The Tensor Product as a Representation
We have classified the irreducible representations of \(\mathfrak{sl}(2,\mathbb{C})\).
For each non-negative integer \(m\) there is exactly one of dimension \(m + 1\) up to
isomorphism, the
irreducible representation
\((\pi_m, V_m)\) obtained from the polynomial representation \(\Pi_m\) of \(SU(2)\).
We have also learned how to
tensor two representations of one group or Lie algebra
into a representation on the tensor-product space.
The natural question now joins these two threads. Given two irreducibles \(V_m\)
and \(V_n\), the tensor product \(V_m \otimes V_n\) is again a representation of
\(\mathfrak{sl}(2,\mathbb{C})\). It is almost never irreducible. How does it
break into irreducible pieces? The answer is the Clebsch-Gordan decomposition,
and it turns out to be governed entirely by a single arithmetic fact. On a tensor
product, the eigenvalues of the diagonal generator add.
The Action and Its Diagonal Generator
We keep the
working basis
\(H, E, F\) of \(\mathfrak{sl}(2,\mathbb{C})\), with \(H\) the generator we diagonalize,
\(E\) raising \(H\)-eigenvalues by \(2\), and \(F\) lowering them by \(2\). For a single
irreducible \((\pi_m, V_m)\) the operator \(\pi_m(H)\) is
diagonalizable
with eigenvalues
\[
m, m - 2, \dots, -m,
\]
each occurring once. We write \(u_m, u_{m-2}, \dots, u_{-m}\) for a corresponding basis of
eigenvectors, so that \(\pi_m(H)\,u_j = j\,u_j\). The eigenvalue \(j\) of a vector under
\(\pi_m(H)\) is its
weight.
On the tensor product \(V_m \otimes V_n\), the Lie algebra acts by the
one-group tensor-product rule
\[
(\pi_m \otimes \pi_n)(X) = \pi_m(X) \otimes I + I \otimes \pi_n(X),
\quad X \in \mathfrak{sl}(2,\mathbb{C}).
\]
The additive form on the right is not a convention but a consequence of differentiating the
group rule \((\Pi_m \otimes \Pi_n)(A) = \Pi_m(A) \otimes \Pi_n(A)\) at the identity, and it
is exactly the structure that makes weights add.
Weights Add
Take eigenvectors \(u_j \in V_m\) and \(v_k \in V_n\), so that \(\pi_m(H)\,u_j = j\,u_j\)
and \(\pi_n(H)\,v_k = k\,v_k\). Applying the diagonal generator to the elementary tensor
\(u_j \otimes v_k\),
\[
\begin{align*}
(\pi_m \otimes \pi_n)(H)\,(u_j \otimes v_k)
&= \bigl(\pi_m(H) \otimes I + I \otimes \pi_n(H)\bigr)(u_j \otimes v_k) \\\\
&= (\pi_m(H)\,u_j) \otimes v_k + u_j \otimes (\pi_n(H)\,v_k) \\\\
&= j\,(u_j \otimes v_k) + k\,(u_j \otimes v_k) \\\\
&= (j + k)\,(u_j \otimes v_k).
\end{align*}
\]
So each elementary tensor \(u_j \otimes v_k\) is again an eigenvector of the diagonal
generator, now with weight \(j + k\). The \((m+1)(n+1)\) tensors \(u_j \otimes v_k\)
form a basis
of \(V_m \otimes V_n\), and they diagonalize \((\pi_m \otimes \pi_n)(H)\) outright.
The entire decomposition problem reduces to bookkeeping. We need only count how many
basis tensors land at each weight.
The same elementary tensors are not, however, in general eigenvectors of the
raising and lowering operators. By the
raising-lowering relations,
\((\pi_m \otimes \pi_n)(E)\) sends a weight-\(w\) vector to weight \(w + 2\) and
\((\pi_m \otimes \pi_n)(F)\) sends it to weight \(w - 2\), just as on a single
irreducible. What changes is that an eigenspace at a given weight may now be more
than one-dimensional, and the raising operator no longer annihilates only the
single top vector. Recovering the irreducible pieces means finding the highest
weight vectors
inside this graded space, the vectors that are
annihilated by \(E\), and then following the chains they generate. That is the content
of the next two sections.
Why a Rotation-Equivariant Network Cares
In a rotation-equivariant network the activations of a layer carry an action of
\(\mathfrak{so}(3) \cong \mathfrak{su}(2)\), and they decompose into irreducible
"type-\(\ell\)" features. These include a type-\(0\) scalar, a type-\(1\) vector, and
higher tensors, each a copy of some \(V_m\). When two such features interact
multiplicatively, the layer forms their tensor product, and the tensor
product of two irreducibles is what we are about to decompose. The arithmetic
just established, that weights add on a tensor product, is the reason the
combined object reorganizes into a predictable list of new type-\(\ell\)
features rather than into an unstructured mixture. The Clebsch-Gordan
decomposition is the rule a layer follows when it multiplies two equivariant
features together.
The Motivating Example: \(V_1 \otimes V_1\)
Before stating the general theorem, we work out the smallest non-trivial case by hand. It
is small enough to see every vector explicitly, yet it already exhibits the full mechanism. We locate the
highest weight, descend with the lowering operator until the chain closes, and
read off the leftover. Let \(V_1 = \mathbb{C}^2\) be the
standard representation
of \(\mathfrak{sl}(2,\mathbb{C})\), for which \(\pi_1(X) = X\). With \(\{e_1, e_2\}\) the
standard basis of \(\mathbb{C}^2\) and \(H = \operatorname{diag}(1, -1)\), we have
\(\pi_1(H)\,e_1 = e_1\) and \(\pi_1(H)\,e_2 = -e_2\), so \(e_1\) and \(e_2\) carry weights
\(+1\) and \(-1\). The representation is irreducible, since an invariant line would be
spanned by an eigenvector of \(\pi_1(H)\), hence by \(e_1\) or \(e_2\), while
\(\pi_1(F)\,e_1 = e_2\) and \(\pi_1(E)\,e_2 = e_1\). By the classification,
\(\mathbb{C}^2\) is therefore a model for \((\pi_1, V_1)\).
The Four Weights
The four elementary tensors \(e_k \otimes e_l\), \(1 \le k, l \le 2\), form a basis of
\(\mathbb{C}^2 \otimes \mathbb{C}^2\). By the additive rule of the previous section, each is
an eigenvector of \((\pi_1 \otimes \pi_1)(H)\) whose weight is the sum of the two factor
weights. The tensor \(e_1 \otimes e_1\) has weight \(+2\), the two tensors
\(e_1 \otimes e_2\) and \(e_2 \otimes e_1\) have weight \(0\), and
\(e_2 \otimes e_2\) has weight \(-2\). The weights run \(2, 0, 0, -2\). The value
\(0\) occurs twice, and the values \(\pm 2\) occur once each. A single
irreducible never repeats a weight, so this four-dimensional space cannot be
irreducible and must split.
Descending from the Top
The largest weight is \(+2\), attained only by \(e_1 \otimes e_1\). Since there is no vector
of weight \(+4\) for it to map to, the raising operator must annihilate it:
\((\pi_1 \otimes \pi_1)(E)\,(e_1 \otimes e_1) = 0\). It is therefore a
highest weight vector,
and the
chain it generates under the lowering operator
spans an irreducible subspace. We follow that chain. Writing
\(L := (\pi_1 \otimes \pi_1)(F) = \pi_1(F) \otimes I + I \otimes \pi_1(F)\) and using
\(\pi_1(F)\,e_1 = e_2\), \(\pi_1(F)\,e_2 = 0\),
\[
\begin{align*}
L\,(e_1 \otimes e_1)
&= (\pi_1(F)\,e_1) \otimes e_1 + e_1 \otimes (\pi_1(F)\,e_1)
= e_2 \otimes e_1 + e_1 \otimes e_2, \\\\
L\,(e_1 \otimes e_2 + e_2 \otimes e_1)
&= 2\,(e_2 \otimes e_2), \\\\
L\,(e_2 \otimes e_2) &= 0.
\end{align*}
\]
The chain closes after three steps. Its three vectors carry weights \(2, 0, -2\) and span a
three-dimensional invariant irreducible subspace, a copy of \(V_2\).
The Leftover
One direction of the four-dimensional space has not been used: the antisymmetric
combination \(e_1 \otimes e_2 - e_2 \otimes e_1\), of weight \(0\). A direct calculation
with the same operators shows that \(\mathfrak{sl}(2,\mathbb{C})\) acts on it as zero,
\[
\begin{align*}
(\pi_1 \otimes \pi_1)(H)\,(e_1 \otimes e_2 - e_2 \otimes e_1) &= 0, \\\\
(\pi_1 \otimes \pi_1)(E)\,(e_1 \otimes e_2 - e_2 \otimes e_1) &= 0, \\\\
(\pi_1 \otimes \pi_1)(F)\,(e_1 \otimes e_2 - e_2 \otimes e_1) &= 0,
\end{align*}
\]
so it spans a one-dimensional invariant subspace on which the action is trivial, a copy
of \(V_0\).
Together with the three-dimensional piece this subspace accounts for all four
dimensions, and the two subspaces meet only in zero, since the four vectors
\(e_1 \otimes e_1\), \(e_1 \otimes e_2 + e_2 \otimes e_1\), \(e_2 \otimes e_2\), and
\(e_1 \otimes e_2 - e_2 \otimes e_1\) are linearly independent. We have shown
\[
V_1 \otimes V_1 \cong V_2 \oplus V_0,
\]
an isomorphism of \(\mathfrak{sl}(2,\mathbb{C})\) representations. The familiar splitting of
a rank-two tensor into its symmetric and antisymmetric parts is exactly this
decomposition. The three symmetric tensors are the \(V_2\), and the single
antisymmetric tensor is the \(V_0\).
Every feature of the general case is already visible here. The weights add. The top
weight is unrepeated and seeds the largest irreducible. Descending with the
lowering operator traces out that irreducible, and what remains assembles into
strictly smaller irreducibles, each appearing once. The next section turns this procedure into a theorem.
The Clebsch-Gordan Theorem
The example generalizes without surprise. The top weight seeds the largest irreducible. One peels it
off, passes to an invariant complement, and repeats, each step lowering the top
weight by two, until nothing is left. We state the result and then carry out exactly this
induction.
Theorem (Clebsch-Gordan Decomposition)
Let \(m\) and \(n\) be non-negative integers with \(m \ge n\). As a representation of
\(\mathfrak{sl}(2,\mathbb{C})\),
\[
V_m \otimes V_n \cong V_{m+n} \oplus V_{m+n-2} \oplus \cdots \oplus V_{m-n}
= \bigoplus_{k=0}^{n} V_{\,m+n-2k}.
\]
Every irreducible appearing on the right occurs exactly once. The decomposition is
multiplicity-free. Here \(V_p\) denotes the irreducible
representation of \(\mathfrak{sl}(2,\mathbb{C})\) of dimension \(p + 1\), unique
up to isomorphism.
The dimensions check at once. The right-hand side has dimension
\(\sum_{k=0}^{n} (m + n - 2k + 1) = (n + 1)(m + 1)\), matching
\(\dim(V_m \otimes V_n) = (m+1)(n+1)\). Multiplicity-freeness is not a general
feature. For tensor products of representations of \(\mathfrak{sl}(3,\mathbb{C})\),
for instance, an irreducible can occur more than once.
Weights and Their Multiplicities
Choose weight bases \(u_m, u_{m-2}, \dots, u_{-m}\) of \(V_m\) and
\(v_n, v_{n-2}, \dots, v_{-n}\) of \(V_n\), where \(\pi_m(H)\,u_j = j\,u_j\) and
\(\pi_n(H)\,v_k = k\,v_k\). The products \(u_j \otimes v_k\) are a basis of
\(V_m \otimes V_n\), and by the additive rule each has weight \(j + k\). The weights range
from \(m + n\) down to \(-(m+n)\) in steps of \(2\), and we record how many basis tensors
realize each.
The top weight \(m + n\) is attained only by \(u_m \otimes v_n\), so its multiplicity is
one.
Lowering by \(2\) to weight \(m + n - 2\), the realizations are \(u_{m-2} \otimes v_n\) and
\(u_m \otimes v_{n-2}\), a multiplicity of two, provided \(n \gt 0\). Each further step of \(-2\)
adds one new realization, because the second index \(k\) gains one more admissible value,
until \(k\) saturates at \(-n\). Concretely the multiplicity of weight \(w\) is the number of pairs
\((j, k)\) with \(j + k = w\), \(j \in \{m, m-2, \dots, -m\}\), and
\(k \in \{n, n-2, \dots, -n\}\). Since \(m \ge n\), this count climbs by one at each
step from \(w = m+n\) until it reaches \(n + 1\) at weight \(w = m - n\). It then
holds steady at \(n + 1\) across the plateau \(m - n \ge w \ge -(m-n)\), and
falls symmetrically by one per step down to \(-(m+n)\).
Peeling Off the Irreducibles
Proof:
First we record that \(V_m \otimes V_n\) is
completely reducible,
in the internal sense that it is a direct sum of irreducible invariant
subspaces. The group \(SU(2)\) is
compact,
so its representation \(\Pi_m \otimes \Pi_n\) on \(V_m \otimes V_n\) is completely
reducible, since
every finite-dimensional representation of a compact group is completely reducible.
Write \(V_m \otimes V_n = U_1 \oplus \cdots \oplus U_r\) with each \(U_i\) invariant
and irreducible under \(SU(2)\). We check that the same subspaces are invariant and
irreducible under \(\mathfrak{sl}(2,\mathbb{C})\). By the
outer-product formula
evaluated at the pair \((X, X)\), the representation of \(\mathfrak{su}(2)\)
associated to \(\Pi_m \otimes \Pi_n\) is \(X \mapsto \pi_m(X) \otimes I + I \otimes \pi_n(X)\),
the restriction of \(\pi_m \otimes \pi_n\) to \(\mathfrak{su}(2)\). Each \(U_i\) is
invariant under this restriction, being invariant under every
\((\Pi_m \otimes \Pi_n)(e^{tX})\) and hence under the derivative at \(t = 0\), because
a subspace of a finite-dimensional space is
closed.
Because \(SU(2)\) is connected (sending \(U\) to its first column identifies it with
the unit sphere in \(\mathbb{C}^2\)), the
group-algebra correspondence
(Point 1), applied to the restriction of \(\Pi_m \otimes \Pi_n\) to \(U_i\), makes
each \(U_i\) irreducible under \(\mathfrak{su}(2)\). Finally
\(\mathfrak{sl}(2,\mathbb{C})\) is the
complexification of \(\mathfrak{su}(2)\),
and the hypothesis \(\mathfrak{su}(2) \cap i\,\mathfrak{su}(2) = \{0\}\) holds because
a matrix that is both skew-Hermitian and Hermitian is zero. The
complex-linear extension \(\pi(X + iY) = \pi(X) + i\,\pi(Y)\)
of the restriction is unique, so it is \(\pi_m \otimes \pi_n\) itself. A subspace
invariant under \(\pi(X)\) and \(\pi(Y)\) is invariant under
\(\pi(X) + i\,\pi(Y)\), so each \(U_i\) is invariant under
\(\mathfrak{sl}(2,\mathbb{C})\), and by the same proposition it is irreducible under
\(\mathfrak{sl}(2,\mathbb{C})\) because it is irreducible under \(\mathfrak{su}(2)\).
Thus \(V_m \otimes V_n\) is completely reducible as a representation of
\(\mathfrak{sl}(2,\mathbb{C})\), and we may split off any invariant subspace and
find an
invariant complement that is again completely reducible.
Now run the induction on the top weight. Consider the vector \(u_m \otimes v_n\), of weight
\(m + n\). No weight \(m + n + 2\) exists for its image under the raising
operator \((\pi_m \otimes \pi_n)(E)\) to occupy, so that image is forced to
be zero. Thus \(u_m \otimes v_n\) is a
highest weight vector,
and the
chain it generates under repeated lowering
spans an invariant irreducible subspace \(W\) whose weights are \(m + n, m + n - 2, \dots,
-(m+n)\), each once. That is, \(W \cong V_{m+n}\).
If \(n = 0\), then \(\dim W = m + 1 = \dim(V_m \otimes V_n)\), so \(W\) is the whole
space and the theorem reads \(V_m \otimes V_0 \cong V_m\). Suppose \(n \gt 0\), and
take an invariant complement \(W'\) with \(V_m \otimes V_n = W \oplus W'\), itself
completely reducible. Weight multiplicities split along this decomposition. Writing
\(v = w + w'\) with \(w \in W\) and \(w' \in W'\), both summands being invariant
under \((\pi_m \otimes \pi_n)(H)\), the eigenvalue equation holds for \(v\) if and
only if it holds for \(w\) and for \(w'\) separately. Hence each weight space of
\(V_m \otimes V_n\) is the direct sum of the corresponding weight spaces of \(W\)
and \(W'\). Comparing dimensions, \(W'\) is spanned by its weight vectors. Because
\(W\) contributes multiplicity exactly one to every weight from \(m + n\) down to
\(-(m+n)\), removing it lowers each multiplicity by one. In particular weight
\(m + n\) no longer appears in \(W'\), and the largest weight surviving in \(W'\)
is \(m + n - 2\), now with multiplicity one. A nonzero vector of that weight in
\(W'\) is sent by \(E\) into the weight-\((m+n)\) part of \(W'\), which is zero, so
it is a highest weight vector, and its chain generates an irreducible subspace of
\(W'\) isomorphic to \(V_{m+n-2}\).
Iterate. At each stage we pass to the invariant complement of all irreducibles extracted so
far. This lowers every remaining multiplicity by one and so lowers the top surviving
weight by two. The process extracts \(V_{m+n}, V_{m+n-2}, \dots\) in turn. It halts when the
complement is exhausted, which by the multiplicity profile occurs precisely after
\(V_{m-n}\) is removed. At that point every weight has been accounted for the
correct number of times, and the remaining space is zero. Collecting the extracted irreducibles,
\[
V_m \otimes V_n \cong \bigoplus_{k=0}^{n} V_{\,m+n-2k},
\]
and since each top weight \(m + n, m + n - 2, \dots, m - n\) was extracted once, no
irreducible repeats.
Why the Count Stops at \(V_{m-n}\)
The lower factor \(V_n\) sets the length of the decomposition. There are \(n + 1\)
summands, one for each admissible value of the second weight \(k\), and the smallest
irreducible that appears is \(V_{m-n}\), the difference of the two highest
weights. The list begins at \(V_{m+n}\), the sum of the two highest weights.
The whole decomposition is thus pinned by two numbers, the sum and the
difference of the highest weights, with everything in between filled in by steps of two.
Decomposition in Practice
The theorem is best absorbed through a concrete case larger than the motivating one. Take
\(m = 4\), \(n = 2\), so that \(V_4 \otimes V_2\) is \(15\)-dimensional. The theorem predicts
\[
V_4 \otimes V_2 \cong V_6 \oplus V_4 \oplus V_2,
\]
of dimensions \(7 + 5 + 3 = 15\). We verify it by the weight count alone, with no operator
algebra.
Reading the Decomposition Off the Weights
The first factor contributes weights \(4, 2, 0, -2, -4\), and the second contributes
\(2, 0, -2\). Adding every pair gives the weight multiplicities of the tensor product, which
we tabulate.
| Weight \(w\) |
Multiplicity |
Realizing tensors \(u_j \otimes v_k\) |
| \(6\) | \(1\) | \(u_4 \otimes v_2\) |
| \(4\) | \(2\) | \(u_2 \otimes v_2, u_4 \otimes v_0\) |
| \(2\) | \(3\) | \(u_0 \otimes v_2, u_2 \otimes v_0, u_4 \otimes v_{-2}\) |
| \(0\) | \(3\) | \(u_{-2} \otimes v_2, u_0 \otimes v_0, u_2 \otimes v_{-2}\) |
| \(-2\) | \(3\) | \(u_{-4} \otimes v_2, u_{-2} \otimes v_0, u_0 \otimes v_{-2}\) |
| \(-4\) | \(2\) | \(u_{-4} \otimes v_0, u_{-2} \otimes v_{-2}\) |
| \(-6\) | \(1\) | \(u_{-4} \otimes v_{-2}\) |
The multiplicities are \(1, 2, 3, 3, 3, 2, 1\). Subtracting the weight profile of \(V_6\)
(one copy of each weight \(6, 4, 2, 0, -2, -4, -6\)) leaves \(0, 1, 2, 2, 2, 1, 0\). The largest surviving weight is \(4\), seeding a
\(V_4\). Subtracting the profile of \(V_4\)
(weights \(4, 2, 0, -2, -4\)) leaves \(0, 0, 1, 1, 1, 0, 0\), the weight profile of \(V_2\).
Nothing remains after that, and the count confirms
\(V_4 \otimes V_2 \cong V_6 \oplus V_4 \oplus V_2\) with each summand once. The peeling induction of the proof is exactly this subtraction of
weight profiles, carried out one irreducible at a time from the top.
The Coupling Rule of an Equivariant Layer
A rotation-equivariant network stores its activations as type-\(\ell\) features, each a
copy of an irreducible of \(\mathfrak{so}(3) \cong \mathfrak{su}(2)\). Under the
lifting criterion for \(SO(3)\)-representations
these are exactly the \(V_m\) with \(m\) even, a type-\(\ell\) feature being a copy
of \(V_{2\ell}\), of dimension \(2\ell + 1\). When a layer multiplies a
type-\(\ell_1\) feature by a type-\(\ell_2\) feature, the product lives in
\(V_{2\ell_1} \otimes V_{2\ell_2}\), and the Clebsch-Gordan decomposition is the rule
that reorganizes it back into a sum of admissible type-\(\ell\) features, with
\(\ell\) ranging in integer steps over \(|\ell_1 - \ell_2|, \dots, \ell_1 + \ell_2\).
Multiplicity-freeness means each output type is produced in exactly one way, so the
coupling carries no ambiguity that the architecture would have to resolve by extra
bookkeeping. This is the algebraic content
behind the tensor-product layers that combine directional features in equivariant
models. The decomposition fixes which output channels a product of two input
channels is allowed to populate.
The construction reaches further than the case worked here. The complete reducibility
that opened the proof, combined with
Schur's lemma,
governs how operators relate the irreducible pieces, and pushing in
that direction leads to the selection rules and reduced matrix elements that organize how a
rotationally symmetric system responds to vector quantities. The decomposition of a tensor
product into irreducibles is the first and most basic of these structural rules.