Spin Representations and Descent to Rotations

A representation of $SU(2)$ gives an ordinary representation of $SO(3)$ exactly when the double cover’s kernel acts trivially. It gives an action on all rays exactly when that kernel acts scalarly. This distinction precisely separates integer spin, half-integer spin, and mixtures of the two. The physical introduction and double-cover picture are in SO(3) and SU(2); this page supplies the representation-level descent test.

Required background. Combining Ray Representations explains why different block phases need not preserve a combined ray. Helpful background. SO(3) and SU(2) introduces the covering map.

A quotient representation must forget the kernel

Let $p:SU(2)\to SO(3)$ be the covering homomorphism, with kernel $\{I,-I\}$. Let $D$ be a unitary representation of $SU(2)$.

To define an ordinary representation by $R\mapsto D(u)$ where $p(u)=R$, the answer must be independent of whether the lift is $u$ or $-u$. Since

$$D(-u)=D(-I)D(u),$$

this is equivalent to $D(-I)=I$.

For an action on rays, independence requires only that $D(-I)$ multiply every vector by one common phase. Thus $D(-I)$ must be scalar. Because $(-I)^2=I$, that scalar can only be $+1$ or $-1$.

Necessity is not just a convenient convention. If a linear operator preserves every ray, phase uniqueness forces it to be scalar. A matrix that is $+I$ on one summand and $-I$ on another fails this test on rays with both components nonzero.

The spin label determines the central sign

In the irreducible spin-$j$ representation, a $2\pi$ rotation is represented by $e^{-2\pi iJ_z/\hbar}$. Its magnetic eigenvalues are $m\hbar$ with $m=j,j-1,\ldots,-j$. Therefore every eigenvector acquires the same phase:

$$D_j(-I)=(-1)^{2j}I.$$

Integer-spin representations descend to ordinary representations of $SO(3)$. Half-integer-spin representations give well-defined ray actions of $SO(3)$ while remaining ordinary vector representations of $SU(2)$.

The angular-momentum spectrum used here is supplied by the spin representation theory. The point is its consequence for the kernel, not a second derivation of the ladder construction or the spinor rotation formula.

Reducible representations require a common spin parity

For a finite or Hilbert direct sum of spin types,

$$D=\bigoplus_j D_j\otimes I_{\mathcal M_j},$$

the central element acts as the direct sum of $(-1)^{2j}$ on the occupied components. It is scalar if and only if all nonzero components have the same integer-versus-half-integer parity.

For example, $D_0\oplus D_1$ descends to an ordinary rotation representation. The representation $D_{1/2}\oplus D_{3/2}$ descends to a projective rotation action on all rays. In contrast, $D_0\oplus D_{1/2}$ has

$$D(-I)=I_1\oplus(-I_2).$$

A vector with components in both summands changes its relative sign under the two possible lifts of one rotation, so it does not define an unambiguous full ray action of $SO(3)$ through this construction.

This is a group-choice statement. The direct sum is a perfectly valid ordinary representation of $SU(2)$. Inferring a physical superselection rule requires additionally insisting on the unextended rotation group’s action and specifying the physical state and observable structure. Rotation representation theory alone is not the relativistic spin-statistics theorem.

Tensor products multiply the central signs

For two spin factors,

$$(D_{j_1}\otimes D_{j_2})(-I) =(-1)^{2j_1+2j_2}I.$$

Two half-integer factors therefore have integer central parity. Their total-spin decomposition contains only integer spins, consistently with the ordinary $SO(3)$ action. One integer and one half-integer factor retain half-integer parity.

This is a quick global consistency check on angular-momentum addition. The Clebsch–Gordan coefficients determine the detailed decomposition, but the central element already constrains which parity can occur.

Direct sums and tensor products answer different physical questions. One compares alternative transformation sectors; the other combines simultaneous subsystems. Treating both as “adding spin representations” can hide this distinction.

A global obstruction can be invisible locally

The Lie algebras of $SO(3)$ and $SU(2)$ are isomorphic. Their local commutator relations therefore do not decide whether a vector representation descends through the double cover. The central sign records information obtained by following a full rotation loop.

Nor can one remove the half-integer obstruction by shifting the generator of every rotation axis independently. An isolated circle subgroup admits phase conventions that change its endpoint phase, but these choices must also satisfy the relations connecting rotations about different axes.

The physical meaning of comparing the spinor sign with an interferometer reference is treated in Spinors and 2π Rotations. A scalar sign on one entire state and a relative sign between controlled branches are different experimental operations.

Exercise: classify three constructions

Which of $D_1\oplus D_{3/2}$, $D_{1/2}\otimes D_{1/2}$, and $D_{1/2}\oplus D_{5/2}$ descends to an ordinary $SO(3)$ representation? Which gives a projective action on all rays?

Solution

The first has mixed central signs and gives no full ray descent through the stated quotient. The tensor product has sign $(-1)(-1)=+1$ and descends ordinarily. The last direct sum has the common sign $-1$, so its ray action descends but its given vector representation does not descend ordinarily.

References

• B. C. Hall, Lie Groups, Lie Algebras, and Representations: An Elementary Introduction, 2nd ed., Graduate Texts in Mathematics 222, Springer, 2015 — $SU(2)$ representations, covering groups, and descent.
• K. Landsman, Foundations of Quantum Theory: From Classical Concepts to Operator Algebras, Fundamental Theories of Physics 188, Springer, 2017, Chapter 5 — projective rotation symmetry and spin.