Covering groups and global forms

A representation of a Lie algebra first integrates to its simply connected group. It is a representation of a chosen quotient group precisely when the quotient kernel acts trivially. This is why the same infinitesimal generators can admit different sets of matter multiplets.

The objects here are connected real Lie groups and finite-dimensional, continuous complex representations in characteristic zero. For compact groups these representations can be made unitary. A complex vector space can still carry a representation of real or quaternionic type; that is a separate question. Read Lie algebras and global groups first if the local/global distinction is new.

Start with the universal cover

For a connected Lie group $G$, there is a simply connected Lie group $\widetilde G$ and a covering homomorphism

$$1\longrightarrow\Gamma\longrightarrow\widetilde G \overset{p}{\longrightarrow}G\longrightarrow1, \qquad G\simeq\widetilde G/\Gamma.$$

Its differential identifies the Lie algebras. The kernel $\Gamma$ is discrete and central, and is naturally isomorphic to $\pi_1(G)$ with identity basepoints. Centrality has a useful elementary explanation: for fixed $\gamma\in\Gamma$, the map $g\mapsto g\gamma g^{-1}$ is continuous from a connected group to the discrete kernel, so it is constantly $\gamma$. These statements and integration of Lie-algebra representations are Etingof, Proposition 3.5 and Corollary 11.2, printed pp. 26 and 62.

The simply connected hypothesis matters for the fundamental group. If $H\to H/\Gamma$ is merely a connected covering group, then instead

$$1\longrightarrow\pi_1(H)\longrightarrow\pi_1(H/\Gamma) \longrightarrow\Gamma\longrightarrow1.$$

The last map lifts a loop and records its endpoint in the kernel. Thus $\pi_1(H/\Gamma)$ need not equal $\Gamma$. Conrad, Fundamental groups of Lie groups, pp. 1–2 proves this exact sequence. For instance, $z\mapsto z^k$ covers a circle by a circle: the sequence is $0\to\mathbb Z\overset{\times k}{\to}\mathbb Z \to\mathbb Z_k\to0$, not a finite fundamental group for the target.

Derive the descent test

Let $\widetilde\rho:\widetilde G\to GL(V)$ be a representation. The proposed action on a coset is $\rho(g\Gamma)=\widetilde\rho(g)$. Replacing $g$ by $g\gamma$ gives the same matrix for all $g$ exactly when

$$\boxed{\quad\widetilde\rho(\gamma)=I_V\quad \text{for every }\gamma\in\Gamma.\quad}$$

This is both necessary and sufficient for descent. It holds for reducible representations too. In an irreducible complex representation, Schur’s lemma expresses every central action as a scalar:

$$\widetilde\rho(z)=\chi_V(z)I_V,\qquad \chi_V:Z(\widetilde G)\longrightarrow\mathbb C^\times.$$

The central character $\chi_V$ must therefore be trivial on the specified subgroup $\Gamma$. It is not the trace character $\operatorname{tr}\rho(g)$: on a central element that trace is $\dim(V)\chi_V(z)$. For a direct sum, check every summand; for a tensor product, central characters multiply. This explains how two individually obstructed representations can have a tensor product that descends.

For a compact semisimple Lie algebra, its simply connected compact group has finite center. All connected global forms are quotients by subgroups of that center. With a chosen maximal torus and positive roots, let $P$ be the weight lattice and $Q$ the root lattice. The allowed weight lattice of $G=\widetilde G/\Gamma$ is

$$P_G=\{\lambda\in P:\chi_\lambda|_\Gamma=1\}, \qquad Q\subseteq P_G\subseteq P.$$

Its dominant weights label the irreducible representations of $G$; the adjoint form $\widetilde G/Z(\widetilde G)$ has $P_G=Q$. Etingof, §43.1, Proposition 43.2 and Corollary 43.5, printed pp. 230–232, gives the classification and central-character criterion. The roots and weights lesson develops the lattices.

SU(2) and SO(3): the sign of a full rotation

The covering map $SU(2)\to SO(3)$ acts by conjugation on the real space of traceless Hermitian $2\times2$ matrices. It preserves $\tfrac12\operatorname{tr}(XY)$ and has kernel $\{I,-I\}$. In particular, the cover, quotient and algebra are three different objects: SU(2), SO(3) and $\mathfrak{su}(2)$. The construction is Etingof, Exercise 3.9, p. 26.

Write the highest weight as the nonnegative integer $n=2j$. The irrep is $V_n=\operatorname{Sym}^n\mathbb C^2$ with dimension $n+1$. The central matrix $-I$ changes the sign of each fundamental vector, so on a symmetric product of $n$ such vectors it acts as $(-1)^n$. Consequently

$$V_n\text{ descends to }SO(3) \quad\Longleftrightarrow\quad n\text{ is even} \quad\Longleftrightarrow\quad j\in\mathbb Z_{\ge0}.$$

Highest weight $n$	Spin $j$	Complex dimension	Action of $-I$	SO(3) representation?
$0$	$0$	$1$	$+1$	Yes
$1$	$\tfrac12$	$2$	$-1$	No
$2$	$1$	$3$	$+1$	Yes
$3$	$\tfrac32$	$4$	$-1$	No

The symmetric-power realization is Conrad–Landesman, Example 28.5 and Lemma 28.6, printed pp. 124–125. It makes the sign calculation explicit. A spin-$\tfrac12$ doublet gives an ordinary representation of $SU(2)$ and a projective action of $SO(3)$, rather than an ordinary two-dimensional $SO(3)$ representation.

Try the obstructed doublet and allowed triplet. For a further check, $V_1\otimes V_1=V_0\oplus V_2$: the product of two minus signs is plus, and both resulting summands descend. The tensor-product workbench reproduces it.

SU(3) and PSU(3): dimension does not decide descent

Use the ordered $A_2$ Dynkin labels $(p,q)$, with $(1,0)=\mathbf3$ and $(0,1)=\overline{\mathbf3}$. Put $\omega=e^{2\pi i/3}$ and $z=\omega I_3$. The fundamental carries scalar $\omega$; its dual carries $\omega^{-1}=\omega^2$. A highest-weight $(p,q)$ summand occurs in $\mathbf3^{\otimes p}\otimes\overline{\mathbf3}^{\otimes q}$, on which $z$ acts everywhere by $\omega^{p+2q}$. Therefore

$$\chi_{(p,q)}(z)=\omega^{p+2q},\qquad (p,q)\text{ descends to }PSU(3)\iff p+2q\equiv0\pmod3.$$

This uses the scalar-matrix embedding of the center in Conrad, Centers of simply connected semisimple groups, pp. 1–2. The following dimensions follow from $\dim(p,q)=(p+1)(q+1)(p+q+2)/2$; the center calculation is independent of that dimension formula.

Irrep	$(p,q)$	Dimension	$p+2q\bmod3$	PSU(3) representation?
Fundamental	$(1,0)$	$3$	$1$	No
Antifundamental	$(0,1)$	$3$	$2$	No
Symmetric square	$(2,0)$	$6$	$2$	No
Adjoint	$(1,1)$	$8$	$0$	Yes
Symmetric cube	$(3,0)$	$10$	$0$	Yes

The adjoint action is conjugation, so scalar matrices cancel directly: $zXz^{-1}=X$. The ten-dimensional representation supplies another useful check: three fundamental factors contribute $\omega^3=1$, yet its conjugate is $(0,3)$, a different irrep. Trivial central charge does not imply real type. Compare the adjoint calculation and symmetric cube, then consult the PSU(3) record.

The general SU(N)/Zₖ rule

Let $N\ge2$, let $k$ divide $N$, and specify the subgroup

$$z=e^{2\pi i/N}I_N,\qquad \Gamma_k=\langle z^{N/k}\rangle\subset SU(N).$$

Order the $A_{N-1}$ nodes so that $\omega_r$ labels $\bigwedge^r\mathbb C^N$. The scalar $z$ acts on that exterior power by $e^{2\pi i r/N}$. Taking tensor products of these fundamental representations shows that highest weight $\lambda=\sum_{i=1}^{N-1}a_i\omega_i$ has center charge

$$b=\sum_{i=1}^{N-1}i\,a_i\pmod N,\qquad \chi_\lambda(z^{N/k})=e^{2\pi i b/k}.$$

Hence descent requires exactly $k\mid b$. The tensor construction is explained in Conrad–Landesman, Example 28.10, printed p. 126. Since the cover is simply connected, $\pi_1(SU(N)/\Gamma_k)\simeq\mathbb Z_k$; its center is $\mathbb Z_N/\Gamma_k\simeq\mathbb Z_{N/k}$. These are different invariants.

Worked exercise. The antisymmetric two-index representation of $SU(4)$ has labels $(0,1,0)$, dimension $\binom42=6$ and $b=2$. It descends through $\Gamma_2=\{I,-I\}$ but not through $\Gamma_4$: the respective generator phases are $e^{2\pi i}=1$ and $e^{\pi i}=-1$. With only this matter irrep, the compatible quotient orders are $1$ and $2$. Reproduce the result in Matter & global gauge structure using $N=4$, labels 0,1,0, and quotient orders $2$ and $4$ in turn.

Specify the embedding, and keep the hypotheses

An abstract label such as $\mathbb Z_2$ does not specify a quotient of a product group. In $SU(2)\times SU(2)$, the kernels generated by $(-I,I)$ and $(-I,-I)$ impose, respectively, even $n_L$ and even $n_L+n_R$ on $V_{n_L}\otimes V_{n_R}$. Thus $(n_L,n_R)=(1,1)$ descends only for the second choice. This follows by multiplying the two central signs; no dimension test can replace the map. The product-group workbench retains such embedded generators explicitly.

Two hypotheses delimit the finite-center classification:

• Semisimplicity: $U(1)$ is compact but its universal cover is $\mathbb R\to U(1)$, $t\mapsto e^{it}$, with kernel $2\pi\mathbb Z$. A unitary character $t\mapsto e^{iqt}$, $q\in\mathbb R$, descends exactly for $q\in\mathbb Z$. A compact product containing a circle is not its own universal cover.
• Compactness of the real form: polar decomposition retracts $SL(2,\mathbb R)$ onto $SO(2)$, so $\pi_1\simeq\mathbb Z$. Its universal cover has infinite cyclic center. If $c$ denotes the lift of a rotation from $0$ to $\pi$, then projection to $SL(2,\mathbb R)$ kills $\langle c^2\rangle$, whereas projection to $PSL(2,\mathbb R)$ kills $\langle c\rangle$. The two targets are both non-simply-connected despite having different centers.

The infinite center is recorded by Brodzki–Niblo–Plymen–Wright, §1, printed p. 1 (PDF p. 2); the rotation construction identifies the two kernels. Every finite-dimensional complex representation of this universal cover factors through $SL(2,\mathbb R)$, so none is faithful. This is Etingof, Remark 36.7, printed p. 196. Infinite-dimensional unitary representations constitute a different category; the compact highest-weight tools do not classify them. See the universal-cover record.

Finally, the algebra sees local structure near the identity. For a disconnected group, its other components and their action are additional data. Likewise, representation descent checks whether matter is an honest representation of a proposed gauge group; it does not select magnetic/dyonic line spectra, theta parameters, or establish anomaly cancellation. The distinct line choices are discussed in Aharony–Seiberg–Tachikawa, §§1–2.

Continue with the Lie groups and topology path, the global-form exercises, or the physics of the global gauge group.