Construction of Nahm data and BPS monopoles with continuous symmetries
Benoit Charbonneau, Anuk Dayaprema, C. J. Lang, ?kos Nagy, Haoyang Yu
CCONSTRUCTION OF NAHM DATA AND BPS MONOPOLES WITH CONTINUOUSSYMMETRIES
BENOIT CHARBONNEAU, ANUK DAYAPREMA, C. J. LANG, ÁKOS NAGY, AND HAOYANG YU
Abstract.
We study solutions to Nahm’s equations with continuous symmetries and, undercertain (mild) hypotheses, we classify the corresponding Ansätze. Using our classification,we construct novel Nahm data, and prescribe methods for generating further solutions.Finally, we use these results to construct new BPS monopoles with spherical symmetry.
Contents
1. Background and motivation 12. Nahm’s equations and their symmetries 43. Axially symmetric solutions 94. Spherically symmetric solutions 145. The Nahm transform 28Appendix A. The coe ffi cients of the Higgs fields 32References 331. Background and motivation
When one studies a particular equation of interest and is confronted with the desireto prove the existence of solutions, the trick favored amongst all others is to imposesymmetry to simplify the equation, in the hope that the reduced equation is more tractable.In gauge theory, this trick has manifested itself over and over again, and this paper is anew manifestation of it.Of all the gauge theory equations, one of longest lived is the
Bogomolny equationexpressing the relationship F ∇ = ∗ d ∇ Φ between a connection ∇ of curvature F ∇ over a vectorbundle E , and an endomorphism Φ of E called the Higgs field . The study of symmetricmonopoles on R goes back to the very first attempts to produce monopoles back in1974–1975, when some spherically symmetric monopoles were explicitly calculated in Date : February 23, 2021.2020
Mathematics Subject Classification.
Key words and phrases.
BPS monopoles, Nahm equation, Nahm transform, symmetric Ansätze. a r X i v : . [ m a t h - ph ] F e b
32, 34, 41]. The study of axial symmetry came a few years later with [24, 43]. Sincethen, spherical symmetry was explored further in [5, 12, 13, 33, 44, 45], and cylindricalin [12, 13, 28]. The bulk of the research on symmetric monopoles however concernedmonopoles with discrete group of symmetries in [7–9, 18–22, 31, 37–40]. Everything citedso far takes place on the Euclidean space, but symmetries should be useful wherever thereare some, and have been exploited for studying monopoles on the hyperbolic 3-space in[4, 11, 27, 30, 36], and for studying monopole chains, that is monopoles on R × S in [14].Using symmetry is of course not just about proving existence. Because of the explicitnature of the solutions, it allows one to test other ideas. For instance, the recent work [15]used the known examples to compute explicit tails of monopoles. And in [18], Hitchin,Manton, and Murray prove that the charge k cyclic SU(2) monopoles form a geodesicsubmanifold of the total moduli space, hence this can be used to illustrate some of themonopole dynamics.The tools used to study monopoles are many, and include twistor spaces constructions,rational maps, spectral curves, Nahm transform; see [35] for a comprehensive book thatalso includes a lot of the physics. Our choice of tool is the Nahm transform and through itNahm’s equations. Nahm’s equations were first introduced by Nahm in [29]. They form asystem of ordinary di ff erential equations on a triple of matrix-valued functions, definedon an open subset of R . Such a triple T = ( T , T , T ) satisfies Nahm’s equations if˙ T = [ T , T ] , (Nahm1)˙ T = [ T , T ] , (Nahm2)˙ T = [ T , T ] , (Nahm3)or, in a more compact form, involving the antisymmetric Levi-Civita (cid:15) -tensor: ∀ i ∈ { , , } : ˙ T i = 12 (cid:88) j,k =1 (cid:15) ijk [ T j , T k ] . (1.2)In practice, solving Nahm’s equations is di ffi cult as it is equivalent to a purely quadraticsystem of ordinary di ff erential equations, that is a system of equations on functions x ( t ) , . . . , x n ( t ) of the form ∀ a ∈ { , . . . , n } : ˙ x a = n (cid:88) b,c =1 C abc x b x c , for some C abc ∈ R . Such equations are not well-understood even in the n = 2 case. Tocircumvent these issues, we can use two things to our advantage: First of all, Nahm’sequations can be formulated as a Lax pair, thus it has many conserved quantities. Second, e only search for axially and spherically symmetric solutions which further reduces thecomplexity of the system.Nahm’s equations corresponding to monopoles with maximal symmetry breaking havebeen understood for a long time, dating back to Nahm in [29] and then Hitchin forstructure group SU(2) in [16, 17] and Hurtubise–Murray for arbitrary classical groups in[25]. For arbitrary symmetry breaking, apart from low charge rank 2 minimal symmetrybreaking work of Dancer [12] and the work of Houghton–Weinberg [23] extending it,nothing comprehensive about the Nahm transform and the behavior of the Nahm datacorresponding to monopoles with arbitrary symmetry breaking was known prior to [1, 2,10]. The eigenvalues of the Higgs field at infinity correspond to singular points on theinterval of definition of the solution to Nahm’s equations, and at those singular points the T i have residues forming representations of su (2). Unlike in the maximal symmetry breakingcase, where [25] showed that as one approaches a singular point from the side of highestrank there are continuing and terminating components in the T i and only a continuingcomponent from the other side, here terminating components can arise on both sides.In this paper, we do not explicitly consider the full possibilities of solutions to Nahm’sequations where multiple adjoining intervals are permitted, and only produce explicitsolutions on a single interval, corresponding to monopoles whose Higgs field has only twodistinct eigenvalues at infinity. While our method could potentially produce sphericallysymmetric monopoles whose Higgs fields have more than two distinct eigenvalues atinfinity but finding such examples is beyond the scope of this paper. Thus, for our currentpurpose, it su ffi ces to describe the behavior of the Nahm data at the two ends of thisinterval: the residues of the T i form a (possibly reducible) representation of su (3).The main result of our paper, Theorem 4.4, is a structure theorem for sphericallysymmetric solutions to Nahm’s equations under certain reasonable conditions. Using thesimplest version of this structure theorem, we provide in Theorem 4.9 Ansätze for familiesof spherically symmetric solutions based on a long chain of representations V n +2 k ⊕ · · · ⊕ V n .We capitalize on this Ansatz for a chain of length 3 in Theorem 4.12 to produce Nahm datafor a spherically symmetric monopole with symmetry breaking S(U(3) × U(3)). Chainsof length 2 are fully explored in Theorems 4.11 and 4.13 to contrast an infinite familyof solutions to Nahm’s equations, indexed by n ∈ N + , that are defined on ( − ,
1) andwhose residues correspond to the representation V n +1 ⊕ V n +1 on one end and V ⊕ ( n +1)2 on theother. The corresponding monopoles are spherically symmetric SU( n + 3) with symmetrybreaking S(U(2) × U( n + 1)), so their symmetry breaking is neither minimal nor maximal. rganization of the paper. The paper is organized as follows: In Section 2, we recallsome relevant properties of Nahm’s equations, with an emphasis on the two canonicalgroup actions, rotations and gauges. We then introduce the notions of axially and spher-ically symmetric solutions. In Section 3, we prove an infinitesimal version of the axialsymmetry condition and classify axially symmetric solutions. In Section 4, we prove aninfinitesimal version of the spherical symmetry condition and we classify (under certainextra hypotheses) and construct novel spherically symmetric solutions. We also provide amethod for constructing further spherically symmetric Nahm data. Finally, in Section 5,we use the spherically symmetric solutions found in Section 4 to construct novel BPSmonopoles using the (ADHM–)Nahm transform.
Acknowledgment.
BC is supported by NSERC Discovery Grant. Portion of this work wasdone while CJL received NSERC USRA support. CJL was supported by NSERC CGS-D. AD,ÁN, and HY thank the organizers of DOmath 2019 at Duke University, where a large portion ofthis work was completed. Nahm’s equations and their symmetries
Let us begin with three well-known facts about Nahm’s equations (Nahm1) to (Nahm3):(1) Since the map M n × n ( C ) ⊕ → M n × n ( C ) ⊕ ; ( A, B, C ) (cid:55)→ ([ B, C ] , [ C, A ] , [ A, B ]) , is analytic (in fact polynomial), solutions to Nahm’s equations are (real) analyticand unique, for any initial condition; see, for example, [42, Theorem 4.2].(2) Since su ( n ) is a matrix Lie-algebra, it is closed under commutators. Thus if asolution to Nahm’s equations that is defined on a connected interval ( a, b ) and forall i ∈ { , , } and for some t ∈ ( a, b ) satisfies that T i ( t ) ∈ su ( n ), then it must alsosatisfy that T i ( t ) ∈ su ( n ) for all t ∈ ( a, b ) and, again, for all i ∈ { , , } . In fact, su ( n )can be replaced by an arbitrary Lie algebra.
3) Any solution to Nahm’s equations have the following five conserved quantities: C = tr( T T ) , (2.1a) C = tr( T T ) , (2.1b) C = tr( T T ) , (2.1c) C = tr (cid:16) T − T (cid:17) , (2.1d) C = tr (cid:16) T − T (cid:17) . (2.1e)Solutions to Nahm’s equations typically develop singularities. In fact, in all known,nonconstant examples develop at least one, simple (that is, first order) pole. It is notknown whether all nonconstant solutions develop singularities, and when they do, whattype of singularities can occur. The lemma below—which is well-known in the literatureand thus we state it without proof—describes the behavior of a solution at a simple pole.For the remainder of this paper, let ( X , X , X ) the standard basis of so (3), that is, forall i, j, k ∈ { , , } , we have ( X i ) jk = − (cid:15) ijk . Lemma 2.1.
Let T be a solution to Nahm’s equations (Nahm1) to (Nahm3) , such that it isdefined on an open interval ( t , t ) ⊆ R , and T has a simple pole at t . Then the map so (3) → su ( n ); a X + a X + a X (cid:55)→ a Res( T , t ) + a Res( T , t ) + a Res( T , t ) , is a Lie algebra homomorphism (that is, a representation of so (3) ). There are two canonical group actions on solutions to Nahm’s equations.
Definition 2.2.
Let T = ( T , T , T ) ∈ M n × n ( C ) ⊕ .(1) For each A ∈ SO(3) , define A T = ( A T , A T , A T ) via ∀ i ∈ { , , } : A T i (cid:66) (cid:88) j =1 A ij T j . (Rotation) (2) For each U ∈ SU( n ) , define U T = ( U T , U T , U T ) via ∀ i ∈ { , , } : U T i (cid:66) U T i U − . (Gauge) We denote the induced actions of X ∈ so (3) and Y ∈ su ( n ) on T as X T and [ Y , T ] , respectively. Remark 2.3.
The Nahm transform (explained in Section 5) provides a correspondence betweenmonopoles and some solutions to Nahm’s equations. The action defined in (Rotation) correspondson the monopole side to pulling back via the same rotation on R . he following proposition is straightforward, hence we state it without proof. Proposition 2.4.
The actions defined by equations (Rotation) and (Gauge) define smooth groupactions. Furthermore, the two actions commute and preserve su ( n ) ⊕ , that is ∀ A ∈ SO(3) : ∀ U ∈ SU( n ) : A ( U T ) = U ( A T ) (cid:67) AU T . (2.2) and if T ∈ su ( n ) ⊕ , then AU T ∈ su ( n ) ⊕ . Let now T be a smooth function T : ( a, b ) → su ( n ) ⊕ ; t (cid:55)→ T ( t ) = ( T ( t ) , T ( t ) , T ( t )) . By an abuse of notation, let AU T be the function defined as AU T : ( a, b ) → su ( n ) ⊕ ; t (cid:55)→ AU ( T ( t )) . Then we have the following lemma for solutions to Nahm’s equations.
Lemma 2.5.
Let T a solution to Nahm’s equations (Nahm1) to (Nahm3) on some connected,open interval I ⊆ R . Then for all A ∈ SO(3) and for all U ∈ SU( n ) , the function AU T also solvesNahm’s equations.Furthermore, let (cid:101) T be another solution to Nahm’s equations. If there exists t ∈ I ∩ dom (cid:16) (cid:101) T (cid:17) , A ∈ SO(3) , and U ∈ SU( n ) , such that (cid:101) T ( t ) = AU T ( t ) , then (cid:101) T can be extended to all of I , and AU T | I ∩ dom ( (cid:101) T ) = (cid:101) T | I ∩ dom ( (cid:101) T ) .Proof. Since the actions of SO(3) and SU( n ) commute, by equation (2.2), it is enough toverify the first claim separately for rotations and gauges.Let T be a solution to Nahm’s equations (Nahm1) to (Nahm3) and A ∈ SO(3). Let( a ij ) i,j =1 be the components of A . Recall the formula ∀ A ∈ SO(3) : (cid:88) l,m,n =1 (cid:15) lmn a il a jm a kn = (cid:15) ijk . (2.3)Furthermore, note that for all i ∈ { , , } , T i = (cid:80) l =1 a li A T l . Using these, we can computethe right hand side of equation (1.2) for A T , and getdd t ( A T ) i = (cid:88) l =1 a il ˙ T l = (cid:88) l =1 a il (cid:88) m,n =1 (cid:15) lmn [ T m , T n ] (cid:88) i,j,k,l,m,n =1 a il (cid:15) lmn a jm a kn [( A T ) j , ( A T ) k ]= 12 (cid:88) i,j,k =1 (cid:15) ijk [( A T ) j , ( A T ) k ] , which completes the proof of the first claim for rotations.Since [Ad( U )( T ) , Ad( U )( S )] = Ad( U )([ T , S ]), it is clear that T is a solution to Nahm’sequations, if and only if U T is a solution, which completes the proof of the first claim forgauges.The second claim follows from the uniqueness of solutions to the initial value problemcorresponding to Nahm’s equations. (cid:3) The following is then immediate.
Corollary 2.6.
Let A ∈ SO(3) , U ∈ SU( n ) , and T ∈ su ( n ) ⊕ , such that AU T = T . If T : ( t − (cid:15), t + (cid:15) ) → su ( n ) ⊕ , is the (unique) solution to Nahm’s equations (Nahm1) to (Nahm3) with the initial condition T ( t ) = T , then AU T = T , that is ∀ t ∈ ( t − (cid:15), t + (cid:15) ) : AU ( T ( t )) = T ( t ) . Remark 2.7. If A ∈ O(3) , then A T can still be defined via (Rotation) . Then equation (2.3) becomes (cid:88) l,m,n =1 (cid:15) lmn a il a jm a kn = det( A ) (cid:15) ijk . In particular, when A (cid:60) SO(3) , the only thing that changes in the proof of Lemma 2.5 is the signin equation (2.3) . Thus if T , is a solution to Nahm’s equations (Nahm1) to (Nahm3) , then ( S , S , S ) (cid:66) A T is a solution to the anti-Nahm equations : ∀ i ∈ { , , } : ˙ S i = − (cid:88) j,k =1 (cid:15) ijk [ S j , S k ] . Definition 2.8.
Fix A ∈ SO(3) and let I ⊆ R be a nonempty, connected, and open interval. Wecall a solution T : I → su ( n ) ⊕ to Nahm’s equations (Nahm1) to (Nahm3) A -equivariant , ifthere exists U A ∈ SU( n ) , such that A T = U − A T . (2.4) Similarly, if H ⊆ SO(3) , then T is H -equivariant , if for all A ∈ H , T is A -equivariant. emark 2.9. Equation (2.4) can be written as AU A T = T , thus T is a fixed by the simultaneousactions of A and U A .Note also that when T is H -equivariant (for some H ⊆ SO(3) ), we did not require anythingelse from the map H → SU( n ); A (cid:55)→ U A , in particular it need not be homomorphic or smooth, and may not even be unique.Nonetheless, using equation (2.4) , we see that if T is equivariant for both A and A , withcorresponding gauge transformations U A and U A , respectively, then A A T = A ( A T ) = A ( U − A T ) = U − A ( A T ) = U − A ( U − A T ) = U − A U − A T = ( U A U A ) − T , so T is A A -equivariant, and we can choose the corresponding gauge transformation to be U A A = U A U A . Thus the set H T (cid:66) { A ∈ SO(3) | T is A -equivariant } , is a subgroup of SO(3) . The goal of this paper is to study and construct H -invariant solutions to Nahm’s equa-tions, with H being a connected, nontrivial Lie subgroup of SO(3), thus either H (cid:39) SO(2),or H = SO(3). Motivated by spatial geometry, we make the following definitions: Definition 2.10.
When H (cid:39) SO(2) , the solution is called axially symmetric . Similarly, when H = SO(3) , the solution is called spherically symmetric . Remark 2.11.
The conserved quantities in equations (2.1a) to (2.1e) transform according to the5-dimensional irreducible representation of SO(3) under the action of
SO(3) , and are invariantunder the action of
SU( n ) ; see for instance [13, Equation (21)] .More precisely, let T be a solution to Nahm’s equation and let us redefine the correspondingconserved quantities as ∀ i, j ∈ { , , } : C ij ( T ) (cid:66) tr (cid:16) T i T j (cid:17) − (cid:88) k =1 tr (cid:16) T k (cid:17) . (2.5) Note that equation (2.5) defines a 3-by-3, real, symmetric, and traceless matrix, and its 5independent parameters can be chosen to be the conserved quantities in equations (2.1a) to (2.1e) .Let us denote this matrix by C ( T ) . Then for all A ∈ SO(3) and U ∈ SU( n ) we have C (cid:16) AU T (cid:17) = A C ( T ) A − . In particular, if T is spherically symmetric, then all conserved quantities in equations (2.1a) to (2.1e) must vanish by Schur’s Lemma. n either of the above cases, H -invariance implies the existence of a function, U : H → SU( n ), such that ∀ A ∈ H : A T = U ( A ) − T . However, as in Remark 2.9, this function need not be homomorphic or smooth, andmoreover, it need not be unique. Indeed, if the stabilizer subgroup of T S T (cid:66) { U ∈ SU( n ) | U T = T } , is nontrivial, then one can choose a function H → S T and “twist”. Hence, U may not beunique, and moreover, even if U was homomorphic or smooth, the twisted version maynot have these properties.The case in which U can be chosen to have some regularity, at least around the identity,is easier to handle. For this reason, our main theorems require that U is either once (in theaxially symmetric case), or twice (in the spherically symmetric case) di ff erentiable at theidentity. 3. Axially symmetric solutions
In this section, we consider axially symmetric solutions, that is, when H is isomorphicto SO(2) in Definition 2.10.All subgroups of SO(3) that are isomorphic to SO(2) are maximal tori of SO(3) andthus are conjugate to each other. Moreover, they can be viewed as rotations around agiven, oriented axis (that is, an oriented line through the origin). Hence, without loss ofgenerality, it is enough to study one of them. Let A : R → SO(3); θ (cid:55)→ A ( θ ) (cid:66) cos( θ ) − sin( θ ) 0sin( θ ) cos( θ ) 00 0 1 . (3.1)Then our choice of such a subgroup is H (cid:66) { A ( θ ) | θ ∈ [0 , π ) } . This subgroup is the group of rotations around the third axis. Moreover, equation (3.1)provides a global parametrization of H .Our first theorem gives an infinitesimal version of axial symmetry. Theorem 3.1. If T = ( T , T , T ) is axially symmetric around the third axis and the correspond-ing U function in (2.4) can be chosen to be continuously di ff erentiable at the identity of SO(3) , hen there exists a Y ∈ su ( n ) such that T = [ T , Y ] , (3.2a) T = [ Y , T ] , (3.2b)0 = [ Y , T ] . (3.2c) Conversely, if equations (3.2a) to (3.2c) are satisfied, then T is axially symmetric around thethird axis.More generally, T is axially symmetric around some axis if there exists B ∈ SO(3) such that B T satisfies equations (3.2a) to (3.2c) . Definition 3.2.
We call Y the generator of the axially symmetry for T .Proof. Assume that T = ( T , T , T ) is axially symmetric around the third axis and thecorresponding U function in (2.4) can be chosen so that it is continuously di ff erentiable atthe identity of SO(3), and let Y (cid:66) dd θ (cid:16) U A ( θ ) (cid:17)(cid:12)(cid:12)(cid:12) θ =0 . Then equations (3.2a) to (3.2c) are justthe linearizations of A ( θ ) U A ( θ ) T = T , at θ = 0.On the other hand, if equations (3.2a) to (3.2c) hold for some Y ∈ su ( n ), then, for all θ ∈ R , let A ( θ ) be defined via equation (3.1) and let U ( θ ) (cid:66) exp( θY ) ∈ SU( n ) . (3.3)Now equations (3.2a) to (3.2c) are equivalent to ˙ A (0) T + [ Y , T ] = 0 . Now simple computation, using equations (3.1) and (3.3), shows that for all θ ,dd θ (cid:18) A ( θ ) U ( θ ) T (cid:19) = ˙ A ( θ ) U ( θ ) T + A ( θ )˙ U ( θ ) T = ˙ A (0) (cid:18) A ( θ ) U ( θ ) T (cid:19) + (cid:20) Y , A ( θ ) U ( θ ) T (cid:21) = A ( θ ) U ( θ ) (cid:16) ˙ A (0) T + [ Y , T ] (cid:17) = 0 . Hence ∀ θ ∈ R : T = A (0) U (0) T = A ( θ ) U ( θ ) T , which is equivalent to T being axially symmetric around the third axis.The last claim follows from the discussion in the beginning of this section. (cid:3) The moral of Theorem 3.1 is that, up to rotation and gauge, axially symmetric solutions toNahm’s equations are labeled by elements Y ∈ su ( n ). Note that after a gauge transformationof T , the corresponding Y changes by the adjoint action of SU( n ). Thus, we only need toconsider the adjoint orbits in su ( n ). It is easy to find canonical representatives in every rbit: In every orbit there is a unique Y of the form Y = i · diag( α , α , . . . , α n ) with α k ∈ R , α k (cid:62) α k +1 . Of course, (cid:80) ni =1 α i = 0 has to also hold, as elements of su ( n ) are traceless.However, there are adjoint orbits that cannot carry a nontrivial Nahm datum satisfyingequations (3.2a) to (3.2c), as shown in the next lemma. Lemma 3.3. If T is a nonconstant, axially symmetric solution to Nahm’s equations (Nahm1) to (Nahm3) , and Y ∈ su ( n ) is the generator of the axially symmetry for T , then − ∈ Spec (cid:16) ad Y (cid:17) .Conversely, if − ∈ Spec (cid:16) ad Y (cid:17) , then there is a nonconstant, axially symmetric solution toNahm’s equations, whose generator of the axial symmetry is Y . Remark 3.4.
Classical computation shows that − is in the spectrum of ad Y , if and only if α j = α i + 1 , for some (cid:54) i < j (cid:54) n .Proof of Lemma 3.3. Equations (3.2a) and (3.2b) imply thatad Y ( T ) = − T , and ad Y ( T ) = − T . This proves the first claim.For the converse, we first show that equations (3.2a) to (3.2c) can be satisfied at a point,call t . Note that ad Y always has a nontrivial kernel, because 0 (cid:44) Y ∈ ker(ad Y ). Pickany nonzero element T ∈ ker(ad Y ) and let T ( t ) = T . As − ∈ Spec (cid:16) ad Y (cid:17) , we can choose T ( t ) to be a ( − T ( t ) = ad Y ( T ( t )). The Picard–Lindelöf Theoremguarantees the existence of a (local) solution with these initial values. As dd t T is nonzero at t = t , T is necessarily nonconstant. This concludes the proof. (cid:3) Using equation (3.2b), T can be eliminated from Nahm’s equations. Since [[ Y , T ] , T ] =[ Y , [ T , T ]], the reduced system of ordinary di ff erential equations, which we call the axiallysymmetric Nahm’s equations , has the form:˙ T = [ Y , [ T , T ]] , (3.4a)˙ T = [ T , [ Y , T ]] . (3.4b)3.1. Axially symmetric solutions to the SU ( ) Nahm’s equations.
We now illustrate howTheorem 3.1 and Lemma 3.3 can be used to find axially symmetric solutions to Nahm’sequations through the n = 3 case. Now Y ∈ su (3) and thus, after a change of basis, it canalways be brought to the form Y α,β (cid:66) αi βi
00 0 − ( α + β ) i , with α, β ∈ R , and α (cid:62) β (cid:62) − ( α + β ) . y Lemma 3.3, to get nonconstant solutions, we need to have β = α − , or β = (1 − α ) , or β = 1 − α. Let us write a generic element of su (3), say T , as T = ai z z − z bi z − z − z − ( a + b ) i , with a, b ∈ R , and z , z , z ∈ C . Then we get ad Y α,β ( T ) = − ( α − β ) z − (2 α + β ) z ( α − β ) z − ( α + 2 β ) z (2 α + β ) z ( α + 2 β ) z . (3.5)Thus, equation (3.5) imply that, when no two of the diagonal elements of Y α,β are equal,then the kernel is spanned by elements of the form i · diag( a, b, − ( a + b )), and thus this isthe most general form of T , in this case. When two of the diagonal elements are equal,then the kernel is 4-dimensional and isomorphic (as a Lie algebra) to R ⊕ so (3).The table below summarizes the cases in which the ( − α (cid:62) β (cid:62) − ( α + β )).case α, β ) example(s) dim R (cid:16) ker (cid:16) ad Y α,β + su ( n ) (cid:17)(cid:17)
1. ( α, α − , α ∈ (cid:16) , (cid:17) ∪ (1 , ∞ ) (cid:16) , − (cid:17) , (2 ,
1) 22. ( α, − α ) , < α < (cid:16) , (cid:17) (cid:16) α, (1 − α ) (cid:17) , α ∈ (cid:16) , (cid:17) ∪ (1 , ∞ ) (cid:16) , (cid:17) , (3 , −
1) 24. (1 ,
0) (1 ,
0) 45. (cid:16) , − (cid:17) (cid:16) , − (cid:17) (cid:16) , (cid:17) (cid:16) , (cid:17) ff erentialequation on four real functions. In the cases 4., 5., and 6., the Ansätze have 4 + 4 = 8parameters, and the axially symmetric Nahm’s equations (3.4a) and (3.4b) reduce to asystem of eight ordinary di ff erential equation on eight real functions.We end this section by computing the solutions explicitly in a particular case. Example: The ( α, β ) = (cid:16) , (cid:17) case. In this example we show how our technique recovers theresults of [13, Proposition 3.10]. et Y (cid:66) Y , , that is Y = i / − i / . Now both the kernel and the ( − Y are 2-dimensional. More concretely,we can write our Ansatz as T = z − z & T = ai − ( a + b ) i
00 0 bi , where z is a complex function and, a and b are real functions. Using the residual gaugesymmetry, we can assume, without any loss of generality, that z is, in fact, real. The axiallysymmetric Nahm’s equations (3.4a) and (3.4b) then become˙ z = ( a − b ) z, (3.6a)˙ a = 2 z , (3.6b)˙ b = − z . (3.6c)Note that if one knows a and b , then z can be computed via equation (3.6a). The conservedquantities in equations (2.1a) to (2.1c) are automatically zero. The other two, given byequations (2.1d) and (2.1e), are related and satisfy C = C = a + b + ( a + b ) − z . Furthermore, equations (3.6b) and (3.6c) imply that the quantities k (cid:66) a + b, and k (cid:66) a − k a − z , are also conserved. Using equation (3.6b) we get˙ a = 2 (cid:16) a − k a − k (cid:17) = 2 (cid:18) a − k (cid:19) − (cid:32) k + k (cid:33) . (3.7)Let A (cid:66) a − k = a − b, and K (cid:66) − k − k . Then equation (3.7) is equivalent to ˙ A = A + K . (3.8) et c ∈ R be the constant of integration, and then the solutions of equation (3.8) are A ( t ) = −√ K cot (cid:16) √ K ( t − c ) (cid:17) K > , − ( t − c ) − K = 0 , −√− K coth (cid:16) √− K ( t − c ) (cid:17) K < . (3.9)From this the functions a, b , and z , and thus the corresponding axially symmetric solutionof Nahm’s equations can easily be reconstructed. Remark 3.5.
In each case of equation (3.9) , the solutions develop singularities. When
K > ,the singularities are at the points (cid:110) c + πK k (cid:12)(cid:12)(cid:12) k ∈ Z (cid:111) . When K (cid:54) , the only singularity is at t = c .According to Lemma 2.1 these singularities induce a representation of so (3) . In all of thethree cases of equation (3.9) this representation is the (unique, up to isomorphism) irreducible,3-dimensional representation. Spherically symmetric solutions
In this section we consider spherically symmetric solutions to Nahm’s equations. Thatis, solutions T , such that for all A ∈ SO(3) there exists U A ∈ SU( n ) such that AU A T = T .The next theorem is analogous to Theorem 3.1 as it gives an infinitesimal version ofspherical symmetry. Theorem 4.1. If T = ( T , T , T ) is spherically symmetric and the corresponding U function in (2.4) can be chosen to be twice continuously di ff erentiable at the identity of SO(3) , then thereexists a triple, ( Y , Y , Y ) ∈ su ( n ) , such that for all i, j ∈ { , , } [ Y i , T j ] = (cid:88) k =1 (cid:15) ijk T k , (4.1a)[[ Y i , Y j ] , T ] = (cid:88) k =1 (cid:15) ijk [ Y k , T ] . (4.1b) Conversely, if equations (4.1a) and (4.1b) are satisfied, then T is spherically symmetric. Definition 4.2.
We call elements of the triple ( Y , Y , Y ) the generators of the sphericallysymmetry for T .Proof. Assume first that T is spherically symmetric and the corresponding U function in(2.4) can be chosen so that it is twice continuously di ff erentiable at the identity of SO(3). n this case, T is, in particular, axially symmetric around all of the three coordinate axesand the corresponding U functions in (2.4) can be chosen so that they are continuouslydi ff erentiable at the identity of SO(3), thus the first part of Theorem 3.1 can be applied.For each i ∈ { , , } , let X i ∈ so (3) be the infinitesimal generator of the (positively oriented)rotation around the i th axis, that is ( X i ) jk = − (cid:15) ijk , and let Y i be the corresponding generatorof axial symmetry. Then the nine equations in (4.1a) are exactly the three triples ofequations that one gets from these three di ff erent axial symmetries.To prove equation (4.1b), let for all i ∈ { , , } and for all t ∈ R let A i ( t ) (cid:66) exp( tX i ) , that is, the rotation around the i th axis, and let U i ( t ) ∈ SU( n ) be such that A i ( t ) U i ( t ) T = T , (4.2)and U i is twice di ff erentiable at t = 0 and (necessarily) ˙ U i (0) = Y i . By equation (4.2), for all i, j ∈ { , , } and for all ( s, t ) ∈ R we get that A i ( s ) A j ( t ) A i ( − s ) A j ( − t ) U i ( s ) U j ( t ) U i ( − s ) U j ( − t ) T = T . The quadratic Taylor series expansion in ( s, t ) ∈ R around the origin gives us0 = A i ( s ) A j ( t ) A i ( − s ) A j ( − t ) U i ( s ) U j ( t ) U i ( − s ) U j ( − t ) T − T = (cid:16) [ X i ,X j ] T + [[ Y j , Y i ] , T ] (cid:17) st + O (cid:16) | ( s, t ) | (cid:17) . Using the vanishing of the st -terms and the commutator relations of ( X , X , X ), we getequation (4.1b). This concludes the proof of the “if” direction.Now assume that equations (4.1a) and (4.1b) hold and let A ∈ SO(3). Due to thesurjectivity of the exponential map exp : so (3) → SO(3), there exists X = a X + a X + a X ∈ so (3), such that A = exp( X ). Let A ( s ) (cid:66) exp( sX ) ∈ SO(3), Y (cid:66) a Y + a Y + a Y ∈ su ( n ), and U ( s ) (cid:66) exp( sY ). Clearly, A (1) = A , A (0) = , and U (0) = n . Thus A (0) U (0) T = T .Next we show that A ( s ) (cid:66) A ( s ) U ( s ) T − T is independent of s . Usingdd s A ( s ) = XA ( s ) = A ( s ) X, and dd s U ( s ) = Y U ( s ) = U ( s ) Y , we get dd s A ( s ) = A ( s ) U ( s ) (cid:16) X T + [ Y , T ] (cid:17) . Now the vanishing of the s -independent quantity in the parentheses is equivalent toequation (4.1a). Thus, A is constant and since A (0) = 0, we have A (1) = 0, which isequivalent to T being A -equivariant. Since A ∈ SO(3) was arbitrary, this concludes theproof. (cid:3) .1. Structure theorem for the spherically symmetric Ansatz.
In this section we prove astructure theorem for spherically symmetric solutions of Nahm’s equations, under certainhypotheses. This structure theorem classifies spherically symmetric Anätze throughrepresentation theoretic means. In order to set up the stage for this, let us begin with aremark.
Remark 4.3.
If a triple ( Y , Y , Y ) ∈ su ( n ) ⊕ satisfies the so (3) commutator relations ∀ i ∈ { , , } : [ Y i , Y j ] = (cid:88) k =1 (cid:15) ijk Y k , (4.3) then equation (4.1b) holds independent of T . A compact way to rephrase equation (4.3) canbe given as follows: consider the linear map from so (3) to su ( n ) that sends X i to Y i . Byequation (4.3) , this map gives a representation of so (3) . Let us call it ( C n , ρ ) . Let ( V k , ρ k ) be theirreducible, k -dimensional complex representation of so (3) and let (cid:16) ˆ V , ˆ ρ (cid:17) (cid:66) ( C n , ρ ) ⊗ (( C n ) ∗ , ρ ∗ ) ⊗ ( V , ρ ) . The values of T can be viewed as an endomorphism on V and the action of any X ∈ so (3) on T is given by ˆ ρ ( X )( T ) (cid:66) X T + [ ρ ( X ) , T ] . If T is spherically symmetric with generators ( Y , Y , Y ) , then equation (4.1a) is equivalent to ∀ X ∈ so (3) : ˆ ρ ( X )( T ) = 0 , (4.4) or, in other words, the values of T lies in the trivial component of ( V , ˆ ρ ) . Since only representa-tion of so (3) that is both trivial and irreducible is the 1-dimensional one, Theorem 4.1 impliesthe following: When equation (4.3) holds, T is spherically symmetric exactly if T takes valuesin the direct sum of the 1-dimensional irreducible components of (cid:16) ˆ V , ˆ ρ (cid:17) . Theorem 4.1 and Remark 4.3 tell us that certain spherically symmetric solutions toNahm’s equations are labeled by representations of so (3).Let us recall the Clebsch–Gordan Theorem: For all positive integers, m (cid:62) n , we have thefollowing decomposition of representations( V m , ρ m ) ⊗ ( V n , ρ n ) (cid:39) n (cid:77) k =1 ( V m + n +1 − k , ρ m + n +1 − k ) . (4.5)The above equation (4.5) implies that the representation ( V m , ρ m ) ⊗ ( V n , ρ n ) ⊗ ( V , ρ ) hasa single 1-dimensional irreducible summand when m = n (cid:62)
2, or m = n + 2, and noneotherwise. When m = n + 2, let us call B n the unit length generator of the unique 1-dimensional representation in ( V n +2 , ρ n +2 ) ⊗ ( V n , ρ n ) ⊗ ( V , ρ ), which is well-defined, up to U(1) factor. Note, that B n can be viewed as an so (3)-invariant triple of maps ( B n , B n , B n )from the n -dimensional irreducible representation to the ( n + 2)-dimensional one. Withthis in mind, let us state and prove our structure theorem. Theorem 4.4 (Structure theorem for spherically symmetric solutions to Nahm’s equations) . Let T be a spherically symmetric solution to Nahm’s equations (Nahm1) to (Nahm3) such thatthe generators induce the representation ( C n , ρ ) (as in Remark 4.3), and write the decompositionof ( C n , ρ ) into irreducible summands as ( C n , ρ ) (cid:39) k (cid:77) a =1 (cid:16) V n a , ρ n a (cid:17) . (4.6) We can assume, without any loss of generality, that for all a ∈ { , . . . , k − } , we have n a (cid:62) n a +1 .For all i ∈ { , , } and a ∈ { , . . . , k } let Y i,a (cid:66) ρ n a ( X i ) ,Y i (cid:66) ρ ( X i ) = diag (cid:0) Y i, , Y i, , . . . , Y i,k (cid:1) . Fix t in the domain of T and for all i ∈ { , , } , write T i ( t ) in a block matrix form at any pointin the domain of T , according to equation (4.6): T i ( t ) = ( T i ) ( T i ) . . . ( T i ) k ( T i ) ( T i ) ...... . . . ( T i ) k . . . ( T i ) kk , (4.7) with ( T i ) ab = − ( T i ) ∗ ba ∈ V n a ⊗ V ∗ n b . Then, up to a ρ -invariant gauge and for all a, b ∈ { , , . . . , k } ,we have that(1) There exist c a ∈ R , such that for all i ∈ { , , } , we have ( T i ) aa = c a Y i,a .(2) If a (cid:44) b and | n a − n b | (cid:44) , then ( T i ) ab = 0 .(3) If n a = n b + 2 , then there exists c ab ∈ C , such that for all i ∈ { , , } , we have ( T i ) ab = c ab B n a i . Remark 4.5.
An important question in any gauge theory is the reducibility of solutions. If T isa spherically symmetric Nahm datum, then its decomposition into irreducible components iseasy to read from equation (4.7) : I (cid:66) { , . . . , k } , and define a relation on I via a ∼ b exactlywhen there exists i ∈ { , , } , such that ( T i ) ab (cid:44) . Note that ∼ is reflexive and symmetric(but need not be transitive). Thus it generates an equivalence relation on I which we denoteby ∼ . The set of irreducible components of T is then labeled by J (cid:66) I / ∼ and the component orresponding to { a , a , . . . , a l } ∈ J is given by (cid:16) T { a ,a ,...,a l } (cid:17) i ( t ) (cid:66) ( T i ) a a ( T i ) a a . . . ( T i ) a a l ( T i ) a a ( T i ) a a ...... . . . ( T i ) a l a . . . ( T i ) a l a l . In particular, we have the following:(1) The third bullet point in Theorem 4.4 implies that if the set { n , n , . . . , n k } contains botheven and odd numbers, then T is reducible.(2) If k = 2 , n = n + 2 , and for some i ∈ { , , } , ( T i ) (cid:44) , then T is irreducible. Remark 4.6.
The k = 1 case in Theorem 4.4, that is when ( C n , ρ ) is irreducible, was studied,albeit for low ranks only, by Dancer in [12] . This solution is also equivalent to the one found inthe K = 0 case of equation (3.9) .Proof of Theorem 4.4. Let us begin with the k = 1 case, that is when ( C n , ρ ) is irreducible.By Remark 4.3 the value of T at any point is an element in the trivial component of (cid:16) ˆ V , ˆ ρ (cid:17) .Then the Clebsch–Gordan decomposition, equation (4.5), tells us that there is a unique1-dimensional trivial summand in (cid:16) ˆ V , ˆ ρ (cid:17) , thus if we find one Ansatz, then it is the mostgeneral one. Let ρ ( X i ) (cid:66) Y i . Note that if for all i ∈ { , , } we have T i = f Y i at any point ofthe domain of T , then we have that for all, and it is a spherically symmetric Ansatz. Thiscompletes the proof in k = 1 case.Let us turn to the general case. Regard T as an element of ˆ V via evaluating it at t .Recall that so (3) acts on V via (the extension of)ˆ ρ : so (3) ⊗ ˆ V → ˆ V ; X ⊗ R (cid:55)→ X R + [ ρ ( X ) , R ] . By Theorem 4.1 and Remark 4.3, T is spherically symmetric if equation (4.4) holds, or, inother words, if T takes values in the 1-dimensional irreducible summands of (cid:16) ˆ V , ˆ ρ (cid:17) . Wecan write the decomposition of (cid:16) ˆ V , ˆ ρ (cid:17) into irreducible components as (cid:16) ˆ V , ˆ ρ (cid:17) = k (cid:77) a =1 (cid:16) su (cid:16) V n a , ρ n a (cid:17) ⊗ ( V , ρ ) (cid:17) (cid:77) k − (cid:77) b =1 k (cid:77) c = b +1 (cid:16) V n b ⊗ V n c ⊗ V , ρ n b ⊗ ρ n c ⊗ ρ (cid:17) = diagonal ⊕ o ff -diagonalLet (cid:16) ˆ V , ˆ ρ (cid:17) be the direct sum of 1-dimensional irreducible components in (cid:16) ˆ V , ˆ ρ (cid:17) . By theproof of the k = 1 case, each summand of the diagonal terms has a unique copy of ( V , ρ ), roving the first bullet point. Furthermore, each o ff -diagonal term also has unique copy, ifand only if | n b − n c | = 2, proving the second bullet point. Finally, one can easily verify thatwhen n a = n b for some a (cid:44) b , then there exists U ∈ SU( n ), such that [ U , Y ] = 0, U only actsnontrivially on the (cid:16) V n a ⊕ V n b , ρ n a ⊕ ρ n b (cid:17) summand, and the ab -component of U T is zero.This completes the proof. (cid:3) Remark 4.7.
A weaker version of Theorem 4.4 was stated, without rigorous proof, in [6] . Next we analyze the o ff -diagonal terms in equation (4.7). As before, let ( X , X , X ) be astandard basis of so (3), that is for all i, j, k ∈ { , , } , we have ( X i ) jk = − (cid:15) ijk . Theorem 4.8.
Fix n ∈ N + , and let ( V n +2 , ρ n +2 ) and ( V n , ρ n ) be as above. For all i ∈ { , , } , let Y + i (cid:66) ρ n +2 ( X i ) , Y − i (cid:66) ρ n ( X i ) , and Y i (cid:66) diag (cid:16) Y + i , Y − i (cid:17) . Let (cid:16) B n , B n , B n (cid:17) be the (unique up to a U(1) factor) triple in the (unique) 1-dimensional irreducible component of (cid:16) ˆ V , ˆ ρ (cid:17) (cid:66) ( V n +2 ⊗ V n ⊗ V , ρ n +2 ⊗ ρ n ⊗ ρ ) . Then, after a potential rescaling , we have the following identities (for all j, k ∈ { , , } , where itapplies): Y + j B nk − B nk Y − j = (cid:88) i =1 (cid:15) ijk B ni , (4.8a) (cid:88) i =1 B ni (cid:16) B ni (cid:17) ∗ = n +2 , (4.8b) (cid:88) i =1 (cid:16) B ni (cid:17) ∗ B ni = n + 2 n n , (4.8c) B nj (cid:16) B nk (cid:17) ∗ − B nk (cid:16) B nj (cid:17) ∗ = − n + 1 (cid:88) i =1 (cid:15) ijk Y + i , (4.8d) (cid:16) B nj (cid:17) ∗ B nk − (cid:16) B nk (cid:17) ∗ B nj = 2( n + 2) n ( n + 1) (cid:88) i =1 (cid:15) ijk Y − i , (4.8e) Y + j B nk − Y + k B nj = n + 32 (cid:88) i =1 (cid:15) ijk B ni , (4.8f) B nj Y − k − B nk Y − j = − n − (cid:88) i =1 (cid:15) ijk B ni . (4.8g) That is, replacing B n with λB n , for some λ ∈ C − { } . roof. Equation (4.8a) is the defining equation for (cid:16) B n , B n , B n (cid:17) , hence needs no proof.Using equation (4.8a), for all j ∈ { , , } we get that Y + j , (cid:88) i =1 B ni (cid:16) B ni (cid:17) ∗ = 0 , Y − j , (cid:88) i =1 (cid:16) B ni (cid:17) ∗ B ni = 0 , thus, by Schur’s Lemma, we have there are real numbers, say a n and b n , such that (cid:88) i =1 B ni (cid:16) B ni (cid:17) ∗ = a n n +2 , (cid:88) i =1 (cid:16) B ni (cid:17) ∗ B ni = b n n . By taking traces we can conclude that both a n and b n are positive. Since (cid:16) B n , B n , B n (cid:17) can berescaled by a nonzero constant, we can achieve a n = 1. Again by the properties of the trace,we get b n = n +2 n . This proves equations (4.8b) and (4.8c).For all i ∈ { , , } let (cid:101) Y + i (cid:66) (cid:88) j,k =1 (cid:15) ijk B nj (cid:16) B nk (cid:17) ∗ , (4.9) (cid:101) Y − i (cid:66) (cid:88) j,k =1 (cid:15) ijk (cid:16) B nj (cid:17) ∗ B nk , (4.10) (cid:101) B + i (cid:66) (cid:88) j,k =1 (cid:15) ijk Y + j B nk , (4.11) (cid:101) B − i (cid:66) (cid:88) j,k =1 (cid:15) ijk B nj Y − k . (4.12)Using equation (4.8a), we get that for all i, j ∈ { , , } (cid:104) Y + i , (cid:101) Y + j (cid:105) = (cid:88) k =1 (cid:15) ijk (cid:101) Y + k , (cid:104) Y − i , (cid:101) Y − j (cid:105) = (cid:88) k =1 (cid:15) ijk (cid:101) Y − k , + i (cid:101) B ± j − (cid:101) B ± j Y − i = (cid:88) k =1 (cid:15) ijk (cid:101) B ± k . Thus, by the proof of the k = 1 case in Theorem 4.4 and the uniqueness (up to scale) of (cid:16) B n , B n , B n (cid:17) , we get that that there are real numbers α ± n and β ± n , such that for all i ∈ { , , } (cid:101) Y ± i = α ± n Y ± i , (cid:101) B ± i = β ± n B ni . Next we four, independent, linear equations on ( α + n , α − n , β + n , β − n ) ∈ R to be able to (uniquely)solve for them. Let C + n n +2 (cid:66) (cid:88) i =1 (cid:16) Y + i (cid:17) = − ( n + 1)( n + 3)4 n +2 , (4.13) C − n n (cid:66) (cid:88) i =1 (cid:16) Y − i (cid:17) = − n − n , (4.14)be the Casimir operators corresponding to the two irreducible representations. Usingequations (4.9), (4.10), (4.13), and (4.14), we get α + n C + n n +2 = (cid:88) i =1 Y + i ( α + n Y + i ) = (cid:88) i =1 Y + i (cid:101) Y + i = (cid:88) i,j,k =1 (cid:15) ijk Y + i B nj (cid:16) B nk (cid:17) ∗ = (cid:88) k =1 (cid:101) B + k (cid:16) B nk (cid:17) ∗ = β + n n +2 , which gives α + n C + n − β + n = 0 . (4.15)Similarly α − n C − n n = (cid:88) k =1 ( α − n Y − k ) Y − k = (cid:88) k =1 (cid:101) Y − k Y − k = (cid:88) i,j,k =1 (cid:15) ijk (cid:16) B ni (cid:17) ∗ B nj Y − k = (cid:88) i =1 (cid:16) B ni (cid:17) ∗ (cid:101) B − i = β − n n + 2 n n , so, we get α − n C − n − n + 2 n β − n = 0 . (4.16)Next, adding up equations (4.8a), (4.11), and (4.12), and using equation (4.8a) yields β + n + β − n = 2 . (4.17)Let us define B (cid:66) (cid:88) i =1 Y + i B ni = (cid:88) i =1 B ni Y − i . (4.18) sing equation (4.8a), we get Y + i B − BY − i = 0 for all i ∈ { , , } , and thus, by Schur’s Lemma, B = 0. Using this and equation (4.11), for all i ∈ { , , } we have( β + n ) B ni = (cid:88) j,k =1 (cid:15) ijk Y + j (cid:16) β + n B nk (cid:17) = (cid:88) j,k,l,m =1 (cid:15) ijk (cid:15) klm Y + j Y + l B nm = (cid:88) j,l,m =1 (cid:16) δ il δ jm − δ im δ jl (cid:17) Y + j Y + l B nm = (cid:88) j =1 (cid:16) Y + j Y + i B nj − Y + j Y + j B ni (cid:17) = (cid:88) j,k =1 (cid:15) jik Y + k B nj + Y + i (cid:88) j =1 Y + j B nj − C + n B ni = β + n B i + 0 − C + n , thus we get ( β + n ) = β + n − C + n . (4.19)An analogous computation gives ( β − n ) = β − n − C − n . (4.20)Subtracting equation (4.20) from equation (4.19), and using equation (4.17) gives( β + n ) − ( β − n ) = ( β + n + β − n )( β − n − β − n ) = 2( β − n − β − n ) = ( β − n − β − n ) − C + n + C − n , and thus β + n − β − n = C − n − C + n = n + 4 n + 3 − ( n − n + 2) . (4.21)The equations (4.15) to (4.17) and (4.21) form a set of four independent linear equationsfor four unknowns, so the solution is unique: α + n = − n + 1 , α − n = 2( n + 2) n ( n + 1) , β + n = n + 32 , β − n = − n − . This completes the proof of equations (4.8d) to (4.8g). (cid:3)
Ansätze and solutions.
In the next theorem, using Theorem 4.8 we investigate ageneralization of the irreducible solutions defined in the second bullet point of Remark 4.5. heorem 4.9. Let n, k (cid:62) , and suppose that ( V , ρ ) decomposes as ( V , ρ ) (cid:39) k (cid:77) a =0 (cid:16) V n +2 k − a , ρ n +2 k − a − (cid:17) = ( V n +2 k , ρ n +2 k ) ⊕ . . . ⊕ ( V n +2 , ρ n +2 ) ⊕ ( V n , ρ n ) . (4.22) For all m (cid:62) and i ∈ { , , } , let Y mi (cid:66) ρ m ( X i ) and (cid:16) B m , B m , B m (cid:17) as in Theorem 4.8.Then the spherically symmetric Ansatz given in equation (4.7) takes the following form: thereexists real, analytic functions f , f , . . . , f k , g , g , . . . , g k − on the domain of T such that for all i ∈ { , , } we have T i = f k Y n +2 ki g k − B n +2( k − i . . . − g k − (cid:18) B n +2( k − i (cid:19) ∗ f k − Y n +2( k − i g k − B n +2( k − i ... − g k − (cid:18) B n +2( k − i (cid:19) ∗ . . . ... . . . g B ni . . . − g (cid:16) B ni (cid:17) ∗ f Y ni . (4.23) By convention, let g − = g k ≡ .When n = 1 , then (cid:16) Y , Y , Y (cid:17) = (0 , , , and thus f can be omitted. The rest of the functionsabove satisfy the following equations: ∀ a ∈ { , , . . . , k } : ˙ f a = f a + 1 a g a − − a + 3(2 a + 1)( a + 1) g a , (4.24a) ∀ a ∈ { , , . . . , k − } : ˙ g a = (( a + 2) f a +1 − af a ) g a . (4.24b) When n (cid:62) , then the functions above satisfy the following equations: ∀ a ∈ { , , . . . , k } : ˙ f a = f a + 2 n + 2 a − g a − − n + 2 a + 2)( n + 2 a )( n + 2 a + 1) g a , (4.25a) ∀ a ∈ { , , . . . , k − } : ˙ g a = (cid:18) n + 2 a + 32 f a +1 − n + 2 a − f a (cid:19) g a . (4.25b)Note that Equations (4.24a) and (4.24b) are simply Equations (4.25a) and (4.25b) with n = 1, with the exception that the di ff erential equation for f is omitted. Proof.
When (
V , ρ ) decomposes as in equation (4.22), equation (4.7) in Theorem 4.4 reducesto equation (4.23). The equations (4.24a), (4.24b), (4.25a), and (4.25b) follow usingequations (4.8a) and (4.8d) to (4.8g). (cid:3)
Remark 4.10.
Let T = ( T , T , T ) be a solution to Nahm’s equations (Nahm1) to (Nahm3) onan open, connected interval I ⊆ R , and f be a nonconstant a ffi ne function on R , that is for all ∈ R , f ( t ) = at + b , for some a, b ∈ R with a (cid:44) . We define the pull-back of T via f as t (cid:55)→ f ∗ ( T )( t ) (cid:66) ( aT ( f ( t )) , aT ( f ( t )) , aT ( f ( t ))) , Then f ∗ ( T ) is also a solution to Nahm’s equations on the interval f − ( I ) .In particular, if T has domain I = ( α, β ) , then choosing f ( t ) = β − α t + α + β yields that f ∗ ( T ) has domain ( − , . Furthermore, the residues of f ∗ ( T ) at α (resp. at β ) are exactly a times theresidues of T at α (resp. at β ).Thus, we can assume that the domain of T is ( − , , albeit at a price of losing two degrees offreedom in the general solution. In the following three theorems we use Theorem 4.9 in the cases when n = 1 and k ∈ { , } and when n (cid:62) k = 1, to find solutions to Nahm’s equations. Theorem 4.11.
Under the hypotheses and notation of Theorem 4.9 let n = 1 and k = 1 .Thus ( V , ρ ) (cid:39) ( V , ρ ) ⊕ ( V , ρ ) . Let T be a spherically symmetric solution to Nahm’s equa-tions (Nahm1) to (Nahm3) , with representation induced by ( V , ρ ) . Let the domain of T be aconnected, open interval, I .In this case, choose ρ to be the identity and (cid:16) B , B , B (cid:17) to be the standard (orthonormal andoriented) basis of R . Then, in some gauge, there are real, analytic functions, f = f and g = g ,on I , such that for all i ∈ { , , } T i = f X i gB i − g (cid:16) B i (cid:17) ∗ , (4.26) and f and g satisfy the following equations: ˙ f = f + g , (4.27a)˙ g = 2 f g. (4.27b) Any maximally extended, irreducible solution to equations (4.27a) and (4.27b) develops polesin both direction, thus the domain of such a solution is necessarily a bounded, open interval,and the only maximally extended, irreducible solution to equations (4.27a) and (4.27b) withthe boundary conditions that the residues are exactly at t = ± is f ( t ) = t − t , (4.28a) g ( t ) = 11 − t , (4.28b) There is another solution with g ≡ K = 0 case of equation (3.9) and a trivial solution. ith residues given by Res( f , ±
1) = − , (4.29a)Res( g, ±
1) = − . (4.29b) The spherically symmetric Nahm datum given by equations (4.26) , (4.28a) , and (4.28b) , inducesrepresentations that isomorphic to ( V , ρ ) ⊕ at both poles.Proof. First of all, equation (4.26) is a special case of equation (4.23) and equations (4.27a)and (4.27b) are special cases of equations (4.24a) and (4.24b), with n = 1, k = 1, f = f , and g = g . Simple commutations give equations (4.28a) and (4.28b) and the equations (4.29a)and (4.29b). The claim about the irreducibility follows from equation (4.34c) and thesecond bullet point in Remark 4.5.The decomposition of the poles can be verified using the Casimir operatorsˆ C ± (cid:66) − (cid:88) i =1 Res( T i , ± . (4.30)Substitute f and g from equations (4.29a) and (4.29b) into equation (4.30) to getˆ C ± = 2 − . Both Casimir operators are proportional to the identity, thus the representations are eitherirreducible, or reducible to isomorphic summands. By inspecting the dimensions and thefactors of proportionality we find that the latter is true. In fact, we get that the inducedrepresentations are isomorphic to ( V , ρ ) ⊕ at both poles, which completes the proof. (cid:3) To demonstrate the same idea in higher rank cases, we present, without proof (as theclaim is easy to check), the case of n = 1 and k = 2 in the following theorem. Theorem 4.12.
Under the hypotheses and notation of Theorem 4.9), let ( V , ρ ) (cid:39) ( V , ρ ) ⊕ ( V , ρ ) ⊕ ( V , ρ ) . Let T be a spherically symmetric solution to Nahm’s equations (Nahm1) to (Nahm3) , with representation induced by ( V , ρ ) .Then, in some gauge, there are real, analytic functions, f , f , g , and g , on the domain of T ,such that for all i ∈ { , , } T i = f Y i g B i − g (cid:16) B i (cid:17) ∗ f Y i g B i − g (cid:16) B i (cid:17) ∗ , For example, equations (4.24a) and (4.24b) can be decoupled via the substitution F ± = f ± g . nd f , f , g , and g satisfy the following equations ˙ f = f + g − g , ˙ f = f + 12 g , ˙ g = 2 f g , ˙ g = (3 f − f ) g . The above equations have the following maximally extended, irreducible solution on ( − , : f ( t ) = f ( t ) = − tt − , g ( t ) = √ √ t − , g ( t ) = √ t − , with residues given by Res( f , ±
1) = Res( f , ±
1) = − , Res( g , ±
1) = ± (cid:114) , Res( g , ±
1) = ± √ . The above Nahm datum induces representations that are isomorphic to ]( V , ρ ) ⊕ at both poles. Finally, investigate the case of n (cid:62) k = 1. Theorem 4.13.
Under the hypotheses and notation of Theorem 4.9 let n (cid:62) and k = 1 , thus ( V , ρ ) (cid:39) ( V n +2 , ρ n +2 ) ⊕ ( V n , ρ n ) , and let T be a spherically symmetric solution to Nahm’s equa-tions (Nahm1) to (Nahm3) , with representation induced by ( V , ρ ) . Let the domain of T be aconnected, open interval, I . Finally, let (cid:16) Y ± , Y ± , Y ± (cid:17) and ( B , B , B ) as in Theorem 4.4.Then, in some gauge, there are real, analytic functions, f ± and g , on I , such that for all i ∈ { , , } T i = f + Y + i gB i − gB ∗ i f − Y − i , (4.31) and they satisfy the following system of ordinary di ff erential equations ˙ f + = ( f + ) + 2 n + 1 g , (4.32a)˙ g = (cid:18) n + 32 f + − n − f − (cid:19) g, (4.32b)˙ f − = ( f − ) − n + 2) n ( n + 1) g . (4.32c) The above equations have the following maximally extended, irreducible solution on ( − , : f + ( t ) = − ( n + 1) t + ( n − n + 1)( t − , (4.33a) f − ( t ) = − ( n + 1) t + ( n + 3)( n + 1)( t − , (4.33b) However this solution need not be (and probably is not) unique, in any sense. However this solution need not be (and probably is not) unique, in any sense. ( t ) = (cid:18) nn + 1 (cid:19) t − , (4.33c) with residues given by Res( f + , ±
1) = (2 − n ) ∓ ( n + 2)2( n + 1) , (4.34a)Res( f − , ±
1) = 12 ∓ n + 32( n + 1) , (4.34b)Res( g, ±
1) = ± (cid:114) n n + 1) . (4.34c) The spherically symmetric Nahm datum given by equations (4.31) and (4.33a) to (4.33c) ,induces representations that isomorphic to ( V n +1 , ρ n +1 ) ⊕ at t = 1 and to ( V , ρ ) ⊕ ( n +1) at t = − .Proof. First of all, equation (4.31) is a special case of equation (4.23) and equations (4.31)and (4.32a) to (4.32c) are special cases of equations (4.24a) and (4.24b), with n (cid:62) k = 1, f = f − , f = f + , and g (cid:66) g . Checking equations (4.33a) to (4.33c) and (4.34a) to (4.34c)then is straightforward. The claim about the irreducibility follows from equation (4.34c)and the second bullet point in Remark 4.5.In order to find the decomposition of the induced representations of so (3) at t = ±
1, wecompute the corresponding Casimir operators:ˆ C ± (cid:66) − (cid:88) i =1 Res( T i , ± . For each i ∈ { , , } and t ∈ ( − , T i ( t ) = f + ( t ) (cid:16) Y + i (cid:17) − g ( t ) B i B ∗ i f + ( t ) g ( t ) Y + i B i + f − ( t ) g ( t ) B i Y − i f + ( t ) g ( t ) B ∗ i Y + i + f − ( t ) g ( t ) Y − i B ∗ i − g ( t ) B ∗ i B i + f − ( t ) (cid:16) Y − i (cid:17) . (4.35)Using equations (4.8b), (4.8c), (4.13), (4.14), (4.18), and (4.35) we get − (cid:88) i =1 T i ( t ) = (cid:16) ( n +1)( n +3)4 f + ( t ) + g ( t ) (cid:17) n +2 (cid:16) n − f − ( t ) + n +2 n g ( t ) (cid:17) n . (4.36)Now substituting f + , f − , and g from equations (4.34a) to (4.34c) into equation (4.36) andtaking residues, we getˆ C + = ( n + 1) − n +2 , and ˆ C − = 2 − n +2 . Both Casimir operators are proportional to the identity, thus the representations are eitherirreducible, or reducible to isomorphic summands. By inspecting the dimensions and the actors of proportionality we find that the latter is true. In fact, we get that the inducedrepresentations are isomorphic to ( V n +1 , ρ n +1 ) ⊕ at t = 1 and to ( V , ρ ) ⊕ ( n +1) at t = − (cid:3) The Nahm transform
In this section we use the solutions to Nahm’s equations that found in the previoussections, to construct solutions to the BPS monopole equation, using the
Nahm transform ,which was was introduced by Nahm in [29], and it is a version of the Atiyah–Drinfeld–Hitchin–Manin construction; cf. [3].5.1.
Brief summary of the technique.
We outline the Nahm transform of Nahm datadefined on a single interval. Let T = ( T , T , T ) be a rank N Nahm data that is analyticon the nonempty, open interval ( a, b ), and satisfies certain boundary conditions at a and b . The precise form of these boundary conditions can be found in [25] in the maximalsymmetry breaking case, and in [10] in the general case. We only consider the case when T has a pole at both boundary points, and the residue forms (potentially reducible) N -dimensional representation of so (3). Let e , e , and e represent multiplication by thestandard unit quaternions on C (cid:27) H , and let L , (cid:16) ( a, b ); C N ⊗ H (cid:17) be the Hilbert spaceof L functions with Dirichlet boundary values. For each x ∈ R , we define a first-orderdi ff erential operator Λ T x , called the Nahm–Dirac operator , to be Λ T x : L , (cid:16) ( a, b ); C N ⊗ H (cid:17) → L (cid:16) ( a, b ); C N ⊗ H (cid:17) ; f (cid:55)→ i d f d t + (cid:88) a =1 (( iT a − x a C N ) ⊗ e a ) f . Note that Λ T x is a Fredholm operator. Moreover, (cid:110) Λ T x (cid:12)(cid:12)(cid:12) x ∈ R (cid:111) is a norm-continuousfamily of Fredholm operators; in particular, the index is independent of x ∈ R .Nahm’s main observation in [29] was that if T is a solution to Nahm’s equations, thenfor all x ∈ R , the kernel of Λ T x is trivial. Thus the index of Λ T x is nonpositive, andthe cokernel-bundle, defined via E T x (cid:66) coker (cid:16) Λ T x (cid:17) , is a Hermitian vector bundle over R , with a distinguished connection induced by the product connection on the trivialHilbert-bundle R × L (cid:16) ( a, b ); C N ⊗ H (cid:17) . Let us denote this connection by ∇ T , let Π E T x be he orthogonal projection from L , (cid:16) ( a, b ); C N ⊗ H (cid:17) to E T x , and let m : L , (cid:16) ( a, b ); C N ⊗ H (cid:17) → L , (cid:16) ( a, b ); C N ⊗ H (cid:17) ; f (cid:55)→ ( t (cid:55)→ itf ( t )) . Finally, let us define Φ T ∈ End (cid:16) E T (cid:17) as ∀ x ∈ R : Φ T x (cid:66) Π T x ◦ m − tr E T ( Π T x ◦ m ) rk ( E T ) E T . Note that Φ T ∈ su (cid:16) E T (cid:17) . The main results of [10, 25, 29] is that the pair ( ∇ T , Φ T ) is a (finiteenergy) BPS monopole on E T . Moreover, the asymptotic behavior of this monopole isgoverned by the poles of T at a and b .5.2. Construction of spherically symmetric BPS monopoles.
In this section we use The-orems 4.11 and 4.13 to generate spherically symmetric monopoles. We employ Maple (see[26] for details) to carry out the Nahm transform. More precisely, we solve the followingordinary di ff erential equation for all r ∈ R + : (cid:16) Λ T ( r, , (cid:17) ∗ u = 0 , and reduce the general solution to only the square integrable ones. Finally, we use thesesolutions to compute the Higgs field. Remark 5.1.
Before we begin, let us make a few important remarks:(1) We only carry out the computation in the cases of found in Theorem 4.11 and in the n = 2 case of Theorem 4.13. While we also found an explicit, irreducible solution toNahm’s equations in Theorem 4.12, we could not perform the Nahm transform of itwith our current technique. Nonetheless, general theory (cf. [10] ) tells us that thereis monopole with nonmaximal symmetry breaking corresponding to this Nahm datum,whose Higgs field, in some gauge, has the following asymptotic expansion Φ ( r ) = i · diag(1 , , , − , − , − − i r diag(3 , , , − , − , −
3) + O (cid:16) r − (cid:17) . In more geometric terms, this monopole (topologically) decomposes the trivial
SU(6) bundle over the “sphere at infinity” into two, nontrivial
U(3) bundles, both of which arefurther decomposed (holomorphically) into line bundles with Chern numbers ± .(2) By the second bullet point in Remark 4.5, the Nahm data found in Theorems 4.11 to 4.13are all irreducible, hence so are the corresponding BPS monopoles. Explicit solutions for any n ∈ N + can be found similarly.
3) For the computations below, one needs to know explicit formulae for (cid:16) Y n , Y n , Y n (cid:17) and (cid:16) B n , B n , B n (cid:17) . The former are well-known in the literature, while the latter can be com-puted uniquely (up to a U(1) factor) via equations (4.8a) and (4.8b) .The ( V , ρ ) (cid:39) ( V , ρ ) ⊕ ( V , ρ ) case: We have the following Nahm data from Theorem 4.11 : T = g f − f g , T = − f
00 0 0 − gf − g , T = − f f g g ,f = − tt − , g = it − . After the Nahm transform we get a Higgs field of the form Φ ( r ) = i · diag( F ( r ) , G ( r ) , − F ( r ) , − G ( r )) , where the functions F and G can be found in equation (A.1) in Appendix A.1. Theasymptotic expansion of this Higgs field is Φ ( r ) = i · diag(1 , , − , − − i r diag(2 , , − , −
2) + O (cid:16) r − (cid:17) . In more geometric terms, this monopole (topologically) decomposes the trivial SU(4)bundle over the “sphere at infinity” into two, nontrivial U(2) bundles, both of which arefurther decomposed (holomorphically) into line bundles with Chern numbers ± r ) of this SU(4) monopole. Note that | Φ | has a single zero at the origin. Figure 1.
Norm squared of the Higgs field (solid) and energy density(dashed) of the spherically symmetric monopole corresponding to the solu-tion to Nahm’s equations found in Theorem 4.11.The representation given by the generators of the spherical symmetry satis-fies (
V , ρ ) (cid:39) ( V , ρ ) ⊕ ( V , ρ ). We use a slightly di ff erent (but gauge equivalent) version of the solutions found in Theorem 4.11. he ( V , ρ ) (cid:39) ( V , ρ ) ⊕ ( V , ρ ) case: We the following Nahm data from Theorem 4.13: T = i f + i f + (cid:113) g
00 0 − i f + (cid:113) g − i f + − (cid:113) g − i f −
00 0 − (cid:113) g i f − ,T = √ f + i √ g − √ f + f + i √ g − f + √ f + i √ g
00 0 − √ f + i √ g i √ g i √ g − f − i √ g i √ g f − ,T = i √ f + − √ g i √ f + if + − √ g if + i √ f + 1 √ g
00 0 i √ f + √ g √ g − √ g − i f − √ g − √ g − i f − ,f + = − t + 13( t − , f − = − t + 53( t − , g = 2 √ t − . After the Nahm transform we get a Higgs field of the form Φ ( r ) = i · diag( F ( r ) , G ( r ) , − ( F ( r ) + F ( − r ) + G ( r ) + G ( − r )) , F ( − r ) , G ( − r )) , where the function F and G can be found in equation (A.2) in Appendix A.2. While thisHiggs field is not traceless, and hence structure group of the monopole is only U(5), thetraceless part of Φ , call Φ , together with the same connection forms an SU(5) monopole.The asymptotic expansion of this SU(5) Higgs field is Φ ( r ) = i · diag (cid:18) , , , − , − (cid:19) − i r · diag(2 , , , − , −
3) + O (cid:16) r − (cid:17) . n more geometric terms, this monopole (topologically) decomposes the trivial SU(5)bundle over the “sphere at infinity” into a direct sum of a U(3) and a U(3) bundle, both ofwhich are further decomposed (holomorphically) into line bundles with Chern numbers 2and −
3, respectively.Figure 2 shows the pointwise norm squared of this Higgs field and energy density (as afunction of r ) of this SU(5) monopole. Note that | Φ | has a single zero at the origin. Figure 2.
Norm squared of the Higgs field (solid) and energy density(dashed) of the spherically symmetric monopole corresponding to the solu-tion to Nahm’s equations found in Theorem 4.13.The representation given by the generators of the spherical symmetry satis-fies (
V , ρ ) (cid:39) ( V , ρ ) ⊕ ( V , ρ ). Appendix A. The coefficients of the Higgs fields
A.1.
The ( V , ρ ) (cid:39) ( V , ρ ) ⊕ ( V , ρ ) case: F ( r ) = e r p ( r ) − p ( − r ) e r p ( r ) + p ( − r ) ,G ( r ) = e r p ( r ) + e r p ( r ) − e r p ( − r ) − p ( − r ) − e r p ( r ) + e r p ( r ) + e r p ( − r ) − p ( − r ) .p ( r ) = 4 r − r + 3 ,p ( r ) = 4 r − r,p ( r ) = 4 r − r + 1 ,p ( r ) = 64 r − r + 36 r + 6 r − ,p ( r ) = 64 r − r + 4 r − r. (A.1) .2. The ( V , ρ ) (cid:39) ( V , ρ ) ⊕ ( V , ρ ) case: F ( r ) = e r p ( r ) + p ( r ) e r p ( r ) − p ( r ) ,G ( r ) = e r p ( r ) − e r p ( r ) − e r p ( r ) + p ( r ) e r p ( r ) + e r p ( r ) − e r p ( r ) − p ( r ) .p ( r ) = 8 r − r + 15 r − ,p ( r ) = 4 r + 9 r + 6 ,p ( r ) = 8 r − r + 3 r,p ( r ) = 4 r + 3 r,p ( r ) = 64 r − r + 224 r − r + 27 r − ,p ( r ) = 256 r − r + 256 r − r − r + 45 r − ,p ( r ) = 128 r + 128 r + 112 r − r − r + 9 ,p ( r ) = 4 r + 9 r + 3 ,p ( r ) = 64 r − r + 96 r − r + 3 r,p ( r ) = 256 r − r + 64 r − r + 60 r − r,p ( r ) = 128 r + 128 r + 48 r + 24 r − r. (A.2) References [1]
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Department of Pure Mathematics, Department of Physics and Astronomy,University of Waterloo, Ontario, Canada
Email address : [email protected] URL : (Anuk Dayaprema) Duke University
Email address : [email protected] (C. J. Lang) University of Waterloo
Email address : [email protected] (Ákos Nagy) University of California, Santa Barbara
Email address : [email protected] URL : akosnagy.com (Haoyang Yu) Duke University
Email address : [email protected]@duke.edu