I'm teaching using A. Zee's book, Group Theory in a Nutshell for Physicists. I'm thinking about assigning exercise 3 in chapter IV.7 because it looks like a good way for the students to practice working with subgroups, but the way Zee phrases the exercise it sounds like students are supposed to notice something deeper and I can't figure out what it is.

The problem concerns the local isomorphism between $SO(4)$ and $SU(2)\times SU(2)$. Zee makes this isomorphism explicit by writing,

\begin{equation} W\rightarrow U^\dagger W V \end{equation}

where $U$ and $V$ are $SU(2)$ matrices and $W=t I+i \vec{x}\cdot\vec{\sigma}$.

Zee talks about the diagonal subgroup $(V,V)$ in the text, and shows that it corresponds to the $SO(3)$ that rotates $\vec{x}$ and leaves $t$ fixed. Then, in the exercise he asks,

What does the $SU(2)$ subgroup of $SO(4)$, consisting of the elements $(V^\dagger,V)$, correspond to? Verify that these transformations do change $t$.

Does anyone have any idea what this subgroup is supposed to correspond to? It's straightforward enough to work out a generic group element (and confirm that $t$ is transformed), but it doesn't seem very illuminating.


1 Answer 1


I would like to stay out of transduction of pettifoggery, but here is what I would teach my students, in your place. For U,V,W elements of SU(2), and likewise for their adjoints, \begin{equation} W\rightarrow U^\dagger W V \end{equation} is, indeed, an π‘†π‘ˆ(2)Γ—π‘†π‘ˆ(2) transformation. To the extent the Us and the Vs are unlinked to each other, they close into the left and right SU(2), respectively.

Now, as he says, the diagonal restriction U=V is an SU(2) subgroup, all right, $$ W\rightarrow V_1^\dagger W V_1 \rightarrow V_2^\dagger V_1^\dagger W V_1 V_2 = (V_1 V_2)^\dagger W (V_1 V_2), $$ and, in current algebra, is referred to as the Vector (isospin) subgroup.

That is, your students must see how the sum of the left and right generators of the Us and V, respectively, dubbed $\vec v\cdot \vec\sigma$, commute among themselves to a $i(\vec v_1\times \vec v_2)\cdot\vec\sigma$. The above map is called "adjoint action", and rotates the triplet $\vec x$, even though, superficially, they might think the Οƒs can only rotate doublets.

However, for $$ W\rightarrow V_1 W V_1 \rightarrow V_2 V_1 W V_1 V_2 \neq (V_1 V_2) W (V_1 V_2), $$ so these SU(2) group elements do not close to a subgroup of the chiral π‘†π‘ˆ(2)Γ—π‘†π‘ˆ(2). (They are in the π‘†π‘ˆ(2)Γ—π‘†π‘ˆ(2)/π‘†π‘ˆ(2) coset space, but no matter.)

Lie-algebraically, they are generated by left-minus-right generators, and are called the axial generators, $\vec a\cdot\vec\sigma$. The commutator of two such axials, as they should expect, is not an axial, but, instead, a vector generator, $i(\vec a_1\times \vec a_2))\cdot\vec\sigma$, the conceptual cornerstone of chiral dynamics. The fact they transform (shift) t is the reason they are said to realize the axials nonlinearly (~break chiral symmetry spontaneously).

So, if they do the problem, the good ones will balk at the specious subgroup thing.

  • 1
    $\begingroup$ Ah, I had the vague inkling this couldn't possibly be a subgroup but kept thinking "oh, but Zee directly says it is...", should have realized it would be an un-caught error. $\endgroup$
    – Matt
    Commented Feb 18, 2023 at 8:06
  • 1
    $\begingroup$ ...been catching those in Tony's lectures since Spring of '74... $\endgroup$ Commented Feb 18, 2023 at 8:34

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