{i\chi _{g}} h(x)^{-1}$is invertible, we see that

$$ \bar{\partial} \chi_{g} = 0 $$

(C.5)

Since χ _g are real valued, it is a constant. □

As a result of the the proposition, it suffices for us to look for gauge invariant (ψ,A) under actions of $H(\mathcal {L})$ whose gauge factor $h_{g}(x)=e^{i \chi _{g}}$ is a constant. Hence we consider spaces of the form

$$ \left\{ \psi\left( R_{\xi}^{-1} i x\right) = \eta \psi(x),\ R_{\xi} A\left( R_{\xi}^{-1} x\right) = \eta^{\prime} A(x)\right \} $$

(C.6)

where $\eta , \eta ^{\prime } \in \mathbb {C}$. One realises that such space corresponds to irreducible representations of $H(\mathcal {L})$.

D Table of C ₆-Equivariant Theta Functions

Vortex number	Value of r	Theta functions that span Vn,6,r
n = 2	0	𝜃 0
	2	𝜃 2
n = 4	0	${\theta _{0}^{2}}$
	1	𝜃 4
	2	𝜃 0 𝜃 4
	4	${\theta _{2}^{2}}$
n = 6	0	${\theta _{0}^{3}}, \ {\theta _{2}^{3}}, {\theta _{4}^{2}} \theta _{2}^{-1}$
	1	𝜃 0 𝜃 4
	2	${\theta _{0}^{2}} \theta _{2}$
	3	𝜃 4 𝜃 2
	4	$\theta _{0} {\theta _{2}^{2}}$
n = 8	0	${\theta _{0}^{4}}, \ \theta _{0} {\theta _{2}^{3}}$
	1	${\theta _{0}^{2}} \theta _{4}$
	2	${\theta _{2}^{4}}, \ {\theta _{4}^{2}}, \ {\theta _{0}^{3}} \theta _{2}$
	3	𝜃 0 𝜃 4 𝜃 2
	4	${\theta _{0}^{2}} {\theta _{2}^{2}}$
	5	$\theta _{4} {\theta _{2}^{2}}$
n = 10	0	${\theta _{0}^{5}},\ {\theta _{0}^{2}} {\theta _{2}^{3}}$
	1	${\theta _{0}^{3}} \theta _{4}, \ \theta _{4} {\theta _{2}^{3}}$
	2	${\theta _{0}^{4}} \theta _{2},\ \theta _{0} {\theta _{2}^{4}}, \ \theta _{0} {\theta _{4}^{2}}$
	3	${\theta _{0}^{2}} \theta _{4} \theta _{2}$
	4	${\theta _{0}^{3}} {\theta _{2}^{2}} , \ {\theta _{2}^{5}}, \ {\theta _{4}^{2}} \theta _{2}$
	5	$\theta _{0} \theta _{4} {\theta _{2}^{2}}$