THE VARIATIONAL PROBLEM AND BACKGROUND FIELD IN RENORMALIZATION GROUP METHOD FOR NON-LINEAR SIGMA MODELS

. We study the variational problem as described by Balaban in his renormalization group method for Yang-Mills theories in d = 3 , 4 and adapt it to a class of Non-Linear Sigma Models in d = 2. The result of the variational problem is a minimal conﬁguration which serves as a classical background ﬁeld in the renormalization group analysis.


Introduction
Balaban with his renormalization group method has managed to establish the ultraviolet stability of the lattice Yang-Mills theories in d = 3, 4 [1][2][3][4][5][6][7][8][9][10][11][12].His work in four dimensional Yang-Mills theories is described as a qualitative breakthrough [13].However, Balaban's renormalization group method has remained largely inaccessible to readers as the analysis is difficult to follow and is spread over several papers [14].The goal of our program is to apply Balaban's renormalization group method to study the ultraviolet problem in Euclidean Non-Linear Sigma Models in d = 2.The Non-Linear Sigma Models in two dimensions are both critical and asymptotically free just like the Yang-Mills theories in four dimensions.
Here we focus on understanding the variational problem in the renormalization group method as discussed by Balaban in [7].We study it for Non-Linear Sigma Models.The result of the variational problem is a classical background field.The fluctuations about this background field configuration denote the quantum corrections.
The background field method is also widely used in physics especially while quantizing gauge theories, see for example, [15,16] for applications.This is because the method preserves the symmetries of the theory while computing quantum corrections.
E-mail address: abhishek.goswami@amu.edu.pl.Date: September 23, 2022.where M and N are large positive integers while L is a large positive odd integer.The lattice notation we use here is due to Dimock [17].From this torus we construct a (scaled) lattice with spacing L −k as T −k M+N−k .We also consider a unit lattice (the lattice of centers of unit blocks) given by and an L lattice (the lattice of centers of L × L blocks) given by Let η = L −k and denote T η ≡ T −k M+N−k .Consider the field configurations U U : where G is a Lie group SU (2) in which case we have a O(4) Non-Linear Sigma Model.
More generally we take G as SU(n) and then we have a class of Non-Linear Sigma Models known as Non-Linear Chiral Models.
Next we rescale to the unit lattice T 0 M+N .The field configurations U L k : As the field configurations take values on the group manifold and not on the lattice, rescaling of the lattice has no effect on them.This property of the field configurations to be constrained on the group manifold is what makes the model non-linear.
Let x, x + ηe µ and x, x − ηe µ denote oriented bonds starting from x on the lattice T η .Define Action.We work with Wilson-like action as remarked by Balaban (Eq.0.32 in [8]) for Non-Linear Chiral Models.Let b be an oriented bond between two sites x and y on T η .Denote the boundary of b as ∂b = xy .The action functional is with d = 2 where Note that as d = 2 the functional ( 6) is independent of the lattice spacing.

Symmetries of A. First rewrite A as
Let v, w ∈ SU(n).Consider a global transformation U → vUw.Then where we have used the cyclic property of the trace.
We are interested in the integrals of the form where g is the coupling constant which is dimensionless in d = 2, E is the vacuum energy and dU = x∈Tη dU(x) is the product Haar measure of the group G.The integral is over the group manifold.We intend to study such integrals using the method of steepest descents.This is how the variational problem enters into the analysis.To do so we will have to define the integral over a linear vector space instead of the group manifold.We do that by transforming to the local coordinates of U given by Lie algebra variables; A, as U = e iA .First we give an overview of the method of steepest descents as applied in our analysis.
1.2.Steepest descent.This discussion follows the one in [18].Let A be an element of some finite domain in a linear vector space.Consider the integral where g is finite and E is a regular function of g and is sufficiently small.We assume that f has a unique critical point in the domain which is a minima.Denote the minima by A 0 .We look to expand f about A 0 and hence A 0 is called a background.Let A − A 0 = gA ′ denote the scaled fluctuation variable.Then the integral I becomes Here V is a function that includes higher order O(A ′ 3 ) terms in the expansion of f and d 2 (A 0 )f denotes the quadratic form also known as Hessian of f .The integral in ( 12) is then studied using cluster expansion.The output of the cluster expansion is a function of the background A 0 which is regular in g and also sufficiently small.
Our goal here is to understand the background configuration which is the minimum of the functional as written above.The cluster expansion and analysis of the coupling constant will be carried out in the future work.
Before we can formally state the variational problem we need few more definitions.
1.3.Block averaging.Consider the L −k lattice T η and a L −k+1 lattice given by T −k+1 M+N−k .For a site y ∈ T −k+1 M+N−k define a block centered on y as Note that since L is odd the boundary of the block lies between the two neighboring sites.
We define a one step block averaging operation that maps the field configurations on Note that if U(x) = e iA(x) and U(y) = e iA(y) with A small then using Baker-Campbell-Hausdorff formula: log [e iA(x) e −iA(y) ] = iA(x) − iA(y) Let v, w ∈ SU(n).Then under a global transformation U → vUw, first note that has same eigenvalues as U(x)U −1 (y) and hence is unitary equivalent to U(x)U −1 (y).Then by the spectral orthogonal projection representation (Eq.22 in [3]), their logarithms are also unitary equivalent Thus, we have the last step follows from the taylor expansion of the exponential function since for a matrix X, (vXv A composition of averaging operations k times is same as averaging over a block of side length L k and volume L 2k .Let y ∈ T 0 M+N−k .Define a block centered on y B k (y) is a block with L k sites on each side.Then 1.4.Spaces of regular configurations.Let M = L m be a very large positive integer (M is not same as the vol.parameter in (1)).Then the lattice of the centers of M × M blocks given by We take Ω k = T η .Let U be a field configuration on the lattice T η .Let ε 0 > 0. Define a space of the field configurations by the conditions The norms are operator norms for a matrix X, "the operator norm Define another space B k,1 (ε 1 ) as Balaban's notation.On the lattice T η Balaban [7] considers a sequence of domains • For any domain Ω ⊂ T η , Ω (j) = Ω ∩ (L j ηZ) 2 .Thus, Ω (j) are centers of L j blocks in Ω which are points on the lattice T M+N−k .The sequence admits the case where some domains are equal to the lattice T η .Balaban also considers the sets of the form Ω (j) Here we work with the case where all Ω j equal to the entire lattice T η .Hence, we work with a single scale determined by Ω k for spaces (21), (22).This is sufficient for the first treatment of the model as we work in a small field approximation.For the full model we have to consider the multiscale geometry.
Note that on the scale determined by j, where a bond variable as in (19) has a bond length of L j η, we have Since V is fixed, any global transformation must keep (23) invariant and hence, As any global transformation is independent of the lattice points, note that if V is any constant configuration and u is any When V is not a constant configuration, the transformations that satisfy (24) are of the form uVu −1 .Such transformations commute with V; [u, V] = 0 and thus, uV = Vu imply uVu −1 = V.Denote the subgroup formed by the transformations satisfying (24) as For any group G = SU(n), the only element that commutes with every other element is the identity.Thus, for a general V, the subgroup G(V) is just the identity.
We can similarly define a space B k,1 (B k,1 , W) for configurations U satisfying (23) but for (k + 1) instead of (k) with W instead of V and on B k,1 .
1.4.2.New space.Consider a set of two new field configurations U 0 and U = U ′ U 0 .Here we define a space whose arbitrary configuration is of the form U ′ U 0 .Let a 0 , a 1 > 0. We assume that Note that by (23) and the first line in (26), ( Then the variational problem is defined as Theorem 1 Let a 0 , a 1 be strictly positive constants.Then for 0 < ε 0 a 0 and 0 < ε 1 a 1 let V ∈ B k (ε 1 ) be fixed.There exists a constant B 3 > 0 obeying B 3 ε 1 ε 0 such that a unique minimal configuration of the functional A (k) (U) exists in the space Remark 2 In the k th step of the actual renormalization group construction, the action used is the effective action for scale k, which includes, in addition to the A (k) of Theorem 1, both contributions from the previous fluctuation integrals and the effects of renormalization.A (k) is only the classical contribution to the effective action and hence the background configuration is classical as well.Note that only the pure small fields, i.e.U with |∂U − 1| sufficiently small, are being considered in this paper.In the actual renormalization group construction all U ′ s must be considered.
Outline In Sect. 2 we first understand the above statement of the variational problem in the context of the renormalization group (RG) transformation.Then the proof of the Theorem 1 follows in the Sect.3. In the final Sect. 4 we discuss the analyticity of the background configuration.

RG transformation
Here we give a brief overview of the renormalization group (RG) transformation with the aim to understand the variational problem and the role of critical configuration in a single iteration.RG iteration has three steps: integration, rescaling and extension.
The first step introduces a split between the short distance (or high momenta) degrees of freedom and long distance (or low momenta) degrees of freedom.The unit lattice configurations here correspond to the high momenta degrees of freedom while the L lattice configurations correspond to the low momenta degrees of freedom.
We analyze the above integral by the method of steepest descent.This is where we require a minimal configuration of the functional A(U) with delta function constraint.The minimal configuration is the high momenta degree of freedom which serves as a background field.We integrate out the fluctuation about this high momenta background and then scale back to the unit lattice for convenience.
This procedure is carried out until we reach a unit scale denoted as L k ǫ = 1, where ǫ = L −K .Since we are rescaling back to the unit lattice after each integration, we are working with Note that the unit lattice at the start of the renormalization is T 0 M+N .Then the unit lattice at the end of k RG iterations is For our model, rescaling does not introduce any canonical dimension dependent changes to the field configurations as shown in (4).
The last step extension will be mentioned shortly.
2.1.Defining the variational problem.We first discuss the role of a critical configuration in a RG iteration.After that we define the variational problem in a manner such that its solution "that is the existence of a unique critical configuration" would mean that the RG iteration can be started.k = 1 At the start of the first iteration we have The action functional A(U) is functional of only one variable U.The critical configuration U 0 (V) belongs to the space U 0 (B 3 ε 1 ) ∩ B 0,1 (B 0,1 , V).We take critical configuration U 0 (V) as a background configuration and expand the functional A(U) around it.Note that as U 0 (V) is defined on a unit lattice, its fluctuation is also defined on a unit lattice which we integrate out.This completes the first step of the iteration.
Next we rescale back to the unit lattice so that The final step extension corresponds to simply recognizing that U 1 (V) is already in the space At the end of the first iteration the functional of two variables U and V is written as A (1) (U 1 (V)).It is the minimal value of the functional A (1) in variable U. Thus, at the end of the first iteration the variational problem is defined by . Thus, showing the existence of U 1 (V) implies that the first iteration can be started as described above.It provides the background configuration U 0 (V).k = 2 At the start of the second iteration we have A critical configuration of the functional A (1) (U 1 (V)) is critical in both the variables U and V.We already have configuration U 1 (V) which is critical in variable U. Denote a critical configuration in V of the functional A (1) (U 1 (V)) with constraint V = W as V (1) (W) ∈ B 1 .
After integrating out the fluctuation, we do the rescaling so that Extension corresponds to simply recognizing that the background configuration U 2 (W) is already in the space Thus, at the end of the second iteration the functional is A (2) (U 2 (W)) and the variational problem is defined by . Thus, once again showing the existence of a unique critical configuration U 2 (W) would imply that the second RG iteration can be started as described above.
After k iterations We have a functional of two variables U and V as The variational problem is defined as

Proof
We start by recalling definitions of some operators from [4].After that we discuss a mapping transformation from [6].We use the mapping transformation to reduce the proof of Theorem 1 to proving a Proposition 4. We then first give a detailed proof of the Proposition 4 in Subsect.3.3 using perturbative methods.Proof of Theorem 1 follows in the next Subsect.3.4.

3.1.
Minimizers.The minimizer is a solution of the linear variational problem defined as (Eqs.3.109, 3.110 [4]) where the constraint Q is linear.Both Q and ∆ depend on the background.The minimizer is a linear operator H and has a representation (Eq.1.103 in [1] or Eq.3.126 [4]) We now construct a transformation u satisfying (43) such that h = U ′ u −1 satisfies the following where Q k (U 0 , A) is given by .
Theorem 2 For arbitrary configurations U 0 and U ′ U 0 satisfying (26) there exists a constant c 1 such that if a 0 > 0 and a 1 > 0 obey a 0 + a 1 c 1 then there is exactly one transformation u satisfying (43) on B k such that the conditions (45) hold for the configuration h = U ′ u −1 .
Proof The proof of the Theorem is by induction.First note that the conditions U 0 ∈ U k (a 0 ) and ) have an inductive property that they hold for k − 1 if they are true for k.Whereas the condition does not have the same inductive property.To apply induction, we follow Balaban and use an equivalent condition (Eq.1.66, [6]) where . Now we assume that the Theorem is true for k − 1.This means we have a desired transformation u 1 satisfying R 0 u 1 (k−1) = 1.Then inductively we have R 0 u 1 (k) = 1 and we construct the desired transformation for k th step as a perturbation Then solving Q ′ (u 1 , λ) = 0 gives us λ.Note that we do not need the Landau gauge condition of Balaban (Eq.1.80 [6]).By Proposition 5 [6], we get a unique λ satisfying We get the desired configuration as Since U 1 is constructed inductively from the solution of the previous step, we have U 1 = e iA ′ with |A ′ | < O(1)(a 0 + a 1 ).Then from the bounds on A ′ and λ it follows that To use perturbative methods we make use of the Theorem 2. Set a 0 = ε 0 and a ) by Theorem 2 there exists exactly one transformation u satisfying (43) such that h = U ′ u −1 satisfies the conditions (52).k = 1.Note that k = 1 defines the variational problem at the end of the first iteration.We take the given configuration V ∈ B 2 (ε 1 ) and construct a configuration V 0 ∈ B 1 (ε 1 ) as (53).We take U 0 = V 0 such that U 0 ∈ U 1 (ε 1 ) and consider the functional A (1) (U) on the space U 1 (ε 0 ) ∩ B 1 (B 1 , V 0 ) with U = U ′ U 0 (and U 0 on a larger space U 1 (B 3 ε 1 )).Then by definition (26) we have | Ū0 − V 0 | < a 1 on B 1 .Now we apply Theorem 2 as discussed above.
This completes the proof of Lemma 3.
Proposition 4 For h = e iA with variables A in the space (52) there exist variables A 1 and a map D from A to the variables on B k such that the unique critical configuration of the functional A (k) (hU 0 ) in the space |A| < ε 2 is given by the configuration where |A 1 | < 2ε 3 with ε 2 , ε 3 sufficiently small and for some constant C 2 .

3.3.
Proof of Proposition 4. We start by an expansion of A (k) (e iA U 0 ) about the background U 0 as (Eq.3.189 [4]) where V 0 (A) denotes O(A 3 ), ∆ is given by (Eq.3.191 [4]) is a small perturbation of the Laplacian that depends on the background and A, J is defined as Then we linearize the constraint Q k by a suitable change of variables.We denote the new variables as A ′ and substitute them in (57).After some minor readjustments including another change of variables A ′ → A 1 we get a functional whose critical configuration is the one we are looking for.We make use of contraction mapping theorem to show the existence of a unique critical configuration.
3.3.1.Linearizing transformation.We follow Sect.C of [7] on Ω k .The constraint is It is a non-linear function of A. From the definition of the block averaging ( 14) and using Baker-Campbell-Hausdorff the linear term is Q k A = Ā+Af (U 0 ), where f (U 0 ) is a function of the background configuration.Then Here we will linearize the constraint so that Q k is linear.
The linearizing transformation is constructed as where D will be a mapping defined on configurations A and taking values in configurations on B k .In fact we will see that D is related to C k .
Proof The constraint is linear if We study this fixed point using contraction mapping theorem.Let Then the map (64) is contractive if (see the discussion below Eq. 54 in [7]) Identifying X with D(A ′ ) in (65), we get |D(A ′ )| < B −1 0 ε 3 .Since ε 3 was arbitrary, we can take it close to and D (n) be homogeneous polynomials of n th order, then write and therefore, This completes the proof of the Lemma 5.
Locality of D(A ′ ).Note that D (2) (A ′ ) is ultra localized at a site whereas D (m) (A ′ ) for m > 2, contains the operator H which has an exponential decaying kernel.Thus, we can say that functional derivative of D(A ′ ) decays exponentially.Hence, D(A) is localized.
Proof D(A ′ ) is a power series in A ′ as (69) and the bound (68) implies that the transformation Similarly, using and from (66) that the transformation A = A ′ − HD(A ′ ) is defined for ε 3 (18C 2 B 0 ) −1 on Ω k .Now see Eqns.59-62 in [7] in addition to our discussion.This completes the proof of the Proposition 6.
Denote the linearized constraint Q k (A) as Q.

3.3.2.
Equations for a solution of the variational problem.Make change of variables as A = A ′ − HD(A ′ ) in (57) and consider the functional on the space of configurations A ′ satisfying Note that in the functional F (A ′ ) the zeroth order and the first order terms in A ′ are A (k) (U 0 ) and A ′ , J respectively.Next we separate the quadratic terms.Write Proof See Sect.E in [7].

3.4.
Proof of Theorem 1.The proof of Theorem 1 is by induction.For k = 1, a fixed configuration V 0 ∈ B 1 (ε 1 ) is given.We define U 0 = V 0 such that U 0 ∈ U 1 (ε 1 ).Then we consider the functional A 1 (U) on the space U 1 (ε 0 ) ∩ B 1 (B 1 , V 0 ).The definition (26) gives us freedom to consider U 0 on a larger space bounded by ε 0 .Thus, we can have a large constant B 3 obeying B 3 ε 1 ε 0 and take U 0 ∈ U 1 (B 3 ε 1 ) such that for B 3 = 1, we have Ū0 = V 0 on B 1 .Now take U = U ′ U 0 .Next we make use of the mapping transformation (Theorem 2) and get the functional A 1 (hU 0 ) = A 1 (e iA U 0 ) to solve the critical problem perturbatively as shown in Lemma 3. Then Proposition 4 gives us a unique critical configuration of A 1 (e iA U 0 ) in the space |A| < ε 2 < O(1)ε 0 .
To understand what B 3 is like let denote a M − block and ˜ denote a block of linear size 3M consisting of nearest neighbors of .Denote ′ ∈ c ∩ ˜ .Let y 1 ∈ .We follow Sect.F (Eqs. 162, 163) in [7] and write and assume that B 3 e −κ(2M ) 1 2 .This completes the proof for k = 1.For each inductive step k > 1, the critical configuration for k − 1 serves as U 0 ∈ U k (B 3 ε 1 ) for the k th step, where B 3 is provided by the inductive hypothesis on k − 1.Then we consider the functional A k (U) on the space U k (ε 0 ) ∩ B k (B k , V) and solve the critical problem perturbatively as in Lemma 3 and Proposition 4. To see that U k is a minimum we apply the whole procedure again with U k instead of U 0 .We get a functional F (A ′ ) as (80) for which the critical configuration is A ′ = 0. Then from (81), (82), (83) together with A ′ = 0 we conclude that δA ′ , J = 0 for all δA ′ : QδA ′ = 0 (99) But due to the shift to U k we also have QA = 0 (the constraint is automatically satisfied since Ū(k) k = V).Hence the configuration A ′ satisfies the same conditions as δA ′ and we