Constructions on Representations

Direct Sums

We now have a working vocabulary — a representation of a matrix Lie group, its associated Lie algebra representation, the notion of an irreducible representation, and the intertwining maps that compare two of them. The constructions of this page are, in a precise sense, not new. The direct sum of vector spaces, the tensor product, and the dual space are operations of linear algebra we have already built; nothing about them required a group. What this page adds is the group. Once a group acts on the spaces being combined, each of these three operations carries the action along with it, and the result is again a representation. The constructions were waiting in the linear algebra for a symmetry to turn them into representation theory.

We take them in turn — direct sum, tensor product, dual — beginning with the simplest. The direct sum makes precise the idea of running several representations side by side without interaction.

That these formulas define representations is a direct check. For the group case, each factor is a homomorphism, so for \(A, B \in G\), \[ (\Pi_1 \oplus \cdots \oplus \Pi_k)(AB)(v_1, \ldots, v_k) = \big( \Pi_1(AB) v_1, \ldots \big) = \big( \Pi_1(A)\Pi_1(B) v_1, \ldots \big), \] which is exactly \((\Pi_1 \oplus \cdots \oplus \Pi_k)(A)\) applied to \((\Pi_1 \oplus \cdots \oplus \Pi_k)(B)(v_1, \ldots, v_k)\); continuity and \((\Pi_1 \oplus \cdots \oplus \Pi_k)(I) = I\) are immediate. The Lie algebra version is the same computation with the bracket in place of the product.

The construction is transparent in matrix terms. Choosing a basis of each \(V_j\) and concatenating them into a basis of \(V_1 \oplus \cdots \oplus V_k\) puts every operator \((\Pi_1 \oplus \cdots \oplus \Pi_k)(A)\) into block-diagonal form, \[ (\Pi_1 \oplus \cdots \oplus \Pi_k)(A) = \begin{pmatrix} \Pi_1(A) & & \\ & \ddots & \\ & & \Pi_k(A) \end{pmatrix}, \] with no off-diagonal coupling between the summands. The blocks evolve independently: the group acts within each \(V_j\) and never mixes one summand into another.

This independence is precisely why the direct sum is the most transparent source of reducible representations. Each summand \(V_j\), sitting inside \(V_1 \oplus \cdots \oplus V_k\) as the subspace whose other coordinates vanish, is invariant under every \((\Pi_1 \oplus \cdots \oplus \Pi_k)(A)\); as soon as there is more than one summand of positive dimension, this is a nontrivial invariant subspace, so a direct sum of two or more nonzero representations is never irreducible. The reverse direction — whether a given representation can be broken apart as a direct sum of irreducible pieces — is the substantive question of complete reducibility, and it is governed by the intertwining maps between the pieces. The remaining two constructions of this page, by contrast, build spaces on which the group acts in a genuinely entangled way.

Tensor Products of Spaces

The second construction rests on the tensor product of vector spaces, which we have already met. Before building representations on it we recall the underlying linear algebra and fix the form in which we will use it. Given finite-dimensional vector spaces \(U\) and \(V\), the tensor product \(U \otimes V\) is a new space built from formal products \(u \otimes v\) of an element \(u \in U\) with an element \(v \in V\); a general element is a linear combination \[ a_1\, u_1 \otimes v_1 + a_2\, u_2 \otimes v_2 + \cdots + a_n\, u_n \otimes v_n, \] and the symbol \(\otimes\) is bilinear, so that \[ \begin{align*} (u_1 + a\, u_2) \otimes v &= u_1 \otimes v + a\, u_2 \otimes v, \\\\ u \otimes (v_1 + a\, v_2) &= u \otimes v_1 + a\, u \otimes v_2. \end{align*} \] The product is not commutative: even when \(U = V\), the elements \(u \otimes v\) and \(v \otimes u\) are in general distinct.

Although the spaces of our primary interest are complex, the construction is identical over \(\mathbb{R}\) and over \(\mathbb{C}\); the basis and dimension count below hold verbatim in either case, and we apply them to complex representations without further comment. The decisive structural fact is that this construction converts bilinear maps into linear maps, a property that determines \(U \otimes V\) up to canonical isomorphism without reference to any basis.

Concretely, if \(e_1, \ldots, e_n\) is a basis of \(U\) and \(f_1, \ldots, f_m\) is a basis of \(V\), then bilinearity lets every element be written in terms of the products \(e_j \otimes f_k\), and these form a basis of \(U \otimes V\). In particular \[ \dim(U \otimes V) = (\dim U)(\dim V). \] The basis-free counterpart of this fact is the universal property: there is a bilinear map \((u, v) \mapsto u \otimes v\) from \(U \times V\) into \(U \otimes V\) through which every bilinear map out of \(U \times V\) factors uniquely as a linear map out of \(U \otimes V\). That is, for any bilinear \(\psi : U \times V \to X\) there is a unique linear \(\tilde{\psi} : U \otimes V \to X\) with \(\tilde{\psi}(u \otimes v) = \psi(u, v)\). This is the working principle behind every construction in this section: to define a linear map on \(U \otimes V\), it suffices to give a bilinear recipe on pairs \((u, v)\), and the universal property supplies a unique, well-defined linear extension.

The point is subtle enough to be worth stating plainly, since we will lean on it repeatedly. To pin down a linear map on \(U \otimes V\) one could instead write down its values on the basis \(\{ e_j \otimes f_k \}\), but then its independence of the chosen basis would need a separate check. One could try defining it directly on the products \(u \otimes v\), but a single element of \(U \otimes V\) can be written as such a combination in many ways, so one would have to verify that the value does not depend on which decomposition is used. The universal property dispatches both concerns at once: a bilinear recipe on pairs is exactly the data that extends to one well-defined linear map, with no basis chosen and no consistency check left to the reader. Each operator we build below is defined this way, so its well-definedness will never be in question.

Operators on a Tensor Product

The universal property delivers operators on \(U \otimes V\) from operators on the factors. If \(A : U \to U\) and \(B : V \to V\) are linear, the assignment \((u, v) \mapsto (Au) \otimes (Bv)\) is bilinear, so it extends to a unique linear operator \(A \otimes B\) on \(U \otimes V\) characterized by \[ (A \otimes B)(u \otimes v) = (Au) \otimes (Bv). \] The same recipe applies to operators between different spaces. A short computation on elementary tensors — which span \(U \otimes V\) — gives the composition rule \[ (A_1 \otimes B_1)(A_2 \otimes B_2) = (A_1 A_2) \otimes (B_1 B_2). \]

In coordinates this operator is concrete. Relative to the basis \(\{ e_j \otimes f_k \}\), the matrix of \(A \otimes B\) is precisely the Kronecker product of the matrices of \(A\) and \(B\), and the composition rule above is the mixed-product property of the Kronecker product. The abstract operator and its matrix are two views of the same object: the basis-free definition fixes what \(A \otimes B\) is, while the Kronecker product is how it is computed once bases are chosen.

Tensor Products of Representations

With the tensor product of spaces and of operators in hand, we can let a group act on a tensor product. There are two ways to do this, both standard, and the distinction between them is the source of most of the bookkeeping in the subject. The first takes representations of two possibly different groups and produces a representation of their direct product; the second takes two representations of the same group and produces another representation of that same group.

The Outer Tensor Product

That this is a representation follows at once from the composition rule for tensor products of operators: for \((A_1, B_1), (A_2, B_2) \in G \times H\), \[ (\Pi_1 \otimes \Pi_2)\big((A_1, B_1)(A_2, B_2)\big) = \Pi_1(A_1 A_2) \otimes \Pi_2(B_1 B_2) = \big(\Pi_1(A_1)\Pi_1(A_2)\big) \otimes \big(\Pi_2(B_1)\Pi_2(B_2)\big), \] and by that composition rule this equals \(\big(\Pi_1(A_1) \otimes \Pi_2(B_1)\big)\big(\Pi_1(A_2) \otimes \Pi_2(B_2)\big)\), which is \((\Pi_1 \otimes \Pi_2)(A_1, B_1)\,(\Pi_1 \otimes \Pi_2)(A_2, B_2)\). The product group \(G \times H\) is itself a matrix Lie group — realized as block-diagonal pairs inside \(GL(n + m; \mathbb{C})\) — and its Lie algebra is the direct sum \(\mathfrak{g} \oplus \mathfrak{h}\). The corresponding Lie algebra representation has a form that is worth deriving in full, because the answer is not the one a first guess would suggest.

The additive form is forced, and the alternative is not a valid choice. The naive guess \((\pi_1 \otimes \pi_2)(X, Y) = \pi_1(X) \otimes \pi_2(Y)\) fails to be a representation of \(\mathfrak{g} \oplus \mathfrak{h}\) for a structural reason: a Lie algebra representation must be a linear map on \(\mathfrak{g} \oplus \mathfrak{h}\), and \((X, Y) \mapsto \pi_1(X) \otimes \pi_2(Y)\) is bilinear in the pair, not linear — doubling \(X\) and tripling \(Y\) multiplies it by six, whereas a linear map would scale the two contributions separately. Differentiation resolves the matter automatically: the product rule converts the multiplicative pairing at the group level into the additive pairing at the algebra level, with the identity operator marking the factor that is, momentarily, held fixed.

The Inner Tensor Product

The second construction takes two representations of a single group \(G\) and regards their tensor product as a representation of \(G\) itself, rather than of \(G \times G\).

This is the previous construction composed with a diagonal restriction. At the group level one restricts \(\Pi_1 \otimes \Pi_2\) from \(G \times G\) to the diagonal copy of \(G\) sitting inside it as the pairs \((A, A)\); at the algebra level the diagonal \(\mathfrak{g} \hookrightarrow \mathfrak{g} \oplus \mathfrak{g}\), \(X \mapsto (X, X)\), turns the additive form of the previous proposition into \(\pi_1(X) \otimes I + I \otimes \pi_2(X)\). The notation \(\Pi_1 \otimes \Pi_2\) is therefore genuinely ambiguous: the same symbol denotes a representation of \(G \times G\) or of \(G\) according to which construction is meant, and one must say which. In practice the reading is fixed by the argument the representation is fed: the outer product accepts a pair \((A, B)\) of group elements and reads off \(\Pi_1(A) \otimes \Pi_2(B)\), whereas the inner product accepts a single element \(A\) and reads off \(\Pi_1(A) \otimes \Pi_2(A)\). The two agree precisely along the diagonal, which is what the restriction above records. The distinction also surfaces in the algebra formulas: the outer product carries an independent pair \((X, Y)\) and the inner product the single \(X\) substituted into both slots.

The inner tensor product is where the subject acquires its arithmetic. If \(\Pi_1\) and \(\Pi_2\) are irreducible, their tensor product as a representation of \(G\) is in general not irreducible, and the problem of decomposing it as a direct sum of irreducibles is a definite computational question. For the rotation group this decomposition is the rule for combining angular momenta in quantum mechanics; in the representation-theoretic description of equivariant networks on geometric data it is the operation that fuses two feature channels of given types into channels of new types. We take up that decomposition, for the representations of \(SU(2)\) built earlier, in the next page.

Dual Representations

The third construction acts not on a product of spaces but on the dual space \(V^*\), the space of linear functionals on \(V\). A group acting on \(V\) induces an action on functionals, and the natural way to transport a linear operator to the dual is the dual map. For a linear operator \(A\) on \(V\), its dual (or transpose) operator \(A^\top\) on \(V^*\) is defined by \[ (A^\top \varphi)(v) = \varphi(A v), \qquad \varphi \in V^*,\; v \in V. \] Relative to a basis of \(V\) and its dual basis, the matrix of \(A^\top\) is the ordinary transpose of the matrix of \(A\) — the plain transpose, not the conjugate transpose, even when \(V\) is complex. A short check from the definition gives the order-reversing composition law \[ (AB)^\top = B^\top A^\top. \]

This order reversal is the whole difficulty. A representation must preserve the order of multiplication, so the assignment \(g \mapsto \Pi(g)^\top\) is not a representation: by the law above it reverses products. The remedy is to compose with the inverse, which restores the order, and this is exactly the definition.

Both formulas define representations, and in each the corrective ingredient is essential. At the group level, the inverse converts the order-reversing transpose into an order-preserving map: using \((AB)^\top = B^\top A^\top\), \[ \Pi^*(g) \Pi^*(h) = \big[\Pi(g^{-1})\big]^\top \big[\Pi(h^{-1})\big]^\top = \big[\Pi(h^{-1}) \Pi(g^{-1})\big]^\top = \big[\Pi\big((gh)^{-1}\big)\big]^\top = \Pi^*(gh), \] where the third equality uses \(\Pi(h^{-1})\Pi(g^{-1}) = \Pi(h^{-1} g^{-1}) = \Pi((gh)^{-1})\). At the algebra level the minus sign plays the same role: differentiating \(\Pi^*(e^{tX}) = [\Pi(e^{-tX})]^\top\) at \(t = 0\) brings down a factor of \(-1\) from the inverse in the exponent and the transpose from the dual, giving \[ \left. \frac{d}{dt} \right|_{t=0} \Pi^*(e^{tX}) = -\,\pi(X)^\top, \] which is \(\pi^*(X)\). One verifies directly that \(\pi^*\) so defined respects the bracket: \(\pi^*([X, Y]) = [\pi^*(X), \pi^*(Y)]\), the two transposes and two minus signs combining to reproduce the bracket without a leftover sign. Dropping either the inverse in \(\Pi^*\) or the minus sign in \(\pi^*\) destroys the homomorphism property.

Two structural facts complete the picture. First, taking the dual twice returns to the original representation: the canonical isomorphism \(V \cong V^{**}\), which identifies \(v \in V\) with evaluation at \(v\), intertwines \(\Pi\) with \((\Pi^*)^*\), so that \((\Pi^*)^* \cong \Pi\). Second, the dual of an irreducible representation is again irreducible: an invariant subspace of \(V^*\) under \(\Pi^*\) is the annihilator of an invariant subspace of \(V\) under \(\Pi\), and the correspondence between a subspace and its annihilator is inclusion-reversing and dimension-complementary, so a proper nonzero invariant subspace on one side would force one on the other. Consequently \(\Pi^*\) is irreducible exactly when \(\Pi\) is.

For some groups the dual carries no new information. When a representation is unitary — \(\Pi(g)\) preserving a Hermitian inner product, as every representation of a compact group can be arranged to be — the dual representation is isomorphic to the complex conjugate representation \(\overline{\Pi}\), because for unitary operators the inverse-transpose equals the entrywise conjugate. Whether \(\Pi^*\) is then isomorphic to \(\Pi\) itself depends on the group: it holds for every representation of \(SU(2)\), where each irreducible representation is self-dual, but fails for \(SU(3)\), whose standard representation on \(\mathbb{C}^3\) and its dual are inequivalent — the distinction physicists record by calling them the quark and antiquark representations. The three constructions of this page — direct sum, tensor product, and dual — together generate, from a handful of basic representations, the full library on which the decomposition theory of the next pages operates.