The Gribov horizon and ghost interactions in Euclidean gauge theories

may appear again. We show ghost condensation happens in the nonlinear gauge, and the zero mode repetition is avoided. ...................................................................................................................


Introduction
A perturbative calculation in gauge theories requires gauge fixing. However, in non-Abelian gauge theories, there is a problem of gauge copies (Ref. [1]). Gribov showed that gauge-equivalent copies exist in the Landau gauge (1.1) PTEP 2017, 023B04 H. Sawayanagi and the FP operator has zero modes. These zero modes can cause a problem in proving the gauge equivalence (Ref. [12]). Thus physical effects of the horizon ∂ are worth studying.
In this paper, we study the effect of these zero modes. In the next section, we show that a pair of zero modes in the Landau gauge can yield additional ghost interactions. If we require BRS invariance, an effective Lagrangian becomes a Lagrangian in a nonlinear gauge. In Sect. 3, the gauge ∂ μ A μ = 0 is considered. If there is a pair of zero modes, the nonlinear gauge is realized as well. We also show that the partition function does not vanish even if the FP operator yields a single zero mode. In Sect. 4, the effect of a single zero mode is discussed in the Landau gauge. In the low energy region, ghost condensation appears in the nonlinear gauge. The effect of the zero modes under the condensation is discussed in Sect. 5. Section 6 is devoted to a summary. In Appendix A, examples of zero modes in the Coulomb gauge are given in three dimensional space-time. In Appendix B, the effective Lagrangian in Sect. 2 is derived by the use of a source term. The nonlinear gauge has two gauge parameters. Renormalization group equations for these parameters are presented in Appendix C. In Appendix D, symmetries in the nonlinear gauge are discussed.

Effect of ghost zero modes in the Landau gauge
We consider the SU(2) gauge theory with structure constants f ABC . Using the notation a partition function in the Landau gauge is Z L = Z α=0 with where ic · ∂ μ D μ c = ic A ∂ μ (∂ μ + gA μ ×) AB c B . The gauge condition (Eq. (1.1)) leads to the relations Namely, ∂ μ D μ is hermitian, and its eigenvalues are real. The eigenfunction u n with the eigenvalue λ n satisfies −∂ μ D μ u n (x) = λ n u n (x). (2.4) When A μ is on the first Gribov horizon, the lowest eigenvalue is λ 0 = 0 and u 0 (x) is a zero mode. If we can make u 0 (x) complex, as Eq. (2.4) leads to is also a zero mode. We assume a pair of zero modes (u 0 (x), u * 0 (x)) exists. Some examples of a zero-mode pair (u 0 (x), u * 0 (x)) are presented in Appendix A. If u 0 is real, it may be a single zero mode. An example of such a zero mode is given in Appendix A, and its effect is discussed in Sect. 4. Now we expand the ghost c as 1 1 We assume that eigenfunctions of the FP operator form an orthonormal complete set. Strictly speaking, to ensure it, spaces and/or configurations of A μ must be restricted. We emphasize that what is important here is that c contains ξ u 0 , ξ † u * 0 andc containsξ u 0 ,ξ † u * 0 .

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holds. We note that if there are some pairs of zero modes (u , respectively. However the discussion below is also applicable. Equations (2.4) and (2.5) imply that the Lagrangian dx ic·∂ μ D μ c does not contain the Grassmann variables ξ , ξ † ,ξ , andξ † . However the measures Dc and Dc contain dξ dξ † and dξ dξ † , respectively. Since a Grassmann variable ζ satisfies the partition function vanishes: We know that fermions in an instanton background have zero modes. These zero modes yield the additional interaction of fermions (Refs. [13,14]). Likewise, the above ghost zero modes may produce additional ghost interactions, because From Eqs. (2.6) and (2.7), we obtain where ABCD = u A 0 u B 0 u * C 0 u * D 0 , and terms denoted by . . . lack some or all of ξ , ξ † ,ξ , andξ † . Therefore Eq. (2.9) leads to where σ [AB][CD] is antisymmetric with respect to A and B, and C and D as well. Thus ghost zero modes produce effective ghost interactions. Now we determine σ [AB] [CD] , and construct effective Lagrangians. The first candidate is  holds. Therefore, as in the instanton case (Ref. [15]), Eq. (2.11) is derived from the nonvanishing partition function where K 1 is a dimensionless constant. Interaction with other fields is also possible. If we use and Eq. (2.10) becomes (2.14) Taking account of Eq. (2.12), we find Eq. (2.14) is derived from we obtain If we set K 2 = − i g K 1 = igα 2 , we get the BRS-invariant effective Lagrangian Thus the Gribov horizon yields the Lagrangian in the nonlinear gauge L NL (Refs. [16][17][18]).

α = 0 gauge
In the α = 0 gauge, as ∂ μ A μ = 0 and the operator ∂ μ D μ is not hermitian. We assume that the operator ∂ μ D μ has a pair of zero modes (u 0 , u * 0 ) and a real single zero mode v 0 . Then c is expanded as where ξ , ξ † , and ζ are independent Grassmann variables. Although the Lagrangian Eq. (2.2) does not contain ξ , ξ † , and ζ , the measure Dc contains dξ dξ † dζ . Thus we find However Eq. (3.2) contradicts ghost number conservation. To avoid this problem, a pair of zero modes (ū 0 ,ū * 0 ) and a real single zero modev 0 of the operator D μ ∂ μ must exist, 3 andc is expanded asc , andv 0 = v 0 . Now we consider the effect of the zero-mode pairs (u 0 , u * 0 ) and (ū 0 ,ū * 0 ). Since the Lagrangian does not contain ξ , ξ † ,ξ , andξ † , and the measure contains dξ dξ † dξ dξ † , to obtain a nonzero partition function, we must repeat the consideration in Sect. 2. Namely the zero-mode pairs give rise to the effective Lagrangian L eff , and the nonlinear gauge is realized.
H. Sawayanagi Next we study the terms ζ v 0 in Eq. (3.1) andζv 0 in Eq. (3.3). The Lagrangian L eff has the term igα 2 B · (c × c). Although this term is necessary to ensure BRS symmetry, as the partition function does not vanish even if DcDc contains dζ dζ . Thus, when α = 0, the partition function changes from Eq. (2.1) to Eq. (2.18), if the FP operator ∂ μ D μ has a pair of zero modes. This result is unchanged even if this operator has a single zero mode.

Renormalization group flow of α
We return to the gauge α = 0, and assume ∂ μ D μ has a single zero To avoid this difficulty, we first construct the partition function Z NL α = 0, and then take the limit α → 0, i.e., lim α→0 Z NL α . From the Lagrangian L NL , the equation of motion for B is So, when α → 0, the term −igα 2 (c × c) must be taken into account. In this section, treating the interactions perturbatively at the one-loop level, we study the behavior of α.
In Appendix C, we derive the renormalization group (RG) equations which coincide with the results in Refs. [20] and [21]. 4 We emphasize that the equation for α 1 does not contain α 2 , and vice versa. From Eq. (4.1), α = α 1 + α 2 satisfies When |α| 1, Eq. (4.2) becomes Therefore, when α 2 = 0, α increases as μ decreases. The quartic ghost interaction makes α = 0, and the situation in Sect. 3 is realized. Even if a single zero mode v 0 exists, the partition function does not vanish. Eq. (4.1) shows that (α 1 , α 2 ) = (0, 0) is an infrared fixed point. Does this fact imply that the Landau gauge (Eq. (1.1)) is retrieved as μ → 0? Does the process in Sect. 2 repeat again? In the next section, we show that such a problem does not happen.

Summary
In the Landau gauge α = 0, the FP operator −∂ μ D μ has zero modes on the Gribov horizon. As the ghost c and the antighostc are Grassmann variables, it is natural to expect that these zero modes yield effective ghost interactions. We have shown that the quartic ghost interaction is produced by a pair of zero modes. If we impose BRS invariance, the Lagrangian in the nonlinear gauge is obtained. Thus the Landau gauge changes to the nonlinear gauge. In the α = 0 gauge, the same result is obtained as well.
The effect of a single zero mode was also studied. Although there is no trouble in the α = 0 gauge, the partition function Z may vanish in the α = 0 gauge. We can avoid this problem by taking the limit α → 0.
Usually, when det ∂ μ D μ = 0 for some configuration A μ , we can avoid the Z = 0 problem by choosing another gauge (locally) (Ref. [25]). In this paper, we have shown that such a configuration changes the gauge to the nonlinear gauge automatically.
The partition functions in the Landau gauge and the nonlinear gauge are equivalent perturbatively. In the nonlinear gauge, (α 1 , α 2 ) = (0, 0) is an infrared fixed point at the one-loop level. In this case, the Landau gauge is retrieved and the zero-mode problem appears again. However, this scenario is not true. The nonlinear gauge yield the ghost condensation below the energy scale μ 0 , and the zero-mode problem no longer happens.

Appendix A. Examples of zero modes in the Coulomb gauge
In this appendix, choosing the gauge ∂ j A j = 0, we study the eigenvalue equation in three-dimensional space-time.

A.1. A pair of zero modes
If the eigenfunction has the form u A = e is w A with gA j × (∂ j w) = 0, Eq. (A1) becomes Since H is a real antisymmetric 3 × 3 matrix, its eigenvalues are pure imaginary or 0, i.e., The last equation of Eq. (A3) means that the effect of A C j disappears and w 0 does not become a zero mode. From Eqs. (A2) and (A3), we obtain h(x)e ±is w A ± = ( + λ)e ±is w A ± .
Thus we find the two functions u ± = e ±is w ± become a zero-mode pair, if A.1.1. Three-torus T 3 Gribov copies in the three-torus T 3 are studied in Ref. [5]. The constant configuration is on the first Gribov horizon, where L is the size of the torus. Setting s = 2πx 1 /L, we find Eq. (A4) is satisfied by a zero-mode pair

A.1.2. Axially symmetric configuration in R 3
Next we consider the configuration where (r, θ, φ) are spherical coordinates. Using the angular momentum operatorL j = −i jkl x k ∂ l , we find Then it is natural to set e is = e imφ and where l and m are integers, and We note e imφ Now, following Henyey (Ref. [26]), we substitute the functions into Eq. (A7), where K, r 0 , d, ρ, κ, and σ are constants. Equation (A7) is satisfied by Thus we obtain the Abelian configuration and the corresponding zero-mode pairs as ≥ 1, m = 1, 2, . . . , l).

A.2. A single zero mode
In Ref. [27], a single zero mode was found in an instanton background. Here we give an example in R 3 . Generalizing Eqs. (A5) and (A6), we choose the configuration First we solve the equation where iα is an eigenvalue of . We substitute the expansion and, for simplicity, choose l = 1. Then we find that the eigenvalues are α = 2, 1, and −1, and the numbers of eigenfunctions are 1, 3, and 5, respectively. We choose the real eigenfunctions Next we determine R α . From Eq. (A11) with λ = 0 and Eq. (A12), R α satisfies As in the previous subsection, we substitute Eq. (A8) into Eq. (A13). Then we find Two real zero modes are replaced by a pair of zero modes. So one real zero mode remains for each value of α.

Appendix B. Derivation of the Lagrangians Eqs. (2.19) and (5.1) by the use of "source"
In the instanton case, the fermion determinant does not vanish if fermion sources exist (Refs. [13,14]). Following this case, we introduce a field ϕ(x), and replace ic · ∂ μ D μ c with The eigenvalue equation is We treat the term gϕ× as perturbation, and perform the expansion where (0) n = λ n , and w Using the normalization dx u * n · u n = 1 and dx u n · u n = 0, we obtain for u n and for u * n , where f ABC u A n ϕ B u C n = 0 has been used. Therefore, if ∂ μ D μ has a pair of zero modes (u 0 , u * 0 ), Eq. (B1) gives rise to the determinant where k n is the number of eigenfunctions that have the eigenvalue n or * n . Thus, although Since gives the determinant Eq. (B2). To derive Eq. (2.19), we multiply Eq. (B3) by exp[− dx (ϕ + α 2 B) 2 /(2α 2 )], and integrate with respect to ϕ: After the ϕ integration, we obtain Eq. (2.19).
We note that to derive Eq. (5.1), then Eq. (B3) must be multiplied by exp where w is a constant determined later.

Appendix C. Derivation of the RG equations (4.1) and (5.2)
In Sect. C.1, using L NL , we derive the RG equations (4.1). In Sect. C.2, the RG equation ( We define the renormalization constant Z 4 by wherec 0 =Z 1/2 3c and c 0 =Z 1/2 3 c. First we consider the ghost self-energy. Although L NL gives additional one-loop diagrams, the divergence of them cancels out. Thus we obtain, as usual, where ε = (4−D)/2, and C 2 (G) = 2 is inserted. We note the gauge parameter in L NL is α = α 1 +α 2 .
Next we study Z 4 . Using the notation of Fig. C1, one-loop diagrams that contribute to Z 4 come from the diagrams in Figs. C2 and C3. However Fig. C2(b) does not yield divergence, and the divergences of Figs. C2(c1)-(c3) cancel out. Furthermore some of the diagrams derived from Fig. C3       Then performing the replacement g → gμ −ε or 1/ε → 2 ln /μ in Eqs. (C2) and (C3), and using the RG equation Then L NL gives the counter terms The first counter term cancels the divergence of Fig. C5(a), and we obtain As the gauge parameter in L NL is α, the constant Z as usual. Using these results, Z The divergence of Fig. C5(b) is canceled by the second counter term, i.e.,
Appendix D. Symmetries of the Lagrangian L ϕ in Eq. (5.1)

D.1. BRS symmetry
It is easy to check that L ϕ is invariant under the BRS transformation The constant w is determined in such a way as to conserve this symmetry. From the partition function we can derive the equation of motion for B as where = 1 Z ϕ Dμ exp − dx (L inv + L ϕ ) .
In addition, we must set w = ϕ 0 = 0 to maintain BRS symmetry. Asδ B L ϕ = −g(B ×c) · w, the Lagrangian does not respect the anti-BRS symmetry. 16
Namely, because of BRS symmetry, Z ϕ remains invariant under this symmetry.
In the same way, we can show that the breaking by w cannot be observed in any function ( ) , if ( ) is BRS invariant. To show this, we consider the function which appears in δ θ ( ) . Using δ θ L ϕ = −i(w × θ) · δ Bc and δ B ( ) = 0, we find Eq. (D3) vanishes. Thus BRS-invariant Green functions aren't broken by δ θ L ϕ .