Two Groups, One Algebra
Every representation of a matrix Lie group differentiates to a representation of
its Lie algebra. This is the content of the
induced Lie algebra homomorphism.
The reverse question is more delicate. Given a representation of the Lie algebra,
does it integrate back to a representation of the group? For a
simply connected
group the answer is always yes, and the correspondence between group and algebra
representations is exact. When the group is not simply connected, the answer can be
no, and the failure is governed entirely by the group's fundamental group. This
section sets up the sharpest instance of the phenomenon, and the rest of the page
resolves it completely.
The two groups in question share a single Lie algebra. We established that
\(\mathfrak{su}(2) \cong \mathfrak{so}(3)\). The two are the same three-dimensional
real Lie algebra, written in two matrix guises. Yet the groups they sit inside are
topologically different. The group \(SU(2)\) is homeomorphic to the three-sphere
\(S^3\) and is simply connected. The group \(SO(3)\) is homeomorphic to
\(\mathbb{RP}^3\) and has
fundamental group \(\mathbb{Z}/2\mathbb{Z}\).
They are joined by the
double cover
\(\Phi : SU(2) \to SO(3)\), a surjective two-to-one homomorphism with kernel
\(\{I, -I\}\), whose derivative at the identity is precisely the isomorphism
\(\phi : \mathfrak{su}(2) \to \mathfrak{so}(3)\).
Where the Asymmetry Comes From
Because \(SU(2)\) is simply connected, every representation of \(\mathfrak{su}(2)\)
lifts to a representation of \(SU(2)\). We have, moreover, already classified the
representations of \(\mathfrak{su}(2)\). Complexifying, a representation of
\(\mathfrak{su}(2)\) extends complex-linearly to a representation of its
complexification
\(\mathfrak{sl}(2;\mathbb{C})\), and the
classification of irreducible \(\mathfrak{sl}(2;\mathbb{C})\)-representations
shows that the irreducible ones are exactly the
polynomial representations
\(\pi_m\), one in each dimension \(m + 1\) for integer \(m \geq 0\). Each \(\pi_m\)
is irreducible and arises by differentiating a representation of the group
\(SU(2)\). So on the \(SU(2)\) side there is no obstruction at all. Every
irreducible Lie algebra representation comes from the group.
Transporting through the isomorphism \(\phi\), the irreducible representations of
\(\mathfrak{so}(3)\) are exactly the maps
\[
\sigma_m = \pi_m \circ \phi^{-1}, \quad m = 0, 1, 2, \dots,
\]
one in each dimension \(m + 1\). The question that remains is asymmetric to the
\(SU(2)\) case. For which \(m\) does \(\sigma_m\) lift to a representation
\(\Sigma_m\) of the group \(SO(3)\), that is, to an honest homomorphism on
\(SO(3)\) satisfying \(\Sigma_m(e^{X}) = e^{\sigma_m(X)}\) for every
\(X \in \mathfrak{so}(3)\)? The next section gives the answer in a single line, and
it is not "all \(m\)."
The Lifting Criterion
The obstruction is a parity. An irreducible representation of \(\mathfrak{so}(3)\)
lifts to the group \(SO(3)\) exactly when its highest weight \(m\) is even,
equivalently exactly when its dimension is odd.
Theorem (Lifting Criterion for \(SO(3)\)-Representations)
Let \(\sigma_m = \pi_m \circ \phi^{-1}\) be the irreducible complex representation of
\(\mathfrak{so}(3)\) of dimension \(m + 1\), where \(m \geq 0\) is an integer. If
\(m\) is even, there is a representation \(\Sigma_m\) of the group \(SO(3)\) such that
\(\Sigma_m(e^{X}) = e^{\sigma_m(X)}\) for every \(X \in \mathfrak{so}(3)\). If
\(m\) is odd, no such representation of \(SO(3)\) exists.
The condition that \(m\) be even is equivalent to the condition that the dimension
\(\dim V_m = m + 1\) be odd. Thus it is the odd-dimensional irreducible
representations of \(\mathfrak{so}(3)\) that descend from group representations of
\(SO(3)\), while the even-dimensional ones live only on the cover \(SU(2)\). The
smallest excluded case is \(m = 1\), the two-dimensional representation. It is a
perfectly good representation of \(\mathfrak{so}(3)\), and of \(SU(2)\), but it
does not exist as a representation of \(SO(3)\).
Both halves of the criterion trace to the kernel \(\{I, -I\}\) of the double cover.
Lifting \(\sigma_m\) to \(SO(3)\) means defining a value on each rotation \(R\),
and each \(R\) is the image of a pair \(\{U, -U\}\) in \(SU(2)\). A well-defined
lift requires the representation built on \(SU(2)\) to assign \(U\) and \(-U\) the
same operator, that is, to send the kernel element \(-I\) to the identity. Whether
it does is decided by the parity of \(m\), and the proof is the computation of a
single matrix exponential.
Proof
The entire argument turns on one rotation and one eigenvalue computation. Choose the
rotation about the third axis by angle \(2\pi\). In the
standard basis of \(\mathfrak{so}(3)\)
its generator is \(E_3\), the skew-symmetric matrix
\[
E_3 = \begin{pmatrix} 0 & -1 & 0 \\\\ 1 & 0 & 0 \\\\ 0 & 0 & 0 \end{pmatrix},
\quad e^{2\pi E_3} = I.
\]
A full turn returns to the identity in the group. This is the single global fact
about \(SO(3)\) that the proof needs. We track what a candidate lift \(\Sigma_m\)
would have to do to it.
The Eigenvalue Computation
We must evaluate \(\sigma_m(E_3)\). Recall that
\(\sigma_m = \pi_m \circ \phi^{-1}\), where
\(\phi : \mathfrak{su}(2) \to \mathfrak{so}(3)\) is the
isomorphism \(\mathfrak{su}(2) \cong \mathfrak{so}(3)\)
sending \(\tfrac{1}{2}F_k \mapsto E_k\). Hence
\(\phi^{-1}(E_3) = \tfrac{1}{2}F_3\), the diagonal generator of
\(\mathfrak{su}(2)\). Since \(\pi_m\) was analyzed on the complexified algebra in
the first place, we pass to the
complexified basis,
where \(F_3 = -iH\), and obtain
\[
\phi^{-1}(E_3) = \tfrac{1}{2}F_3 = -\tfrac{i}{2}H,
\]
so that
\[
\sigma_m(E_3) = -\tfrac{i}{2}\,\pi_m(H).
\]
The operator \(\pi_m(H)\) is the one we diagonalized in classifying these
representations. In the
weight basis
\(u_0, u_1, \dots, u_m\) it is diagonal with eigenvalues \(m, m-2, \dots, -m\), the
eigenvalue on \(u_j\) being \(m - 2j\). Therefore \(\sigma_m(E_3)\) is diagonal in
the same basis, with eigenvalue \(-\tfrac{i}{2}(m - 2j)\) on \(u_j\), and the
operator \(e^{2\pi\sigma_m(E_3)}\) is diagonal with eigenvalues
\[
\begin{align*}
e^{2\pi \cdot \left(-\frac{i}{2}(m - 2j)\right)}
&= e^{-\pi i (m - 2j)} \\\\
&= (-1)^{m - 2j} = (-1)^{m}.
\end{align*}
\]
Every diagonal entry is the same number \((-1)^m\), independent of \(j\), so
\[
e^{2\pi\sigma_m(E_3)} = (-1)^m\, I.
\]
This is the crux. A half-turn's worth of phase, \((-1)^m\), records the parity of
\(m\) and nothing else.
Odd \(m\): No Lift Exists
Proof (odd case):
Suppose \(m\) is odd and a representation \(\Sigma_m\) of \(SO(3)\) existed with
\(\Sigma_m(e^X) = e^{\sigma_m(X)}\) for every \(X \in \mathfrak{so}(3)\). Apply it to
\(X = 2\pi E_3\). On one hand \(e^{2\pi E_3} = I\), and a homomorphism must send the
identity to the identity:
\[
\Sigma_m\bigl(e^{2\pi E_3}\bigr) = \Sigma_m(I) = I.
\]
On the other hand, the defining relation gives
\[
\Sigma_m\bigl(e^{2\pi E_3}\bigr) = e^{2\pi \sigma_m(E_3)} = (-1)^m\, I = -I,
\]
since \(m\) is odd. Thus \(I = -I\) on the representation space \(V_m\), which
is false, since \(V_m\) is nonzero and \(2I \neq 0\) on it. The contradiction
shows no such \(\Sigma_m\) can exist.
Even \(m\): A Lift Exists
For even \(m\) the same computation removes the obstruction rather than creating one. The
representation \(\pi_m\) of \(\mathfrak{su}(2)\) is the differential of the group
representation \(\Pi_m\) of \(SU(2)\) on the
degree-\(m\) polynomials.
We define \(\Sigma_m\) on \(SO(3)\) by pushing \(\Pi_m\) down through the double cover,
and the computation above is exactly what makes this well-defined.
Proof (even case):
The kernel of the
double cover
\(\Phi : SU(2) \to SO(3)\) is \(\{I, -I\}\). The nontrivial kernel element is
\(-I \in SU(2)\), and it is itself an exponential: in \(\mathfrak{su}(2)\),
\[
\begin{align*}
e^{2\pi \cdot \frac{1}{2}F_3} = e^{\pi F_3}
&= e^{\pi \cdot (-iH)} \\\\
&= \mathrm{diag}\bigl(e^{-\pi i}, e^{\pi i}\bigr) = -I.
\end{align*}
\]
Applying the group representation \(\Pi_m\) and using that its differential is
\(\pi_m\),
\[
\begin{align*}
\Pi_m(-I) = \Pi_m\bigl(e^{\pi F_3}\bigr)
&= e^{\pi\, \pi_m(F_3)} = e^{2\pi \cdot \frac{1}{2}\pi_m(F_3)} \\\\
&= e^{2\pi \sigma_m(E_3)} = (-1)^m\, I.
\end{align*}
\]
Because \(m\) is even, \(\Pi_m(-I) = I\). Now each rotation \(R \in SO(3)\) is the
image under \(\Phi\) of a pair \(\{U, -U\} \subset SU(2)\), and
\[
\Pi_m(-U) = \Pi_m(-I)\,\Pi_m(U) = \Pi_m(U).
\]
The two preimages carry the same operator, so the rule
\[
\Sigma_m(R) := \Pi_m(U), \quad R = \Phi(U),
\]
does not depend on which preimage is chosen, and \(\Sigma_m\) is therefore
well-defined on \(SO(3)\). It is a homomorphism because \(\Pi_m\) and \(\Phi\)
are, and it satisfies \(\Sigma_m(e^X) = e^{\sigma_m(X)}\) by construction,
since differentiating \(\Sigma_m \circ \Phi = \Pi_m\) at the identity recovers
\(\sigma_m \circ \phi = \pi_m\). Thus \(\Sigma_m\) is the desired
representation of \(SO(3)\).
Integer and Half-Integer Spin
The parity that decides liftability has a name in physics. Label each irreducible
representation not by its highest weight \(m\) but by the spin \(\ell = m/2\), so
that the dimension is \(2\ell + 1\). The criterion then reads as follows. A
representation of \(\mathfrak{so}(3)\) comes from a representation of the group
\(SO(3)\) exactly when \(\ell\) is an integer. The integer-spin representations,
with \(\ell = 0, 1, 2, \dots\) and dimensions \(1, 3, 5, \dots\), are honest
representations of rotations. The half-integer ones, with
\(\ell = \tfrac{1}{2}, \tfrac{3}{2}, \dots\) and dimensions \(2, 4, \dots\), are
not. They live only on the double cover \(SU(2)\) and are called
"spin" representations precisely because they have no realization on \(SO(3)\)
itself.
The smallest spin representation, \(\ell = \tfrac{1}{2}\) (highest weight
\(m = 1\), dimension \(2\)), is the state space of the electron. Our odd-case
computation is the mathematical content of a famous physical statement. Rotating an
electron's wavefunction by a full \(2\pi\) does not return it to itself but
multiplies it by \(-1\). That sign is exactly the value \((-1)^m = -1\) we found
for \(e^{2\pi\sigma_1(E_3)}\). A rotation by \(4\pi\), twice around, multiplies by
\((-1)^2 = 1\) and restores the state, which is the group-theoretic shadow of the
fact that the \(4\pi\) loop is contractible in \(SO(3)\) while the \(2\pi\) loop is
not. The
order-two fundamental group
of \(SO(3)\) and the sign \((-1)^m\) are the same fact seen twice.
Which Features a Rotation-Equivariant Network Can Carry
A network built to respect 3D rotations stores its activations in channels that
transform under irreducible representations of the rotation group. A
"type-\(\ell\)" feature occupies a \((2\ell+1)\)-dimensional channel and
rotates by the corresponding representation when the input is rotated. The
architectures used in practice, on molecules, point clouds, and physical
fields, act on data living in ordinary space, where the symmetry that
physically applies is \(SO(3)\). A rotation is a rotation, with no \(\pm\)
bookkeeping attached to the input. The lifting criterion is therefore the
statement of which feature types such a network can use directly. Integer-spin
features (scalars \(\ell = 0\), vectors \(\ell = 1\), and higher tensors)
transform under genuine \(SO(3)\) representations and are exactly the
type-\(\ell\) features these networks are assembled from.
Half-integer-spin features have no consistent \(SO(3)\) transformation rule at
all. The would-be assignment is two-valued, sending a single rotation to a
matrix and its negative. A layer that tried to carry one could not decide which
value to use, and the failure is not an engineering inconvenience but the same
obstruction this page proved, namely the kernel \(\{I, -I\}\) of the double
cover refusing to map to the identity. Spinor features enter only when the
relevant symmetry is genuinely \(SU(2)\), as in the quantum-mechanical setting
where the \(\pm\) sign carries physical meaning. For data in plain Euclidean
space, the representations are pinned down with no freedom: one feature type in
each odd dimension, their weights laid out by the classification, their
liftability settled by a single parity.