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The Unified Standard Model

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Abstract

The aim of this work is to find a simple mathematical framework for our established description of particle physics. We demonstrate that the particular gauge structure, group representations and charge assignments of the Standard Model particles are all captured by the algebra M(8, \(\mathbb {C})\) of complex \(8 \times 8\) matrices. This algebra is well motivated by its close relation to the normed division algebra of octonions. (Anti-)particle states are identified with basis elements of the vector space M(8, \(\mathbb {C})\). Gauge transformations are simply described by the algebra acting on itself. Our result shows that all particles and gauge structures of the Standard Model are contained in the tensor product of all four normed division algebras, with the quaternions providing the Lorentz representations. Interestingly, the space M(8, \(\mathbb {C})\) contains two additional elements independent of the Standard Model particles, hinting at a minimal amount of new physics.

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Notes

  1. This GUT is often referred to as “SO(10)” by physicists. However, the Lie group used in the model is indeed Spin(10), the double cover of SO(10).

  2. Note that the real numbers are also taken into account since a tensor product with \(\mathbb {R}\) does not alter the structure of the algebra and thus \(\mathbb {D}= \mathbb {R}\otimes \mathbb {C}\otimes \mathbb {H}\otimes \mathbb {O}\).

  3. In fact, complex conjugation will also flip the spin, such that the complex conjugate of a right-handed Weyl spinor with spin up is a left-handed Weyl spinor with spin down.

  4. The associative algebra \(\mathbb {H}\), is isomorphic to the algebra of maps \(\overleftarrow{\mathbb {H}}\cong \mathbb {H}\). Both \(\overleftarrow{\mathbb {O}}\) and \(\mathbb {H}\) are thus subalgebras of \(\overleftarrow{\mathbb {D}}\cong \mathbb {C}\otimes \mathbb {H}\otimes \overleftarrow{\mathbb {O}}\). Moreover, \(\overleftarrow{\mathbb {O}}\) is equivalent to the algebra of adjoint actions on the octonions [18]. It follows that \(\overleftarrow{\mathbb {D}}\) is the algebra of adjoint actions on \(\mathbb {D}\).

  5. Note that there is an overall factor of \(\mathbb {C}\) in this decomposition. By (1) the same complex field enters into the Lorentz representations discussed in Appendix B. However, we are here only focusing on a subalgebra of this tensor product, namely \(\mathbb {C}\otimes \mathbb {O}\). As such, we cannot currently comment on any physical significance of the overall factor of \(\mathbb {C}\) in the decomposition. It is known [20], that for vector fields such an overall \(\mathbb {C}\) corresponds to the combination of hermitian and anti-hermitian vectors in \(\mathbb {C}\otimes \mathbb {H}\).

  6. These basis vectors also satisfy \({R_I}^{\bar{*}}\equiv \begin{pmatrix} 0 &{}\eta \\ \eta &{}0 \end{pmatrix} {R_I}^* =R_{I+4}\), where \(\eta =\mathrm {diag}(1,-1,-1,-1)\). The \(\bar{*}\) is an operation in M(8,\(\mathbb {C}\)) that corresponds to complex conjugation in \(\mathbb {C}\otimes \overleftarrow{\mathbb {O}}\), as described in appendix A.2.

  7. Additionally, we will require that \((V^\pm _a)^{\bar{*}}=V^\mp _a\) which implies \(\left( a_{aI}^+\right) ^* = a_{a(I+4)}^-\).

  8. In fact, in the standard formulation of particle physics, such an action from the right is not even well defined: Particles in the fundamental representation of the gauge group are represented by column vectors and can only be acted upon by matrices from the left. Here, the situation is different since all our particles correspond to elements of the matrix algebra. Hence they can be multiplied by matrices both from the left and from the right.

  9. The complex conjugation here refers to complex conjugation in the Clifford algebra and hence to the \(\bar{*}\) operator on M(8,\(\mathbb {C}\)). This is discussed in detail in appendix A.2.

  10. If the symmetric matrix has has rank 1, it only has 1 non-vanishing eigenvalue and can therefore not be traceless. For the antisymmetric matrix \(m_A\), this can also be seen by noting that a rank-1 matrix can be written in terms of 2 vectors u and v as \(m_A=uv^\mathrm {T}\). But then antisymmetry implies \(uu^\mathrm {T}=0\) and thus \(u=0\).

  11. Even though \(\eta \) is the same as the Minkowski metric in Cartesian coordinates, this is just an artefact of how we chose to represent our basis elements, and not related to Lorentz transformations.

  12. Note that this is also the case for the \(\mathfrak {su}\)(3) generators. However since SU(3) transformations act from the left and commute with the projectors \(\mathcal {R}\) and \(\bar{\mathcal {R}}\) this would imply that defining similar operators \(\hat{\lambda }_I := \mathcal {R}\bar{\lambda }_I|1 - \bar{\mathcal {R}}\bar{\lambda }_I^{\bar{*}}|1 = \left( \bar{\lambda }_I-\bar{\lambda }^{\bar{*}}_I\right) |1=\lambda _I|1\), where we have used \(\bar{\lambda }_I\) as defined in (32).

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Acknowledgements

This work is supported by a grant from the Max-Planck-Society.

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Correspondence to Brage Gording.

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Appendices

Appendix A: The Matrix Algebra \(\text {M(8,}\mathbb {C})\)

1.1 A.1. The Clifford Algebra \(\mathbb {C}l(6)\)

We briefly review basic properties of the 6-dimensional Clifford algebra. All basis elements of the space \(\mathbb {C}l(6)\) can be generated by two sets of vectors \(\{\alpha _i\}_{i=1}^3\) and \(\{\alpha _i^\dagger \}_{i=1}^3\), where hermitian conjugation \({}^\dagger \) is an anti-automorphism, implying \(i^*=-i\) and \(\alpha ^* = -\alpha ^\dagger \). These generating vectors satisfy the anti-commutation relations,

$$\begin{aligned} \{\alpha _i,\alpha _j^\dagger \} = \delta _{ij},\quad \{\alpha _i,\alpha _j\}=0,\quad \{\alpha _i^\dagger ,\alpha _j^\dagger \}=0. \end{aligned}$$
(63)

We will here not work with the elements of the generating space, but rather with the full set of basis elements which span the space \(\mathbb {C}l(6)\). To this end we define \(\omega :=\alpha _1\alpha _2\alpha _3\) and the projectors

$$\begin{aligned} P_0 := \omega ^\dagger \omega ,\quad P_i:= \alpha _i\omega ^\dagger \omega \alpha _i^\dagger . \end{aligned}$$
(64)

Let moreover \(\bar{P}_a := P_a^*\) for all \(a\in \{0,1,2,3\}\). It follows from (63) that \(\omega \) is annihilated by right or left action of any \(\alpha _i\). The projector \(P_0\) was used in [19] to find the Standard Model structure associated to one generation of fermions, but to our knowledge this is the first use of the projections \(P_i\).

The above 8 projectors are linearly independent and split our space into 8 complex linearly independent subspaces. Specifically, each one of these projectors will define a left ideal. The space \(\overleftarrow{\mathbb {C}\otimes \mathbb {O}}\,P_b\) can then be spanned by 8 linearly independent basis vectors,

$$\begin{aligned} B_{ab} := \alpha _a\omega ^\dagger \omega \alpha _b^\dagger ;\quad A_{ab} := \alpha _a^\dagger \omega \alpha _b^\dagger \quad \quad a \in \{0,1,2,3\}. \end{aligned}$$
(65)

The basis vectors which span \(\overleftarrow{\mathbb {C}\otimes \mathbb {O}}P_b^*\) are found by taking the complex conjugate of (65), which we will denote by

$$\begin{aligned} \bar{B}_{ab} := \alpha _a^\dagger \omega \omega ^\dagger \alpha _b\quad \bar{A}_{ab} := \alpha _a\omega ^\dagger \alpha _b\quad \quad a \in \{0,1,2,3\}. \end{aligned}$$
(66)

This provides a compact way of writing all basis elements of \(\mathbb {C}l(6)\) in terms of \(B_{ab},\bar{B}_{ab},A_{ab},\bar{A}_{ab}\).

1.2 A.2. \(\mathbb {C}l(6)\cong \text {M(8,}\mathbb {C})\)

We will now demonstrate that \(\mathbb {C}l(6)\) is isomorphic to the algebra of \(8\times 8\) complex matrices \(\text {M}(8,\mathbb {C})\). Clearly the vector spaces over which the two algebras are defined are isomorphic by virtue of having the same dimension. We thus only need to show that the Clifford product in \(\mathbb {C}l(6)\) is identified with the matrix product in \(\text {M}(8,\mathbb {C})\).

Let \(\{M_{IJ}\}\) be a basis of \(\text {M}(8,\mathbb {C})\). We take one of their matrix entries to be equal to 1 (in the Ith row and Jth column), while all other entries are zero. A general matrix F in \(\text {M}(8,\mathbb {C})\) can then be written as \(F=\sum _{I,J} F^{IJ}M_{IJ}\). The matrix product expressed in this basis reads,

$$\begin{aligned} FH=\sum _{I,L}\left( \sum _JF^{IJ}H^{JL}\right) M_{IL}. \end{aligned}$$
(67)

Next we identify the basis \(M_{IJ}\) with the basis elements of \(\mathbb {C}l(6)\) via

$$\begin{aligned} M_{IJ} \longleftrightarrow \left\{ \begin{matrix} B_{(I-1)(J-1)} &{} \text {for } I,J\in \{1,2,3,4\} \\ A_{(I-5)(J-1)} &{}\ \qquad \qquad \text { }\text { for } I\in \{5,6,7,8\}\text {, }J\in \{1,2,3,4\} \\ \bar{A}_{(I-1)(J-5)} &{}\qquad \qquad \quad \text { }\text { for } I\in \{1,2,3,4\}\text {, }J\in \{5,6,7,8\} \\ \bar{B}_{(I-5)(J-5)} &{}\text {for } I,J\in \{5,6,7,8\} \end{matrix} \right. \end{aligned}$$
(68)

Under this identification, we can evaluate the Clifford algebra product of two basis elements and obtain,

$$\begin{aligned} M_{IJ} M_{KL} = \delta _{JK} M_{IL}. \end{aligned}$$
(69)

This reproduces precisely the standard matrix product of M(8,\(\mathbb {C}\)) in (67).

Hermitian conjugation of elements of \(\mathbb {C}l(6)\) correspond to the usual hermitian conjugation of matrices in \(\text {M}(8,\mathbb {C})\). Thus we will not distinguish between hermitian conjugation in the two algebras and label both the operations by \({}^\dagger \). Specifically, for any \(M\in \) M(8,\(\mathbb {C}\)) we have that \(M^\dagger := \left( M^*\right) ^\mathrm {T}\) where \({}^\mathrm {T}\) is the matrix transpose. The situation is different for the operation of complex conjugation. Due to the Clifford algebra property \(\alpha ^* = -\alpha ^\dagger \), complex conjugation acts on the \(\mathbb {C}l(6)\) basis elements (65) and (66) as,

$$\begin{aligned} \left( B_{ab}\right) ^* = \sum _{c,d} \eta _{ac}\bar{B}_{cd}\eta _{db},\quad \left( A_{ab}\right) ^* = \sum _{c,d} \eta _{ac}\bar{A}_{cd}\eta _{db}, \end{aligned}$$
(70)

where \(\eta \) is a diagonal matrix with entries \(\{1,-1,-1,-1\}\).Footnote 11

From (68) and (70), it is clear that, in the matrix representation, complex conjugation necessarily affects the index structure. Namely, matrices \(M\in \text {M}(8,\mathbb {C})\) satisfy,

$$\begin{aligned} M^{\bar{*}} := \begin{pmatrix} 0 &{}\eta \\ \eta &{}0 \end{pmatrix} M^* \begin{pmatrix} 0 &{}\eta \\ \eta &{}0 \end{pmatrix}, \end{aligned}$$
(71)

where, in order to distinguish complex conjugation in the two algebras, we have introduced the symbol \(\bar{*}\) to denote \(\mathbb {C}l(6)\) complex conjugation in the matrix representations. We continue to use \(^*\) to denote the conjugation of complex numbers.

As our matrix space can be written as the outer product of two vector spaces \(\mathbb {C}^{8}\), this implies that, for any \(V\in \mathbb {C}^8\), we have that,

$$\begin{aligned} V^{\bar{*}} \equiv \begin{pmatrix} 0 &{}\eta \\ \eta &{}0 \end{pmatrix} V^*. \end{aligned}$$
(72)

In assigning basis elements of \(\mathbb {C}l(6)\) to Standard Model particle types we use the matrix representation \(\text {M}(8,\mathbb {C})\). This simplifies the analysis of linear independence and makes the paper more accessible to readers less familiar with Clifford algebras. However, we stress that we need properties, like the complex conjugation \(\bar{*}\), associated to the complex Clifford algebra \(\mathbb {C}l(6)\). The latter is isomorphic to the more fundamental structure \(\mathbb {C}\otimes \overleftarrow{\mathbb {O}}\).

Appendix B: Lorentz Structures in \(\mathbb {C}\otimes \mathbb {H}\)

We briefly outline in which way the complex quaternions \(\overleftarrow{\mathbb {C}\otimes \mathbb {H}}= \mathbb {C}\otimes \mathbb {H}\) contain the Lorentz representations of the Standard Model. For details, see Ref. [19]. The basis elements of this algebra are \(\{1,i,\varepsilon _x,i\varepsilon _x,\varepsilon _y,i\varepsilon _y,\varepsilon _z,i\varepsilon _z\}\), where \(\{\varepsilon _x,\varepsilon _y,\varepsilon _z\}\) anti-commute and satisfy

$$\begin{aligned} \varepsilon _x\varepsilon _y = \varepsilon _z, \quad \varepsilon _y\varepsilon _z=\varepsilon _x, \quad \varepsilon _z\varepsilon _x=\varepsilon _y , \quad \varepsilon _i^2=-1. \end{aligned}$$
(73)

The unit scalar, 1, and the unit imaginary, i, commute with all other elements of the algebra. Hermitian conjugation \({}^\dagger \) can be defined on the algebra as an anti-automorphism which maps \(\varepsilon _j\rightarrow -\varepsilon _j\) and \(i\rightarrow -i\).

The above basis is useful for showing that \(\mathbb {C}\otimes \mathbb {H}\) contains the Lie algebra \(\mathfrak {so}(1,3)\) of Lorentz transformations. \(\varepsilon _j\) generates rotations and \(i\varepsilon _j\) generates boosts. The same algebra contains Lorentz spinors and their transformations are described by letting the algebra act on itself. To see this, it is convenient to work with the basis vectors,

$$\begin{aligned} \varepsilon _{\uparrow \uparrow }:= & {} \frac{1}{2}\left( 1-i\varepsilon _z\right) , \quad \varepsilon _{\downarrow \downarrow } := \frac{1}{2}\left( 1+i\varepsilon _z\right) , \nonumber \\ \varepsilon _{\downarrow \uparrow }:= & {} \frac{1}{2}\left( \varepsilon _y+i\varepsilon _x\right) ,\quad \varepsilon _{\uparrow \downarrow } := \frac{1}{2}\left( -\varepsilon _y+i\varepsilon _x\right) . \end{aligned}$$
(74)

This basis satisfies,

$$\begin{aligned} \varepsilon _{\uparrow \uparrow }^2=\varepsilon _{\uparrow \uparrow },\quad \varepsilon _{\downarrow \downarrow }^2=\varepsilon _{\downarrow \downarrow },\quad \varepsilon _{\uparrow \downarrow }^2=\varepsilon _{\downarrow \uparrow }^2=0. \end{aligned}$$
(75)

Defining \(P := \epsilon _{\uparrow \uparrow }\), left and right handed spinors correspond to the following subspaces,

$$\begin{aligned} \Psi _L\in (\mathbb {C}\otimes \mathbb {H})P, \quad \Psi _R\in (\mathbb {C}\otimes \mathbb {H})P^*. \end{aligned}$$
(76)

Hence, \(\epsilon _{\uparrow \uparrow }\) and \(\epsilon _{\downarrow \downarrow }\) are projectors related by complex conjugation. Left and right handed Lorentz transformations of these spinors are defined as,

$$\begin{aligned} \Psi _L \rightarrow e^{is}\Psi _L, \quad \Psi _R \rightarrow e^{-is^*}\Psi _R, \end{aligned}$$
(77)

for \(s\in \mathfrak {so}(1,3)\) spanned by the elements \(\varepsilon _j\) and \(i\varepsilon _j\). Then a general spinor behaves under Lorentz transformations as,

$$\begin{aligned} \Psi \rightarrow e^{is}\Psi P + e^{-is^*}\Psi P^*. \end{aligned}$$
(78)

Additionally within the algebra one can find scalar, vector, and field strength representations, as further elaborated on in Ref. [19].

Appendix C: Gauge Transformations

1.1 C.1. Commuting Transformations

Here we discuss general conditions for gauge transformations acting on states such as (11) and (12) to commute. Let us consider two gauge groups G and \(G'\) and denote their gauge transformations by the operators \(\mathcal {O}_G\) and \(\mathcal {O}_{G'}\). Then let some element \(\Psi \) transform under these gauge groups as,

$$\begin{aligned} \Psi \rightarrow \mathcal {O}_G\Psi ,\quad \Psi \rightarrow \mathcal {O}_{G'}\Psi . \end{aligned}$$
(79)

If both of the operators are represented by matrix multiplication from the left, then,

$$\begin{aligned} \mathcal {O}_G\Psi =e^{i\lambda }\Psi ,\quad \mathcal {O}_{G'}\Psi =e^{i\lambda '}\Psi , \end{aligned}$$
(80)

where \(\lambda \) and \(\lambda '\) are generators in the Lie algebras of G and \(G'\). When these transformations arise from entirely independent gauge groups, they should commute,

$$\begin{aligned} e^{i\lambda }e^{i\lambda '}\Psi \overset{!}{=} e^{i\lambda '}e^{i\lambda }\Psi . \end{aligned}$$
(81)

This holds if and only if the generators \(\lambda \) and \(\lambda '\) commute. For instance, in our present setup, the generators of \(\text {U}_Y(1)\) and SU(3) commute, and so do their gauge transformations on all particle states.

On the other hand, suppose now that the transformations corresponding to the group G are represented by matrix multiplication from the left and those of \(G'\) are represented by matrix multiplication from the right,

$$\begin{aligned} \mathcal {O}_G\Psi =e^{i\lambda }K\quad \mathcal {O}_{G'}\Psi =\Psi e^{i\lambda '}. \end{aligned}$$
(82)

In this case, as long as one works with an associative algebra, it does not matter which action is carried out first since,

$$\begin{aligned} e^{i\lambda }\Big (\Psi e^{i\lambda '}\Big )= \Big (e^{i\lambda }\Psi \Big )e^{i\lambda '}, \end{aligned}$$
(83)

always holds true. Hence, requiring gauge transformations to commute puts no constraints on the generators \(\lambda \) and \(\lambda '\).

1.2 C.2: Transformations of Generators

Next let us see how gauge transformations act on generators that do not commute with each other. For definiteness, let us take the gauge group to be \(G=\) SU(N). Let \(\Psi \) be a state in the fundamental representation of G. A generator \(\lambda \in \,\mathfrak {su}(N)\) then takes \(\Psi \) to a different state \(\tilde{\Psi }\) via the infinitesimal transformation,

$$\begin{aligned} (1+i\lambda ) \Psi = \tilde{\Psi }. \end{aligned}$$
(84)

Consider then also \(\Psi '\) and \(\tilde{\Psi }'\) obtained via another, finite transformation generated by \(\lambda '\),

$$\begin{aligned} \Psi ' = e^{i\lambda '}\Psi , \quad \tilde{\Psi }' = e^{i\lambda '}\tilde{\Psi }. \end{aligned}$$
(85)

We can now derive the generator \(\tilde{\lambda }\) which infinitesimally relates the states \(\Psi '\) and \(\tilde{\Psi }'\). Combining (84) and (85), one arrives at,

$$\begin{aligned} \tilde{\Psi }'-\Psi ' =i\tilde{\lambda }\Psi '=i\left( e^{i\lambda '}\lambda e^{-i\lambda '}\right) \Psi ' . \end{aligned}$$
(86)

Of course this derivation is well-known and shows that elements \(\lambda \) of the Lie algebra \(\mathfrak {su}(N)\) transform in the adjoint representation of the gauge group.

We will use the same method to derive how the SU(2) generators \(T_i\) in our setup (which do not commute with the elements \(\lambda _I\) of \(\mathfrak {su}(3)\)) transform under the SU(3) transformations. A particle state \(\Psi \) is infinitesimally related to \(\tilde{\Psi }\) with different SU(3) and SU(2) charges via,

$$\begin{aligned} \Psi +i\lambda _I \Psi + i\Psi T_i = \tilde{\Psi }, \end{aligned}$$
(87)

Consider then a finite SU(3) transformation generated by \(\lambda '\) that produces the states \(\Psi '=e^{i\lambda '}\Psi \) and \(\tilde{\Psi }'=e^{i\lambda '}\tilde{\Psi }\), Following the same procedure as above, we find,

$$\begin{aligned} \tilde{\Psi }'-\Psi '=i e^{i\lambda '}\lambda _Ie^{-i\lambda '}\Psi '+i\Psi 'T_i . \end{aligned}$$
(88)

This shows that, even though they do not commute with each other, the SU(2) generators do not transform under SU(3) transformations because they are not acting on states from the same side. It is easy to show that by the same argument the SU(3) generators and the \(\text {U}_Y(1)\) generator are invariant under SU(2) transformations. We conclude that all gauge generators transform only under their own gauge group, in the adjoint representation, as expected.

Appendix D: Right Handed Representations

We here show explicitly how one may accommodate right handed gauge representations in the set-up provided here. To do so we first start by discussing the discrimination of the SU(2) transformations on different matrix elements.

The SU(2) transformations presented in (16) and (17) clearly act differently on matrix elements \(R_I(V_a^\pm )^\dagger \) depending on whether \(I\in \{1,2,3,4\}\) or \(I\in \{5,6,7,8\}\) respectively. We could incorporate such a discrimination at the level of the generators by defining projectors that single out these relevant matrix subalgebras. To do so define simultaneous multiplication on M(8,\(\mathbb {C}\)) from the left by X and from the right by Y, for some \(X,Y\in \text {M(8,}\mathbb {C}\)), as the operation X|Y. Explicitly, for any \(K\in \text {M(8,}\mathbb {C}\)), \((X|Y)K:= XKY\). Defining the projectors

$$\begin{aligned} \mathcal {R}:= & {} \sum _{I=1}^{4} R_I (R_I)^\dagger \end{aligned}$$
(89)
$$\begin{aligned} \bar{\mathcal {R}}:= & {} \mathcal {R}^{\bar{*}} = \sum _{I=5}^{8} R_I (R_I)^\dagger \end{aligned}$$
(90)

we can then describe the SU(2) transformations via operators

$$\begin{aligned} \hat{T}_j := \mathcal {R}|T_j-\bar{\mathcal {R}}|T_J^{\bar{*}} \end{aligned}$$
(91)

such that (16) and (17) can collectively be written as

$$\begin{aligned} K\rightarrow e^{i\hat{T}_j}K \end{aligned}$$
(92)

for any \(K\in \)(11).Footnote 12

This idea can be extended further when we consider spatial representations by ensuring that only left handed particles and right handed anti-particles transform under SU(2). In this case, using the projectors P and \(\bar{P}\) introduced in appendix B, we would write operators

$$\begin{aligned} \hat{T}'_j := \left( \mathcal {R}|T_j\right) P-\left( \bar{\mathcal {R}}|T_J^{\bar{*}}\right) P^*, \end{aligned}$$
(93)

and SU(2) transformations via operators

$$\begin{aligned} e^{i\hat{T}'_j}. \end{aligned}$$
(94)

For \(\hbox {U}_Y\)(1) transformations we may also describe the action of left vs. right handed fermions via projectors. Here the story is quite similar, but with a small twist. Note that for left handed fermions, the hypercharge is the average of the electric charges of the SU(2) doublet. As such defining a matrix element

$$\begin{aligned} Q := -R_4(R_4)^\dagger -\frac{1}{3}\sum _{I=1}^3 R_I(R_I)^\dagger -\frac{2}{3}\sum _{I=5}^7 R_I(R_I)^\dagger , \end{aligned}$$
(95)

it is clear that

$$\begin{aligned} Y \equiv Q-Q^{\bar{*}} \end{aligned}$$
(96)

Now, define projectors

$$\begin{aligned} \mathcal {V}:= & {} \sum _a V^-_a(V^-_a)^\dagger \end{aligned}$$
(97)
$$\begin{aligned} \bar{\mathcal {V}}:= & {} \mathcal {V}^{\bar{*}} = \sum _a V^+_a(V^+_a)^\dagger , \end{aligned}$$
(98)

in the same way we defined the projectors \(\mathcal {R}\) and \(\bar{\mathcal {R}}\). It is then straight forward to verify that the operator

$$\begin{aligned} \hat{Y} := \left( (Q-Q^{\bar{*}})|1\right) P + \left( Q|\mathcal {V}-Q^{\bar{*}}|\bar{\mathcal {V}}\right) P^* \end{aligned}$$
(99)

yields the correct hypercharge assignments for both left and right handed fermions, all from application of Q and its complex conjugate. This is similar to how we obtain the correct SU(2) transformations from application of \(T_j\) and their complex conjugates.

While we here detail an approach for how the correct Standard Model charges may be incorporated for both left and right handed fermions, it is clear that the use of projectors as introduced here is ad-hoc and not natural. However, it is the purpose of this paper only to show that the structure of \(\mathbb {C}\otimes \overleftarrow{\mathbb {O}}\) when used as a subset of \(\overleftarrow{\mathbb {D}}\) may describe all Standard Model gauge representations, as presented here. Studies of the full \(\overleftarrow{\mathbb {D}}\) where the Lorentz structures are included explicitly, and thus any questions regarding the natural appearance of the Standard Model structures, is left to future work.

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Gording, B., Schmidt-May, A. The Unified Standard Model. Adv. Appl. Clifford Algebras 30, 55 (2020). https://doi.org/10.1007/s00006-020-01082-8

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