# Journal of Nonlinear Mathematical Physics

Volume 27, Issue 2, January 2020, Pages 199 - 218

# On constant solutions of SU(2) Yang-Mills equations with arbitrary current in Euclidean space ℝn

Authors
Dmitry Shirokov
National Research University Higher School of Economics, Myasnitskaya str. 20, Moscow, 101000, Russia,dm.shirokov@gmail.com
Institute for Information Transmission Problems of Russian Academy of Sciences, Bolshoy Karetny per. 19, Moscow, 127051, Russia
Received 21 July 2019, Accepted 20 August 2019, Available Online 27 January 2020.
DOI
10.1080/14029251.2020.1700625How to use a DOI?
Keywords
Yang-Mills equations; singular value decomposition; cubic equations; constant solutions; SU(2)
Abstract

In this paper, we present all constant solutions of the Yang-Mills equations with SU(2) gauge symmetry for an arbitrary constant non-Abelian current in Euclidean space ℝn of arbitrary finite dimension n. Using the invariance of the Yang-Mills equations under the orthogonal transformations of coordinates and gauge invariance, we choose a specific system of coordinates and a specific gauge fixing for each constant current and obtain all constant solutions of the Yang-Mills equations in this system of coordinates with this gauge fixing, and then in the original system of coordinates with the original gauge fixing. We use the singular value decomposition method and the method of two-sheeted covering of orthogonal group by spin group to do this. We prove that the number (0, 1, or 2) of constant solutions of the Yang-Mills equations in terms of the strength of the Yang-Mills field depends on the singular values of the matrix of current. The explicit form of all solutions and the invariant F2 can always be written using singular values of this matrix. The relevance of the study is explained by the fact that the Yang-Mills equations describe electroweak interactions in the case of the Lie group SU(2). Non-constant solutions of the Yang-Mills equations can be considered in the form of series of perturbation theory. The results of this paper are new and can be used to solve some problems in particle physics, in particular, to describe physical vacuum and to fully understand a quantum gauge theory.

Open Access

## 1. Introduction

Up to now the law of elementary particles physics is given by quantum gauge theories [5]. We need exact solutions of classical Yang-Mills equations to describe the vacuum structure of the theory and to fully understand a quantum gauge theory [15]. During the last 50 years, many scientists have been searching for particular classes of solutions of the Yang-Mills equations. The well-known classes of solutions of the Yang-Mills equations are described in detail in various reviews [1], [22]. Only certain (nontrivial) classes of particular solutions of these equations are known because of their nonlinearity: monopoles [21], [10], [16], instantons [4], [20], [3], merons [2], etc.

The main result of this paper is the presentation of all constant (that do not depend on x ∈ ℝn) solutions of the Yang-Mills equations with SU(2) gauge symmetry for an arbitrary constant non-Abelian current in Euclidean space of arbitrary finite dimension. The relevance of the study is explained by the fact that the Yang-Mills equations describe electroweak interactions in the case of the Lie group SU(2). Note that instantons are solutions in Euclidean space-time (with imaginary time) and thus the Euclidean case (not only Minkowski case) is important for applications. Constant solutions of the Yang-Mills equations are essentially nonlinear solutions and, from this point of view, are particularly interesting for applications.

Constant solutions of the Yang-Mills equations with zero current were considered in [17] and [18]. In [17], Prof. R. Schimming wrote: “The following problems concerning constant Yang-Mills fields are actual ones in our opinion: Is there a gauge- and coordinate-invariant characterization of those Yang-Mills fields which admit constant potentials with respect to some gauge and some coordinate system? Find as many as possible (in the ideal case: all) constant Yang-Mills fields and classify them! . . . ” In the current paper, we give a complete answer to these questions in the case of the Lie group SU(2). Our results for an arbitrary current are consistent with the results of [17] and [18] for zero current (and arbitrary compact Lie algebra).

In this paper, we present the general solution of the special system (system of the SU(2) Yang-Mills equations for constant solutions with arbitrary current) of 3n cubic equations with 3n unknowns and 3n parameters. This algebraic problem is solved using the singular value decomposition method and the method of two-sheeted covering of orthogonal group by spin group. Using the invariance of the Yang-Mills equations under the orthogonal transformations of coordinates and gauge invariance, we choose a specific system of coordinates and a specific gauge fixing for each constant current and obtain all constant solutions of the Yang-Mills equations in this system of coordinates with this gauge fixing, and then in the original system of coordinates with the original gauge fixing.

## 2. The main ideas

Let us consider Euclidean space ℝn of arbitrary finite dimension n. We denote Cartesian coordinates by xμ, μ = 1,...,n and partial derivatives by μ = ∂/∂xμ.

Let us consider the Lie group

G=SU(2)={SMat(2,)|SS=1,detS=1},dimG=3(2.1)
and the corresponding Lie algebra
𝔤=𝔰𝔲(2)={SMat(2,)|S=S,trS=0},dim𝔤=3.(2.2)

Denote by 𝔤Tba a set of tensor fields of ℝn of type (a, b) with values in the Lie algebra 𝔤. The metric tensor of ℝn is given by the identity matrix 1 = diag(1,...,1) = ‖δμν‖ = ‖δμν‖. We can raise or lower indices of components of tensor fields with the aid of the metric tensor. For example, Fμν = δμαδνβFαβ.

Let us consider the Yang-Mills equations

μAννAμ[Aμ,Aν]=Fμν,(2.3)
μFμν[Aμ,Fμν]=Jν,(2.4)
where Aμ ∈ 𝔤T1 is the potential, Jν ∈ 𝔤T1 is the non-Abelian current, Fμν = −Fνμ ∈ 𝔤T2 is the strength of the Yang-Mills field. One suggests that Aμ, Fμν are unknown and Jν is known.

Note that (2.3) can be considered as a definition of the strength

Fμν:=μAννAμ[Aμ,Aν].(2.5)

We can substitute the components of the skew-symmetric tensor Fμν from (2.3) into (2.4) and obtain

μ(μAννAμ[Aμ,Aν])[Aμ,μAννAμ[Aμ,Aν]]=Jν.(2.6)

From this point of view, the system (2.3)(2.4) can be considered as a system for the uknown Aμ and the known Jν. For each potential Aμ, the corresponding strength (2.5) can be calculated. Note that from a physical point of view, strength is important, not potential. Also note that a special case of the system (2.3)(2.4) for the Abelian Lie group G = U(1) (Maxwell’s equations) can be considered only for the unknown Fμν and the known Jν, but in this paper, we consider the case of the non-Abelian Lie group G = SU(2) and need the potential Aμ for calculations.

We may verify that the current (2.4) satisfies the non-Abelian conservation law

νJν[Aν,Jν]=0.(2.7)

The Yang-Mills equations are gauge invariant. Namely, the transformed tensor fields

Aμ=S1AμSS1μS,Fμν=S1FμνS,S=S(x):nG,Jv=S1JνS(2.8)
satisfy the same equations
μAννAμ[Aμ,Aν]=Fμν,μFμν[Aμ,Fμν]=Jν.

One says equations (2.3)(2.4) are gauge invariant w.r.t. the transformations (2.8). The Lie group G is called the gauge group of the Yang-Mills equations.

From (2.6), we obtain the following algebraic system of equations for constant solutions (that do not depend on x ∈ ℝn)

[Aμ,[Aμ,Aν]]=Jν,ν=1,,n,(2.9)
and the following expression for the strength of the Yang-Mills field
Fμν=[Aμ,Aν].

Constant solutions of the Yang-Mills equations with zero current Jμ = 0 were considered in [17] and [18]. In this paper, we give all solutions of (2.9) for an arbitrary constant non-Abelian current Jν, ν = 1,...,n.

Note that we have already studied constant solutions of the Yang-Mills-Proca equations, which generalize the Yang-Mills equations and the Proca equation, in [13] and covariantly constant solutions of the Yang-Mills equations in [12], [14], [19] using the techniques of Clifford algebras. We do not use these results in the current paper.

Our aim is to obtain a general solution Aμ ∈ 𝔰𝔲(2)T1 of (2.9) for any Jμ ∈ 𝔰𝔲(2)T1. If n = 1, then (2.9) transforms into 0 = J1. Therefore, the equation (2.9) has an arbitrary solution A1 ∈ 𝔤T1 for J1 = 0 and it has no solutions for J1 ≠ 0. Note that two dimensional Yang-Mills theory is discussed in [8] and other papers. We consider the case n ≥ 2 further for the sake of completeness.

The Pauli matrices σa, a = 1, 2, 3

σ1=(0110),σ2=(0ii0),σ3=(1001)(2.10)
satisfy
(σa)=σa,trσa=0,{σa,σb}=2δab1,[σa,σb]=2iεabcσc,
where εabc=εabc is the antisymmetric Levi-Civita symbol, ε123 = 1.

We can take the following basis of the Lie algebra 𝔰𝔲(2):

τ1=σ12i,τ2=σ22i,τ3=σ32i(2.11)
with
(τa)=τa,trτa=0,[τa,τb]=εabcτc,(2.12)
i.e. the structural constants of the Lie algebra 𝔰𝔲(2) are the Levi-Civita symbol.

For the potential and the current, we have

Aμ=Aaμτa,Jμ=Jaμτa,Aaμ,Jaμ.(2.13)

Latin indices take values a = 1, 2, 3 and Greek indices take values μ = 1, 2,...,n.

Let us substitute (2.13) into (2.9). We have

AμcAaμAbν[τc,[τa,τb]]=Jaντa,AμcAaμAbνεabd[τc,τd]=Jaντa,AμcAaμAbνεabdεcdkτk=Jaντa,
and, finally,
AμcAaμAbνεabdεcdk=Jkν,ν=1,,n,k=1,2,3.(2.14)

We obtain 3n equations (k = 1, 2, 3, ν = 1, 2,...,n) for 3n expressions Akν and 3n expressions Jkν. We can consider this system of equations as the system of equations for two matrices An×3=Akν and Jn×3=Jkν.

We will give the general solution Akν of the system (2.14) for all Jkν using algebraic methods. In Section 3, we also calculate the strength Fμν (using (2.5)) and the invariant F2 = FμνFμν for each solution Aμ, because they are important from a physical point of view.

We have the following well-known theorem on the singular value decomposition (SVD), see [6], [7].

### Theorem 2.1.

For an arbitrary real matrix An×N of the size n × N, there exist orthogonal matrices Ln×n ∈ O(n) and RN×N ∈ O(N) such that

Ln×nTAn×NRN×N=Dn×N,(2.15)
where
Dn×N=diag(μ1,,μs),s=min(n,N),μ1μ2μs0.

The numbers μ1,..., μs are called the singular values, the columns li of the matrix L are called the left singular vectors, the columns ri of the matrix R are called the right singular vectors. From (2.15), we get AR = LD and ATL = RDT. We obtain the following relation:

AATL=LDRTRDT=LDDT,ATAR=RDTLTLD=RDTD,
i.e. the columns of the matrix L are eigenvectors of the matrix AAT, and the columns of the matrix R are eigenvectors of the matrix ATA. The squares of the singular values are the eigenvalues of the corresponding matrices. From this fact, it follows that singular values are uniquely determined.

### Lemma 2.1.

The system of equations (2.14) is invariant under the transformation

AA=AP,JJ=JP,PSO(3)
and under the transformation
AA^=QA,JJ^=QJ,QO(n).

### Proof.

The system (2.9) is invariant under the transformation

Aμ=S1AμS,Jν=S1JνS,SG=SU(2).(2.16)

It follows from the invariance under (2.8) and the fact that an element S ∈ G = SU(2) does not depend on x now.

Let us use the theorem on the two-sheeted covering of the orthogonal group SO(3) by the spin group Spin(3) ≃ SU(2). For an arbitrary matrix P=pbaSO(3), there exist two matrices ±S ∈ SU(2) such that

S1τaS=pbaτb.

We conclude that the system (2.14) is invariant under the transformation

Aμ=S1AaμτaS=AaμS1τaS=Aaμpbaτb=Abμτb,Abμ=Aaμpba,Jμ=S1JaμτaS=JaμS1τaS=Jaμpbaτb=Jbμτb,Jbμ=Jaμpba.

The Yang-Mills equations are invariant under the orthogonal transformation of coordinates. Namely, let us consider the transformation xμx^μ=qνμxν, where Q=qνμO(n). The system (2.14) is invariant under the transformation

A^μ=qμνAμ=qμνAaμτa=A^aντa,A^aν=qμνAaμ,J^ν=qμνJμ=qμνJaμτa=J^aντa,J^aν=qμνJaμ.

The lemma is proved.

Combining gauge and orthogonal transformations, we conclude that the system (2.14) is invariant under the transformation

AbνA^bν=qμνAaμpba,An×3A^n×3=Qn×nAn×3P3×3,JbνJ^bν=qμνJaμpba,Jn×3J^n×3=Qn×nJn×3P3×3(2.17)
for any P ∈ SO(3) and Q ∈ O(n).

### Theorem 2.2.

Let A=Akν, J=Jkν satisfy the system of 3n cubic equations (2.14). Then there exist matrices P ∈ SO(3) and Q ∈ O(n) such that QAP is diagonal. For all such matrices P and Q, the matrix QJP is diagonal too and the system (2.14) takes the following form under the transformation (2.17):

a1((a2)2+(a3)2)=j1,a2((a1)2+(a3)2)=j2,a3((a1)2+(a2)2)=j3(2.18)
in the case n ≥ 3 and
a1(a2)2=j1,a2(a1)2=j2(2.19)
in the case n = 2.

We denote diagonal elements of the matrix QAP by a1, a2, a3 (or a1, a2) and diagonal elements of the matrix QJP by j1, j2, j3 (or j1, j2).

### Proof.

Let the system (2.14) has some solution Aaμ, Jaμ. Let us synchronize gauge transformation and orthogonal transformation such that A=Aaμ will have a diagonal form. Namely, we take P ∈ SO(3) and Q ∈ O(n) such that QAP is diagonal. Note that we can always find the matrix R ∈ SO(N) from the special orthogonal group in SVD (2.15). If it has the determinant −1, then we can change the sign of the first columns of the matrices L and R and the determinant will be +1.

Let us consider the case n ≥ 3. In (2.14), we must take μ = a = c, ν = b to obtain nonzero summands. Also we need b = k, i.e. ν = k. In this case, the product of two Levi-Civita symbols in (2.14) equals −1. If νk, then the expression on the left side of the equation equals zero. If ν = k, then we obtain the following sum over index μ = a = c

A^kkak(A^aa)2,
where we have only 2 summands in the sum (except the value μ = k because of the Levi-Civita symbols).

Under our transformation the expressions Jaμ are transformed into some new expressions J^aμ. We obtain the following system of 3n equations

A^11((A^22)2+(A^33)2)=J^11A^22((A^11)2+(A^33)2)=J^22,A^33((A^11)2+(A^22)2)=J^33,0=J^kν,νk,ν=1,,n,k=1,2,3.(2.20)

This system of equations has solutions if the matrix J^ is also diagonal.

In the case n = 2, we obtain the system

A^11(A^22)2=J^11,A^22(A^11)2=J^22,0=J^kν,νk,ν=1,2,k=1,2,3(2.21)
instead of the system (2.20) and the proof is similar.

### Remark 2.1.

In Theorem 2.2. we calculate SVD of the matrix A and obtain non-negative singular values a1, a2, a3 (or a1, a2 in the case n = 2). The diagonal elements of the matrix QJP will be non-positive because of the equations (2.18) (or (2.19)). If we want, we can change the matrix Q ∈ O(n) (multiplying by the matrix −1 ∈ O(n)) such that the elements of the new matrix QJP will be non-negative (they will be singular values of the matrix J) and the elements of the new matrix QAP will be non-positive. Multiplying the matrices P and Q by permutation matrices, which are also orthogonal, we can obtain the diagonal elements of the new matrix QJP in decreasing order, the diagonal elements of the new matrix QAP will be in some other order.

### Remark 2.2.

Suppose we have known matrix J and want to obtain all solutions A of the system (2.14). We can always calculate singular values j1, j2, j3 (or j1, j2) of the matrix J and solve the system (2.18) (or (2.19)) using lemmas below. Finally, we obtain all solutions AD = diag(a1, a2, a3) (or AD = diag(a1, a2)) of the system (2.14) but in some other system of coordinates depending on Q ∈ O(n) and with gauge fixing depending on P ∈ SO(3). The matrix

will be solution of the system (2.14) in the original system of coordinates and with the original gauge fixing.

### Remark 2.3.

Note that Q1Q11ADP11P1, for all Q1 ∈ O(n) and P1 ∈ SO(3) such that Q1JDP1 = JD, will be also solutions of the system (2.14) in the original system of coordinates and with the original gauge fixing because of Lemma 2.1. Here we denote JD = diag(j1, j2, j3) (or JD = diag(j1, j2)).

Let us give one example. If the matrix J = 0, then all singular values of this matrix equal zero and we can take Q = P = 1 for its SVD. We solve the system (2.18) (or (2.19)) for j1 = j2 = j3 = 0 (or j1 = j2 = 0) and obtain all solutions AD = diag(a1, a2, a3) (or AD = diag(a1, a2)) of this system. We have Q1JDP1 = JD for JD = 0 and any Q1 ∈ O(n), P1 ∈ SO(3). Therefore, the matrices Q1ADP1 for all Q1 ∈ O(n) and P1 ∈ SO(3) will be solutions of the system (2.14) because of Lemma 2.1.

Let us present a general solution of the systems (2.18) and (2.19) and discuss symmetries of these systems.

The systems (2.18), (2.19) can be rewritten in the following form using bk := −ak, k = 1, 2, 3 or k = 1, 2:

n3:b1(b22+b32)=j1,b2(b12+b32)=j2,b3(b12+b22)=j3,(2.22)
n=2:b1b22=j1,b2b12=j2.(2.23)

In the following lemmas, we assume that j1, j2, j3 are known (parameters), and b1, b2, b3 are unknown. We give general solutions of the corresponding systems of equations.

The system (2.22) has the following symmetry. Suppose that (b1, b2, b3) is a solution of (2.22) for known (j1, j2, j3). If we change the sign of some jk, k = 1, 2, 3, then we must change the sign of the corresponding bk, k = 1, 2, 3. Thus, without loss of generality, we can assume that all expressions bk, jk, k = 1, 2, 3 in (2.22) are non-negative. Similarly for the system (2.23).

### Lemma 2.2.

The system of equations (2.23) has the following general solution:

1. (1)

in the case j1 = j2 = 0, has solutions (b1, 0), (0, b2) for all b1,b2 ∈ ℝ;

2. (2)

in the cases j1 = 0, j2 ≠ 0; j1 ≠ 0, j2 = 0, has no solutions;

3. (3)

in the case j1 ≠ 0, j2 ≠ 0, has a unique solution

b1=j22j13,b2=j12j23.

### Proof.

The proof is by direct calculation.

The system (2.22) has the following symmetry.

### Lemma 2.3.

If the system (2.22) has a solution (b1, b2, b3), where b1 ≠ 0, b2 ≠ 0, b3 ≠ 0, then this system has also a solution (Kb1, Kb2,Kb3), where K=(b1b2b3)23.

### Proof.

Let us substitute (Kb1, Kb2,Kb3) into the first equation. We have

j1=4Kb1(K2b22+K2b32)=K3(b22+b32)b1b22b32.

Using j1=b1(b22+b32), we obtain

K=(b1b2b3)23.

We can verify that the same will be for the other two equations.

For example, let us take j1 = 13, j2 = 20, j3 = 15. Then the system (2.22) has solutions (b1, b2, b3) = (1, 2, 3) and (623, 6232, 6233).

### Lemma 2.4.

The system of equations (2.22) has the following general solution:

1. (1)

in the case j1 = j2 = j3 = 0, has solutions (b1,0,0), (0,b2,0), and (0,0,b3) for all b1,b2,b3 ∈ ℝ;

2. (2)

in the cases j1 = j2 = 0, j3 ≠ 0 (or similar cases with circular permutation), has no solutions;

3. (3)

in the case j1 ≠ 0, j2 ≠ 0, j3 = 0 (or similar cases with circular permutation), has a unique solution

b1=j22j13,b2=j12j23,b3=0;

4. (4)

in the case j1 = j2 = j3 ≠ 0, has a unique solution

b1=b2=b3=j123;

5. (5)

in the case of not all the same j1, j2, j3 > 0 (and we take positive for simplicity), has the following two solutions

(b1+,b2+,b3+),(b1,b2,b3)
with the following expression for K from Lemma 2.3
K:=b1+b1=b2+b2=b3+b3=(b1+b2+b3+)23=(b1b2b3)23:
1. (a)

in the case j1 = j2 > j3 > 0 (or similar cases with circular permutation):

b1±=b2±=j32z±3,b3±=z±b1±,z±=j1±j12j32j3.

Moreover,

z+z=1,K=(j32)23.

2. (b)

in the case j3 > j1 = j2 > 0 (or similar cases with circular permutation):

b1±=1w±b3,b2±=w±b3,b3±=b3=j1s3,w±=s±s242,s=j3+j32+8j122j1.

Moreover,

w+w=1,b1±=b2,K=(j1s)23.

3. (c)

in the case of all different j1, j2, j3 > 0:

b1±=j3t0y±z±3,b2±=y±b1±,b3±=z±b1±,z±=y±(j1j2y±)j2j1y±,y±=t0±t0242,
where t0 > 2 is the solution (it always exists, moreover, it is bigger than j2j1+j1j2) of the cubic equation
j1j2t3(j12+j22+j32)t2+4j32=0.

Moreover,

y+y=1,z+z=1,K=(j3t0)23.

We can use the explicit Vieta or Cardano formulas for t0:

t0=Ω+2Ωcos(13arccos(12βΩ3)),
Ω:α+β3,α:A+1A>2,β:=B2A,A:=j2j1,B:=j3j1,t0=Ω+L+Ω2L,L:=Ω32β+2β(βΩ3)3.

### Proof.

The proof is rather cumbersome, we give it in Appendix A.

## 3. Results for the potential and the strength of the Yang-Mills field

In the case of the constant potential of the Yang-Mills field, we have the following expression for the strength

Fμν=[Aμ,Aν]=[Aaμτa,Abντb]=AaμAbνεabcτc=Fcμντc.(3.1)

If we take a system of coordinates depending on Q ∈ O(n) and a gauge fixing depending on P ∈ SO(3) such that the matrices A=Aaμ and J=Jaμ are diagonal (see Theorem 2.2), then the expressions Fcμν are nonzero only in the case of three different indices μ = a, ν = b, and c, which take the values 1, 2, 3. For each solution, we calculate the invariant F2 = FμνFμν, which is present in the Lagrangian of the Yang-Mills field.

Using results of the previous section for the system (2.14) and the expression (3.1), we obtain the following results for the potential A and strength F of the Yang-Mills field depending on the constant current J. The case n = 2 is much simpler than the case n ≥ 3, we discuss this case for the sake of completeness.

In the case of dimension n = 2:

1. (1)

In the case of zero current J = 0, we have zero potential A = 0 or nonzero potential (see Case 1 of Lemma 2.2)

A=(a00000),a\{0}.

In these cases, we have zero strength F = 0 (Fμν = 0). Note that this fact is already known (see [17], [18]).

2. (2)

In the case rank(J) = 1, we have no constant solutions (see Case 2 of Lemma 2.2).

3. (3)

In the case rank(J) = 2, we have a unique solution (see Case 3 of Lemma 2.2)

A=(a1000a20),a1=j22j13,a2=j12j23.

For the strength, we have the following nonzero components

F12=F21=j1j23τ3(3.2)
using specific system of coordinates and specific gauge fixing, where j1 and j2 are singular values of the matrix J=Jaμ. In this case, we obtain the following expression for the invariant F2 = FμνFμν:
F2=FμνFμν=12(j1j2)2310.(3.3)

In the case of dimension n ≥ 3:

1. (1)

In the case J = 0, we have zero potential A = 0 or nonzero potential (see Case 1 of Lemma 2.4):

A=(a00000000),a\{0}.(3.4)

In these cases, we have zero strength F = 0.

2. (2)

In the case rank(J) = 1, we have no constant solutions (see Case 2 of Lemma 2.4).

3. (3)

In the case rank(J) = 2, we have a unique solution (see Case 3 of Lemma 2.4):

A=(a1000a20000000),a1=j22j13,a2=j12j23.(3.5)

For the strength, we have the following nonzero components (3.2) and again (3.3) using specific system of coordinates and gauge fixing, where j1, j2, and j3 = 0 are singular values of the matrix J.

4. (4)

In the case rank(J) = 3, we have one or two solutions.

In the specific case of all the same singular values j := j1 = j2 = j3 ≠ 0, we have a unique solution (see Case 4 of Lemma 2.4)

A=(a000a000a000000),a=j23.(3.6)

We have the following nonzero components of the strength:

F12=F21=j243τ3,F23=F32=j243τ1,F31=F13=j243τ2.(3.7)

In this case, we have

F2=FμνFμν=32j416310.(3.8)

In the case of not all the same singular values j1, j2, j3 of the matrix J, we have two different solutions

A=(b1±000b2±000b3±000000),(3.9)
where b, k = 1, 2, 3 are from Case 5 of Lemma 2.4.

We have the following nonzero components of the strength:

F±12=F±21=b1±b2±τ3,F±23=F±32=b2±b3±τ1,F±31=F±13=b3±b1±τ2.(3.10)

In this case, we have

F±2=Fμν±F±μν=12((b1±b2±)2+(b2±b3±)2+(b3±b1±)2)10.(3.11)

In the next lemma, we give the explicit form of (3.11).

### Lemma 3.1.

In the case of not all the same j1, j2, j3, (3.11) takes the form:

1. (1)

in the case j1 = j2 > j3 > 0 (or similar cases with circular permutation):

F±2=K2(1+2z±2)2z±431,F+2F2,(3.12)
where z±=j1±j12j32j3d, K=(j32)23.

2. (2)

in the case j3 > j1 = j2 > 0 (or similar cases with circular permutation):

F±2=K2(s21)21,F+2=F2,(3.13)
where s=j3+j32+8j122j1>2, K=(j1s)23.

3. (3)

in the case of all different j1, j2, j3 > 0:

F±2=K2(y±2+z±2+y±2z±2)2(y±z±)431,F+2F2,(3.14)
where K=(j3t0)23, and y±, z±, t0 are from Case (5) - (c) of Lemma 2.4.

In all cases of Lemma, the expression K is the invariant for each pair of solutions (see Lemmas 2.3 and 2.4).

### Proof.

We give the proof in Appendix B.

Note that in Case 2 of Lemma 3.1, we have two constant solutions of the Yang-Mills equations with the same invariant F2=F±2. In each of two Cases 1 and 3, we have two constant solutions of the Yang-Mills equations with different invariants F+2F2.

We summarize the results for the case of arbitrary Euclidean space ℝn, n ≥ 2, in Table 1.

n rank(J) additional conditions rank(A) A F F2
n ≥ 2 0 0 A = 0 F = 0 F2 = 0
n ≥ 2 0 1 see (3.4) F = 0 F2 = 0
n ≥ 2 1
n ≥ 2 2 2 see (3.5) see (3.2) see (3.3)
n ≥ 3 3 j1 = j2 = j3 3 see (3.6) see (3.7) see (3.8)
n ≥ 3 3 j1 = j2 > j3 3 see (3.9) see (3.10) see (3.12)
n ≥ 3 3 j3 > j1 = j2 3 see (3.9) see (3.10) see (3.13)
n ≥ 3 3 all different j1, j2, j3 3 see (3.9) see (3.10) see (3.14)
Table 1.

All constant solutions of SU(2) Yang-Mills equations in ℝn.

## 4. Conclusions

The main result of this paper is the presentation of all constant solutions of the Yang-Mills equations with SU(2) gauge symmetry for an arbitrary constant current in Euclidean space of arbitrary finite dimension. Using the invariance of the Yang-Mills equations under the orthogonal transformations of coordinates and gauge invariance, we choose a specific system of coordinates and a specific gauge fixing for each constant current and obtain all constant solutions of the Yang-Mills equations in this system of coordinates with this gauge fixing, and then in the original system of coordinates with the original gauge fixing (see Remarks 2.2 and 2.3). We prove that the number (0, 1, or 2) of constant solutions of the Yang-Mills equations (solutions of the system (2.14)) in terms of the strength F (3.1) depends on the rank of the matrix J and, sometimes, on the singular values of this matrix (see Section 3). The explicit form of these solutions and the invariant F2 can always be written using singular values of the matrix J.

We plan to solve the same problem as in this paper, but in pseudo-Euclidean space of arbitrary finite dimension, in particular, in the case of Minkowski space. This will allow us to obtain all constant solutions of the Dirac-Yang-Mills equations, which is interesting for applications. Another task is to consider the same problem on curved manifolds. We need another technique to solve the same problem for the case of the Lie group SU(3), which is important for describing strong interactions.

Note that now we can consider nonconstant solutions of Yang-Mills equations in the form of series of perturbation theory using all constant solutions from Lemmas 2.2 and 2.4 as a zeroth approximation. The problem reduces to solving systems of linear partial differential equations. This will allow us to give a local classification of all solutions of the classical SU(2) Yang-Mills equations.

The results of this paper are new and can be used to solve some problems in particle physics, in particular, in describing physical vacuum [1], [9], [11], [15]. In this paper, we discuss mathematical structures and constructions. Relating the proposed mathematical constructions to real world objects goes beyond the scope of this investigation. The explicit formulas for solutions (see the results of Section 3 and Lemmas 2.2, 2.3, and 2.4) are fundamental for the Yang-Mills field and should be interesting for physicists.

## A. The proof of Lemma 2.4

The first four cases of Lemma 2.4 are easily verified.

Let us consider the case of not all the same positive j1, j2, j3. As we mentioned before the lemma, we can assume that jk > 0 and bk > 0 because if we change the sign of jk, then the sign of bk is also changed.

We use the following change of variables

x=b1>0,y=b2b1>0,z=b3b1>0.

We obtain

j1=x3(y2+z2),j2=yx3(1+z2),j3=zx3(1+y2).(A.1)

Using notation

A=j2j1>0,B=j3j1>0,
we get the system for two variables y and z:
y(1+z2)=A(y2+z2),z(1+y2)=B(y2+z2).(A.2)

For the variable x = b1, we have

b1=xj1(y2+z2)3.(A.3)

From the first equation (A.2), we obtain

(Ay)z2=y(1Ay).(A.4)

Let us consider two cases: A = 1 and A ≠ 1.

If A = 1 (and B ≠ 1, because we consider the case of not all the same j1, j2, j3), then we can rewrite (1−y)z2 = y(1−y) in the form (1−y)(yz2) = 0. If y = 1, then we substitute this condition into the second equation (A.2) and obtain 2z = B(1+z2). If 0 < B < 1, then z=1±1B2B. For B > 1, there is no solution of this type. If y = z2, then we substitute this condition into the second equation (A.2) and get z(1 + z4) = B(z2 + z4), z5Bz4Bz2 + z = 0. Dividing both sides by z2 and using notation s=z+1z>0, we get s=B±B2+82. We have z > 0 and s > 2 (if s = 2, then z = y = 1 and we obtain the case j1 = j2 = j3, which is not considered now). Therefore, B > 1 and we have two expressions z=s±s242 for one s=B+B2+82.

The results for the case A = 1 can be summarized as follows. If j1 = j2 > j3 (A = 1, 0 < B < 1), then we have two solutions

b1±=b2±=j11+z±23,b3±=z±b1±,z±=1±1B2B,B=j3j1.

Using z+z = 1 and z±1+z±2=B2, we can verify that K=b1+b1=b2+b2=b3+b3=(b1+b2+b3+)23=(b1b2b3)23=(Bj12)23=(j32)23 and rewrite solutions in the following form

b1±=b2±=j32z±3,b3±=z±b1±,z±=j1±j12j32j3.

If j3 > j1 = j2 (A = 1, B > 1), then we have two solutions

b1±=j1w2+w43,b2±=w±2b1±,b3±=w±b1±,w±=s±s242,s=B+B2+82,B=j3j1.

Using w±1+w±2=1s and w+w = 1, we can verify that K=b1+b1=b2+b2=b3+b3=(b1+b2+b3+)23=(b1b2b3)23=(j1s)23. Moreover, b3+ = b3−, b1+ = b2−, b1− = b2+ and we can rewrite solutions in the following form

b1±=1w±b3,b2±=w±b3,b3±=b3=j1s3,w±=s±s242,s=j3+j32+8j122j1.

Now let us consider the case A ≠ 1. If B = 1, then we can similarly consider this case as previous case (A = 1, B ≠ 1), since the system (A.2) is symmetric w.r.t. the change yz, AB. Therefore, let us consider the remaining case, when A ≠ 1 and B ≠ 1. If A = B, then j2 = j3 and this case is similar to the previous one again. Let us consider the case AB, i.e. all jk, k = 1, 2, 3 are different now.

If A ≠ 1, then Ay. Really, suppose that A = y. Then from (A.4) we obtain A = 1 and a contradiction. Since Ay, we get from (A.4)

z2=y(1Ay)Ay.(A.5)

Because all variables are positive, the second equation (A.2) is equivalent to z2(1 + y2)2 = B2(z2 + y2)2. Substituting (A.5) into this expression, we get

y(1Ay)Ay(1+y2)2=B2(y(1Ay)Ay+y2)2.

Note, that when we will find y from this equation, expression y(1Ay)Ay will be positive. Therefore z=y(1Ay)Ay. We have

y(1Ay)(1+y2)2Ay=B2(yy3)2(Ay)2,(1Ay)(1+y2)2(Ay)=B2y(1y2)2,Ay6(B2+A2+1)y5+3Ay4+2(B2A21)y3+3Ay2(B2+A2+1)y+A=0.

Dividing both sides by y3 and using notation t=y+1y>0(t2=y2+1y2+2,t3=y3+3y+31y+1y3), we get

A(y3+1y3)(B2+A2+1)(y2+1y2)+3A(y+1y)+2(B2A21)=0,A(t33t)(B2+A2+1)(t22)+3At+2(B2A21)=0,At3(B2+A2+1)t2+4B2=0.

We obtain the following cubic equation

f(t):t3(B2A+A+1A)t2+4B2A=0.(A.6)

We are interested only in the positive solutions t > 0 of this equation. We have y2ty + 1 = 0. This equation has positive solutions y=t±t242 only in the case t ≥ 2.

We have f (−∞) = −∞, f(0)=4B2A>0, f(2)=4(A1)2A<0, f (+∞) = +∞. This means that the cubic equation (A.6) has one negative solution, one solution between 0 and 2, and one solution t0 > 2. Moreover, solution t0 is bigger than A+1A, because f(A+1A)=B2(A21)2A3<0. We can calculate t0 using Cardano formulas.

Finally, if A ≠ 1 and B ≠ 1, then we have two solutions

b1±=j1y±2+z±23,b2±=y±b1±,b3±=z±b1±,z±=y±(1Ay±)Ay±,y±=t0±t0242,
where t0 = t0(A, B) > 2 is a solution of the equation (A.6). Using y±z±y±2z±2=By±1+y±2=Bt0, y+y = 1, and z+z = 1, we can verify that K=b1+b1=b2+b2=b3+b3=(b1+b2+b3+)23=(b1b2b3)23=(j3t0)23 and rewrite solutions in the following form
b1±=j3t0y±z±3,b2±=y±b1±,b3±=z±b1±,z±=y±(1Ay±)Ay±,y±=t0±t0242,
where t0 = t0(A, B) > 2 is a solution of the equation (A.6). The lemma is proved.

## B. The proof of Lemma 3.1

1. 1)

In Case (5) - (a) of Lemma 2.4, using

b1±=b2±,b3±=z±b1±,b1±=j32z±3,K=(j32)23,
we obtain
F±2=12((b1±b2±)2+(b2±b3±)2+(b3±b1±)2)1=12b1±4(1+2z±2)1=12(j32z±)43(1+2z±2)1=K2(1+2z±2)2z±431.

Let us prove that F+2F2 in this case. Suppose that we have F+2=F2, i.e.

1+2z+2z+43=1+2z2z43,z43+2z+2z43=z+43+2z2z+43.

Using z+z = 1, we get

z43+2z+23=z+43+2z23,(z231)2=(z+231)2,(z23+z+232)(z23z+23)=0,
which is not possible, because z23+z+232(zz+)13=2 and z+, z do not equal ±1.

2. 2)

In Case (5) - (b) of Lemma 2.4, using

b1±=1w±b3,b2±=w±b3,b3±=b3=j1s3,K=(j1s)23,
we obtain
F±2=12((b1±b2±)2+(b2±b3±)2+(b3±b1±)2)1=12b3±4(1+w±2+1w±2)1=12(j1s)43(1+w±2+1w±2)1=K22(1+w±2+1w±2)1=K22(s21)1.

In the last equality, we used w±+1w±=s, i.e. w±2+1w±2=s22.

We have F+2=F2, because F±2 does not depend on w± in this case.

3. 3)

In Case (5) - (c) of Lemma 2.4, using

b2±=y±b1±,b3±=z±b1±,b1±=j3t0y±z±3,K=(j3t0)23,
we obtain
F±2=12((b1±b2±)2+(b2±b3±)2+(b3±b1±)2)1=12b1±4(y±2+z±2+y±2z±2)1=12(j3t0y±z±)43(y±2+z±2+y±2z±2)1=K2(y±2+z±2+y±2z±2)2(y±z±)431.

Using z±2=y±(1Ay±)Ay±, we also get

F±2=K2(y±2+z±2+y±2z±2)2(y±z±)431=K2(y±2+(1+y±2)y±(1Ay±)Ay±))2y±43(y±(1Ay±)Ay±)231=K2(1Ay±3)2y±(Ay±)13(1Ay±)231.

Let us prove that F+2F2 in this case. Suppose that we have F+2=F2, i.e.

(1Ay+3)3y+3(Ay+)(1Ay+)2=(1Ay3)3y3(Ay)(1Ay)2.

Using y=y+1, we get

y+3(Ay+)(1Ay+)2(1Ay+3)3=(1Ay+3)31y+3(A1y+)(1Ay+)2,(y+3A)3(1Ay+)=(1Ay+3)3(y+A),(A3A)y+10+(1A4)y+9+3(A3A)y+6+3(AA3)y+4+(A41)y++(AA3)=0.

Dividing both sides of the equation by A2 − 1 ≠ 0 and y+50, we obtain

A(y+51y+5)(1+A2)(y+41y+4)+3A(y+1y+)=0.

Dividing both sides of the equation by (y+1y+)0, we obtain

A(y+4+y+2+1+1y+2+1y+4)(1+A2)(y+3+y++1y++1y+3)+3A=0.

Using t=y++1y+=y++y, we have

y+2+1y+2=t22,y+3+1y+3=t33t,y+4+1y+4=t44t2+2
and obtain
At4(1+A3)t33At2+2(1+A2)t+4A=0.

Dividing by t ≠ 0, we get

A(t2+4t2)(1+A2)(t2t)3A=0.

Using d:=t=2t, we have t2+4t2=d2+4 and obtain

If d = A, then

t2At2=0.(B.1)

But it is in a contradiction with

At3(B2+A2+1)t2+4B2=0.(B.2)

Really, multiplying both sides of (B.1) by At, we get

At3A2t22At=0.(B.3)

From (B.2) and (B.3), we obtain

(1+B2)t22At4B2=0.(B.4)

From (B.4) and (B.1), we get

2At+4B21+B2=At+2,tA(1B2)=2(1B2),t=2A,
because B±1. Substituting t=2A into (B.1), we get 4A2=4, i.e. a contradiction, because A±1.

If d = A−1, then

At2t2A=0.(B.5)

But it is in a contradiction with

At3(B2+A2+1)t2+4B2=0.(B.6)

Really, multiplying both sides of (B.5) by t, we get

At3t22t=0.(B.7)

From (B.6) and (B.7), we obtain

(A2+B2)t22At4B2=0.(B.8)

From (B.8) and (B.5), we get

2At+4B2A2+B2=t+2AA,t(A2B2)=2A(A2B2),t=2A,
because A±B. Substituting t = 2A into (B.5), we get 4A(A2 −1) = 0, i.e. a contradiction, because A ≠ 0, A±1.

The lemma is proved.

## Acknowledgements

The author is grateful to N. G. Marchuk for fruitful discussions. The author is grateful to the participants of the 2nd JNMP Conference on Nonlinear Mathematical Physics (Chile, Santiago, 2019) for useful comments. The author is grateful to the reviewers for their careful reading of the paper and helpful remarks.

This work is supported by the Russian Science Foundation (project 18-71-00010).

## References

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[15]J. Nian and Y. Qian. A topological way of finding solutions to Yang-Mills equations, arXiv:1901.06818 [hep-th]
[16]A.M. Polyakov, Isomeric states of quantum fields, Sov.Phys. - JETP, Vol. 41, 1975, pp. 988-995.
[17]R. Schimming, On constant solutions of the Yang-Mills equations, Arch. Math, Vol. 24, No. 2, 1988, pp. 65-73.
[21]T.T. Wu and C.N. Yang, H. Mark and S. Fernbach (editors), in Properties of Matter Under Unusual Conditions, Interscience New York, 1968.
[22]R.Z. Zhdanov and V.I. Lahno, Symmetry and Exact Solutions of the Maxwell and SU(2) Yang-Mills Equations, Adv. Chem. Phys. Modern Nonlinear Optics, Vol. 119, 2001, pp. 269-352. part II
Journal
Journal of Nonlinear Mathematical Physics
Volume-Issue
27 - 2
Pages
199 - 218
Publication Date
2020/01/27
ISSN (Online)
1776-0852
ISSN (Print)
1402-9251
DOI
10.1080/14029251.2020.1700625How to use a DOI?
Open Access

TY  - JOUR
AU  - Dmitry Shirokov
PY  - 2020
DA  - 2020/01/27
TI  - On constant solutions of SU(2) Yang-Mills equations with arbitrary current in Euclidean space ℝⁿ
JO  - Journal of Nonlinear Mathematical Physics
SP  - 199
EP  - 218
VL  - 27
IS  - 2
SN  - 1776-0852
UR  - https://doi.org/10.1080/14029251.2020.1700625
DO  - 10.1080/14029251.2020.1700625
ID  - Shirokov2020
ER  -