Cooking pasta with Lie groups

We extend the (gauged) Skyrme model to the case in which the global isospin group (which usually is taken to be $SU(N)$) is a generic compact connected Lie group $G$. We analyze the corresponding field equations in (3+1) dimensions from a group theory point of view. Several solutions can be constructed analytically and are determined by the embeddings of three dimensional simple Lie groups into $G$, in a generic irreducible representation. These solutions represent the so-called nuclear pasta state configurations of nuclear matter at low energy. We employ the Dynkin explicit classification of all three dimensional Lie subgroups of exceptional Lie group to classify all such solutions in the case $G$ is an exceptional simple Lie group, and give all ingredients to construct them explicitly. As an example, we construct the explicit solutions for $G=G_{2}$. We then extend our ansatz to include the minimal coupling of the Skyrme field to a $U(1)$ gauge field. We extend the definition of the topological charge to this case and then concentrate our attention to the electromagnetic case. After imposing a"free force condition"on the gauge field, the complete set of coupled field equations corresponding to the gauged Skyrme model minimally coupled to an Abelian gauge field is reduced to just one linear ODE keeping alive the topological charge. We discuss the cases in which such ODE belongs to the (Whittaker-)Hill and Mathieu types.


Introduction
Nuclear pasta is a phase of matter that appears organized in some ordered structures when a large number of Baryons is confined in a finite volume [1], [2], [3], [4], [5], [6]. These configurations appear, for instance, in the crust of neutron stars. Such aggregations of Baryons may take the form of tubular structures, called Spaghetti states, or layers having a finite width, called Lasagna states, or even globular shape, the gnocchi. Until very recently, it was always tacitly assumed that nuclear pasta phase is the prototypical situation in which it is impossible to reach a good analytic grasp. This is related to the fact that such structures appear in the low energy limit of Quantum Chromodynamics (QCD) in which perturbation theory does not work and, at a first glance, the strong non-linear interactions prevent any attempt to find exact solutions. Now, the low energy limit of QCD is described by the Skyrme model [7] at the leading order in the 't Hooft expansion (see [8], [9], [10], [11], [12], [13], as well as [14], [15] and references therein). Unsurprisingly, the highly non-linear character of the Skyrme field equations discouraged any mathematical description of this kind of structures. Consequently, as the above references show, numerical methods (which, computationally, are quite demanding) are dominating in this regime. The situation is even worse when one wants to analyze the electromagnetic field generated in the nuclear pasta phase as, when the minimal coupling with the U (1) gauge field is taken into account; even the available numerical methods are not effective.
On the other hand, one may ask: is the mathematical dream of an analytic description of nuclear pasta structure really out of reach? Analytical methods to infer the general dependence of the nuclear pasta phase on relevant physical parameters (such as the Baryon density) not only would greatly improve our understanding of the nuclear pasta phase itself, but they could also shed considerable light on the interactions of dense nuclear matter with the electromagnetic field.
From the mathematical viewpoint, the problem is very deep and yet simple to state: can we find analytic solutions of the (gauged) Skyrme model able to describe typical configurations of the nuclear pasta phase? Despite the fact that this model has been introduced in the early sixties, for several years only numerical solutions had been available (the only exceptions being [16], in which the authors constructed analytic solutions of the Skyrme field equations in a suitable fixed curved background). Nevertheless, the mathematical beauty of the Skyrme model attracted the attention of many leading mathematicians and physicists. In particular, in [17], [18], [19], [20] and [21], the authors were able to disclose the geometrical structures of configurations with two Skyrmions, to analyze the interaction energy of well separated solitons, to establish necessary conditions for the existence of Skyrmionic crystals and so on. All these remarkable results have been obtained without the availability of analytic solutions of the Skyrme field equations. These efforts (together with the comparison with Yang-Mills theory in which explicit solutions representing instantons and non-Abelian monopoles shed considerable light on the mathematical and physical properties of Yang-Mills theory itself) show very clearly the importance to search for new analytic tools to analyze the gauged Skyrme model in sectors with high Baryonic charge.
Quite recently, new methods have been introduced that allowed the construction of explicit analytical solutions of the Skyrme field equations. Such solutions are suitable to describe nuclear Lasagna and Spaghetti states, see [22], [23], [24], [25], [26], [27], [28], [29], [30], [31], [32], and [33]. Let us recall that the Skyrme model is a non-linear field theory for a scalar field U taking values in the SU (N ) Lie group, where N is the flavor number. This theory possesses a conserved topological charge (the third homotopy class) which physically is interpreted as the Baryonic charge of the configuration.
Most of the solutions found so far have been constructed by employing ad hoc ansätze adapted to the properties of the SU (2) group, but soon it has been realized that particular group structures seem to be at the root of the solvability of the Skyrme field equations. For example, the exponentiation of certain linear functions taking value in the Lie algebra lead to Spaghetti-like configurations, while Euler parameterization of the field U , with suitable linear exponents, lead to Lasagna-like solutions. In all these cases, the solutions are also topologically non-trivial with arbitrary Baryonic charge. A proper mathematical understanding and generalization of the strategy devised in [22], [23], [24], [25], [26], [27], [28], [29], [30], [31], [32] and [33], offers the unique opportunity to disclose the deep connections of the nuclear pasta phase with the theory of Lie groups; two topics which (until very recently) could have been considered extremely far from each other. The present paper is devoted to this opportunity: to provide nuclear pasta configurations of Lasagna and Spaghetti types with the mathematical basis of Lie group theory.
A first step in this direction was to link certain properties of the semi-simple Lie group to the possibility of getting explicit solutions of the Skyrme equations in the Lasagna configurations for the case of SU (N ) groups with arbitrary N [33]. More in general, using the methods developed in [26], [27], [28], [30], [31], [33], with the generalization of the Euler angles to SU (N ) of [34], [35], [36], it has been possible to construct non-embedded multi-Baryonic solutions of nuclear Spaghetti and nuclear Lasagna, at least for the case for the SU (N ) groups, see [37].
A fundamental ingredient in the theory of Lie groups with relevant applications in the Skyrme model is the concept of non-embedded solutions introduced in [11] and [12]. These are solutions of the SU (N )-Skyrme model which cannot be written as trivial embeddings of SU (2) in SU (N ). However, the techniques used to get such results, for example in [33], where quite specific of the group SU (N ). In fact, as we will show in the present manuscript, there is a very interesting relation between Lie group theory and such families of solutions, which allows to generalize the above results in a much more general setting and to classify the solutions: this is exactly the main goal of the present paper.

Resume of the results
Firstly, we will prove that, having fixed a compact connected Lie group G with a given irreducible representation (irrep), the solutions are determined in general by deformations of embeddings of three dimensional Lie groups into G.
Secondly, we will prove that inequivalent families of solutions correspond to inequivalent embeddings (not related by conjugation in G). The problem of determine all possible three dimensional subgroups of a simple Lie group has been solved by E. B. Dynkin in [38]. In particular, in that paper, all possible three dimensional subalgebras of the exceptional Lie algebras are written down.
Thirdly, we will show that such classification also classifies the Spaghetti and Lasagna solutions determined via group theory methods. The difference between Spaghetti and Lasagna depends on the realization of the subgroup element of G: if it is generated by the exponentiation of a linear combination of the generators of a three-dimensional subalgebra of g = Lie(G), then we get Spaghetti-like solutions, while if the realization is through Euler parameterization we get Lasagna-like solutions. Then, we will compute explicitly relevant quantities such as the energy of these configurations.
Fourthly, we will extend this classification to the case of the gauged Skyrme model minimally coupled to Maxwell theory. In particular, we will extend the definition of topological (Baryonic) charge to this case. We will reduce the complete set of coupled field equations both in the gauged Lasagna case and in the gauged Spaghetti case to a single linear equation and we will analyze the integrable cases which correspond to Whittaker-Hill and Mathieu types linear differential equations.

Main tools employed in the analysis
In the present work, we will employ abstract techniques and general properties of semi-simple Lie groups in order to investigate their relation with solvability of the Skyrme equations. This allows to extend all the results found in [33] for the special unitary groups to an arbitrary semi-simple compact Lie group. Indeed, all results will be based on the properties of the roots and weights of the associated Lie algebras, while a generalized Euler parameterization of the Skyrme field U , taking values in G, will lead in general to Lasagna configurations. Similarly, the direct exponentiation of the algebra, as discussed above, will lead us to Spaghetti structures extending the results of [37]. In any case, we will compute the energy of such configurations and will show that they have always a non-trivial Baryon (topological) charge. Interestingly enough, a strategy for constructing non-trivial non-SU (2) solutions in the sense of [11] and [12] will result to be strictly related to the classification of all three dimensional groups in any given simple Lie group, provided by Dynkin in his PhD thesis work, see [38]. As an application of our general analysis, we will show how to construct all non-trivial Lasagna and Spaghetti configurations in any exceptional Lie group, making very explicit the case of G = G 2 .
The generalization of our ansätze which allows to include the minimal coupling of the model to a U (1) electromagnetic field will be introduced as follows. As usual, the gauge field will work as a connection making all derivative covariant under the action of the U (1) gauge field, while their dynamics is expressed by the usual Maxwell action (although our methods also work in the Yang-Mills case). The covariant derivatives break the topological nature of the original term expressing the Baryonic charge. Therefore, generalizing the result in [8], we will deform the Baryonic density expression in order to recover topological invariance .
The introduction of the electromagnetic field makes the field equations of the gauged Skyrme model minimally coupled to Maxwell theory extremely more complicated than in the Skyrme case. Nevertheless, quite surprisingly, the equations will be separable (in a suitable sense) and once again solvable, after imposing the free force conditions on the gauge field. This condition appears quite naturally in Plasma physics (see [39], [40], [41], [42], [43] and references therein). Quite interestingly, such condition implies that the gauge field disappears from the gauged Skyrme field equations (without being a trivial gauge field, of course) and therefore, in this way the gauged Skyrme field equations can be solved as in the ungauged case. It is a very non-trivial result that the remaining field equations (which correspond to the Maxwell equations with the source term arising from the gauged Skyrme model) reduce just to one linear equation for a suitable component of the gauge field in which the Skyrmion act as a source-like term. We will analyze the integrable cases in which this last remaining equation takes the form of a Hill equation for the case of Lasagna states, while a Schrödinger equation with a bi-periodic potential of finite type in the Spaghetti case.
Interestingly enough, for the Lasagna case another nice coincidence shows up here: the relevant solutions we need are exactly the periodic solutions whose existence has been investigated in [44], and which explicit form for the case of a Whittaker-Hill equation has been determined in [45].
It is a truly remarkable result that such a complicated phase such as the nuclear pasta phase of the low energy limit of QCD (even taking into account the minimal coupling with Maxwell theory) can be understood so cleanly in terms of the theory of Lie group.

Notations and conventions
Our conventions are as follows. The action of the Skyrme model in (3 + 1) dimensions is where K and λ are positive coupling constants and g is the metric determinant. The Skyrme field U is a map where G is semi-simple compact Lie group, so that where {T i } is a basis for the Lie algebra g = Lie(G).
The system is confined in a box of finite volume with a flat metric. For Lasagna states we will use a metric of the form where the adimensional spatial coordinates have the ranges so that the solitons are confined in a box of volume V = (2π) 3 L r L γ L φ . For nuclear Spaghetti we will use the metric ansatz with adimensional coordinates ranging in and a total volume V = 4π 3 L r L θ L φ . The energy-momentum tensor associated to the Skyrme field is given by The topological charge is defined by (see Proposition 3) where V is the spatial region spanned by the coordinates at any fixed time t, L = U −1 dU and Tr is the trace over the matrix indices.

Lasagna groups
In [46] it has been shown how Lasagna configurations can be determined as solutions of the Skyrme equations realized as Euler parameterizations of three dimensional cycles in SU (N ). Indeed, these cycles result to be suitable deformations of different non-trivial embeddings of SU (2) into SU (N ). Here we want to prove that such construction can be easily extended to any simple Lie group (at least for the case of the undeformed embedding). Recall that in the case of SU (N ) the embedding was defined [33] by the generalized Euler map where σ is a constant, m is an integer, κ a suitable matrix in SU (N ) and h(r) results to be a linear function of r with values in the Cartan algebra H. Indeed, the main trick was to determine a suitable matrix κ able to make everything easily computable and to grant periodicity of e mγκ . The convenient strategy has been composed in two steps: first we have taken a basis of eigenmatrices of the simple roots, λ j , j = 1, . . . , r, where r = N − 1 is the rank of the group, and defined the matrix where † means hermitian conjugate and c j are complex constants. The second step consisted in determining the allowed values for the c j . We want to do the same with a generic simple Lie group G replacing SU (N ). The first problem we ran into is the following. If λ ∈ g C is an eigenmatrix of a root α of the Lie algebra g of G (so it belongs to the complexification g C of g), in general λ † doesn't belong to g C if G = SU (N ) for some N . So in general κ defined above is not a matrix of g.
In order to overcome this problem, we notice that a compact simple Lie group G always contain a split maximal subgroup [47], which is a maximal subgroup K with the property that 2dim(K) + r = dim(G) and that there exists a Cartan subalgebra H of g all contained in p, the orthogonal complement of the Lie algebra k of K in g (w.r.t. the Killing product): Of course k is a subalgebra of g, while p is not, since which says that p is a representation space for G and K is an isotropy group for p. One can easily show that a root matrix λ, associated to a root α, must have the form Then, k − ip also is a root matrix, corresponding to the root −α. We replace the hermitian conjugation with the ∼ conjugation defined by This way, if λ j , j = 1, . . . , r are matrix roots corresponding to the simple roots of g, then Notice that for G = SU (N ) we haveλ = −λ † . If we choose normalizations as in Appendix A, we can use the matrices J k to decompose h(z) = r j=1 y j (z)J j . The properties of the roots can be inferred case by case from the lists in Appendix A. Exactly the same calculations as in [46] show that the field equations for the Skyrme field are equivalent to the system The first equation has solution h ′′ = 0, as a consequence of the strict positivity of the Cartan matrix for each simple group. The second system, using that the λ αj +α k are independent, reduces to the set of equations Since (α j |α k ) = 0 if and only if α j and α k are linked and since there are r − 1 links in a connected Dynkin diagram, these are exactly r − 1 equations. These are independent and assuming a = α 1 (h) = 0 have the general solution α j (h ′ ) = ǫ j a, j = 2, . . . , r, (2.11) where ǫ j are signs. As in [46], we can solve it by writing Applying α k to both hands and defining ǫ 1 = 1 we get where C G is the Cartan matrix associated to G. The Cartan matrix is positive definite and is therefore always invertible, so that (2.14) Therefore, we have proven the following generalization of Proposition 2 in [46].
are given by where a is a real constant and ǫ j are signs, with ǫ 1 = 1.
Now we have to discuss which choices of the coefficients c j are allowed. To this hand, we have that the solution must cover a topological cycle entirely. First, we notice that as a consequence of our normalizations, if we want to get it with r varying in [0, 2π], we must take see [46], Proposition 3. The second step is to grant periodicity of e γκ . This is the difficult part and determines the allowed values for the c j . Notice that κ is diagonalizable (over C). Indeed, since G is compact, in the adjoint representation κ results to be antihermitian and then diagonalizable with imaginary eigenvalues. It follows that it is diagonalizable in any representation with purely imaginary eigenvalues. If N is the dimension of the representation, then the eigenvalues iµ 1 , . . . , iµ N must be in rational ratios, which means that for any µ a = 0 it must exist integers n a = 0 such that µ a n b = µ b n a , (2.19) or, equivalently, that it exists a non-vanishing real number µ and N integers n a ∈ Z, such that µ a = µn a . (2.20) This condition in general will depend on N , G and the constants c j . In [46] this problem has been shown to have a set of solutions for the particular case of G = SU (N ) in the fundamental representation. Here we have to generalize that procedure without exploiting a very explicit realization. Indeed, we can prove that there are solutions with all c j different from zero by following a strategy developed by Dynkin in [38], that we will recall in the next section. Let us choose f in the Cartan subalgebra, such that α j (f ) = b, a positive constant independent on j, so that We can easily determine it as follows. If h j = i[λ j ,λ j ], then set f = r k=1 p k h k . Thus, the above condition is equivalent to from which we immediately get By the properties of the Cartan matrix it follows that p j are all positive. Finally, we set Then, we have the following proposition: If κ is constructed with the above choice of c j , then e κz is periodic with period n 2π b where n may be 1 or 2 depending on the representation, n = 1 for the adjoint representation.
Proof. We first show that periodicity is independent on the phases of c j . If e κz is periodic, then, for any fixed g ∈ G, ge κz g −1 is also periodic with the same period. Since the simple roots α j are linearly independent, for any fixed j we can find an element h j of the Cartan algebra such that α k (h j ) = δ kj . Let us set g = e ψhj . Then, which shows that gκg −1 differs from κ only by the phase of c j . This proves our assert. So, it is sufficient to prove the proposition for ψ j = 0. In this case, T 3 := f, T 1 := κ and Therefore, as before, the periodicity of e κz is equivalent to the periodicity of e f z . But and Now, any given root α is with the n j all non-negative or all non-positive integers. Therefore, All this exponentials are therefore periodic, with the longest period determined by the simple roots, for which e iα(f )z = e ibz , which has period T = 2π/b. But for any root α implies that g = e f T is in the center of the group. Since the center of a simple compact group is finite, this means that g n = I is the unit matrix for some integer n. Now, since κ is not in the Cartan subalgebra, it follows from [48], Section VII, Theorem 8.5 (see also [47]) that n = 1 or n = 2 depending on the specific representation.
This shows that there exist always at least a set of solutions with all non-vanishing c j , parametrized by a torus of phases. In [46] it has been shown that indeed, for the case of SU (N ) in the smallest irreducible representation, there is a further set of deformations that has been called a moduli space. This is a very difficult task to be investigated in general and we will not consider it here.

On the physical meaning of the time-dependence in the ansatz
It is worth to discuss the physical meaning of the time-dependent ansatz in Eq. (2.1) for the Lasagna-type configurations as well as the one in Eqs. (3.1), (3.2) and (3.3) for the spaghetti-type configurations. First of all, despite the time-dependence of the ansatz of the U field, the energy-momentum tensor is still stationary (so that it describes a static distribution of energy and momentum). This approach is inspired by the usual time-dependent ansatz that is used for Bosons stars [58,59] (and generalize it to arbitrary Lie group) in which the U (1) charged scalar field depends on time in such a way to avoid the Derrick theorem (see [60]). Secondly, the peculiar time-dependence is chosen in order to simplify as much as possible the field equations without loosing the topological charge (as, until very recently, the Skyrme field equations have always been considered a very hard nut to crack from the analytic viewpoint). Thirdly (as it will be discussed in the next sections on the minimal coupling with Maxwell), the present ansatz (both for lasagna and spaghetti type configurations) produces U (1) currents associated to the minimal coupling with Maxwell with a manifest superconducting current. Indeed (as it is clear from Eqs. (6.69) and (6.130)), the present U (1) current always has the form where Ω depends on either the Lasagna or the spaghetti profiles (see Eqs. (6.69) and (6.130)) while Φ is a field which is defined modulo 2π. Consequently, the following observations are important.
1) The current does not vanish even when the electromagnetic potential vanishes (A µ = 0).
2) Such a "left over" is maximal where Ω is maximal (and this corresponds to the local maxima of the energy density: see Eqs. (6.69) and (6.70).
3) J (0)µ cannot be turned off continuously. One can try to eliminate J (0)µ either deforming the profiles appearing in Ω integer multiples of π (but this is impossible as such a deformation would kill the topological charge as well) or deforming Φ to a constant (but also this deformation cannot be achieved for the same reason). Moreover, as it is the case in [57], Φ is only defined modulo 2π. Consequently, J (0)µ defined in Eq. (2.33) is a superconducting current supported by the present gauged configurations.
These are the three of the main physical reasons to choose this peculiar time-dependent ansatz. On the other hand, it is worth to emphasize that the peculiar time-dependence we have chosen (for the reasons explained above) prevents one from using the usual techniques (see, for instance, [13]) to "quantize" the present topologically non-trivial solutions. In particular, the typical hypothesis of a static SU (N )-valued field U is violated in our case (since, as it has been already emphasize, the requirement to have a static T µν which describes a stationary distribution of energy and momentum does not imply that U itself is static). Therefore, to estimate the "classical isospin" of the present configurations we will proceed in a different manner in the next sections.

Energy and Baryon number
The energy of these solutions can be easily computed by means of Proposition 6 in Appendix B. We get and where σ depends on the representation and has to be chosen so that the solution correctly covers a cycle when m = 1 and φ varies from 0 to 2π. To specify it, let us investigate the Baryon number integral. To this hand, let us look better at Proposition 2. The fact that n = 1 or 2 obviously distinguishes the SO(3)-type solutions from the SU (2)-type ones (see [46]), since only in the first case the period remains invariant when passing to the adjoint representation. The right ranges are then understood by considering the correct Euler parameterizations for SO(3) and for SU (2). If we write it generically as one finds that, if T is the period of the exponential functions, in both cases z must vary in a period and y in a range of T /4. The difference is in x, which has to vary in a period for SO(3) and half a period for SU (2), for example, see Appendix C in [46]. If we set x = σφ, y = r and z = mγ and we want to normalize the ranges of the coordinates φ, r, γ, so that all vary in [0, 2π], we see that we always have to require where n is an integer. With these conventions we can state the following proposition.

Proposition 3. The Baryonic topological charge is
The proof is exactly the same as in Appendix F of [46], so we omit it. The energy per Baryon g = E/B is therefore

Spaghetti groups
Another kind of configurations is obtained by starting from a different ansatz, which leads to Spaghetti like solutions. Spaghetti can be parameterized by the following ansatz: In the ansatz, T i are matrices of a given representation of the Lie algebra of G and are required to define a three dimensional subalgebra that we can choose to normalize so that and satisfy where I G,ρ is the Dynkin index of su (2) in G (see [38]), that is the coefficient relating the trace product in the representation ρ of Lie(G) to the Killing product of su (2). We also define Together with τ 1 , they satisfy With these rules, we get for L µ = U −1 ∂ µ U : For the other terms, set α = Θ, Φ and using we get and This shows that the expression of the L µ is universal (depend only on the algebra of the τ j ), so the field equations are always the same for any choice of the group. These are What is expected to change is just the topological charge and the energy. Given this universality property, we see immediately that, for any given group G, these kind of solutions are classified by all possible ways of finding a three dimensional simple subalgebra of the lie algebra g. Luckily, we don't need to tackle such a program, since has already been solved by E. B. Dynkin in [38], chapter III. This work as follows.
First, it is convenient to complexify the algebra, recombine and normalize the generators f, e + , e − of the subgroup so that Each complex three dimensional simple algebra is isomorphic to this. However, we must consider as equivalent only the ones which are isomorphic through an automorphism of the group. Let α j , j = 1, . . . , r be simple roots defined from a cartan subalgebra containing f . Then, it results that (α j |f ) must be integer numbers that can assume only the values 0,1,2. The set of numbers d j = α j (f ) are called the Dynkin characteristic of the subgroup. The main result of [38] is that the three dimensional simple subalgebras A are in one to one correspondence with the characteristics and one can indeed classify the characteristics. A subalgebra is said to be regular if its roots are indeed roots of g. The subalgebra A is said to be integral if the projection of the roots of g along the direction of the roots of A are integer multiples of the simple root α A of A. Since α A (f ) = 2, we see that the dual of α A in the Cartan subalgebra H is From this it follows immediately that A is integral if and only if all the numbers of the Dynkin characteristic χ A = (d 1 , . . . , d r ) of A are even (so are 0 and 2). All inequivalent characteristics for the exceptional Lie groups are listed in [38]. Furthermore, given such a characteristic χ = (d 1 , . . . , d r ), there it is explained how to construct explicitly the associated subalgebra.
and choose p j so that α j (f ) = d j . This gives From this we get As usual, C G is the Cartan matrix. In general, the construction of the remaining generators is non-trivial. To do it, one has to consider the subset of the root system Σ defined by Then, all roots are positive. If λ α are the corresponding eigenmatrices (normalized so that Tr(λ α λ α ) = −1), one then has to look for real coefficients k α such that, setting then [e + , e − ] = −if . If χ G is an admissible characteristic, then in general there are infinite solutions, but we know that are all equivalent so it is sufficient to choose one, all the other ones being related to it by conjugation with elements of the group. Notice that the resulting equations are in general where we used that any positive root can be written as with n β,j non-negative integers, and In the particular case when d j = 2 for all j, Σ χG consists of all simple roots and the solution is easily obtained as Finally, we can go back to our real case by taking Notice that this is the same construction we used to get a periodic generator κ for the Lasagna configurations. This also shows that indeed we can construct a κ matrix for each three dimensional subalgebra.

Energy density and Baryon charge
Let us determine the energy density and the Baryon charge. The energy density is defined by the T tt component of the energy-momentum tensor A direct computation gives I G,ρ is the Dynkin index and can be computed as follows. First, observe that a generic root has the form By using (3.22), (3.28) and the definition of Σ χG , we get Therefore, and so The Baryon charge can be written as in which ρ B is the Baryonic density charge Recalling the ranges (1.5) for the coordinates and that q = 2v + 1 and χ(0) = 0, we get The boundary conditions on χ(r) depend on the periodicity of τ 1 , which corresponds to the periodicity of T 3 (T 3 = τ 1 (Θ = π)). We must have χ(2π) = nπT G,ρ , so that where T G,ρ = 1 for representations with even dimension and T G,ρ = 2 for representations with odd dimension.

On the "classical" isospin of these configurations
We have shown in previous sections that the inclusion of a suitable time-dependence in the ansätze, both for lasagna and spaghetti phases (see Eqs. (2.1) and (3.1)), is one of the key ingredients that allows the field equations to be considerably reduced, leading to a single integrable ODE equation for the profiles. This timedependence offers a nice short-cut to estimate the "classical Isospin" of the configurations analyzed in the present paper (a relevant question is whether or not the classical Isospin is large when the Baryonic charge is large). In particular, one may evaluate the "cost" of removing such time-dependence. Such a cost is related to the internal Isospin symmetry of the theory. This is like trying to estimate the angular momentum of a spinning top by evaluating the cost to make the spinning top to stop spinning. In the present case, the time-dependence of the configurations can be removed from the ansätze by introducing a Isospin chemical potential; then the isospin chemical potential needed to remove such time-dependence is a measure of the classical Isospin of the present configurations. We will see how this works for the simplest SU (2) case, where the generators are T j = iσ j , being σ j the Pauli matrices (general group G behave in a similar way).
As it is well known, the effects of the Isospin chemical potential can be taken into account by using the following covariant derivative Now, we will use exactly the same ansatz as before in the spaghetti SU (2) case, but this time without the time dependence: together with the introduction of the Isospin chemical potential in Eq. (3.46) in the theory. One can check directly that the complete set of Skyrme equations can still be reduced to the same ODE for the profile χ (r) in the case of the spaghetti phase in Eq. (3.18) only provided the Isospin chemical potential for the spaghetti phase is given byμ In other word, the cost to eliminate the time-dependence is to introduce an Isospin chemical potential which is large when the Baryonic charge of the spaghetti is large. Something similar happens in the case of the lasagna phase. Let us consider the ansatz in terms of the Euler angles but without the time-dependence for the SU (2) case: Let us introduce the Isospin chemical potential, demanding that the profile h(r) should be the same as before. Then, as in the spaghetti case, the Skyrme field equations with chemical potential can still be satisfied by the very same profile h(r) provided we fix the Isospin chemical potential as At this point it is important to remember that in the SU (2) case the lasagna and spaghetti type solutions have the following values for the topological charges see [25] and [26] for more details. These arguments show that the "classical Isospin" of configurations with high Baryonic charge is large. Finally, it is important to point out that the large Isospin case corresponds to either neutron rich or proton rich matter and due to Coulomb effects (not taken into account in this model), the neutron rich solution is preferred. This fact is very convenient as far as the physics of neutron stars is concerned.

Examples: exceptional pasta
As an example we can consider the "basic exceptional Skyrmions", that are solutions in lowest dimensional representation when G is one of the exceptional Lie groups. There are five cases that we now recall according to the dimension of the group. For each of them we know all inequivalent three dimensional subalgebras, each one determined by the Dynkin characteristic χ I (d 1 , . . . , d r ), where I is the Dynkin index and d j are the coefficients of the characteristic, ordered as the simple root listed in Appendix A.
The smallest exceptional group is G 2 , a 14 dimensional group of rank 2 whose smallest irrep is 7 dimensional. There are four different three dimensional subalgebras. It contains four 3D subalgebras, having characteristics χ 1 and χ 2 are regular but not integral, while χ 4 and χ 28 are not regular but are integral. In particular, the minimal regular subalgebra containing χ 4 is χ 1 ⊕χ 3 , while χ 28 is maximal so that the smallest regular subalgebra containing it is G 2 itself.
As an example, we will finally construct the explicit solutions for G 2 , which we can call "G 2 exceptional pasta".

G 2 exceptional Spaghetti
Here we consider explicit solutions case by case. Our deduction will be quite general and independent on the specific realization in terms of matrices, but just on the chosen representation. Nevertheless, for sake of completeness, in Appendix D we will provide an explicit matrix realization of the subalgebras in the lowest fundamental representation.

χ 28 -Spaghetti
This is the principal case, with χ 28 = (2, 2). Therefore p ≡ (p 1 , p 2 ) = (36,20). We already know the solution in this case. The Spaghetti solution is Because of Proposition 2, we already know that working in the adjoint the solution is of SO(3)-type. In the representation 7 7 7, it is sufficient to verify that for T − = 3λ 1 + √ 5λ 2 , and v the maximal vector of 7 7 7, then the vectors ρ k 7 7 7 (T − )(v), k = 0, . . . , 6 are all linearly independent. Here ρ 7 7 7 : G 2 → End(R 7 ) is the representation map of the algebra. This is proved in Appendix D and proves that R 7 is irreducible under χ 28 . Since it is odd dimensional, it is of SO(3)-type.

G 2 exceptional Lasagna
For the exceptional Lasagna we can use Proposition 2. Since we already know that n = b must be equal to 1, we get that (p 1 , p 2 ) = (18,10), and, if we fix ψ j = 0 for simplicity, then Moreover, from Proposition 1 we get .

Extended ansatz
In order to allow for further generalizations, it is convenient to employ the Euler parameterization in a more general ansatz, after fixing the matrices κ and f . Let us consider the Skyrmionic field 1 where κ is specified in (2.2), and f has the same properties as in (2.21). One of the aim of this generalization is to provide a description of different pasta states without specifying them a priori. This could lead to a comprehensive description of Skyrmions in a finite volume and to an analytical definition of other possible states (such as gnocchi states) and the transitions between them. In this work, we did not analyze all these possibilities and all the limits of this models, but we outline the main properties which characterize them, namely the wave equations, the topological charge and the energy density. If we define then U (t, r, φ, γ) = e −α(t,r,φ,γ)κ e ξ(t,r,φ,γ)κ e χ(t,r,φ,γ)f e ξ(t,r,φ,γ)κ e α(t,r,φ,γ)κ . and f can be normalized so that tr(f 2 ) = tr(κ 2 ). (5.8) This leads to the condition (2.24) and, in particular, |c j | 2 = b 2 p j . Using these conventions we can now write the Skyrme equation explicitly.

Non-linear wave equations
We call wave equations to the field equations for the functions α, ξ and χ. These result to be (5.11)

Baryon charge
The Baryon charge is with ρ B = −12 c 2 ε ijk ∂ i α∂ j ξ∂ k cos(bχ). (5.14) Up to now we have just written local expressions, but in order to compute the Baryonic charge it is necessary to define the ranges of α, ξ and χ. Proposition 2 tells us that the period of e gκ is T κ = ηn 2π b , where η = 1, 2 depending on the representation, while n ∈ Z. Following [46], the ranges must be where σ = 1 for odd-dimensional representations and 1 2 for even-dimensional representations and m, n are both integer. The integration of the density charge leads to We can compute the ratio c 2 b 2 in the following way. From (5.8), we get where the definition f = r j=1 p j h j has been used. Now, we can replace the coefficients p j with (2.23) and Tr(h j f ) = α j (f ) = ib to get The Baryon charge takes the form

Example: the Lasagna case
Let us now compare the results obtained in this section with the previous ones. Our quantities can be written in terms of the Lasagna ansatz as follows Moreover, the profile function depends only on the parameter r (χ = χ(r)). This leads to the following relations With these choices the equations (5.10) and (5.11) are automatically satisfied. The equation (5.9) becomes which leads to the solution where the boundary conditions χ(0) = 0 and χ(2π) = π b have been used. Now, it is easy to compute the energy density, which results The integration over the volume of the box gives the total energy of the Lasagna (5.24)

Coupling with U (1) gauge field
By employing the generalization presented in the previous section, it is now easy to couple the Skyrmion field to an electromagnetic field A µ . To this aim we introduce the action where and the hat stands for the replacement of the partial derivative with a covariant derivative which means thatL Here T is any element of the Lie algebra of the group G, representing the direction of the U (1) gauge field. Later, we will identify T with the generator T 3 . The action (6.1) is now invariant under gauge transformation The gauge invariance appears also in the fact that the theory depends on A µ through the quantity ∂ µ α − A µ , which is invariant for gauge transformations.

Covariant Baryonic charge
As in [49], in order to determine a topological invariant, one is tempted to start directly generalizing (1.7) to the expressionB which, however, is not a topological invariant if the field-strength F is non-vanishing. Nevertheless, a topological invariant can be constructed after a simple subtraction, even for a non-Abelian gauge field. Indeed, we have: Proof. In order to prove the proposition, we have to prove that the first variation ofB w.r.t. U and ω (independently) vanishes at any functional point, that is independently if U and ω are constrained by some equations of motion. Notice that in taking variations, δω is a well defined 1-form on S despite ω could not be. To keep notation compact we will use bold round brackets to indicate a trace ( ( (M) ) ) ≡ Tr(M). Moreover, we first recall the following properties. If a j , j = 1, . . . , k are Lie algebra valued 1-forms then ( ( (a 1 ∧ · · · ∧ a k−1 ∧ a k ) ) ) = (−1) k−1 ( ( (a k ∧ a 1 ∧ · · · ∧ a k−1 ) ) ).
Notice that with our conventions in (6.4), we have to make the identifications  (6.43) where, according to the conventions in [46], the orientation of the coordinates is such that ε rγφ = 1. In our case S is a closed three dimensional manifold in a semisimple compact Lie group G. Since in this case H 2 (G, Q) = 0, we get that the correction to the density does not contribute to the integral and we expectB = B always. It is worth to mention that in the construction of the solutions of the Skyrme equations, however, S is replaced by V that is compact but it is not a closed smooth manifold but a hyperrectangle with boundary. Therefore, the above integral does not define a topological invariant unless we impose suitable boundary conditions. To understand which are the most suitable ones, let us first analyze the case A = 0. In this case the map U maps the hyperrectangle in a closed smooth submanifold of G, soL is the pull-back of a 1-form well defined on a closed compact manifold (indeed, the left-invariant Maurer-Cartan form) and this is the reason we get a topological invariant. This suggest the boundary conditions we are looking for. They have to be imposed so that also A is the pull-back of a well defined 1-form over G (or the image of the hyperrectangle in G). Under these conditions, the quantities A µ are not independent, due to the fact that Φ, Θ and χ defines a map M : R 3+1 → R 3 . Locally, the embedding takes the form

Example: Lasagna states coupled to an electromagnetic field
To be explicit, we now work out the example of Lasagna states. For this case we choose T = κ. The covariant derivative determines the coupling of the gauge field to the Skyrmions, which appears in the definition ofL μ Notice that the introduction of the gauge field in the direction κ causes a shift in L µ given by ∂ µ α → ∂ µ α − A µ . It results that all the quantities we computed in the previous section are shifted by this quantity when the Skyrmions are coupled to a Maxwell field and it is really easy to convert the uncoupled theory with the coupled one. The covariant Baryon density charge now becomeŝ where ρ B is the uncoupled density. The correction to ρ B is a total derivative, so it depends only on the boundary conditions, as discussed above. Differently from [49], our system lives in a box, so, the electromagnetic field is not constrained to zero at the boundaries. Therefore, the Baryonic charge is not necessarily a topological invariant and not even expected to be an integer. As we said above, we can fix this problem by requiring for A to be the pull-back of a well defined potential over the homology cycle of G selected by the map U . This is easily accomplished by looking at the form of the ansatz for the Lasagna states. As t is irrelevant, we fix t = 0 to simplify the expressions: U (r, γ, φ) = e −φσκ e χ(r)f e mγκ = e −φσκ e mγκ(r) e χ(r)f . (6.50) Since bχ(2π) = π, we see thatκ(0) = −κ(2π) = κ, so that, if, for a generic fixed r, U (r, γ, φ) defines a two dimensional surface in G, for r = 0, 2π it collapses down to one dimensional circles: This degeneration means that well defined 1-forms on the whole manifold must have components only along the direction on the degeneration submanifolds, which in our case means A α (r = 0) = 0, (6.53) A ξ (r = 2π) = 0, (6.54) which in the original coordinates becomes 1 2m Also, one between φ and γ has to be identified periodically, while the other one is periodic or "antiperiodic" 3 according to the cases if the cycle is of SO(3) or SU (2), respectively. Therefore, in any case, the 1-forms in the image of the embedding have to be periodically identified so that the integrals at the "boundaries" φ = 0 and φ = 2π cancel out and the same happens for the boundaries at γ = 0 and γ = 2π. So, the only boundaries that may contribute are the ones at r = 0 and r = 2π, which we collectively call ∂ r B. Therefore, the Baryonic charge resultŝ (σA γ (0, γ, φ) + mA φ (0, γ, φ))dγdφ = B (6.57) because of the above boundary conditions, and we used that A α = σA γ + mA φ .

Decoupling of Skyrme equations and free-force conditions
To the Skyrmion equation coupled to a Maxwell field, obtained by shifting ∂ µ α → ∂ µ α − A µ in (5.9), (5.10) and (5.11), we have to add the Maxwell equations, which are given by In the generic Euler parameterization, they become To look for explicit solutions, we aim to decouple the Skyrme equations from the Maxwell field. Since A µ appears in the products ( , we can separate the Skyrme equations from the rest by looking for solutions where these terms are a priori fixed functions Recall that, the quantity ∂ µ α − A µ is gauge invariant.

Example: Lasagna, again
We can use the results of this section in order to study the behavior of Lasagna when coupled to the U (1) gauge field. To simplify the results, we use the free-force conditions and for the gauge field we make the ansatz If we simply shift the gauge field, we can writẽ These conditions are easily solved by using (5.20) (which also apply toÃ µ ) together with (6.72). This leads to ; so, only one gauge field results to be independent, for instance we can takeÃ φ . Thus, the wave equations and the Maxwell equation become The first equation can be rewritten as d dr where M is an integration constant. This determines the boundary values of χ ′ from the ones of χ Vice versa, we can write M in terms of χ ′ (0) We can put this result into the Maxwell equations, getting (6.83) From (6.78) we can locally write r in terms of χ as which leads to a definition ofÃ φ in terms of χ, let us call B(χ) =Ã φ (r(χ)). This way, equation (6.82) can be entirely written in terms of χ where a prime indicates derivative with respect to χ, and the following quantities have been introduced with We can use we have that x must vary in the interval [0, π/2]. Also, we have seen that for the Lasagna states such interval must be one quarter of the period over the cycle, which means that the solution we are looking for must be periodic with period 2π as a function of x. Therefore, it is worth mentioning the following result [44] (see also [50]): Proposition 5. Let y 1 (x) and y 2 (x) be the solutions of the Hill equation where f (x) is an even function of the form for κ an arbitrary constant. The question on the existence of such solutions is investigated in [44]. For example, the existence of solution y (2) is granted if and only if ω takes values for which the determinant of the infinite dimensional matrix vanishes. 4 However, for any practical purposes, such a way is impracticable for looking for explicit solutions in this very general case. Therefore, in place of pursuing this very general analysis, we move now to a particular but more tractable case.

Linear solution of the Skyrme equations
In the particular case when M = 2 b 2 λ , we can find a very simple solution: which is a Whittaker-Hill equation [51,45], see also [52]. It is convenient to introduce the variable changer = r 4 , sor has range 0 ≤r ≤ π 2 . This way, equation (6.102) takes the canonical form We can therefore determine the solutions y (2) and y (3) , following [45]. Using the same notations of that paper, we can identify In particular, η = −ωρ. For the function φ (3) we have to take (see [45]) A n B n cos((2n + 1)x), (6.108) (ρ + 2ji), j > 0, (6.109) and B n solves the recursion relations To find the periodic solution of period 2π, ω and η must be constrained by the following trascendental equation, expressed in terms of a continued fraction (see [45], formula (5.1)) which then gives the solution Finally, B 0 is fixed by the condition φ (3) (0) = 1.
As what concerns the solution κφ (2) (x), we have to consider (ρ + (2j + 1)i), j > 1, (6.116) and D n solves the recursion relations To find the periodic solution of period π, ω and η must be constrained by the following trascendental equation, expressed in terms of a continued fraction (see [45], formula (5.3)) which then gives the solution Since κ is arbitrary, no normalization is required for D 1 . However, notice that κφ (2) (x) can be considered only if equation (6.119) has common solutions with (6.112).

Example: Spaghetti states coupled to an electromagnetic field
In the case of Spaghetti, we do not use a parameterization of Euler type but the exponential parameterization of Section 3. Still, the analysis can be easily extended to this case. Following [53], the gauge field is described by Also in this case, the free-force conditions decouples the Skyrme equations from the Maxwell field. In particular, we take A reasonable explicit form of a gauge field with these properties is given by θ)).  which means that 2χ ′2 λq 2 sin 2 χ 2 + L 2 θ + 4q 2 L 2 r cos χ = 2Z, (6.128) where Z is a constant, which is equivalent to This is a stationary Schrödinger equation with a double periodic potential of finite type. In particular, here one is interested in the zero eigenvalue case. Both the direct and inverse problem for this kind of equation is well studied and much more involuted than the one dimensional case (already highly non-trivial). Here we simply defer the reader to the literature (see [54] and reference therein), and limit ourselves to discuss the boundary conditions. As we discussed in the previous sections, the boundary conditions on A µ are outlined by the behavior of the Skyrme field in the edges of the box, namely (once again, we fix t = 0) This requires the following constraints The contribution of the gauge field to the Baryonic density can be always computed from (6.43). This giveŝ where L r and L θ are specified in (3.9) and (3.16). We easily find U −1 T 3 U = T 3 + sin(qθ)τ 2 − sin(qθ) cos χτ 2 + sin(qθ) sin χτ 3 .

A Roots of simple algebras
Here we list the roots of all simple algebras.

A.1 A N
The corresponding complex algebra is sl(N + 1), while the compact form is su (N + 1). If e a , a = 1, . . . , N + 1, is the canonical basis 5 of R N +1 , then the real linear space generated by the roots is isomorphic to a hyperplane in R N +1 in which all non-vanishing roots are represented by the vectors α j,k = e j − e k , j = k. The simple roots are α j = e j − e j+1 , j = 1, . . . , N . If λ j are the root matrices corresponding to the simple roots, then The split subalgebra is so(N + 1).

A.2 B N
The corresponding compact form is so(2N + 1). The real linear space generated by the roots is isomorphic to R N . If e a , a = 1, . . . , N , is the canonical basis of R N , then all non-vanishing roots are represented by the vectors e j − e k , j = k, ±(e j + e k ), j < k, ±e j . The simple roots are α j = e j − e j+1 , j = 1, . . . , N − 1 and α N = e N . If λ j are the root matrices corresponding to the simple roots, then The split subalgebra is so(N ) ⊕ so(N + 1).

A.3 C N
The corresponding compact form is us p (2N ), the compact symplectic algebra. The real linear space generated by the roots is isomorphic to R N . If e a , a = 1, . . . , N , is the canonical basis of R N , then all non-vanishing roots are represented by the vectors e j − e k , j = k, ±(e j + e k ), j < k, ±2e j . The simple roots are α j = e j − e j+1 , j = 1, . . . , N − 1 and α N = 2e N . If λ j are the root matrices corresponding to the simple roots, then The split subalgebra is u(N ).

A.4 D N
The corresponding compact form is so (2N ). The real linear space generated by the roots is isomorphic to R N . If e a , a = 1, . . . , N , is the canonical basis of R N , then all non-vanishing roots are represented by the vectors e j − e k , j = k, ±(e j + e k ), j < k. The simple roots are α j = e j − e j+1 , j = 1, . . . , N − 1 and α N = e N −1 + e N . If λ j are the root matrices corresponding to the simple roots, the relevant non-vanishing commutators are The split subalgebra is so(N ) ⊕ so(N ).

A.5 G 2
The corresponding compact form is g 2 . The real linear space generated by the roots is isomorphic to a hyperplane in R 3 . If e a , a = 1, . . . , 3, is the canonical basis of R 3 , then all non-vanishing roots are represented by the vectors e j − e k , j = k, and ±(e 1 + e 2 + e 3 − 3e s ), s=1,2,3. The simple roots are α 1 = e 2 − e 3 and α 2 = e 1 − 2e 2 + e 3 . If λ j are the root matrices corresponding to the simple roots, the relevant non-vanishing commutator is [λ 1 , λ 2 ]. The split subalgebra is so(4).

A.6 F 4
The corresponding compact form is f 4 . The real linear space generated by the roots is isomorphic to R 4 . If e a , a = 1, . . . , 4, is the canonical basis of R 4 , then all non-vanishing roots are represented by the vectors e j − e k , j = k, ±(e j + e k ), j < k, ±e j , 1 2 (±e 1 ± e 2 ± e 3 ± e 4 ). The simple roots are α 1 = e 2 − e 3 , α 1 = e 3 − e 4 , α 3 = e 4 , and α 4 = 1 2 (e 1 − e 2 − e 3 − e 4 ). If λ j are the root matrices corresponding to the simple roots, the relevant non-vanishing commutators are The split subalgebra is us p (6) ⊕ us p (2).

A.7 E 6
The corresponding compact form is e 6 . The real linear space generated by the roots is isomorphic to R 6 . If e a , a = 1, . . . , 6, is the canonical basis of R 6 , then all non-vanishing roots are represented by the vectors ±(e j − e k ), j < k < 6, ±(e j + e k ), j < k < 6, where in the parenthesis only an even number of minus signs can appear. The simple roots are If λ j are the root matrices corresponding to the simple roots, the relevant non-vanishing commutators are The split subalgebra is us p (8).

A.8 E 7
The corresponding compact form is e 7 . The real linear space generated by the roots is isomorphic to R 7 . If e a , a = 1, . . . , 7, is the canonical basis of R 7 , then all non-vanishing roots are represented by the vectors ±(e j − e k ), j < k < 7, ±(e j + e k ), j < k < 7, ± √ 2 e 7 , where in the parenthesis only an odd number of minus signs can appear. The simple roots are If λ j are the root matrices corresponding to the simple roots, the relevant non-vanishing commutators are The split subalgebra is su(8).
A.9 E 8 The corresponding compact form is e 8 . The real linear space generated by the roots is isomorphic to R 8 . If e a , a = 1, . . . , 8, is the canonical basis of R 8 , then all non-vanishing roots are represented by the vectors ±(e j − e k ), j < k, ±(e j + e k ), j < k, If λ j are the root matrices corresponding to the simple roots, the relevant non-vanishing commutators are The split subalgebra is so(16).

A.10 Resuming
In conclusion, we see that the commutators we need are strictly related to the Dynkin diagram of the algebra: a commutator between eigenmatrices of two simple roots is non zero only if the roots are linked, that is if the scalar product is not zero. This is simply related to the fact that, with obvious notation, the commutators [λ αi , λ αj ] or is an eigenmatrix for α i + α j , or it vanishes. We also recall here some very well known facts. The fact that Dynkin diagrams have no loops allows to choose the normalization of the matrices λ j so that t.i. [λ αj , λ α k ] = −(α j |α k )λ αj +α k if j < k and with the opposite sign if we change j and k. Here (|) is the scalar product in the space of roots. Notice that also Remember that the trace product is proportional to the Killing product and that the only non-Killing orthogonal root spaces are the ones corresponding to opposite roots. This allows to fix a global normalization so that We also have, for the simple roots α j , where J j are in a Cartan algebra. From the fact that the simple roots are linearly independent, it easily follows that the J j , j = 1, . . . , r, are a basis for the Cartan subalgebra. This is also sufficient to fix the scalar product in the space of roots so that (α j |α k ) = α j (J k ), (A. 13) if the roots are defined as for any h ∈ H. In particular, using ad invariance of the trace product we get Finally, recall that any given simple Lie algebra is characterized by the r × r Cartan matrix With this we can rewrite the normalization conditions as Tr(λ j λ k ) = −δ jk , (A.17) The Cartan matrices of simple Lie groups can be found, for example, in [55], Table 6.

B A proposition
Proposition 6. Let κ = r j=1 (c j λ j + c * jλ j ), and h ∈ H a matrix such that α j (h) = ε j a, where ε j is a sign, j = 1, . . . , r, and set x := e −hz κe hz , z ∈ R. Then and Proof. Let us start with to get where we used that α j (h ′ ) 2 = (ε j a) 2 = a 2 , and that the only non-vanishing traces are Tr(λ m λ n ) = −δ mn . This proves (B. There are different ways of constructing a convenient basis for the Lie algebra of G 2 . We will refer to [56]. In that notation a basis is C J , J = 1, . . . , 14. The only maximal regular subgroup is SO(4) generated by C L , L = 1, 2, 3, 8, 9, 10. The remaining matrices generate p. A convenient Cartan subspace is thus As a basis, we take h 1 = C 11 and h 2 = C 5 . One can easily diagonalize the action of ad(H). If we set To keep contact with our conventions we have to redefine the basis for H as J 1 and J 2 , defined by (notice that λ j is simply the complex conjugate of λ j ) This gives us the geometry of roots, so that We can represent this vectors in the canonical euclidean R 2 as the vectors The corresponding Cartan matrix is with inverse (C G ) −1 = 2 3 1 2 . (D.14) It is also useful to determine the basis for all eigenspaces, in a convenient way, normalized so that Tr(λ α λ α ) = −1 for any root. After setting α 3 = α 1 + α 2 , α 4 = 2α 1 + α 2 , α 5 = 3α 1 + α 2 , α 6 = 3α 1 + 2α 2 , (D. 15) we can state the following proposition.

Proposition 7.
A suitable choice of the eigenmatrices associated to the roots α j , j = 3, 4, 5, 6, is given by Moreover,λ j is the complex conjugate of λ j .
Proof. We know from the general theory that if λ a and λ b are eigenmatrices of the roots α a and α b respectively, and if α a + α b is also root, than the eigenmatrices of α a + α b have the form µ[λ a , λ b ] for any given constant µ. Since λ 1 and λ 2 are eigenmatrices for the fundamental roots α 1 and α 2 , we have that the matrices λ j specified above are surely eigenmatrices for the corresponding roots α j , j = 3, 4, 5, 6. We have only to explain the choices of the constant factors. These are chosen to be real and such that Tr(λ jλj ) = −1. To prove it, first notice that necessarily [λ i ,λ j ] are eigenmatrices for −α i − α j , so we can identifyλ 3 = √ 2[λ 1 ,λ 2 ] and so on. The last part of the proposition then follows from the fact that it is true for λ 1 and λ 2 and that all the coefficient we chosen for defining the remaining λ j are real. Then, using ad-invariance of the trace product, that is Tr( where we used again that [λ 1 ,λ 2 ] = 0, and, therefore, Tr([λ 1 , λ 3 ][λ 1 ,λ 3 ]) = 2(α 2 (J 1 )) 2 Tr(λ 2 λ 2 ) = −2(−1/2) 2 , (D. 22) and putting all together we get Tr(λ 4λ4 ) = −1.
The remaining two cases are proved exactly in the same way.

D.3 Explicit matrix realizations
Here we present the explicit matrix representation of the three dimensional subalgebras in the irrep 7 7 7 of G 2 . These are  6 We could compute it explicitly, but it is not necessary for our purposes.