# A SU(5), 'Standard Model decomposition', direct sum etc.

1. Jul 29, 2016

### Anchovy

This has turned out to be a long question to type out so I apologise, but I don't think it's too hard to follow or read through quickly and I believe the actual question itself may not be too complicated once I get round to asking it. You can possibly skip to the last few paragraphs and still be able to answer.

Anyway, in introductory texts about SU(5) GUT they will often start by showing how the Standard Model fermions are organised into SU(5) multiplets. They use a certain notation involving parentheses and Standard Model quantum numbers that express how each fermion behaves under the three Standard Model gauge groups, $SU(3)_{C} \times SU(2)_{L} \times U(1)_{Y}$.

For example, the up and down quarks are available in 3 colors, so each of these form an SU(3) color triplet, and the left-handed up and down quarks go together in an SU(2) weak isospin doublet. Also the left-handed up quark has a charge of $Q = +\tfrac{2}{3}$ and a weak isospin third component of $T_{3} = +\tfrac{1}{2}$, therefore a hypercharge of $Y = 2(Q - T_{3}) = 2(\tfrac{2}{3} - \tfrac{1}{2}) = 2(\tfrac{4}{6} - \tfrac{3}{6}) = 2(\tfrac{1}{6}) = \tfrac{1}{3}$.

So, the notation I have seen used is as follows: (SU(3) quantum number, SU(2) quantum number, $\tfrac{1}{2} Y$). Thus, the left-handed up and left-handed down quark is assigned the following: $(3, 2, \tfrac{1}{6})$. (I believe this notation has something to do with group theory/group 'representations', which I am shaky on, but I'll leave that till the end of this post.)

Similarly, the left-handed down quark has a charge of $-\tfrac{1}{3}$ and a weak isospin third component of $-\tfrac{1}{2}$, and therefore a hypercharge of $Y = 2(Q - T_{3}) = 2(-\tfrac{1}{3} - -\tfrac{1}{2}) = 2(-\tfrac{2}{6} + \tfrac{3}{6}) = 2(\tfrac{1}{6}) = \tfrac{1}{3}$. So being also an SU(3) triplet, SU(2) doublet and equal to the up quark in hypercharge, it also gets the same assignment as the left-handed up-quark did: $(3, 2, \tfrac{1}{6})$.

Here I will summarise these assignments for the entire first generation of Standard Model fermions, expressed only in terms of left-handed fields just for consistency:

$(u_{r}, u_{g}, u_{b})_{L}, \hspace{2 mm} (d_{r}, d_{g}, d_{b})_{L} \hspace{1 cm}:\hspace{1 cm} (3, 2, \tfrac{1}{6}) \hspace{2 cm} \rightarrow \hspace{1 cm} \text{6 fields} \hspace{0.5 cm} \text{(SU(3) triplet, SU(2) doublet)}$
$\hspace{2.8 cm} (d_{r}, d_{g}, d_{b})^{c}_{L} \hspace{1 cm}:\hspace{1 cm} (3^{*}, 1, \tfrac{1}{3}) \hspace{2 cm} \rightarrow \hspace{1 cm} \text{3 fields} \hspace{0.5 cm} \text{(SU(3) <anti>triplet, SU(2) singlet)}$
$\hspace{2.8 cm} (u_{r}, u_{g}, u_{b})^{c}_{L} \hspace{1 cm}:\hspace{1 cm} (3^{*}, 1, -\tfrac{2}{3}) \hspace{1.5 cm} \rightarrow \hspace{1 cm} \text{3 fields} \hspace{0.5 cm} \text{(SU(3) <anti>triplet, SU(2) singlet)}$
$\hspace{3.5 cm} e^{-}_{L}, \hspace{2 mm} \nu_{eL} \hspace{1 cm}:\hspace{1 cm} (1, 2, -\tfrac{1}{2}) \hspace{2 cm} \rightarrow \hspace{1 cm} \text{2 fields} \hspace{0.5 cm} \text{(SU(3) singlet, SU(2) doublet)}$
$\hspace{4.5 cm} (e^{-}_{L})^{c} \hspace{1 cm}:\hspace{1 cm} (1, 1, 1) \hspace{2 cm} \rightarrow \hspace{1 cm} \text{1 field} \hspace{0.5 cm} \text{(SU(3) singlet, SU(2) singlet)}$

So for one family/generation of fermions we have a total of 15 fields. Now on to my actual question.

The Standard Model fermion fields are arranged into SU(5) multiplets / 'representations' that are named after how many fields they can contain. One of these is called the $5$ (or rather, in terms of left-handed fields they use the $\overline{5}$), and the other is the $10$, so that's enough to neatly accommodate our 5 + 10 = 15 fermions.

When these texts talk about which specific fermions get put into which one of these two multiplets, they talk about using the "Standard Model decomposition". They say that the "Standard Model decomposition" of the $5$ is $(3, 1, -\tfrac{1}{3}) \hspace{1 mm} \oplus \hspace{1 mm} (1,2, \tfrac{1}{2})$ (or for the $\overline{5}$ is $(3, 1, \tfrac{1}{3}) \hspace{1 mm} \oplus \hspace{1 mm} (1,2, -\tfrac{1}{2})$ ).

Now in the summary of these SM quantum number parentheses that I've shown above, I can see that in this 'direct sum', this so-called 'decomposition' of $\overline{5}$, the $(3, 1, \tfrac{1}{3})$ part corresponds to the right-handed, red, green and blue down quark. Also, the $(1,2, -\tfrac{1}{2})$ part corresponds to the left-handed electron and electron neutrino. Thus, we have five fields in total (3 quark, two lepton) that fit neatly into this multiplet. So apparently this assignment is not arbitrary, but rather it has some mathematical reason behind it.

My problem, now that I can finally ask it, is that I would really like to understand how this "decomposition" of $5$ or $\overline{5}$ has actually been calculated/determined. I am very much a novice in my understanding of group theory, group representations etc. but I assume this is about the Standard Model group $SU(3)_{C} \times SU(2)_{L} \times U(1)_{Y}$ being a 'subgroup' of the SU(5) group.

For some further info, I don't understand what the $\oplus$ operation is actually about, I think it means 'direct sum'. From what I've been able to gather, a 'representation' of a group essentially means expressing in matrix form each individual transformation caused by the elements that form the group. So these $(3, 1, \tfrac{1}{3})$, $(1,2, -\tfrac{1}{2})$ objects ('representations' of the $SU(3)_{C} \times SU(2)_{L} \times U(1)_{Y}$ group?) must in some way refer to matrices, and $\oplus$ must mean combining them somehow? Is this the opposite of a 'decomposition'?

I'm just confused about this stuff and if anyone has anything to say, I'd love to hear it. Thanks.

Last edited: Jul 29, 2016
2. Jul 29, 2016

### Orodruin

Staff Emeritus
Yes. It is how the 5 irrep breaks when you break the SU(5) to the SM gauge group. I suggest you read on how to compute the irreps of a subgroup that compose an irrep of the original group.

Yes, it is the direct sum. Given two vector spaces $U$ and $V$, the direct sum is just the vector space containing one element of $U$ and one of $V$ with addition of vectors defined by adding the vectors of $U$ and $V$ separately. A representation of a group is a homomorphism from the group to a set of operators on a vector space, in the case of the decomposition of a representation into irreps, we are showing that the representation is actually a direct sum of representations on irreps, each representation being independent (i.e., acting only on its irrep).

3. Jul 29, 2016

### Anchovy

OK, I'll have to do some more googling. The group theory pdf files I've looked at so far have tended to get incomprehensible to me quite quickly. Lots of unfamiliar words, symbols. "Group theory for dummies"-type searches have been the most fruitful so far, lol.

What is meant by an 'element' of a vector space? And can you give me an example of two vector spaces being added/combined?

Hmm. OK. I'm struggling to visualize things here, because first of all we have this SU(5) irrep, $\overline{5}$:
$$\overline{5} = \begin{pmatrix} \overline{d}_{r} \\ \overline{d}_{g} \\ \overline{d}_{b} \\ e^{-} \\ -\nu_{e} \end{pmatrix}_{L}$$.

So that thing looks like a 5x1 matrix, pretty straightforward. But, then we have these different-looking objects denoted $(3^{*}, 1, \tfrac{1}{3})$ and
$(1, 2, -\tfrac{1}{2})$, which I have also seen written like $(3^{*}, 1)_{\tfrac{1}{3}}$ and $(1, 2)_{-\tfrac{1}{2}}$. So I'm struggling to imagine what happens to get from these two non-matrix/non-vector-looking objects to a 5x1 matrix?

4. Jul 29, 2016

### Orodruin

Staff Emeritus
A vector in that vector space. Let us say $\vec v \in V$ and $\vec u \in U$. Then $(\vec u,\vec v)$ is an element of $U\oplus V$.

If you want to look at it in matrix form, the reduction to SM irreps means that the representations of SM operators are block diagonal. This means that each representation transforms an element in an irrep to another element in that irrep.

5. Jul 30, 2016

### Anchovy

Ok, let me see... I have seen this beginner example involving the symmetry of a triangle being rotated about the origin in 2D space. There are three transformations in such a group; namely: $g_{1}$ (rotating by 120 degrees), $g_{2}$ (240 degrees) or the identity element (360 / 0 degrees). 2D rotations can be expressed in matrix form via

$$\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} cos \theta & sin \theta \\ -sin \theta & cos \theta \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}$$

so these group transformations can be represented by:

$$D(I) = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \hspace{1 cm} D(g_{1}) = \begin{pmatrix} -\tfrac{1}{2} & \tfrac{\sqrt{3}}{2} \\ -\tfrac{\sqrt{3}}{2} & -\tfrac{1}{2} \end{pmatrix}, \hspace{1 cm} D(g_{2}) = \begin{pmatrix} -\tfrac{1}{2} & -\tfrac{\sqrt{3}}{2} \\ \tfrac{\sqrt{3}}{2} & -\tfrac{1}{2} \end{pmatrix}$$
However, this can be extended to 3 dimensions by including a third row & column, but since rotations are around the z-axis, the z-component never changes so it's trivial:
$$D(I) = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \hspace{0.5 cm} D(g_{1}) = \begin{pmatrix} -\tfrac{1}{2} & \tfrac{\sqrt{3}}{2} & 0 \\ -\tfrac{\sqrt{3}}{2} & -\tfrac{1}{2} & 0 \\ 0 & 0 & 1 \end{pmatrix}, \hspace{0.5 cm} D(g_{2}) = \begin{pmatrix} -\tfrac{1}{2} & -\tfrac{\sqrt{3}}{2} & 0 \\ \tfrac{\sqrt{3}}{2} & -\tfrac{1}{2} & 0 \\ 0 & 0 & 1 \end{pmatrix}$$

The upper left 2x2 is irreducible in each case since (aside from the identity matrix) there's supposedly no basis for which they could be turned into a diagonal matrix.
And the lower 1x1 is clearly irreducible too. So I presume this is the sort of thing you mean by having two irreps on a block diagonal.

So... how does this apply in the context of $SU(5)$... what is analogous in that instance to the example above...

The symmetry group is $SU(3) \times SU(2) \times U(1)$, so there should be a 3x3 block for SU(3), a 2x2 block for SU(2), and a 1x1 block for U(1) ?
And... in this case they're not acting in Euclidean space, but transformations in... weak isospin space, color space... for matrices containing... creation/annihilation operators for fermion fields rather than 3D Cartesian vectors? Am I anywhere near the right lines here?

6. Jul 30, 2016

### Orodruin

Staff Emeritus
The blocks are not one per group, there is one block per irrep. The 5 representation of SU(5) is 5-dimensional and splits to one 3-dimensional representation and one 2-dimensional, representing the group elements of the remaining symmetry group with one 3x3 block and one 2x2 block.

7. Jul 30, 2016

### Anchovy

Ah, ok, yes, in that example I posted the 2x2 and 1x1 blocks contained irreps, not groups.

So this thing:

$$\overline{5} = \begin{pmatrix} \overline{d}_{r} \\ \overline{d}_{g} \\ \overline{d}_{b} \\ e^{-} \\ -\nu_{e} \end{pmatrix}_{L}$$.
splits to...
$$\begin{pmatrix} \overline{d}_{r} \\ \overline{d}_{g} \\ \overline{d}_{b} \end{pmatrix}_{L} \hspace{1 cm} \text{and} \hspace{1 cm} \begin{pmatrix} e^{-} \\ -\nu_{e} \end{pmatrix}_{L} \hspace{1 cm}\text{?}$$

I'm unsure of what you mean by the "remaining" symmetry group?

8. Jul 30, 2016

### Orodruin

Staff Emeritus
The SM group is the symmetry group that remains once you break the SU(5) symmetry. Before symmetry breaking, there are elements of SU(5) that mix the blocks.

If you want to have an explicit representation, the generators of SU(3) in the 5 representation have the Gell-Mann matrices in the upper left 3x3 block and are zero everywhere else. The generators of the SU(2) contain the Pauli matrices in the lower left 2x2 block and are zero everywhere else. The U(1) generator is represented by a traceless diagonal matrix which is proportional to the identity (with different proportionality constants) in both the 3x3 and 2x2 block.

9. Jul 30, 2016

### Anchovy

OK. I'll post a screencap of some text regarding unbroken SU(5) and the explicit representation you must be talking about here, and see if I can interpret it...

So, looking at those red equations (1), (2) at the top, that's telling me that, say, the up quark creation operator $u^{\dagger}$ can be 'represented' by $(3, 1)_{\tfrac{1}{3}}$. Straight away I'm confused, because that appears to be talking about $u^{\dagger}$ as though it is an element of a group (since it is assigned a 'representation'). Is this the correct interpretation?

10. Jul 30, 2016

### Orodruin

Staff Emeritus
You need to separate the representation from the vector space the representations act on. Many physics texts are not very explicit about this. The point is that a field transforming according to a particular representation under a symmetry is an element of the vector space the representation of the symmetry acts upon. Therefore it transforms in a particular fashion under the symmetry.

11. Jul 30, 2016

### Anchovy

Is it correct to say that, in the case of the field $u$, it has three colors, so $u_{r}$, $u_{g}$ and $u_{b}$ would be three different vectors in 'color space'? And the transformations that the SU(3) group is comprised of amount to rotations between these vectors?

Secondly, is there a difference between saying "a particular representation under a symmetry" and "a particular representation of a symmetry group element" ?

Thirdly, I'm supposed to be thinking of these representation matrices acting on the vector space itself rather than vectors living in it? Is this the same sort of idea as the difference between active and passive rotations? (ie. active = transform by altering the vector itself, passive = transform by altering the axes)

12. Jul 30, 2016

### Orodruin

Staff Emeritus
They are different directions in color space.

Yes, they span a three-dimensional space transforming under the fundamental representation of SU(3). The leptons also transform under SU(3), just according to the trivial representation. The quark flavor space is not "the" space on which the SU(3) acts, it carries a (fundamental) representation of SU(3). Other fields transform according to other representations.

I have trouble parsing what you would mean by the first statement. You represent an (often abstract) group by a homomorphism to a set of linear operators on a vector space. It is unclear what you would mean by "representation under a symmetry".

The representation is the homomorphism to the linear operators on a vector space. You represent a group element by a linear operator, not by a vector.

13. Jul 30, 2016

### Staff: Mentor

Or quick and dirty: $SU(3) \; \; \begin{array} \text{irrep} \\ \longrightarrow \end{array} \; \; GL(V) = \text{linear operations} \text{ (color space )}$.
(Sorry for interfering.)

14. Jul 30, 2016

### Anchovy

OK, so $u_{r}, u_{g}, u_{b}$ are axes labels for this space. Can the elements of this space (I believe that is to say, the vectors that live on it) be anything other than the unit vectors parallel to these directions?

Hmm. Do I interpret that as, their only transformation under SU(3) is an identity transformation?

How come we've switched from talking about color here to talking about flavor? Is it that this flavor space contains axes/directions for all six quarks, and also three directions per quark, one for each color?

The first statement was something I copied and pasted from your previous post, I was uncertain about it too. From the start of the sentence it was "The point is that a field transforming according to a particular representation under a symmetry is an...".

Hmm, so representations are not the linear operators/matrices themselves, but "homomorphisms" to those operators?

"Homomorphism" is a word that I'm vague on. Take this definition google turns up:

"A homomorphism is a map between two groups which respects the group structure. More formally, let G and H be two group, and f a map from G to H (for every g∈G, f(g)∈H). Then f is a homomorphism if for every g1,g2∈G, f(g1g2)=f(g1)f(g2)."

So that's talking about something between two groups, whereas according to what I've quoted from you here, a homomorphism is to the linear operators on a vector space... which sounds like you're saying it's something different?

This much I am comfortable with. I can see that from my triangle example from earlier.

15. Jul 30, 2016

### Anchovy

This may turn out to be helpful.

16. Jul 30, 2016

### nikkkom

SU(3) acts on the 3-dimensional complex vector space. This is a space of all elements (a,b,c) where a,b,c are complex numbers. There are infinitely many possible vectors in it.

17. Jul 30, 2016

### Staff: Mentor

Yes, as a matrix acts on a vector space: $G(v) = G \circ v$ for $G \in SU(3) \, , \, v \in \mathbb{C}^3$.
But there is no need to define $SU(3)$ as a subgroup of $GL(\mathbb{C}^3)$.
E.g. it's not nearly as obvious when it comes to the $E_8$ in string theory.

The operation of $SU(3)$ via linear operators on a vector space $V$ in the sense of representations, i.e. $\varphi : SU(3) \rightarrow GL(V)$ and $G.v = \varphi(G)(v)$ is something different.
E.g. the dimension of $V$ doesn't need to be three or finite at all, and the representation isn't automatically irreducible.

Part of the confusion that arises, is by mixing the two actions.

18. Jul 30, 2016

### nikkkom

You need to translate this to "English for mere mortals" :D

Are you saying that what's important is how any SU(3) element can be represented? (I.e. any SU(3) element can be represented as a linear composition of 8 Gell-Mann matrices)

19. Jul 30, 2016

### Staff: Mentor

I simply wanted the debate to be clarified. The fact that elements of $SU(3)$ can be written as matrices doesn't matter when we talk about the (irreducible) representations of this symmetry group. It is only one possible representation among many.

I don't see how it helps if you bring in another subject, the infinitesimal representation. The Gell-Mann matrices neither belong to $SU(3)$ nor to a representation $GL(V)$ of it. They aren't invertible and span the Lie algebra $\mathfrak{su}(3)$ of $SU(3)$, its tangent space.
However, there is a representation (conjugation, I think) $SU(3) \rightarrow GL(\mathfrak{su}(3))$, which is again one example, even though an important one. In addition it is one in which the representation space $V= \mathfrak{su}(3)$ isn't three dimensional.

20. Jul 30, 2016

### Orodruin

Staff Emeritus
Yes. Mapping all group elements to the identity is a perfectly valid homomorphism.

Writing too fast on mobile device.

This is just a terminology issue. The representation is the homomorphism, but you can say that under this representation a group element is represented by a particular linear operator (ie., the operator the homomorphism maps it to).

The set of (invertible) linear operators on a vector space form a group. A representation is a homomorphism from an abstract group to such a group of operators.

Note that this is the fundamental (3) representation of SU(3). There are many other representations as well.

You are thinking of the adjoint representation? (Ie, the natural representation of a Lie group on its own Lie algebra) This is indeed an important representation - the gauge fields of Yang-Mills theory transform according to it.

21. Aug 5, 2016

### lpetrich

Nitpick: group-representation matrices are usually specified in complex numbers, and one finds that the rotation group's 2D representations are reducible into two irreducible ones (irreps). In fact, all irreps of abelian groups are one-dimensional. For 2D rotation in general,
$$D_r(\theta) = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} = T \cdot \begin{pmatrix} e^{i\theta} & 0 \\ 0 & e^{-i\theta} \end{pmatrix} \cdot T^{-1}$$
where
$$T = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & -i \\ -i & 1 \end{pmatrix}$$
is selected to be unitary.

To get a dimension-2 irrep, one needs reflections:
$$D_s(\theta) = \begin{pmatrix} \cos\theta & \sin\theta \\ \sin\theta & -\cos\theta \end{pmatrix} = T \cdot \begin{pmatrix} 0 & -i e^{i\theta} \\ i e^{-i\theta} & 0 \end{pmatrix} \cdot T^{-1}$$

Pure 2D rotations are SO(2): Dr(a) for a from 0 to 2*pi. Adding reflections gives O(2): Ds(a) for a from 0 to 2*pi.

The finite order-n subgroup of SO(2) has $D_r(2\pi k/n)$ for k an integer from 0 to n-1. Likewise, the finite order-2n subgroup of O(2) has that and also $D_s(\theta_0 + 2\pi k/n)$ for some θ0 and for k from 0 to n-1.

Note: SO(2) is a one-parameter Lie group: $D_r(\theta) = e^{i L \theta}$ where generator L is
$$L = \begin{pmatrix} 0 & i \\ -i & 0 \end{pmatrix}$$
Lie algebras are sometimes denoted by small letters when one wants to distinguish them from their groups: group SO(2) with algebra so(2).

22. Aug 6, 2016

### lpetrich

As to tensors and representations, I will try to illustrate this by a simple example. Let's start with what I'll call a vector rep, Dij(a) for element a, operating on a vector with indexing j and returning a vector with indexing i. The vectors have some simple basis vectors: e1 = {1} = (1,0,0,...), e2 = {2} = (0,1,0,...), etc. I'm introducing the curly-bracket notation to save myself some typing later on.

Let us now consider a product rep: D(i1i2),(j1j2)(a) = Di1j1(a) * Di2j2(a). Its basis vectors are outer products of the original basic vectors, making them 2-tensors in the original vector space: (e1,e1) = {1,1}, (e1,e2) = {1,2}, etc.

It can be split into symmetric and antisymmetric parts:
Ds,(i1i2),(j1j2)(a) = (1/2) * (Di1j1(a)*Di2j2(a) + Di1j2(a)*Di2j1(a))
Da,(i1i2),(j1j2)(a) = (1/2) * (Di1j1(a)*Di2j2(a) - Di1j2(a)*Di2j1(a))

The symmetric one has symmetric 2-tensor basis vectors: {1,1}, (1/sqrt(2))*({1,2} + {2,1}), (1/sqrt(2))*({1,3} + {3,1}), ..., {2,2}, (1/sqrt(2))*({2,3} + {3,2}), ...

The antisymmetric one has antisymmetric 2-tensor basis vectors: 1/sqrt(2))*({1,2} - {2,1}), (1/sqrt(2))*({1,3} - {3,1}), ..., 1/sqrt(2))*({2,3} - {3,2}), ...

Many of the calculations on Lie-algebra representations are done on their basis sets, because that is usually much easier than working with their representation matrices, and because one can learn much of what one usually wants to know by using basis sets. Things like what irreps are in a rep, product reps, powers of reps with various symmetries ("plethysms"), subgroup reps in a rep, rep reality (real, pseudoreal, complex), etc. I myself have written some code to do that for semisimple Lie algebras.

23. Aug 6, 2016

### lpetrich

Lie-algebra generators one can compose by taking D = I + i*ε*L + O(ε2). One gets for the (anti)symmetric tensor reps,

L(s,a) i1i2,j1j2 = (1/2) * ( (δi2j2*Li1j1 + δi1j1*Li2j2) +- (δi2j1*Li1j2 + δi1j2*Li2j1) )

So once one has one's original L's, one can construct symmetric or antisymmetric 2-tensor L's.

Turning to SU(5), let us see what its Lie-algebra matrices turn into. An easy-to-construct basis for SU(n) in general is (i,j = 1...n):
La,ij = δa,ia,j - δa+1,ia+1,j for a = 1 ... (n-1)
Lab,s,ij = δa,ib,j + δb,ia,j for a = 1 ... (n-1), b = a+1 ... n
Lab,a,ij = - i * (δa,ib,j - δb,ia,j) for a = 1 ... (n-1), b = a+1 ... n
This accounts for all (n2-1) generators of SU(n)

SU(5) breaks down into SU(3) * SU(2) * SU(1) with its algebra matrices becoming:

Nonzero over (1,2,3) * (1,2,3) -- QCD
Nonzero over (4,5) * (4,5) -- weak-isospin particles W
Nonzero over (1,2,3) * (4,5) -- proton-decay leptoquarks
Nonzero over (4,5) * (1,2,3) -- antiparticle of above
diag({2,2,2,-3,-3}) -- weak-hypercharge particle B

24. Aug 8, 2016

### Anchovy

OK thanks, that's a lot to digest, will return to this soon.