# Why do we require gauge symmetries to commute?

My question arises after reading the 87th page of Elementary particle physics in a nutshell by Tully:

which is also given by the following link:

After deriving Eq $$3.83$$, he argued that

Since the $$\tau$$ matrices (which are pauli matrices) don't commute, commutativity of the $$SU(2)$$ and $$U(1)$$ transformations for all $$\vec\alpha$$ and $$\chi$$ is possible only if the $$U(1)$$ charges for the up and down compoments of the $$SU(2)$$ are equal.....And therefore we cannot identify the $$U(1)$$ group in the $$SU(2)\times U(1)$$ direct product group with $$U(1)_{EM}$$

My questions are:

1. why do we require the commutativity of the $$SU(2)$$ group and $$U(1)$$ transformation? And what forbid us to commute them?

2. What are the differences in math between $$U(1)_{EM}$$ and $$U(1)_{Y}$$? Since they are both $$U(1)$$ group I guess there is no difference between them? The only difference is the charge?

3. How is the "charge" defined (hypercharge and electrical charge)? I don't have a backgroud of group theory. As far as I know, it is just a parameter in the lagrangian, is it related to the generator?

4. How does one define hypercharge in experiment? How is it measured?

1. Pretty much by definition. Electroweak theory is built around $$SU(2) \times U(1)$$, which is a Cartesian product of two groups, so it has them as commuting parts by definition.
2. $$U(1)_{\text em}$$ and $$U(1)_Y$$ are different subgroups of $$SU(2) \times U(1)$$ both isomorphic to the abstract $$U(1)$$. The $$U(1)_Y$$ plays a role in the definition of the electroweak model, but because the Higgs acquires a VEV, the $$U(1)_{\text em}$$ plays a role in the broken phase which we happen to be living in.
3. See below.
4. Hypercharge is the $$U(1)_Y$$ gauge charge. In Standard Model, the values of hypercharge are chosen such that the particles have properties consistent with observation. The peculiar pattern of hypercharges is therefore arbitrary in the Standard Model, but its structure has been considered to be a clue into the physics beyond the Standard Model. Quite interestingly most of the GUT theories predict exactly the correct pattern of hypercharges.

To understand gauge charges, a small introduction into the representation theory for semisimple Lie algebra is needed. Keep in mind that I'm leaving a lot behind the curtains here!

An important fact about semisimple Lie algebras is that they admit the Cartan-Weyl basis that splits into a maximal commuting subalgebra called the Cartan subalgebra, and the remaining generators called roots. The number of Cartan subalgebra generators is called the rank of the algebra, and it is the number of gauge charges. Think of the Cartan subalgebra as the space of quantum numbers (the analogy is very close, because both are maximal commuting subspaces of something else, be it the Lie algebra or the configuration space of a system).

All fields (and therefore particles) are classified into finite-dimensional representations of the Lie algebra of $$SU(2) \times U(1)$$, which is $$\mathfrak{su}_2 \oplus \mathfrak{u}_1$$, and each such representation is a direct sum of irreducible representations, or irreps.

Cartan subalgebra generators have simultaneous eigenvalues on the vector space of any irrep, because they all commute. These eigenvalues are called gauge charges.

For example, $$\mathfrak{su}_2$$ has rank 1, and the Cartan-Weyl basis can be chosen as follows:

• Cartan subalgebra is generated by $$J_3$$.
• Two roots are $$J_1 \pm i J_2$$.

Its irreps are labeled by a half-integer $$j$$ called spin (here for abstract $$\mathfrak{su}_2$$, but "spin" also has a meaning in particle physics), and the spin-$$j$$ irrep has dimension $$\dim V_j = 2j + 1.$$

The eigenvalues of $$J_3$$ on $$V_j$$ range from $$-j$$ to $$j$$ with interval of $$1$$: $$-j, -j+1, \dots, j-1, j$$. This constitues a gauge charge that is called isospin (sometimes the term isospin refers to $$j$$, and the eigenvalue of $$J_3$$ is termed the 3-rd projection of isospin).

To give you a real life example, consider a left-chiral duplet $$\left(\begin{array}{c} e_L\\ \nu_{e} \end{array}\right){}$$

It lies in the $$V_{1/2}$$ irrep of $$\mathfrak{su}_2$$, and $$e_L$$ and $$\nu_e$$ are eigenstates of $$J_3$$ with isospin $$\mp 1/2$$ respectively.

Another example is the right-chiral electron $$e_R$$ which is an $$\mathfrak{su}_2$$ singlet, that is, belongs to the 1-dimensional irrep $$V_0$$. Its isospin is $$0$$.

The $$\mathfrak{u}_1$$ subalgebra isn't semisimple and has to be handled separately, but luckily the representation theory of $$\mathfrak{u}_1$$ is pretty straightforward. All irreps of an abelian algebra are 1-dimensional, and are completely parametrized by a choice of a number that corresponds to the single generator $$J$$. For the Lie algebra of $$U(1)_Y$$ that number is called hypercharge, and for $$U(1)_{\text{em}}$$ it is the electric charge.

The question of why electric charge is quantized is open in the Standard Model. GUTs attempt to embed $$\mathfrak{su}_2 \oplus \mathfrak{u}_1$$ into a larger semisimple Lie algebra (e.g $$\mathfrak{su}_5$$) which means that charge quantization (both for hypercharge and electric charge) comes out naturally.

Finally, the formula relating the two $$\mathfrak{u}_1$$ charges is: $$Q = J_3 + \frac{Y}{2},$$ which is of course just the preferred by physicists normalization of $$Q$$ and $$Y$$.

Now for more interesting algebras such as $$\mathfrak{su}_3$$ (for QCD) or $$\mathfrak{su}_5$$ (the simplest GUT model), the ranks are respectively $$2$$ and $$4$$, so irreps form peculiar patterns of gauge charges in 2-dimensional and 4-dimensional spaces respectively.

Garrett Lisi has an awesome browser app called "elementary particle explorer" that maps elementary particles to points on $$\mathbb{R}^n$$ with coordinates corresponding to gauge charges. It works with the Standard Model, several GUT models, as well as with Lisi's pet E8 model (which in its present state isn't well defined).

UPD

I will give you the derivation of $$U(1)_{\text{em}}$$.

The Higgs field is a scalar multiplet with the following transformation properties under the electroweak group $$SU(2) \times U(1)_Y$$:

• Under the $$SU(2)$$ part it transforms as a douplet – the irrep with $$j = 1/2$$. The generators of $$SU(2)$$ in the spin-$$1/2$$ irrep are given by $$J_i = \frac{1}{2} \sigma_i,$$ where $$\sigma_i$$ are the Pauli matrices.
• Under the $$U(1)_Y$$ part it transforms as a singlet with hypercharge $$1$$. The generator of $$U(1)_Y$$, which is of course just $$Y$$, acts on the higgs as $$1$$.

It is known that the Higgs acquires a non-zero vacuum expectation value (VEV) through the dynamics of the electroweak model. This breaks the full electroweak group. The important question is – which part of the group is unbroken (i.e. preserves the VEV)?

The Higgs VEV in the unitary gauge is $$\phi_{0}=\left(\begin{array}{c} 0\\ v \end{array}\right).$$

Acting with $$J_1 \sim \sigma_1$$ or $$J_2 \sim \sigma_2$$ on it mixes up the components, because these matrices aren't diagonal. Therefore no nonzero linear combination of these can preserve the VEV.

Also, the $$U(1)_Y$$ generator which is $$Y$$ doesn't preserve the VEV either. We know that it acts as $$1$$, so

$$Y \phi_0 = \phi_0 \neq 0.$$

However, a linear combination of $$J_3$$ and $$Y$$ does zero out the VEV!

$$J_{3}=\frac{1}{2}\sigma_{3}=\left(\begin{array}{cc} 1/2 & 0\\ 0 & -1/2 \end{array}\right),$$ $$\left(J_{3}+\frac{1}{2}Y\right)\phi_{0}=\left(\begin{array}{cc} 1/2+1/2 & 0\\ 0 & -1/2+1/2 \end{array}\right)\left(\begin{array}{c} 0\\ v \end{array}\right)=\left(\begin{array}{c} 0\\ 0 \end{array}\right).$$

Here we've found the so-called "little group" – the subgroup of the gauge group that preserves the VEV of the Higgs. It turns out that it is the $$U(1)_{\text{em}}$$, generated by $$Q = J_3 + \frac{1}{2} Y.$$

• Thank you so much! I have another question. You mentioned that: $U(1)_{\text em}$ and $U(1)_Y$ are different subgroups of $SU(2) \times U(1)$. How can we see this? – Universe Maintainer May 2 at 16:44
• @UniverseMaintainer the $U(1)_Y$ is the very $U(1)$ in $SU(2) \times U(1)$. It is generated by the hypercharge $Y$: the generic element is written as $\exp(i \varepsilon Y)$. As for $U(1)_{\text{em}}$ – it is generated by $Q = J_3 + Y/2$. It should be evident from this that $U(1)_{\text{em}}$ mixes the direction of $U(1)_Y$ in the full symmetry group with the direction in $SU(2)$ pointed to by $J_3$. As to why that particular direction is physically important – it happens to be the only symmetry that preserves the VEV of the Higgs field. – Solenodon Paradoxus May 2 at 16:50
• @UniverseMaintainer the Higgs field is an $SU(2)$ duplet with hypercharge of $1$. That completely specifies its transformation properties under $SU(2) \times U(1)$. It acquires a VEV (in so-called "unitary gauge") is written as $\phi_{0}=\left(\begin{array}{c} 0\\ v \end{array}\right)$. It's a good exercise to extract a subgroup of $SU(2) \times U(1)$ that preserves such VEV (I urge you to try it, hint – work with the Lie algebra). It turns out that the answer is $U(1)$, but not the $U(1)$ in $SU(2) \times U(1)$, but rather a $U(1)_{em}$ that mixes $U(1)_Y$ with a direction in $SU(2)$. – Solenodon Paradoxus May 2 at 16:55
• Thank you so much for such an amazing answer! It is of great help to me! I have another question perhaps it looks a little bit stupid because I really don't know much about Lie algebra. So seems like the $SU(2)$ group in the standard model is actually $SU(2)_{L}$. What does this mean? any differentce between this group? Do they have the same generators? Also could you please recommand any reference (book or lecture notes) on the topic? Thank you so much! – Universe Maintainer May 2 at 17:04
• @UniverseMaintainer PTAL at the update – I added the derivation of $SU(2)_{\text{em}}$. The $L$ suffix means that $SU(2)$ acts nontrivially only on the left-chiral fermions. In my answer I mentioned that $e_L$ transforms nontrivially under $SU(2)$ (mixes with the neutrino), and $e_R$ is a singlet – it doesn't transform at all. Why particle physics is chiral (i.e. non-left-right-symmetric) is one of the big unanswered questions. – Solenodon Paradoxus May 2 at 17:10