Two Groups, One Algebra
Every representation of a matrix Lie group differentiates to a representation of
its Lie algebra. This is the content of
differentiating a representation.
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. The general theorem behind that answer is stated
here without proof, and the argument below does not use it. When a connected 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
For \(SU(2)\) the general theorem is not needed, because we already know every
irreducible representation of \(\mathfrak{su}(2)\). Every representation of
\(\mathfrak{su}(2)\)
extends uniquely and complex-linearly
to a representation of the
complexification
\(\mathfrak{sl}(2,\mathbb{C})\), and it is irreducible exactly when its extension
is. The hypothesis of that extension,
\(\mathfrak{su}(2) \cap i\,\mathfrak{su}(2) = \{0\}\), holds because a matrix that
is both skew-Hermitian and Hermitian is zero. The
classification of irreducible \(\mathfrak{sl}(2,\mathbb{C})\)-representations
then shows that, up to isomorphism, the irreducible representations of
\(\mathfrak{su}(2)\) are precisely the representations \(\pi_m\) obtained by
differentiating the
polynomial representations \(\Pi_m\)
of \(SU(2)\), one in each dimension \(m + 1\) for integer \(m \geq 0\). 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, up to isomorphism, the maps
\[
\sigma_m = \pi_m \circ \phi^{-1}, \quad m = 0, 1, 2, \dots,
\]
one in each dimension \(m + 1\). Indeed, \(\phi\) is an isomorphism, so composing
with \(\phi^{-1}\) is a bijection between representations of the two algebras that
preserves invariant subspaces. 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 \(SU(2)\). The natural
candidate is the representation \(\Pi_m\) built on \(SU(2)\), pushed down through
\(\Phi\), and this works only if \(\Pi_m\) assigns \(U\) and \(-U\) the same
operator, that is, sends the kernel element \(-I\) to the identity. Whether it does
is decided by the parity of \(m\), and when it does not, the same computation rules
out every other candidate as well. Both facts come from 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 odd case 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. The monomial \(z_2^m\) is a
highest weight vector
of weight \(m\) for \(\pi_m\): by the formulas for \(\pi_m\) on
monomials,
\(\pi_m(E)\) annihilates it and \(\pi_m(H)\) multiplies it by \(m\). Its
chain
\(u_0, u_1, \dots, u_m\) spans an invariant subspace of dimension
\(m + 1 = \dim V_m\), hence all of \(V_m\), and on this basis \(\pi_m(H)\) is
diagonal with eigenvalues \(m, m-2, \dots, -m\), the eigenvalue on \(u_j\) being
\(m - 2j\) (Point 3 of the chain proposition). 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
\(\Pi_m(e^{Z}) = e^{\pi_m(Z)}\) for \(Z \in \mathfrak{su}(2)\), the defining
property of the
differential \(\pi_m\),
\[
\begin{align*}
\Pi_m(-I) = \Pi_m\bigl(e^{\pi F_3}\bigr)
&= e^{\pi\, \pi_m(F_3)} = e^{2\pi \cdot \frac{1}{2}\pi_m(F_3)} \\\\
&= e^{2\pi \sigma_m(E_3)} = (-1)^m\, I.
\end{align*}
\]
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, so \(\Sigma_m\) is well-defined on
\(SO(3)\) and satisfies \(\Sigma_m \circ \Phi = \Pi_m\). It is a homomorphism.
If \(R = \Phi(U)\) and \(R' = \Phi(U')\), then \(RR' = \Phi(UU')\), so
\[
\Sigma_m(RR') = \Pi_m(UU') = \Pi_m(U)\,\Pi_m(U') = \Sigma_m(R)\,\Sigma_m(R').
\]
It is continuous. The group \(SU(2)\) is
compact,
the group \(SO(3)\) is Hausdorff because it is a
metric space
under the distance of \(M_3(\mathbb{R})\), and \(\Phi\) is continuous, since
every entry of \(\Phi_U\) is a polynomial in the entries of \(U\) and their
complex conjugates. By the
quotient-map clause of the
closed map lemma,
a subset of \(SO(3)\) is therefore open exactly when its preimage under the
surjection \(\Phi\) is open. For an open set \(O \subseteq GL(V_m)\) we have
\(\Phi^{-1}(\Sigma_m^{-1}(O)) = \Pi_m^{-1}(O)\), which is open because \(\Pi_m\)
is continuous, so \(\Sigma_m^{-1}(O)\) is open. Hence \(\Sigma_m\) is a
representation
of \(SO(3)\).
Finally, let \(X \in \mathfrak{so}(3)\) and put
\(Y = \phi^{-1}(X) \in \mathfrak{su}(2)\). The map \(\Phi\) takes values in
\(SO(3) \subseteq GL(3, \mathbb{R})\), so it is a real representation of
\(SU(2)\), and its differential is \(\phi\).
Differentiating it
gives \(\Phi(e^{Y}) = e^{\phi(Y)} = e^{X}\), and hence
\[
\begin{align*}
\Sigma_m(e^{X}) = \Sigma_m\bigl(\Phi(e^{Y})\bigr)
&= \Pi_m(e^{Y}) \\\\
&= e^{\pi_m(Y)} = e^{\sigma_m(X)},
\end{align*}
\]
using \(\Pi_m(e^{Y}) = e^{\pi_m(Y)}\) once more and
\(\sigma_m(X) = \pi_m(\phi^{-1}(X)) = \pi_m(Y)\). 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. These half-integer spin representations live only on the double
cover \(SU(2)\) and have no realization on \(SO(3)\) itself.
The smallest half-integer spin representation, \(\ell = \tfrac{1}{2}\) (highest
weight \(m = 1\), dimension \(2\)), is the spin 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
Many networks built to respect 3D rotations store their 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, and the symmetry they are built
to respect contains \(SO(3)\), some of them also tracking reflections. 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 the rotations of data in
plain Euclidean space, the representations are pinned down with no freedom: one
irreducible complex representation, hence one feature type, in each odd
dimension, their weights laid out by the classification, their liftability
settled by a single parity.