Exercise: does the representation descend?

A representation $\rho$ of a group $\widetilde G$ defines a representation of $\widetilde G/\Gamma$ precisely when every element of $\Gamma$ acts as the identity. The exercises below turn this condition into parity, congruence, and topology checks. Work analytically before using the linked tools.

All representations here are finite-dimensional ordinary complex representations of compact connected groups. The Lie-algebra modules integrate uniquely to the stated simply connected compact groups. Descent to a further quotient is an additional requirement; it is not a change in the Lie-algebra action.

For $SU(N)$, $N\geq2$, use the $A_{N-1}$ Dynkin order in which node $r$ labels $\Lambda^r\mathbb C^N$. For labels $(a_1,\ldots,a_{N-1})$, put

$$b=\sum_{r=1}^{N-1}r a_r, \qquad z=e^{2\pi i/N}I_N.$$

Then $z$ acts by $e^{2\pi i b/N}$. The order-$k$ subgroup $\Gamma_k=\langle z^{N/k}\rangle$, for $k\mid N$, acts trivially exactly when $b\equiv0\pmod k$. This follows by testing its generator. The underlying highest-weight and central-quotient results are Etingof, §43.1, Theorem 43.4 and Corollary 43.5, pp. 231–232; the covers and global forms lesson develops the argument.

1. Spin and rotation representations

Let $V_n=\operatorname{Sym}^n\mathbb C^2$ be the $SU(2)$ irrep with highest weight $n$ and spin $j=n/2$.

1. Determine the action of $-I$ on $V_n$ directly from its tensor construction.
2. For $n=1,2,3,4$, give the dimension and decide whether it descends through $SU(2)\to SO(3)$.
3. Does a failed descent mean that the corresponding $\mathfrak{su}(2)$ representation fails to exist?

Solution

The central element contributes one factor of $-1$ on each fundamental tensor factor, so it acts on $V_n$ by $(-1)^n$. The symmetric tensors have basis $x^{n-r}y^r$, $r=0,\ldots,n$, giving dimension $n+1$.

$n$	$j$	Dimension	Central sign	Descends to $SO(3)$?
$1$	$1/2$	$2$	$-1$	No
$2$	$1$	$3$	$+1$	Yes
$3$	$3/2$	$4$	$-1$	No
$4$	$2$	$5$	$+1$	Yes

All four are valid Lie-algebra modules and $SU(2)$ group representations. Only the even-$n$ modules factor through the specified quotient as ordinary linear $SO(3)$ representations. This does not deny the separate projective interpretation of half-integer spin.

Check the doublet and triplet, and inspect the covering map.

2. Triality and reality are different tests

For $SU(3)$, consider labels $(p,q)=(1,0),(1,1),(3,0)$.

1. Compute $p+2q$ modulo $3$ and decide which representations descend to $PSU(3)=SU(3)/\mathbb Z_3$.
2. Compute their dimensions using $\dim(p,q)=(p+1)(q+1)(p+q+2)/2$.
3. Is every representation that descends necessarily self-conjugate?

Solution

Labels	Triality	Dimension	Descends to $PSU(3)$?
$(1,0)$	$1$	$3$	No
$(1,1)$	$0$	$8$	Yes
$(3,0)$	$0$	$10$	Yes

Conjugation exchanges $(p,q)$ with $(q,p)$. Thus the $10$ with labels $(3,0)$ descends, although its conjugate $(0,3)$ is inequivalent. Descent tests a central kernel; self-conjugacy tests equivalence to the conjugate representation. They answer different questions.

The dimension formula and label convention are derived in roots and weights. Check the 10 of PSU(3) and the SU(3) quotient map.

3. An intermediate quotient of SU(4)

Consider matter in the $6=\Lambda^2\mathbb C^4$ with labels $(0,1,0)$ and the adjoint $15$ with labels $(1,0,1)$. The center is generated by $z=iI_4$.

1. Find the subgroup of $Z(SU(4))$ acting trivially on both representations.
2. Which quotients $SU(4)/\Gamma_k$, $k\mid4$, admit this matter content?
3. What are the fundamental group and center of the largest such quotient?
4. What changes if one fundamental $4$ is added?

Solution

The $6$ has $b=2$, so $z$ acts as $-1$ and $z^2=-I_4$ acts as $+1$. This can also be seen without the label formula: a scalar $i$ on each of two exterior factors gives $i^2=-1$. The adjoint has $b=1+3=4$ and trivial center action; directly, scalar conjugation acts trivially on traceless matrices. Their common central kernel is therefore

$$K=\{I_4,-I_4\}=\langle z^2\rangle\cong\mathbb Z_2.$$

The admissible choices are $k=1$ and $k=2$. The $6$ obstructs the full adjoint quotient $PSU(4)$, whose kernel also contains $iI_4$. Matter compatibility permits a choice among these groups; it does not require taking the largest kernel.

Because $SU(4)$ is simply connected,

$$\pi_1(SU(4)/K)\cong K\cong\mathbb Z_2, \qquad Z(SU(4)/K)\cong\mathbb Z_4/K\cong\mathbb Z_2.$$

Here the surviving center is generated by the class of $iI_4$, and it acts nontrivially on the descended $6$. A representation can descend without being trivial on the remaining center. To see that no extra center appears in this quotient, a central coset representative $g$ has commutators $ghg^{-1}h^{-1}\in K$ for all $h\in SU(4)$; continuity and connectedness force that discrete-valued map to be the identity, so $g$ was already central.

Adding a fundamental $4$ adds $b=1$, leaving only the identity in the common kernel. The dimensions independently check as $\binom42=6$ and $4^2-1=15$. Enter rank $N=4$ with labels 0,1,0;1,0,1 in matter-compatible quotients, then add 1,0,0.

4. The order of a product kernel is not enough

In $\widetilde G=SU(2)_L\times SU(2)_R$, compare

$$\Gamma_L=\langle(-I,I)\rangle, \qquad \Gamma_{\mathrm{diag}}=\langle(-I,-I)\rangle.$$

Both are central subgroups of order $2$. Does the representation $(\mathbf2,\mathbf2)$ descend through each? What about $(\mathbf2,\mathbf1)$? Explain why the quotient fundamental group alone cannot decide this.

Solution

On $V_{n_L}\otimes V_{n_R}$ the two generators act as

$$\rho(-I,I)=(-1)^{n_L}, \qquad \rho(-I,-I)=(-1)^{n_L+n_R}.$$

The $(\mathbf2,\mathbf2)$ has $(n_L,n_R)=(1,1)$. It fails the first test and passes the second. The $(\mathbf2,\mathbf1)$ has $(1,0)$ and fails both. Thus including both multiplets obstructs either displayed order-$2$ quotient.

Each displayed quotient has fundamental group isomorphic to $\mathbb Z_2$, since the covering product is simply connected. The abstract isomorphism type of the kernel loses its position inside the product center. Descent retains that embedding. The product gauge-group tool uses explicit center generators for this reason. For a descent check there, choose the two $SU(2)$ factors, zero matter charges under its required spectator $U(1)$, and zero $U(1)$ phase in the quotient generator. The spectator contributes an extra $\mathbb Z$ to the tool’s fundamental group; that is not part of the two-factor problem above.

5. Same orbit, different boundary map

Use the two circle inclusions

$$U(1)\to SU(2):\ z\mapsto\operatorname{diag}(z,z^{-1}), \qquad SO(2)\to SO(3):\ R(\varphi)\mapsto\operatorname{diag}(R(\varphi),1).$$

Each source circle parameter has period $2\pi$. Identify both cosets with $S^2$ and choose compatible sphere generators as in the topology lesson.

1. Compute the image of a unit sphere winding under $\partial:\pi_2(G/H)\to\pi_1(H)$ in each case.
2. How many principal $SU(2)$ and principal $SO(3)$ bundle classes exist on $S^2$?
3. Can an odd $SO(2)$ clutching loop arise from a reduction on the sphere of a smooth $SO(3)$ bundle over a ball?

Solution

The $SU(2)$ circle loop is contractible in $SU(2)$; the $SO(3)$ circle loop of winding $m$ maps to $m\bmod2$. Since both ambient groups have vanishing $\pi_2$, exactness gives boundary images $\mathbb Z$ and $2\mathbb Z$, respectively. With the compatible sign convention,

$$\partial_{\mathrm{SU}}(1)=1, \qquad \partial_{\mathrm{SO}}(1)=2.$$

The spin cover maps the first circle to the second with degree $2$ while identifying the orbit spheres, which checks this normalization. Nevertheless each orbit has $\pi_2(S^2)=\mathbb Z$.

Principal bundles over $S^2$ instead use $\pi_2(BG)=\pi_1(G)$ for connected $G$. There is one $SU(2)$ bundle class and two $SO(3)$ bundle classes. An odd $SO(2)$ clutching loop induces the nontrivial $SO(3)$ class; it cannot extend over a ball, where every principal bundle is trivial. This is a topological obstruction under the smooth extension assumption, not a claim about a monopole solution’s energy or stability. Compare both cases in the topology workbench.

6. Find the missing covering hypothesis

A proposed argument says: “The kernel of $U(1)\to U(1)$, $z\mapsto z^k$, is $\mathbb Z_k$, so the target has $\pi_1=\mathbb Z_k$.” Identify the error for $k\geq2$ and write the correct sequence.

Solution

The source circle is not simply connected. The map induces winding multiplication by $k$, so the correct exact sequence is

$$0\longrightarrow\mathbb Z\xrightarrow{\times k}\mathbb Z \xrightarrow{\bmod k}\mathbb Z_k\longrightarrow0.$$

The kernel records the quotient of the target fundamental group by the image of the source fundamental group. Only a universal covering removes the source term. See Conrad, Fundamental groups of Lie groups, pp. 1–2 and the topology formula reference.