Hyperbolic monopoles with continuous symmetries

We provide a framework to classify hyperbolic monopoles with continuous symmetries and find a Structure Theorem, greatly simplifying the construction of all those with spherically symmetry. In doing so, we reduce the problem of finding spherically symmetric hyperbolic monopoles to a problem in representation theory. Additionally, we determine constraints on the structure groups of such monopoles. Using these results, we construct novel spherically symmetric $\mathrm{Sp}(n)$ hyperbolic monopoles.


Introduction
Let M be a 3-manifold, E → M a vector bundle over M with structure group Sp(n), equipped with a connection A, of curvature F A , and a section Φ of End(E), called the Higgs field.Monopoles are solutions to the Bogomolny equations F A = ⋆D A Φ, with finite energy ε = 1 2π M |F A | 2 dvol M .If M has constant sectional curvature, then twistor methods can be used.When M = R 3 , we have Euclidean monopoles.When M = H 3 , we have hyperbolic monopoles, which have not received as much study as their Euclidean counterparts.This is seen below, as much of the research on hyperbolic monopoles, including this work, has arisen in an effort to find hyperbolic analogues to results about Euclidean monopoles.When M = S 3 , in fact whenever M is compact, both sides of the Bogomolny equation must vanish.As such, the monopoles are flat and this case is not as deep as the previous two.In addition to their deep mathematical structure, hyperbolic monopoles are interesting due to their connection with monopoles in Anti de-Sitter space and Skyrmions [AS05, MS90,Sut11].
The main difference between hyperbolic monopoles and their Euclidean counterparts is the idea of mass, the eigenvalues of the Higgs field at infinity.For Euclidean monopoles, as long as the limit of one eigenvalue is non-zero, we can always change the length scale such that the limit has modulus 1.For hyperbolic monopoles, there is already a length scale, determined by the curvature of the underlying hyperbolic space.Thus, when changing the length scale to modify the mass, we change the curvature of the space.In this paper, we consider hyperbolic space to have constant sectional curvature −1 (the ball model with radius one).Another difference between the two types of monopoles is their motion.Indeed, the natural L 2 metric used to study the motion of Euclidean monopoles is infinite in the hyperbolic case.
The study of hyperbolic monopoles started when Atiyah used the conformal equivalence between S 4 \ S 2 and S 1 × H 3 to associate hyperbolic monopoles with integral mass with circleinvariant instantons on R 4 , leading to a correspondence between these monopoles and based rational maps [Ati84a,Ati84b].Given a vector bundle E → M over a 4-manifold M, equipped with a connection A, an instanton is a solution to the self-dual equations ⋆F A = F A , with finite action M |F A | 2 dvol M .With regards to hyperbolic monopoles, we are only interested in instantons on M = R 4 , equivalently instantons on M = S 4 .The integral mass condition above ensures that the corresponding instanton on S 4 \ S 2 extends to all of S 4 ; otherwise, we get non-trivial holonomy around S 2 ⊆ S 4 [Ati84b].
In this paper, we only investigate hyperbolic monopoles with integral mass, due to their correspondence with circle-invariant instantons.Nonetheless, we review the literature for arbitrary mass hyperbolic monopoles.Because the zero curvature limit of hyperbolic space is Euclidean space, Atiyah conjectured that hyperbolic monopoles correspond to Euclidean monopoles in the zero curvature limit [Ati84a,Ati84b].Chakrabarti quickly gave explicit examples of hyperbolic monopoles, providing evidence to support Atiyah's curvature conjecture [Cha86].Jarvis and Norbury would later confirm this conjecture in general [JN97].
The ADHM transform provides a correspondence between instantons on R 4 (equivalently S 4 ) and ADHM data A, the set of quaternionic matrices M := L M such that M is symmetric and M satisfies the following non-linear constraint.Let x ∈ H ≃ R 4 and define ∆(x) := L M − Ix .Then ∆(x) † ∆(x) must be symmetric and non-singular for all x ∈ H [AHDM78].
In this paper, we look at a subset of A whose instantons are invariant under a particular circle action, giving us hyperbolic monopoles.
Given the similarities between Euclidean and hyperbolic monopoles, Braam and Austin sought to find a hyperbolic analogue to the ADHMN transform, which is a correspondence between Euclidean monopoles and Nahm data, solutions to the Nahm equation, an ordinary differential equation.They succeeded, finding a correspondence between SU(2) hyperbolic monopoles with integral mass and solutions to their discrete Nahm equation: a matrix-valued difference equation [BA90].They also discovered a difference between hyperbolic and Euclidean monopoles reminiscent of the AdS-CFT correspondence: SU(2) hyperbolic monopoles with integral mass are determined by their boundary values at infinity.From this we got a new viewpoint for hyperbolic monopoles: holomorphic spheres, which are holomorphic embeddings of the Riemann sphere in projective space.The discrete Nahm equation was later generalized by Chan to produce SU(N ) hyperbolic monopoles [Cha18].Just as in the Euclidean case, the discrete Nahm equations were shown to be integrable [War99].
At this point, we have multiple ways of looking at hyperbolic monopoles with integral mass: solutions to the Bogomolny equation, rational maps, spectral curves, discrete Nahm data, and holomorphic spheres.The first four are hyperbolic analogues to Euclidean monopoles, however, no analogue of holomorphic spheres exists for Euclidean monopoles, as Euclidean monopoles are not determined by their value at infinity.There are hyperbolic monopoles with non-integral mass.Indeed, Nash used the JNR ansatz (a generalization of 't Hooft's ansatz for instantons) to give charge one spherically symmetric hyperbolic monopoles with arbitrary mass [Nas86].This result was further improved by Sibner and Sibner, who used Taubes' gluing argument to show that hyperbolic monopoles exist with arbitrary charge and mass [SS12].Thus, the study of hyperbolic monopoles turned to finding relationships between these viewpoints for arbitrary mass [MS96, MNS03, MS00, Nor01, Nor04, NR07].However, while the holomorphic sphere viewpoint was generalized to arbitrary mass and charge, spectral curves were found in general for arbitrary mass for charge one and two hyperbolic monopoles as well as for SU(N ) monopoles satisfying specific boundary conditions.Additionally, substantial work has been done analyzing the spectral curves of hyperbolic monopoles with charge three and four [NR07].
As many had focused on the theoretical aspects of hyperbolic monopoles, there were very few explicit examples of these objects, with or without integral mass.Using Jarvis' construction of Euclidean monopoles from rational curves, Ioannidou and Sutcliffe generated some spherically symmetric SU(N ) hyperbolic monopoles, for N = 3, 4 that descended to spherically symmetric Euclidean monopoles in the zero curvature limit [IS99].Harland created spherically symmetric SU(2) hyperbolic monopoles with arbitrary mass by first constructing SO(3)-symmetric hyperbolic calorons (instantons on S 1 × H 3 ) and taking a limit, shrinking the circle [Har08].Oliveira found that given any non-parabolic, spherically symmetric metric on R 3 , there is a one-parameter family of spherically symmetric SU(2) monopoles with arbitrary mass, all of which vanish at the origin [Oli14, Appendix A].In particular, this is true for hyperbolic and Euclidean space.In contrast to this result, in this paper, Proposition 7 gives us a spherically symmetric, hyperbolic Sp(4) monopole that vanishes nowhere.Cockburn used the discrete Nahm equation to generate hyperbolic monopoles with integral mass and axial symmetry and deformed these monopoles to replace axial symmetry with dihedral symmetry [Coc14].Franchetti and Maldonado found examples of hyperbolic monopoles constructed from solutions to the Helmholtz equation and from vortices [FM16,Mal17].
The circle action used by Braam and Austin to generate their discrete Nahm equation led to using the upper-half-space model of hyperbolic space when considering the corresponding monopoles.While axial symmetries are easy to see in this space, others are difficult, such as Platonic symmetries.To that end, Manton and Sutcliffe used a different circle action to end up with the ball model, where rotational symmetries are very easy to see [MS14].They found a set of quaternionic ADHM data that always give a SU(2) ≃ Sp(1) hyperbolic monopole with unit mass [MS14].In this paper, we generalize this set of data, finding a set of ADHM data that always gives Sp(n) hyperbolic monopoles with unit mass.In particular, we provide a framework to classify all spherically symmetric hyperbolic monopoles that have ADHM data in our set.To generate Platonic monopoles, Manton and Sutcliffe used the JNR ansatz, which would generate a circle invariant instanton if all poles were on the boundary of hyperbolic space in R 4 .In addition, they used ADHM data that was previously used to find Platonic instantons to find Platonic monopoles.Using this approach, Bolognesi et al. found exact examples of hyperbolic magnetic bags [BHS15].
Given the importance of the JNR ansatz and Manton and Sutcliffe's ADHM data to generating examples of hyperbolic monopoles, work was done to find formulae for computing the spectral curve and rational map of monopoles constructed from this data [BCS14,Sut21].Additionally, work was done to compute the holomorphic sphere and energy density for monopoles constructed from the JNR ansatz [MN21].
In earlier work, we created a Structure Theorem that generates spherically symmetric Euclidean monopoles [CDL + 22].However, this theorem has some hypotheses, meaning it is not known if it generates all such Euclidean monopoles.Just like others before, given the connection between Euclidean and hyperbolic monopoles, I sought an analogue of this theorem for the hyperbolic case, allowing me to add to the small, but growing list of examples of hyperbolic monopoles.By giving a more abstract proof of the Structure Theorem, I am able to remove the aforementioned hypotheses not only for the case of hyperbolic monopoles, but also Euclidean monopoles.This means that not only do the Structure Theorems provide novel examples of spherically symmetric monopoles with higher rank structure groups, but they generate all spherically symmetric Euclidean monopoles and all spherically symmetric hyperbolic monopoles that can be obtained from the subset of ADHM data I identify below.
The main results of this paper are Theorem 1, Theorem 2, and Theorem 3, the Structure Theorem.The first two theorems are able to linearize the equations of symmetry by working with Lie algebras.They do so for axial and spherical symmetry, respectively.These theorems reduce the problem of finding symmetric hyperbolic monopoles to a problem in representation theory.The Structure Theorem outlines all solutions to the equation of spherical symmetry, solving a key step in finding spherically symmetric hyperbolic monopoles.Using this theorem, we generate many infinite families of spherically symmetric, hyperbolic monopoles.
In Section 2, I introduce the ADHM data that we study, as well as what it means for it to be equivariant under a rotation.In Section 3, we investigate hyperbolic monopoles with axial symmetry.In Section 4, we investigate hyperbolic monopoles with spherical symmetry and prove the Structure Theorem, which greatly simplifies the construction of such monopoles.We then provide examples of spherically symmetric hyperbolic monopoles and discuss constraints on the structure group of these monopoles, providing a framework for the classification of spherically symmetric hyperbolic monopoles.

Monopole data
Recall that ADHM data corresponds to instantons on R 4 (equivalently S 4 ) via the ADHM transform.Additionally, Atiyah's work gave a relationship between circle-invariant instantons and hyperbolic monopole with integral mass.In this section, following Manton and Sutcliffe, we investigate a set of ADHM data whose instantons are invariant under a specific circle action, giving us hyperbolic monopoles [MS14, Section 4].We then examine what it means for such data to be equivariant under a rotation.
In particular, the hyperbolic monopoles corresponding to data in M n,k has structure group Sp(n) and instanton number k, which is the second Chern number of the instanton.
Note 1.Throughout this paper, k indicates both the instanton number and the standard quaternion (k = ij) and its use at any given time is contextual.
When n = 1, LL † is a 1 × 1 matrix.Moreover, it is a non-negative real number.Thus, the condition that LL † is positive definite just becomes L is non-zero, just as in the definition of Manton and Sutcliffe [MS14, (4.9)-(4.11)].Such a condition is natural, as requiring LL † to be positive-definite only rules out those instantons which are trivial embeddings of smaller rank instantons.That is, those whose connection matrices have the form A ν (x) = Ãν (x) ⊕ 0. The final condition is required to use the ADHM transform.
Note 2. The condition LL † being positive definite implies that n ≤ k, as the rows of L must be linearly independent.
Manton and Sutcliffe's definition of M n,k (where n = 1) includes a left eigenvalue µ [MS14, (4.11)].The existence of µ is guaranteed by the first three conditions for M n,k .
Lemma 1.Given M satisfying the first three conditions of M n,k .Then there exists a unique µ ∈ sp(n) such that µL = LM .Specifically, That µ is unique comes from the invertibility of LL † .For the last identity, note that LM 2 = µ 2 L. As L † L − M 2 = I k , we multiply by L on the left and L † on the right to get As LL † is invertible, we obtain the desired identity.
After identifying R 4 ≃ H, Manton and Sutcliffe introduce the following circle action on R 4 : Note that while this circle action is not well-defined on R 4 , we can extend it to a well-defined circle action on S 4 = R 4 ∪{∞}.Moreover, note that instantons on R 4 correspond to instantons on S 4 .The following was proven for the n = 1 case [MS14, Section 4], though the proof works for arbitrary n.
Proposition 1.All M ∈ M n,k correspond to instantons invariant under this circle action.Hence, all of our ADHM data corresponds to hyperbolic monopoles.
Using the conformal equivalence of R 4 \ R 2 ≡ S 1 × H 3 , the model of hyperbolic space produced by this circle action is the ball model: The kernel of ∆(X) † is n-dimensional.Let ψ(X) be a (n + k) × n quaternionic matrix whose columns give an orthonormal basis for the kernel of ∆(X) † .That is, ψ(X) † ∆(X) = 0 and ψ(X) † ψ(X) = I n .Then the Higgs field is given by [MS14, (4.21)] The connection A = A 1 dX 1 + A 2 dX 2 + A 3 dX 3 is given by, for i ∈ {1, 2, 3} [MS14, (4.18)], Lemma 2. If M satisfied the first three conditions of M n,k in Definition 1 and satisfies the final condition for all points in H That is, we can relax the final condition to just points in H 3 .

Rotating monopole data
It is well known that if a non-trivial monopole is invariant under some group of isometries, then a change of coordinates takes the family to a subgroup of O(3) [Lan, Appendix].Otherwise, the energy of the monopole is infinite.As we are interested in investigating monopoles with continuous symmetries-those monopoles whose group of symmetries is a connected Lie group-we are only interested in subgroups of SO (3), that is rotations.Now that we understand the relationship between hyperbolic monopoles and elements in M n,k , we need to understand what it means for an element of M n,k to be equivariant under a rotation.Definition 2. We define the gauge action of Sp(n)×O(k) on M n,k as follows.For q ∈ Sp(n), Q ∈ O(k), and M ∈ M n,k , let Definition 3. We define the rotation action of Sp(1) on M n,k as follows.For p ∈ Sp(1) and M ∈ M n,k , let The above actions are named as such to reflect what is happening to the corresponding monopoles.
(2) For any M ∈ M n,k and p ∈ Sp(1), let R p ∈ SO(3) be the image of p under the double cover map Sp(1) → SO (3).Then the monopole corresponding to p. M is the pull-back of the monopole corresponding to M by R T p , the inverse of R p . Proof.
(2) Let M ∈ M n,k and p ∈ Sp(1).Consider the monopole (Φ, A) associated to M and the monopole (Φ ′ , A ′ ) associated to M ′ := p. M .Let X ∈ H 3 and consider ψ(X) satisfying ψ(X) † ∆(X) = 0 and ψ(X) † ψ(X) = I n .Let ψ ′ (X) := pψ(p † Xp).We see that Thus, we can use ψ ′ (X) to compute the monopole.By the definition of µ, under the rotation transformation, µ maps to pµp † .Note that the double cover Sp(1) → SO(3) takes p → R p , which acts on H 3 via R p X = pXp † .The inverse R T p of R p , acts on R 3 via R T p X = p † Xp.We see that Thus, we have that for l ∈ {1, 2, 3}, Similarly, we have Hence, Therefore, we see that this action just rotates the monopole (in the opposite direction).
Note 3. The gauge action does not only give a gauge-equivalent monopole, it gives the exact same monopole.Thus, while it does not correspond to a gauge transformation on the monopole, it represents a gauge freedom on the ADHM data.
There is another gauge freedom we have, which comes from our choice of ψ(x) when constructing the monopole.Indeed, taking any smooth Sp(n)-valued function g, we see that if ψ(x) † ∆(x) = 0 and ψ(x) † ψ(x) = I n , then so too does ψ ′ (x) := ψ(x)g(x).Under this choice, the monopole changes as Φ(X) → g(X) † Φ(X)g(X), and That is, we get a gauge-equivalent monopole.However, we note that multiplying on the right by g corresponds to a choice of the orthonormal basis of ker∆(x) † and has no effect on M .Note 4. The rotation action descends to an action on In particular, we see that the gauge and rotation actions do not commute on the L part of the actions, but they do commute on the M part.
M .This means that rotation by p has no effect on the monopole, as it is just the result of gauging by (q, Q).
It turns out that M determines L, up to a Sp(n) factor.The choice of this factor is just a choice of gauge, so has no effect on the monopole.Lemma 3. Suppose that M ∈ M n,k .Then L is uniquely determined by M , up to multiplication by some q ∈ Sp(n).Conversely, multiplying L by any q ∈ Sp(n) gives another element of M n,k .Moreover, all such data produces the same monopole.
. Therefore, the two matrices are related via a gauge transformation, so they are both in M n,k and they correspond to the same monopole.
Just as M determines L, if we satisfy the M part of equivariance, then Q determines q such that we have full equivariance.
Lemma 4. Let M ∈ M n,k .Suppose that there exists p ∈ Sp(1) and Q ∈ O(k) such that QM Q T = pM p † .Then there is a unique q ∈ Sp(n) such that qLQ T = pLp † , which is given by q Proof.That q is unique comes from the invertibility of LL † .Indeed, suppose q and q satisfy the desired equation.Then (q − q)LQ T = 0.The invertibility of Q and LL † give q = q.Define q as above.We show that q ∈ Sp(n) and satisfies the desired equation.As In light of the previous lemmas, we see that when considering monopoles equivariant under some rotation, we need only consider the M part.
Definition 5. Let H M ⊆ Sp(1) be the set of unit quaternions under which M is equivariant.Because we are dealing with group actions, H M is a subgroup of Sp(1).
The only non-zero, connected Lie subgroups of Sp(1) are S 1 and Sp(1) itself.
Fact 1.The maximal tori of Sp(1) are S 1 .They are all conjugate to The set H z is the double cover of the set of rotations about the z-axis.Each maximal torus of Sp(1) is the double cover of the set of rotations about a fixed axis.Definition 6.When H M contains a subgroup conjugate to H z , we say that M is axially symmetric.When H M = Sp(1), we say that M is spherically symmetric.

Axial symmetry
In this section, we determine when a monopole is axially symmetric and use this to search for examples.
. Note that M is said to be axially symmetric about the z-axis if it is equivariant under every element of H z .Then M is axially symmetric about the z-axis if and only if there exists Y ∈ so(k) such that The matrix Y ∈ so(k) is called the generator of axial symmetry for M .
Note 6.For axial symmetry around a general axis in H 3 , note that this axis is given by a unit vector û ∈ ∂H 3 and let υ := û 2 ∈ sp(1).Then M is spherically symmetric about û if and only if there exists Y ∈ so(k) such that (7) holds.
Proof.Let X be the smooth manifold comprised of all (n + k) × k quaternionic matrices.The rotation and gauge actions can easily be expanded from M n,k to smooth actions on X .Just as with M n,k , the rotation action on X descends to an action on X /(Sp(n) × O(k)).
Suppose that M ∈ M n,k is axially symmetric about the z-axis, so [ M ] is fixed by S 1 ≃ H z ⊆ Sp(1).Let S ⊆ S 1 × O(k) be the stabilizer group of M restricted to axial rotations.That is With that in mind, we can write S as As M is axially symmetric, clearly the map is compact and its universal cover R is simply connected.
Our goal is to find a smooth map that, when composed with π 1 | S gives the covering map π : R → S 1 taking x → e 2υx .This smooth map will allow us to differentiate the equation for equivariance and arrive at (7).
We first show that S is a Lie group.Let f : As S is a closed subgroup of a Lie group, it is a Lie group, by the Closed-subgroup Theorem.Moreover, as S 1 × O(k) is compact, S must be as well, as it is closed.
Let e := (1, I) be the identity of S ⊆ S 1 × O(k).Consider the Lie algebra homomorphism ϕ := d e (π 1 | S ) : Lie(S) → R. Note that ker(ϕ) ⊆ Lie(S) is more than a Lie subalgebra, it is an ideal of Lie(S).Indeed, for any x ∈ ker(ϕ) and y ∈ Lie(S), As S is compact, it has a bi-invariant metric, which corresponds with a Ad(S)-invariant inner product ⟨•, •⟩ on Lie(S), satisfying for all x, y, z ∈ Lie(S), Let C ⊆ Lie(S) be the orthogonal complement to ker(ϕ).We show that C is an ideal and C ⊕ ker(ϕ) = Lie(S) as Lie algebras (the bracket is zero between the two ideals).Suppose that c ∈ C and s ∈ Lie(S).We see that for all x ∈ ker(ϕ) By the isomorphism theorems, we know that as im(π Let ψ : R → C be the isomorphism. Let Y ⊆ S be the unique connected Lie subgroup corresponding to the Lie algebra C ⊆ Lie(S), whose existence is guaranteed by the Subgroups-subalgebras Theorem.Note that R is a Lie group whose Lie algebra is R.As R is simply connected, the Homomorphisms Theorem tells us that there is a unique Lie group homomorphism Ψ : R → Y such that We show that this map is an isomorphism and we use this to find our desired smooth map.Indeed, if ϕ(ψ(x)) = 0, then ψ(x) ∈ C ∩ ker(ϕ) = {0}, so x = 0, as ψ is an isomorphism.Furthermore, consider y ∈ R = im(ϕ).Then there is some x ∈ Lie(S) such that ϕ(x) = y.We can uniquely write x = c + z for c ∈ C and z ∈ ker(ϕ).But then ϕ(c) = y.As ψ is an isomorphism, there is some w ∈ R such that ψ(w) = c.Therefore, ϕ • ψ(w) = ϕ(c) = y.We have proved that ϕ • ψ is a Lie algebra isomorphism.Call the inverse of this map g.
By the Homomorphisms Theorem, we know that there is a unique Lie group homomorphism G : R → R such that d 0 G = g.We show that Ψ • G is the smooth map that we are searching for.Indeed, we have that But the covering map π : R → S 1 is a Lie group homomorphism whose pushforward at the identity is the identity.By the Homomorphisms Theorem, the two maps must be equal, so As ψ • g is a Lie algebra homomorphism, we know that ψ • g(x) = (h(x), ρ(x)) for some Lie algebra homomorphisms h : R → R and ρ : R → so(k).As Note that as ρ is a Lie algebra homomorphism into so(k), it is a real Lie algebra representation.
From the expression for ψ • g(x) above and as Moving the first factor to the other side, we differentiate and evaluate at θ = 0, obtaining Focusing on the bottom row, let Y := ρ 1 2 ∈ so(k).Then, we see that [υ, M ] = [M, Y ].Conversely, suppose the equations are true for some As real matrices and quaternions commute, we see . By Lemma 4, M axially symmetric about the z-axis.
For axially symmetric monopoles, we do not need to check the final condition of M n,k in Definition 1 at every point in H 3 .
Lemma 5. Suppose that M satisfies (7) for some Y ∈ so(k) as well as the first three conditions of M n,k in Definition 1.If the final condition is satisfied at all

Spherical symmetry
In this section, we determine when a monopole is spherically symmetric, prove the Structure Theorem, which greatly simplifies the construction of such monopoles, and examine novel examples of hyperbolic monopoles.We also find a constraint on the structure groups of spherically symmetric monopoles, providing a framework for classifying these monopoles.We start by proving an analogue to Theorem 1.
Theorem 2. Let M ∈ M n,k .Recall that M is spherically symmetric when it is equivariant under all p ∈ Sp(1).Then M is spherically symmetric if and only if there exists real representation (R k , ρ) with ρ : sp(1) → so(k), such that for all υ ∈ sp(1), The induced representation is said to generate the spherically symmetric monopole corresponding with M .
Proof.We follow the proof of Theorem 1, making modifications when relevant.Suppose that M ∈ M n,k is spherically symmetric, so [ M ] is fixed by Sp(1).Let S ⊆ Sp(1) × O(k) be the stabilizer group of M .As M is spherically symmetric, clearly the map π 1 | S : S ⊆ Sp(1)×O(k) → Sp(1) is surjective.Note that Sp(1) is simply connected and compact.Because Sp(1) is simply connected, instead of worrying about universal covers and covering maps, our goal is just to find a smooth right inverse to π 1 | S , which will allow us to differentiate the equation for equivariance and arrive at (10).Just as in the axial case, S is a compact Lie group.Let e := (1, I) be the identity of S ⊆ Sp(1) × O(k).Consider the Lie algebra homomorphism ϕ := d e (π 1 | S ) : Lie(S) → sp(1).Just as before, C, the orthogonal complement to ker(ϕ), intersects ker(ϕ) trivially and Lie(S) = C ⊕ ker(ϕ).Moreover, by the isomorphism theorems, we know that as im(π Let ψ : sp(1) → C be the isomorphism.
Let Y ⊆ S be the unique connected Lie subgroup corresponding to the Lie algebra C ⊆ Lie(S).As Sp(1) is simply connected, there is a unique Lie group homomorphism Ψ : as before, is an isomorphism, whose inverse we denote g.
There is a unique Lie group homomorphism But id Sp(1) is a Lie group homomorphism whose pushforward at the identity is the identity.By the Homomorphisms Theorem, the two maps must be equal, so Similar to the axial case, we know that there is a Lie algebra homomorphism ρ : sp(1) → so(k) such that ψ • g(x) = (x, ρ(x)).As ρ is a Lie algebra homomorphism whose image consists of real matrices, ρ gives us a real Lie algebra representation.
Therefore, we can remove the conditions on differentiability from that paper, meaning a Euclidean monopole is spherically symmetric if and only if there are matrices Y i , inducing a representation of so(3), satisfying the spherical symmetry equations.As we always get a representation of so(3) if a monopole is spherically symmetric, the Structure Theorem tells us what all spherically symmetric Euclidean monopoles look like.
For spherically symmetric monopoles, we only have to check the final condition of M n,k in Definition 1 along a ray from the origin of H 3 .Lemma 6. Suppose that M satisfies (11) for some Y 1 , Y 2 , Y 3 ∈ so(k) as well as the first three conditions of M n,k in Definition 1.If the final condition is satisfied at Proof.We follow the same proof as Lemma 5, but note for any
The definition of ( V , ρ) is well-motivated.Firstly, note that M ∈ V .Secondly, given the connection between the action of ρ(υ) and (10), we immediately obtain the following corollary.
Note 10.Suppose M ∈ M n,k is spherically symmetric.Corollary 1 tells us that there is some representation (V, ρ) with Y i := ρ(υ i ) ∈ so(k) such that span(M ) ⊆ V is an invariant subspace, which is acted on trivially.
As sp(1) is semi-simple, the representation ( V , ρ) decomposes into irreducible representations.We explore the case M = 0 below, so suppose M ̸ = 0.As span(M ) is a one-dimensional invariant subspace, acted on trivially, (span(M ), 0) is a summand of the representation.So the decomposition of ( V , ρ) must contain trivial summands.Moreover, M is in the direct sum of these trivial summands.Trivially, if M = 0, then it is in the direct sum of trivial summands as well.Moreover, for all X = X 1 i+X 2 j+X 3 k ∈ H 3 , let R := X 2 1 + X 2 2 + X 2 3 .Also, let the norm of an element A ∈ sp(k) be given by |A| 2 := −Tr(A 2 ).Then we have that the corresponding monopole's Higgs field satisfies |Φ(X)| = √ k R 1+R 2 .Also, up to gauge, we have that Proof.Suppose M = 0, so µ = 0. Recall from Note 2 that n ≤ k.As rank(L † L) ≤ n and We can gauge the data without changing the monopole, so we may gauge by L † , to obtain L := L † L = I k .Also, note that M = 0 satisfies the spherical symmetry conditions for any representation ρ.
We now wish to construct the corresponding monopole.Note that ∆(X) = , by (2) we have, up to gauge, We see that such a Higgs field has norm The irreducible real representations of sp(1) are indexed by the dimension of the vector space acted upon; every irreducible real k-representation of sp(1) is isomorphic.However, irreducible real representations only exist for k ∈ N + odd or divisible by four.The tensor product of real representations is simplified by using complex representations, so we shall look at these too.
Like the irreducible real representations, irreducible complex representations of sp(1) are indexed by the dimension of the vector space acted upon.Unlike the irreducible real representations, a unique irreducible complex representation exists for all k ∈ N + , up to isomorphism.Note that the dimension of the vector space acted on k and the highest weight of the representation w are related via w = k−1 2 .Definition 8. Let (V k , ρ k ) be the irreducible complex k-representation of sp(1).Let (R k , ϱ k ) be the irreducible real k-representation of sp(1).
The irreducible complex and real representations are related as follows.For k odd or divisible by four, ϱ k can be complexified and made to act on ⊕2 .Thus, for k odd, the Y i matrices that induce (V k , ρ k ) can be chosen such that they are real matrices.
In particular, we note the following representations.The adjoint representation (sp(1), ad) is the irreducible 3-dimensional representation (V 3 , ρ 3 ).Let F : sp(1) → su(2) be the Lie algebra isomorphism given by The fundamental representation (C 2 , F ) is the irreducible 2-representation (V 2 , ρ 2 ).Finally, the trivial representation (C, 0) is the irreducible 1-representation (V 1 , ρ 1 ).As mentioned earlier, as sp(1) is a semisimple Lie algebra, all representations of it decompose as the direct sum of irreducible representations.The Clebsch-Gordon Decomposition tells us how to decompose tensor products of complex representations: for m ≥ n ≥ 1, As discussed earlier, given a spherically symmetric monopole with ADHM data M ∈ M n,k , M is found in the trivial summands of ( V , ρ).The following lemmas tell us where these trivial summands are, given a decomposition of (V, ρ).First, we see where the trivial summands are when decomposed as a complex representation, providing us with the information needed for the real decomposition.
) has a single trivial summand when m = n ≥ 2 or m = n + 2. Otherwise, it does not have any such summands.
Proof.We first see what representations give a trivial summand when taken as a tensor product with (V 3 , ρ 3 ).We see As we have we see that there is a single (V 3 , ρ 3 ) summand exactly when m + n + 1 − 2k = 3 for some k ∈ {1, . . ., n}.Thus, m + n must be even, so they must have the same parity.Furthermore, as m However, as m + n + 1 − 2k ≤ m + n − 1, we must have m + n − 1 ≥ 3. Thus, m + n ≥ 4. We see that if m = n + 2, then m + n ≥ 4, but if m = n, then we must have m = n ≥ 2. In each case, we get a single (V 3 , ρ 3 ) summand, so the final product has a single (V 1 , ρ 1 ) summand.Otherwise, we get no such summand.
We first investigate the trivial summand present in ( Definition 9. Let n ∈ N + and let B n be the generator of the unique trivial summand in the tensor product ).This generator is well-defined, up to a C * factor, the choice of a scale leaves a U(1) factor.The generator can be viewed as an sp(1)-invariant triple (B n 1 , B n 2 , B n 3 ) of complex (n + 2) × n matrices.

Charbonneau et al. prove the following, giving us some useful identities for these maps [CDL
After potential rescaling, reducing the factor to U(1), we have the following identities (for all i, j ∈ {1, 2, 3}, where it applies) Above we determined where the trivial summands of ( V , ρ) would occur if we decomposed it as a complex representation.We use this information to do the same for the real decomposition.
) has a trivial summand when m = n ≥ 2, both odd, or m = n + 2, both odd, and four trivial summands when m = n ≥ 4, both divisible by four, or m = n + 4, both divisible by four.Otherwise, there are no trivial summands.
Proof.Note that if m and n are both odd, then the complexification of the real representation is isomorphic, as a complex representation, to (V m , ρ m ) ⊗ C (V n , ρ n ) ⊗ C (V 3 , ρ 3 ).By Lemma 7, we know that this tensor product of complex representations has a single trivial summand when m = n ≥ 2 or m = n + 2 and none otherwise.Furthermore, as this tensor product only contains odd dimensional summands, the representation is real, meaning that the real decomposition is the same as the complex.Thus, we have proven the first part of the lemma.
Suppose that m and n are both divisible by four.Then, as a complex representation, the complexification of the real representation is isomorphic to . By Lemma 7, we know that this has four trivial summands when m 2 = n 2 ≥ 2 or m 2 = n 2 + 2 and none otherwise.Just as above, this decomposition contains only odd dimensional summands, so the representation is real and the real and complex decompositions coincide.Thus, we have proven the second part of the lemma.
Finally, suppose that one of n and m is divisible by four and the other is odd.Then, decomposing the complexification of the real representation as a complex representation and expanding the tensor product, we find that the tensor product contains only even dimensional summands, meaning there are no trivial summands, proving the lemma.
Then there is a unique choice of U(1) factor, up to a choice of sign, such that the B n i matrices are real.That is, given this choice of factor, B n spans the unique trivial summand of (R As the complexification of (R n+4 , ϱ n+4 ) is isomorphic to V (n+4)/2 , ρ (n+4)/2 ⊕2 as complex representations, and similarly for (R n , ϱ n ), there exists U ± ∈ SU n+2±2 2 such that We can find four linearly independent triples of real matrices spanning the four trivial summands of (R m , ϱ m ) ⊗ R (R n , ϱ n ) ⊗ R (R 3 , ϱ 3 ).For some choices of α, β, γ, δ ∈ C, these linearly independent triples of real matrices are given by Proof.Suppose that n is odd.Note that Y ± i ∈ su(n + 1 ± 1).We are looking for a non-zero triple (C n 1 , C n 2 , C n 3 ) of real matrices that satisfy (14a), replacing B n i by C n i .Such a triple exists.Indeed, from Lemma 8, we see that the tensor product ) contains a single trivial summand.Any element that spans this real summand satisfies (14a).Hence, any such element is in the span of B n , as the solution space to (14a) is a one dimensional complex vector space.Therefore, the U(1) factor that determines B n i can be chosen such that the B n i matrices are all real.Suppose instead that n is divisible by four.Note that for α, β, γ, δ ∈ C, By Lemma 8, there are four linearly independent triples of real matrices satisfying (14a).That the solution space is four dimensional comes from counting the trivial summands of ).After complexifying this tensor product, we see that it is isomorphic, as a complex representation, to 16), we see that the direct sum of the four trivial summands of the above complex tensor product are given by As the four linearly independent triples of real matrices above satisfy the same equation as the matrices in E, they belong to E. Thus, for some choices of the complex parameters in E, we can find four linearly independent triples of real matrices.
We now investigate the rest of the trivial summands of ( V , ρ).
If n is divisible by four, Lemma 8 tells us that there are four trivial summands of (R The trivial summands are spanned by the following triples of real matrices ). Below, we identify exactly what matrices commute with real irreducible representations.These provide us with exactly enough matrices to span the rest of the trivial summands of ( V , ρ).
Let g be a Lie algebra and (V, ρ) an irreducible representation, over a field F. An endomorphism of the representation is a morphism f : V → V with f a linear map such that for all x ∈ g and v ∈ V , Schur's Lemma tells us that the only endomorphisms are the zero morphism or an isomorphism [BD85, §II, Lemma 1.10].This gives the endomorphism ring of the representation the structure of a division algebra over F. Let f be an endomorphism.As f and ρ(x) are endomorphism, we can associate them to matrices.As matrices, [f, ρ(x)] = 0 for all x ∈ g.That is, the endomorphism ring is exactly the set of matrices that commute with the representation, and we will think of endomorphisms as these matrices going forward.Let ( Ṽ , ρ) : Thus, f is in the endomorphism ring if and only if f is in a trivial summand of (V, ρ) ⊗ F (V * , ρ * ).We are interested in the case g = sp(1) and F = R.
The only real division algebras are R, C, and H.In the case n is odd, we have that (V, ρ)⊗ R (V * , ρ * ) contains a single trivial summand, as the complexification of (V, ρ) is irreducible as a complex representation.Thus, the endomorphism ring is isomorphic to R. As the isomorphism sends I k to 1, the endomorphism ring is just the span of the identity matrix.
In the case n is divisible by four, we have that (V, ρ) ⊗ R (V * , ρ * ) contains four trivial summands, as the complexification of (V, ρ) is isomorphic to V n/2 , ρ n/2 ⊕2 .Thus, the endomorphism ring is isomorphic to H. We show that if X is a n × n real matrix commuting with ρ(x) for all x ∈ sp(1), then X − Tr(X) n I n ∈ so(n).
We have that there is some ring isomorphism ψ : End(V, ρ) → H. Let ψ 0 (X) be the real part of ψ(X).Then ψ(X − ψ 0 (X)I n ) ∈ sp(1), so We know that the ring of n × n real matrices decomposes as Thus, given X, there is a unique antisymmetric matrix A, a unique traceless, symmetric matrix B, and a unique β ∈ R such that X − ψ 0 (X)I n = βI n + A + B. As ψ(X − ψ 0 (X)I n ) ∈ sp(1), we know that β = 0, as A and B give us something in sp(1).Thus, X − ψ 0 (X) Note that [Y i , X] = 0 for all i.Thus, ].Note that as A and Y i are anti-symmetric, the left-hand-side is anti-symmetric.As Y i is antisymmetric and B is symmetric, the right-hand-side is symmetric.But the only symmetric and anti-symmetric matrix is the zero matrix.Therefore, [Y i , A] = 0 = [Y i , B].Thus, A and B correspond to endomorphisms of (V, ρ) themselves, just like X.We show that B must vanish.
Thus, just as we did above for X − ψ 0 (X)I n , B 2 = −γI n for some γ ≥ 0. Note that Taking the trace, we know that ψ 0 (X) = Tr(X) n .We now show exactly what X looks like.Note that for all a, b, c, d ∈ C and i ∈ {1, 2, 3} As the space of such matrices is four-dimensional, some choice of these parameters gives us From above, we know that X − Tr(X) n I n ∈ so(n).Hence, As this matrix is to be traceless, we see that n 2 (a + d) = Tr(X).Additionally, as the matrix is skew-adjoint, Hence, we see that c = −b † , a † + a = 2 Tr(X) n , and d † + d = 2 Tr(X) n .Combining these equations, we see that Re(a) = Re(d) = Tr(X) n and Im(a) = −Im(d).Therefore, we have that d = a † .Thus, Hence, X has the desired form.Moreover, we see that we have two complex parameters, equivalently four real parameters, so all such matrices of this form are real and commute with the representation.Now that we know exactly what spans the trivial summands of ( V , ρ), the following theorem tells us exactly what form M takes if M is spherically symmetric.
Theorem 3 (Structure Theorem).Let M ∈ M n,k be spherically symmetric.Theorem 2 tells us that the monopole corresponding with M is generated by a real representation (V, ρ) of sp(1), which we can decompose as Without loss of generality, we may assume that n a ≥ n a+1 , for all a ∈ {1, . . ., m − 1}.
Let Y i,a := ϱ na (υ i ) ∈ so(n a ) and Then Y i induces (V, ρ).For n a is divisible by four, let y i,a := ρ na 2 (υ i ) ∈ su na 2 induce the irreducible na 2 complex representation.Then there is some Using the decomposition of (V, ρ) given in (19), we have Then, up to a ρ-invariant gauge and for all a, b ∈ {1, . . ., m} we have that (1) if a = b with n a divisible by four, then ∃κ a,1 , κ a,2 , κ a,3 ∈ R such that (2) if a < b and n a = n b is divisible by four, then ∃κ a,b,0 , κ a,b,1 , κ a,b,2 , κ a,b,3 ∈ R such that (3) if n a = n b + 4 are both divisible by four, then ∃κ a,b,0 , κ a,b,1 , κ a,b,2 , κ a,b,3 ∈ C such that and these blocks are real; Hence, κ b,a,0 = κ a,b,0 , κ b,a,1 = κ a,b,2 , κ b,a,2 = κ a,b,1 , and κ b,a,3 = κ a,b,3 .In summary, we see that Mi is of the desired form.We see that if M has the form in the statement of the theorem for some representation (V, ρ) with Y i := ρ(υ i ) ∈ so(k), then M satisfies the spherical symmetry conditions, so M is spherically symmetric, by Theorem 2.

Novel examples of spherically symmetric monopoles
Here we use the Structure Theorem to construct novel examples of spherically symmetric hyperbolic monopoles.We start with irreducible representations.
Proposition 5. Suppose (V, ρ) is an irreducible real representation with dimension k that generates a spherically symmetric hyperbolic monopole with M ∈ M n,k .
(1) If k is odd, then M = 0, which we investigated in Proposition 3.
(2) If k is divisible by four, let Y i and U have the same meaning as in the Structure Theorem.
Then there is some κ ≥ 0 such that M i is gauge equivalent to Having identified all M generated by irreducible real representations, we construct a family of spherically symmetric hyperbolic monopoles.If k is divisible by four, let 0 < κ < 4 k+2 and let Note that as 0 < κ < 4 k+2 , β > 0, so L is well-defined.Using the M i from the second point above, we have M ∈ M k,k and the corresponding Sp(k) monopole is spherically symmetric.
Proof.(1) The Structure Theorem tells us that M i = 0.
(2) The Structure Theorem tells us that there are some κ 1 , κ 2 , κ 3 ∈ R such that We now gauge the data as follows.
Let κ := κ 2 1 + κ 2 2 + κ 2 3 ≥ 0. If κ 2 = κ 3 = 0 and κ 1 ≥ 0, then we have the desired form.If κ 2 = κ 3 = 0 and κ 1 < 0, then we take Q := U † 0 1 −1 0 U .From Lemma 10, we know that Q ∈ so(k).In fact, we see that QQ T = QQ † = I k , so Q ∈ O(k) and QM i Q T has the desired form, as Otherwise, let By Lemma 10, we know that Q ∈ O(k) and Q commutes with the Y i .Hence, Therefore, up to gauge, we may assume that κ 2 = κ 3 = 0 and κ 1 ≥ 0. Now we prove the final part of the proposition.Suppose that we have M as in the statement of the proposition.We show that M ∈ M k,k .Such data is spherically symmetric by the Structure Theorem.Then Simplifying, we see L † L − M 2 = I k .We just need to verify that LL † and ∆(X 3 k) † ∆(X 3 k) are positive definite, the latter for all Suppose that this matrix is not positive definite for some X 3 ∈ [0, 1].Then there is some unit vector v such that ∆(X 3 k) † ∆(X 3 k)v = 0. Hence, Multiply each side on its left by its conjugate transpose, obtaining As the Y i induce the irreducible real k-representation, whose complexification is isomorphic , the largest modulus of an eigenvalue of Finally, suppose that LL † = L 2 is not positive definite.Then there is some unit vector v such that As L 2 − M 2 = I k , we can rewrite this as Multiplying both sides on the left by their conjugate transpose, we get Substituting (22), we see Simplifying, we have (16 But this equation has no roots on 0, 4 k+2 , contradiction!Thus, M ∈ M k,k corresponds with a spherically symmetric Sp(k) monopole.
Note 13.We have found a family of spherically symmetric Sp(k) monopoles, with ADHM data in M k,k , when k is divisible by four.However, there may be other monopoles with ADHM in M n,k (for some n < k) whose corresponding ADHM data shares the M portion of the data with the aforementioned family.Now that we have investigated irreducible representations, let us move on to some reducible ones.
In addition to the above families of spherically symmetric monopoles, the following M ∈ M 2,4 corresponds with a spherically symmetric Sp(2) monopole: Proof.The Structure Theorem tells us that for a spherically symmetric monopole generated by (V, ρ), there is a constant λ 1,2 ∈ R such that the ADHM data M ∈ M m,2n+2 satisfies where the U(1) factor for the B 1 i is chosen so they are all real matrices.Denoting a := λ 1,2 , M has the form given in the statement of the proposition.Furthermore, any data with M i of this form is spherically symmetric by the Structure Theorem.
To see that, up to gauge, we can take a ≥ 0, consider Q := The M in parts two and three of the proposition have M given as above.We show that in both cases, if M satisfies the first three conditions in Definition 1, then the final condition is satisfied.Assuming the first three conditions are satisfied, by Lemma 6, we need only check the final condition at X = X 3 k for X 3 ∈ [0, 1].For such X, we find ∆(X) † ∆(X) = (1 + X 2 3 )I − 2X 3 M 3 .Suppose that ∆(X) † ∆(X) is not positive definite for some X 3 ∈ [0, 1].Then there is some unit vector v such that ∆(X) † ∆(X)v = 0. Thus, Multiplying both sides on the left by their conjugate transpose, we find Substituting M i and using Proposition 4, we see that The largest eigenvalue of Now we prove the second part of the proposition.Suppose 0 < a < n+1 2(n+2) .Let M be as in the statement.Let Substituting our expressions, we see that L † L − M 2 = I 2n+2 .Finally, we show that LL † = L 2 is positive definite.Suppose not, then there is some unit vector v such that L 2 v = 0.As Multiply both sides on the left by . Note that this matrix commutes with the matrix on the right-hand side.Thus, we can use (25) to obtain Simplifying the right-hand side, we note that Thus, we use (25) to find corresponds with a spherically symmetric Sp(2n + 2) monopole.Now we prove the final part of the proposition.Consider the additional M given in the statement.Let a := 1 √ 3 and n := 1.The following matrices induce (R 3 , ϱ 3 ) ⊕ (R 1 , ϱ 1 ): Y i := y i 0 0 0 , where The B i matrices for this case are given, up to a sign, by Indeed, these matrices satisfy (14a) to (14g).Instead of solving these equations on a caseby-case basis, one can solve them in general [Lan, Appendix].As above, consider . Using these generators, we see that Hence, our M is spherically symmetric, assuming it belongs to M 2,4 , which we now verify; indeed, L † L − M 2 = I 4 LL † = I 2 and LL † = I 2 .Therefore, we have M ∈ M 2,4 corresponds with a spherically symmetric Sp(2) monopole.
Note 14.We compute the hyperbolic monopole corresponding to the additional ADHM data above in (24).While the hyperbolic monopole corresponding to the aforementioned ADHM data is spherically symmetric, in a given gauge, the Higgs field varies on spheres of constant radius.However, up to gauge, the Higgs monopole depends only on the radius r from the origin.Hence, along any ray emanating from the origin, up to gauge, the Higgs field Φ is given by Note that the eigenvalues of Φ(r) as r → 1 are ± i 2 .Then we have The energy density of any hyperbolic monopole is given by ε 135r 16 + 840r 14 + 5252r 12 + 13304r 10 + 18282r 8 + 13304r 6 + 5252r 4 + 840r 2 + 135 (3r 4 + 2r 2 + 3) 4 . (28) In Figure 1, we see the norm of the Higgs field squared as well as the energy density of the monopole.From the figure, we see that the Higgs field only vanishes at the origin and the monopole looks like a point-particle, in that the energy density has a global maximum at the origin.
Up to gauge there is some a ≥ 0 such that Note that if n = 1, then M = 0, which we have already covered.
In addition to the above families of spherically symmetric monopoles, the following M ∈ M 4,6 corresponds with a spherically symmetric Sp(4) monopole: Proof.The Structure Theorem tells us that for a spherically symmetric monopole generated by (V, ρ), there is a constant λ 1,2 ∈ R such that the ADHM data M ∈ M m,2n satisfies Denoting a := λ 12 ∈ R, M has the form given in the statement of the proposition.Furthermore, any data with M i of this form is spherically symmetric by the Structure Theorem.
To see that, up to gauge, we can take a ≥ 0, consider Q : The M in parts two and three of the proposition have M given as above.We show that in both cases, if M satisfies the first three conditions in Definition 1, then the final condition is satisfied.Assuming the first three conditions are satisfied, by Lemma 6, we need only check the final condition at Suppose that ∆(X) † ∆(X) is not positive definite for some X 3 ∈ [0, 1].Then there is some unit vector v such that ∆(X) † ∆(X)v = 0. Thus, Multiplying both sides by their conjugate transpose, we find Substituting M i , we see that as 0 < a < 2 n+1 , Hence, (1 − X 2 3 ) 2 < 0, contradiction!Thus, ∆(X) † ∆(X) is positive definite.Now we prove the second part of the proposition.Suppose 0 < a < 2 n+1 .Let M be as in the statement.Then Substituting our expressions, we see that L † L − M 2 = I 2n .Suppose that LL † = L 2 is not positive definite.Then there is some unit vector v such that Multiplying by 1 − a 2 n 2 −1 4 and using (31), we see that corresponds with a spherically symmetric Sp(2n) monopole.Now we prove the final part of the proposition.Consider the additional M given in the statement.Let a := 1 2 and n := 3. Recall from the proof of Proposition 6 that the following matrices induce (R 3 , ϱ 3 ): As above, consider M i = a 0 y i −y i 0 .Using these generators, we see that Y := y 1 i + y 2 j + y 3 k and M are given by Hence, our M is spherically symmetric, assuming it belongs to M 4,6 , which we now verify; indeed, L † L − M 2 = I 6 and LL † = 3 4 I 4 .Therefore, we have M ∈ M 4,6 corresponds with a spherically symmetric Sp(4) monopole.
Note 15.We compute the hyperbolic monopole corresponding to the additional ADHM data above in (30).The spherically symmetric, hyperbolic monopole corresponding to the aforementioned ADHM data has Higgs field Φ satisfying |Φ| 2 = r 12 + 9r 10 + 33r 8 + 58r 6 + 33r 4 + 9r 2 + 1 4(r 4 + r 2 + 1) The energy density of this monopole is given by ε 5r 16 + 70r 14 + 381r 12 + 942r 10 + 1260r 8 + 942r 6 + 381r 4 + 70r 2 + 5 (r 4 + r 2 + 1) 4 .(33) In Figure 2, we see the norm of the Higgs field squared as well as the energy density of the monopole.From the figure, we see that the Higgs field never vanishes and the monopole looks like a shell with a non-zero size, in that the energy density has a global maximum away from the origin.This matrix has eigenvalues i 2 , − i 2 with multiplicity 3 and 1, respectively.

Constrained structure groups
In this section, we prove that there is a constraint on the structure groups of spherically symmetric monopoles generated by a representation of sp(1).Given a real representation (V, ρ) of sp(1), the Structure Theorem helps us find a vector space in which the M component of ADHM data corresponding to spherically symmetric monopoles lies.Lemma 3 tells us that in M n,k , L is uniquely determined by M , up to multiplication by a Sp(n) factor, which does not affect the corresponding monopole.Thus, for a given n, the Structure Theorem helps us find the unique family of spherically symmetric monopoles generated by (V, ρ).However, if n is allowed to vary, the same vector space may give rise to monopoles with different structure groups.For example, see Proposition 6 or Proposition 7. In these examples, the exact M matrices differ when looking at different Proof.As M is spherically symmetric, we have (12).Recall that the Y i := ρ(υ i ).Then we see that for υ = 3 i=1 a i υ i ∈ sp(1), λ(υ)(L) = Proof.As L ̸ = 0, we know span(L) ⊆ Ŵ is an invariant 1-dimensional subspace that is acted on trivially, so (span(L), 0) is a trivial summand of the representation ( Ŵ , λ).
As M has Sp(1) as a structure group, after restricting the scalars of (W, λ), we get a complex 2-representation.In order for ( Ŵ , λ) to have a trivial summand and for the previously mentioned complex representation to be a 2-representation, we must have (W, λ) ≃ (H, ι), the fundamental representation.Hence, there is some p ∈ Sp(1) such that y i = pυ i p † and pυ i p † L − LY i − Lυ i = 0.
Taking L := p † L, we see [υ i , L] = LY i .Such a choice has no effect on the monopole.
Using the Y i matrices from the proof of Proposition 6 and solving these equations, we find there is some c, d ∈ R such that L = ck cj ci d .Recall that with these Y i generators, we have Hence, Hence, a 2 + c 2 = 0 and a 2 + c 2 = 1.Contradiction! Thus, (V, ρ) does not generate any spherically symmetric Sp(1) hyperbolic monopoles with ADHM data in M 1,4 .
Restricting the scalars, this representation is isomorphic, as a complex representation, to (V 3 , ρ 3 ) ⊕2 ⊕ (V 1 , ρ 1 ) ⊕2 .Suppose that (V, ρ) generates a spherically symmetric Sp(1) hyperbolic monopole with ADHM data M ∈ M 1,4 .Then the complex representation obtained by restricting the scalars of (W, λ) is a 2-representation.In order for ( Ŵ , λ) to have a trivial summand and for the previously mentioned complex representation to be a 2-representation, we must have that the complex representation obtained by restricting the scalars of (W, λ) is isomorphic to (V 1 , ρ 1 ) ⊕2 .Hence, y i = 0, so L(Y i + υ i I 4 ) = 0.
Suppose that (V, ρ) generates a spherically symmetric Sp(2) hyperbolic monopole with ADHM data M ∈ M 2,4 .Then the complex representation obtained by restricting the scalars of (W, λ) is a 4-representation.In order for ( Ŵ , λ) to have a trivial summand and for the previously mentioned complex representation to be a 4-representation, we must have the complex representation obtained by restricting the scalars of (W, λ) to be isomorphic to (V 1 , ρ 1 ) ⊕4 , (V 2 , ρ 2 )⊕(V 1 , ρ 1 ) ⊕2 , or (V 3 , ρ 3 )⊕(V 1 , ρ 1 ).Note that we can ignore the final case, as this can not

Note 8 .
We can use a similar proof for symmetric Euclidean monopoles, improving upon previous work [CDL + 22, Theorems 3.1 & 4.1].We fill in the gaps for the spherical symmetry case, but the axial case is similar.In this previous work, the chosen gauge group is SU(n) and there is a rotation action of SO(3) (though we can get an action of Sp(1) using the double cover) [CDL + 22, Definition 2.2].In this case, we let X = Mat(n, n, C) ⊕3 , and let S be the stabilizer group S = {(A, U ) ∈ SO(3) × SU(n) | A U T = T }, where we use the notations for the gauge and rotation actions [CDL + 22, Definition 2.2].
Theorem 2 tells us how to search for spherically symmetric monopoles: use a real k-representation of sp(1) to narrow down the possible M .We are then only left with finding L that gives us M ∈ M n,k .We now investigate what representations generate spherically symmetric monopoles and what the corresponding ADHM data looks like.Definition 7. Let ad : sp(1) → gl(sp(1)) be the Lie algebra homomorphism ad(x)(z) := [x, z].Denote the adjoint representation of sp(1) by (sp(1), ad) and let V := R k .Given a real representation (V, ρ) of sp(1), we define the induced real representation

Proposition 3 (
The M = 0 case).Suppose M = 0 k and M ∈ M n,k .Then n = k and L ∈ Sp(k).Furthermore, gauging by L † gives us L = I k .Such a M is spherically symmetric.
2 .Note 11.Proposition 3 tells us that Sp(k) monopoles with M = 0 have ADHM data in M k,k and are reducible, being constructed by applying a homomorphism Sp(1) → Sp(k) to the basic spherically symmetric instanton number one Sp(1) monopole.

Figure 1 :
Figure 1: The norm of the Higgs field squared and the energy density for the additional ADHM data provided in Proposition 6.The solid line is ε with the left vertical axis and the dashed line is |Φ| 2 with the right vertical axis.Proposition 7. Let n ∈ N + be odd.Suppose (V, ρ) ≃ (R n , ϱ n ) ⊕ (R n , ϱ n ) generates a spherically symmetric hyperbolic monopole with ADHM data M ∈ M m,2n for some m ∈ N + .Let y i := ϱ n (υ i ) ∈ so(n).Up to gauge there is some a ≥ 0 such that

Figure 2 :
Figure 2: The norm of the Higgs field squared and the energy density for the additional ADHM data provided in Proposition 7. The solid line is ε with the left vertical axis and the dashed line is |Φ| 2 with the right vertical axis.