A constraint on the geometry of Yang-Mills theories

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  • JGP - Vol.4, a. 3, 1987

    A constraint on the geometryof Yang-Mills theories

    SIMON DAVISLyman Laboratory of Physics

    Harvard UniversityCambridge, MA 02138


    Abstract. The connection form of a fibre bundle may be identified with the gaugepotential of a Yang-Mills theory only if its dependence on the coordinates of thefibres can be eliminated. To maintain this property under a change of bundlecoordinates, the standard fibre must admit a globally integrable parallelism. Thenecessity of principal bundles in thegeometricalformulation of Yang-Mills theoriescan then be deduced.

    1. INTRODUCTIONThe classical action for a pure Yang-Mills theory can be regarded as a func-

    tional of the curvature form of a principal fibre bundle that is invariant underautornorphisms of the bundle which act trivially on the base space [1) [2]. Whilethe set of gauge transformations determine the structure group of the bundle,there seems to be no similar constraint on the standard fibre. The defining pro-perty of a principal bundle, that the standard fibre coincide with the structuregroup [3], is one that could apparently be eliminated. However, attempts togeneralize pure Yang-Mills theories by allowing the standard fibre to be a non-group manifold have been unsatisfactory [41[5]. The aim of this paper is todemonstrate that the geometry is strongly constrained by physical requirements.Of particular importance is the condition that the fields and transformationrules are defined entirely on the base space; otherwise, any gauge invariance of

    Keywords: Principal connections, Yang-Mills theories.1980 MSC: 53C05, 53 C80.


    the field theory cannot be regarded as an internal symmetry. It will be shownthat the independence of the gauge potential with respect to the fibre coordinatescan be maintained under gauge transformations only if the standard fibre is a Liegroup. One consequence of this result is that the potential must transform underthe adjoint representation of the gauge group. A similar conclusion may bereached by considering the consistency of nonlinear self-couplings of masslessspin-one fields [6].

    2. THE CONNECTION FORMA connection in a principal bundle (F, M, ir, G) (1) provides a decomposition

    of the tangent space at p E P into a horizontal subspace H,~,and a vertical sub-space Vt,, tangent to the fibres (7}. The horizontal subspace H~depends smoothlyon p, and it is required to be invariant under right translations Rg*Hp = Hp~gThe diffeomorphisni between the fibre passing through p and the group G inducesan isomorphism between V~and the Lie algebra of G. Thus, given a connection,a Lie-algebra-valued connection form can be defmed on P by projecting anytangent vector in T~(P) to V~and then mapping the vertical component to theLie algebra. For each local section u : U C M -+ ir(U), this connection formcan be pulled back to the base space by Ft~= o*F. In local coordinates for theneighbourhood U, F~ A~~dx~,where A~0)can be identified as a Lie algebra-valued gauge potential, since, under a change of section, a(x)-+ o(x) = c(x)gl~x),it transforms asA~~= + g1 ~3~g[1].

    Consider now a general fibre bundle (E, M, ir). Given a bundle atlas {U,~.,~1i~I,every point x E M has a neighbourhood Un such that i,l/~is a diffeomorphismfrom ir 1(U) to U x F for some standard fibre F. If x E Un fl U~,y E F, then

    o i~i~~(x,y) = (x, ~~3(x, y)), where ~ ) EDiff(F)are transition functions.

    The trivializations ~i, ~ determine two local sections Un(X) = ~.1(x, y

    0),u~(x)= ~ ~, y0). If U C Un fl U~,an arbitrary trivialization ~i ir

    1(U) ~ U xFwill map a(x), o~(x)to s~oa(x) = (x, y(x)), ~,1io~(x)= (x, y(x)). Let us define

    U x F - F by ~jx, y) ~(x)~ ~1~(y) = y, where E Diff(F).Then the tangent spaces to the two sections are related by

    (1) ~ ~ T~,!i~. ~ Ti4i~~(~~,U0))

    Any bundle (E, M, ir) for which the total space E is a paracompact manifoldadmits a cr connection, 0 ~ r ~ oo [8]. The tangent bundle is then the direct sum

    (1) P is the total space, M is the base space, i~isa smooth projection of P onto M, and Gis the structure group.


    of the vertical and horizontal subbundles, TE = VE n HE. The horizontal pro-jection of the tangent vectors 00* ~ . ~ may be mapped onto TU x TF,with the components along TF denoted by C, C~respectively. The horizontalsubspaces IP*HO(X), P*HO~(x)are spanned by vectors of the form (~,C(y)),~ C~(y)),as the components along TF must independent of ~. Thus, thediffeomorphism ~ induces a mapping from ~]I*HO(X) to ~Li$HQ(X), so that

    (2) C~(y)=TIJ1~ C(y)The connection forms F, F1 are obtained by subtracting C0, C1 from v,V1. The transformation from F0 to N is given by

    = ~

    (3) ~ V0(~~,y)TlJ1~0.C(~~,y)

    = Tl/1~0(~~)+TI/1~0. F(~~,y)The transformation rule of the connection form for the bundle (E, M, ir) generallyinvolves the fibre coordinate.


    To identify the connection forms with gauge potentials depending only on thecoordinates of the base space M, one must extend ~ y) E T~(F)andFa(~~,y) E T~(F)to global vector fields on F. For any y EF, define a mapping

    from a vector space V to the tangent space T~(F),such that ~ y) ==~F


    PROPOSITION 1. The dependence on the fibre coordinate in the transformationrule for the connection form can be eliminated only if the fibre F admits aparallelism.

    Proof It follows from (3) that(4) ~ .f(~)The relation between F(~~)and can be interpreted as a gauge transfor-mation if the fibre coordinate can be eliminated from (4). As both terms on theright-hand side of (4) can be expressed as

    1,,X, X E V, 0), should be a vectorspace isomorphism which depends smoothly on y. Thus, ~ : V x F -~ TF, (.,y) == ~,represents a parallelism on F, with V being the reduced tangent space[9].


    Now consider the second term on the right-hand side of (4). To eliminate thefibre coordinate, we require

    (5)where A(x) : V -+ V is a linear isomorphism depending only on the ~ and notthe fibre coordinate y. For a Yang-Mills theory associated with a principal bundle(F, M, ir, G) V is the Lie algebra of G and A(x) = Ad(g1(x)), g(x) E G. WhenA(x) = Jd~,equation (5) becomes T~ = fr~, . t~.The parallelism ~ defines adifferential system, whose solution is a local diffeomorphism mapping y to y.The parallelism is defined to be integrable if a solution exists for any pair ofpoints y, y EF. When every solution can be extended to a global diffeomorphismon F, the parallelism is globally integrable, and Fis a Lie group [9].

    Even if A(x) ~ Id~,,it can nevertheless be demonstrated that the parallelismmust be globally integrable.

    PROPOSITION 2. For any pair of points ~ y~E F, there exists a global dif-feomorphism iIi~ : F - F, ~ (y

    0) = y~,satisfying Ts,Li~0 . = . A for someconstant matrix A if and only if the parallelism on F is globally integrable.

    Pro of The global diffeomorphism sIi~ is strongly constrained by (5). Indeed,by Frobenius theorem, a local diffeomorphism obeying (5) will exist if andonly if

    (6) [~~A . X, ~ Y]~ = c5~A1. [X, Y]~

    for any two vector fields X, V defined in a neighbourhood of y0. The parallelism

    induces a set of smooth, linearly independent vector fields {~(y)= ~~e1} wherethe {e~} form a basis for the vector space V. Expanding the commutator of twovectors ~, ~ at y in the basis {~(y)},[~, ~ = ciJk~.)~kU~),and noting that(7) ~~A~~1(y) = cb~Ae1=Akf~k(y)the integrability condition (6) becomes(8) Ci.k(YO)Ak = Cklm(YO)AkiAljThis identity holds for all points in a neighbourhood of y0 when is a localdiffeomorphism. If we require ~ to be a global diffeomorphism obeying (5).then (8) must be satisfied at every point on the fibre. It will now be shown thatno constant matrix A can satisfy (8) when the coefficients Cilk depend on thefibre coordinate j. This is a consequence of (8) leading to an infinite number ofindependent constraints on the constant matrix.

    Suppose the contrary. Then for any pair of points (y, j~).there is a globaldiffeornorphism (3 such that T(3 . = . B


    (9) Cf/k (~3~Wmk= Cklm(Y)BkjBlIIf y is chosen to be different from y, cf/k(YY) * cf)k(y), it may be assumed thatB does not obey (8). Let us define the map 7 by ~i~(y) = i.ji~ . (3(y)

    ~Ji~(y) = 7(y). Then Ty .4,,, = C, C = ABA. From (8) and (9),Cf/k (~~O)Amk= Cklm(I~I/~(YO))AkiAII

    (10) = [ClflO(YO)(B~)li(B~)n/BkO}Amk r ~ ~C~ C~ C 1A A tCpqrkYoA pk~ ~ql mr~ kt~l/

    If [A, B] = 0, then B = C and (10) becomes(11) [c

    1~0(y0)A ko ] (B 1),1(B )n/Bmk = [c~~~(y0)ApkAqlI~1~ki~)ijCmrwhich is equivalent to (8). Otherwise (10) represents an independent conditionon A.

    Let F be a smooth n-dimensional manifold, so that it may be covered by openpatches which are homeomorphic to open sets in R~.Since the tangent space atany point on F is isomorphic to R, the reduced tangent space V can be identi-fied with R~.If there is a basis of V, (e

    1, . . . e,~),in which A is a diagonal matrixwith n distinct eigenvalues, the matrices commuting with A are also diagonal.If the eigenvalues of A are not all different, then there will be matrices with off-diagonal elements commuting with A. However, these matrices will not neces-sarily commute with each other, and, in fact, the maximal commuting subgroupin this set is again the group of diagonal matrices. When A has complex eigen-values and cannot be diagonalized over Ri, the argument above can be adaptedto show that the maximal abelian group of linear transformations of V com-muting with A is still n-dimensional.

    However, not all matrices in this group can satisfy an identity such as (9).This is demonstrated in the following lemma.LEMMA 1. Suppose there is a diffeomorphism (3(A) such that T(3(X) . 4, =

    = AB, and

    (12) Ci/k(Y)XBmk = Cklm(Y(A))XBkiXBJ/Then A must equal].

    Proof It follows from (9) and (12) that Cklm(Y) = XCklm(Y(A)). The coefficientsCklm are also constrained by the following property. If X, Y, Z are three parallelvector fields on F, so that their components with respect to the basis {~(y)}are constant, then [X, Y], Z] is also a parallel vector field [FO].Thus for anyy, y EF


    I3~ FI~ ~ 1 l~1 . ~.1iii- 1 1 l~1. i ~ Vy t[cf~c/J~ckJ~

    or equivalently,

    (14) Ci/lCY)Cklm(Y)~ ~k(~Ci/mXY) = Cj/1(y)C~J~(Y)~k~fjrnXY)In particular, forj~,5~(X)=(3(X) ~3~(j~), (14) gives(15) X

    2Cif(5~)C~1~(j~) X~~k(Ci/m)(5~)= ~i.l(y)Ckl ~k~ijmXY)

    There are two solutions to this equation, but by again considering (14) for thepair y, ((3(X) . (3

    1)~~~2(y),we see that A = 1 is the only acceptable one.From this lemma, it follows that the set of matrices commuting with A and

    satisfying (9), to be denoted as MA , has maximum dimension n 1. Now fix apointy

    0 E F. If T13(y0, y) . 41 = 41 B~,then j ~ B~is a mapping ofF ontoMA . This mapping is surjective, but we will now show that its domain cannotinclude all of F. Consider two points )~= f3~.(j),i = 1, 2, such that T$1 . 41 == ~,. . B.. When B1 = B2, T@2 . (3i) =

    4,Y2 which implies Cf/k(y1) == ci/k(Y2). The diffeomorphism ~2 (3~1is known as a translation, and the setof translations forms a pseudogroup [9]. If the pseudogroup is transitive on F,then Cf/k is constant and the parallelism is integrable. By our original assumption,however, the coefficients Cf/k are not constant and the set G

    3, = y E F Cilk 0) == Cf/k(Y1)} is strictly contained in F. It will now be shown that this assumptioncannot hold.

    LEMMA 2. ThesetC~ containsall of F.

    Proof Suppose initially that C~ is a continuous curve passing through y1. LetX~ be the tangent vector at y1 so that X(Cjj~)(y,) = Xk~k(Ci/m)(}~l)= 0. Bythe identity (14), X(C~j~)(y)= 0 for all y on the curve C~ . Thus C,, is an inte-gral curve of the parallel vector field X.

    The derivative of Cf/k in the direction of an arbitrary tangent vector V alsotakes the same value at every point on C~ . Thus, if (4~}is the one-parameterfamily of diffeomorphisms generated by the vector field Y, the translation ofC3, by a distance t along the integral curves of V gives a curve of constant Cf/k,C . The tangent vector to C,~ is


    (16) = X t[X, Y] + [X, Y], Y]

    Associated to each vector field V is a two-dimensional surface S~,spanned bythe integral curves of V passing through C~,. It can be shown that there willexist different surfaces S,~,S,, that intersect at a non-zero distance from C,,


    Suppose Y1, . . ., Y,~_1 are independent vector fields such that the integralcurves of linear combinations of the V1 do not intersect and sweep out an n-di-mensional neighbourhood of y1. If V = E a~Y1and V = aX + bY, the surfaces

    S~,S~ initially coincide at C,, . However, upon substituting Y in (16), one seesthat the tangent spaces to Si,, S,, will differ elsewhere as they are spanned bylinearly independent vectors. Since S,, does not coincide with S,, everywhere,it must intersect another surface S~,,. Let y3 be a point in this intersection.The set C,, consists of a curve in S~, with tangent vector X t[X, Y~I+

    [X, V]. V] . . . and curve in ~ with tangent vector X t[X, V ] +

    + [X, V], Y] . . . where t t are parameters for the integral curves ofV, Y respectively. These vectors are equal if they agree to each order in theinfinitesimally small parameters t, t, which is only possible if V is a multipleof Y. Since V * cY, the tangent vectors do not coincide, and by linearity,they generate a two-dimensional surface of constant Cf/k. By translating thissurface back to y~,one now finds that C,, is two-dimensional. Upon repetitionof the above argument sufficiently many times, it can be concluded that CUk isconstant on an n-dimensional neighbourhood of y1, U,,. Since F can be coveredby overlapping neighbourhoods U3,, Cf/k must be constant on all of F.

    Since {~1(y)} are smooth vector fields, the coefficients CjJk are smooth func-tions and the only remaining possibility is that C,, is a set of isolated points.However, if C,, is zero-dimensional and MA has maximum dimension n 1,there are infinitely many points in F which cannot be mapped onto MA. Cor-responding to each of these points is an independent constraint on A. As A is aconstant matrix, not all of these constraints can be satisfied. Therefore, C,, isn-dimensional and contains all of F.

    As a consequence of Lemma 2, it follows that the coefficients Cilk are constantand the parallelism on F must be globally integrable. The fibre F is then of theform GID, where G is a Lie group and D is a discrete subgroup [10]. AssumingF is simply connected, it will be a Lie group.

    Conversely, when the parallelism is globally integrable and F is a Lie group,the coefficients C/k are constant, and equation (8) will have solutions. Thesimplest solution is Amk = ~mk~More generally, one may note that the reducedtangent space V can be identified with the Lie algebra, and given an arbitrarygroup element g = exp(t~,),A = Ad(g) is an automorphism of V(17) Ad(g)~~= exp(t~/)~fexp( t~,)=where


    (18) Amf = ~mf + t...


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