6  Operations on Representations

Classifying the representations of a group, up to isomorphism, amounts to classifying collections of matrices satisfying the group’s relations up to simultaneous conjugacy. Having an invariant, i.e., a way of attaching an algebraic object to each square matrix that can detect if a pair of matrices are not conjugate, would be invaluable in this endeavor. So: what conjugacy invariants of square matrices are there?

As we have seen, spectral theory plays a special role in studying representations—a natural choice of invariant, then, might be associating a matrix \(A\) to its set of eigenvalues. Remember that the eigenvalues are themselves determined by its characteristic polynomial \[ p_A(t) = t^n + \alpha_1 t^{n-1} + \cdots + \alpha_n, \] another conjugacy invariant. Among the coefficients are \[ \begin{aligned} \alpha_1 & = -(\lambda_1+\cdots+\lambda_n) = -\operatorname{Tr}A, \\ \alpha_n & = (-1)^n \lambda_1 \cdots \lambda_n = (-1)^n \det A, \end{aligned} \] where the \(\lambda_i\) are the solutions of \(p_A(t)\) recorded with multiplicity.

A fundamental object of study in the sequel is the character of a representation \(\rho\), i.e., the function sending elements \(g \in G\) to the trace of the transformation \(\rho(g)\). In some sense, this is because knowing the trace of a matrix \(A\) and its powers is equivalent to knowing its characteristic polynomial. Indeed, since \(A x = \lambda x\) means that \(A^k x = \lambda^k x\), we always have \[ \operatorname{Tr}(A^k) = \lambda_1^k + \cdots + \lambda_n^k. \] Hence the generating function¹ for the sequence \[ \operatorname{Tr}(\mathbb{I}), \operatorname{Tr}(A), \operatorname{Tr}(A^2), \dots, \operatorname{Tr}(A^k), \dots \] is given by \[ \sum_{k=0}^\infty t^k \operatorname{Tr}(A^k) = \sum_{k=0}^\infty \left( \sum_{i=1}^n \lambda_i^k \right) t^k = \sum_{i=1}^n \left( \sum_{k=0}^\infty (\lambda_i t)^k \right) = \sum_{i=1}^n \frac{1}{1-\lambda_i t}. \tag{6.1}\] This formula allows us to recover the set of eigenvalues of \(A\) from only the traces of its powers!²

Example 6.1 Consider the standard representation \(V\) of \(\mathcal{S}_{3}\). As in Example 4.16, we have \[ (1\ 2) \mapsto \begin{pmatrix} -1 & 1 \\ 0 & 1 \\ \end{pmatrix} \qquad \text{ and } \qquad (1\ 2\ 3) \mapsto \begin{pmatrix} 0 & -1 \\ 1 & -1 \\ \end{pmatrix}. \] Call the leftmost matrix \(A\) and the rightmost \(B\), so that \(\operatorname{Tr}A = 0\) and \(\operatorname{Tr}B = -1\). Since \[ A^2 = B^3 = \mathbb{I}\quad \text{ and } \quad B^2 = \begin{pmatrix} -1 & 1 \\ -1 & 0 \\ \end{pmatrix}, \] we have the general formulas \[ \operatorname{Tr}(A^k) = \left\{ \begin{array}{ll} 2 & k \text{ even } \\ 0 & k \text{ odd } \\ \end{array} \right. \quad \text{ and } \quad \operatorname{Tr}(B^k) = \left\{ \begin{array}{rl} 2 & k \equiv 0 \bmod 3 \\ -1 & \text{otherwise}. \\ \end{array} \right. \] We can then compute \[ \begin{split} 2(1+t^2+t^4+\cdots) & = \frac{2}{1-t^2} \\ & = \frac{1}{1-t}+\frac{1}{1+t} \\ \end{split} \] and \[ \begin{split} 2-t-t^2+2t^3-t^4-\cdots & = (1-t)(t+2)(1+t^3+t^6+\cdots) \\ & = \frac{(1-t)(t+2)}{1-t^3} \\ & = \frac{t+2}{t^2+t+1}. \\ & = \frac{t+2}{(t-\zeta)(t-\zeta^2)} \\ & = \frac{t+2}{(\zeta^2 t - 1)(\zeta t - 1)} \\ & = \frac{1}{1-\zeta t} + \frac{1}{1-\zeta^2 t}, \end{split} \] where \(\zeta \coloneqq e^{2\pi i/3}\) satisfies \(\zeta^2+\zeta+1=0\). Hence we conclude that the set of eigenvalues are \(\{1, -1\}\) for \(A\) and \(\{\zeta,\zeta^2\}\) for \(B\), without having to even compute a determinant!

With these notions in mind, we will explore how constructions in linear algebra can be equipped with group actions: given \(G\)-actions on \(V\) and \(W\), what should we make of \(V \oplus W\), \(V^*\), and other associated spaces?

Of special interest is the space of linear maps, \(\operatorname{Hom}(V,W)\); we have seen (as in Example 5.6) that understanding the subspace \(\operatorname{Hom}_G(V,W)\) teaches us about irreducible subrepresentations common to \(V\) and \(W\). Therefore, it would be particularly beneficial to have an action of \(G\) on \(\operatorname{Hom}(V,W)\) that picks out, as its invariants, the subspace of intertwiners (i.e., we want a \(G\)-action so that \(\operatorname{Hom}_G(V,W) = \operatorname{Hom}(V,W)^G\)). This would mean that we could bring to bear ideas like those in Theorem 5.2 to systematize our study of representation theory.³ All constructions described here are in service of this objective.

6.1 Direct Sums

This case is the most clear—indeed, we have already implicitly used this notion in the case of so-called internal direct sums, as in Theorem 5.1, when \(G\) is acting in some coherent way on a larger vector space containing both \(V\) and \(W\). When dealing with abstract vector spaces \(V\) and \(W\) which have no relationship to one another, what group action should we prescribe to the external sum \(V \oplus W\)?

Recall that \(V \oplus W\) is just \(V \times W\) with a linear structure: instead of \((x,y) \in V \times W\), we think of every element in \(V \oplus W\) as a uniquely expressible sum \(x+y\). Given transformations \(A: V \to V\) and \(B: W \to W\), we can define \[ A \oplus B: V \oplus W \to V \oplus W \quad \text{ via } \quad (A \oplus B)(x+y) = Ax + By. \]

Proposition 6.1 If \(A: V \to V\) and \(B: W \to W\) are linear maps, then \[ \operatorname{Tr}(A \oplus B) = \operatorname{Tr}(A) + \operatorname{Tr}(B). \]

Proof. If \(\mathscr{A} \subset V\) and \(\mathscr{B} \subset W\) are bases, then \[ \begin{split} \operatorname{Tr}(A \oplus B) & = \operatorname{Tr}([A \oplus B]_{\mathscr{A} \sqcup \mathscr{B}}) \\ & = \operatorname{Tr}\begin{pmatrix} [A]_{\mathscr{A}} & 0 \\ 0 & [B]_{\mathscr{B}} \end{pmatrix} \\ & = \operatorname{Tr}([A]_{\mathscr{A}}) + \operatorname{Tr}([B]_{\mathscr{B}}) \\ & = \operatorname{Tr}(A) + \operatorname{Tr}(B). \end{split} \]

6.2 Dual Spaces

Remember that, over a field \(F\), the dual space \(V^*\) is the space of linear maps \(\operatorname{Hom}(V,F)\). More casually, every \(f \in V^*\) knows how to gobble up a vector \(x \in V\) and produce a scalar \(f(x) \in F\) in a linear way. Given a basis for \(V\) and the dual basis for \(V^*\), we can think of the vectors in \(V\) and functionals in \(V^*\) as column vectors with their pairing expressed conveniently via the (non-Hermitian!) dot product: \[ f(x) = f^\top x. \]

In deciding how \(G\) should act on \(V^*\) given an existing action on \(V\), this pairing is the fundamental relationship that we should attend to. We would like the action we construct on \(V^*\) to respect this pairing: \[ (g \cdot f)(g \cdot x) = f(x). \] Hence, given \(\rho: G \to \operatorname{GL}(V)\), we will define \[ g \cdot f \coloneqq \rho(g^{-1})^\top f \tag{6.2}\] so that \[ (g \cdot f)(g \cdot x) = \left(\rho(g^{-1})^\top f\right)^\top (\rho(g) x) = f^\top \rho(g^{-1}) \rho(g) x = f^\top x. \] Note that many authors often think of elements in \(V^*\) as row vectors—again, with respect to a dual basis for \(V\)—so that \(f(x)\) is just the matrix multiplication \(fx\), in which case \(g \cdot f = f \rho(g^{-1})\). In any formulation, we have \[ (g \cdot f)(x) \coloneqq f( g^{-1} \cdot x ). \tag{6.3}\]

While the trace of the transpose is the same as the original matrix, and \(\lambda\) is an eigenvalue of \(A\) if and only if \(\tfrac{1}{\lambda}\) is an eigenvalue of \(A^{-1}\), there is not in general a direct formula for the trace of \(A^{-1}\) in terms of \(\operatorname{Tr}A\). In light of Remark 4.3, however, we can think of complex representations of finite groups as maps \(\rho: G \to \operatorname{U}(n)\) given the right basis. In particular, all the eigenvalues \(\lambda\) of the linear maps in the image of our representations have norm \(1\) (Corollary 4.1), which means \(\lambda^{-1} = \overline{\lambda}\). Hence we have the handsome formula: \[ \operatorname{Tr}\left((Q^{-1})^\top\right) = \overline{\operatorname{Tr}Q}. \tag{6.4}\]

We will see in Homework (e.g., Exercise E.3, Exercise F.3) how other constructions involving the dual space allow us to study polynomial functions on \(V\) as \(G\)-representations.

6.3 Tensor Products

Tensors are objects designed to formalize the study of multilinearity, a concept that pervades mathematics, science, and engineering. Unfortunately the pedagogy surrounding tensors often begins with an often unmotivated set of universal properties, emphasizing theorems of existence and uniqueness without any grounding context. Of course it is something of an anachronism to assume that the early creators of these objects started with such universal properties in mind, rather than repeated encounters with some associated phenomena that mathematicians subsequently built up a framework to encompass. As a result, students of tensors can often find the subject intimidating or needlessly abstract.

Aside on multilinear maps

Note

This section is written in response to many students, especially those in physics and chemistry, who have asked for motivation regarding the tensor product. The material in this digression is entirely optional, and students are free to skip straight to Section 6.3.2.

While it is hopeless to succinctly answer the general question, “When does multilinearity come up naturally?” we will focus on a particular historical example from electromagnetism. In particular, we show how the need to “multiply” vectors arises.

Following a discussion from the infamous Classical Electrodynamics (Jackson 1998), mirroring the notation where vectors are written in bold, the change in the mechanical momentum of a multiparticle system in a domain \(D\) is the total Lorentz force: \[ \frac{\mathop{}\!\mathrm{d}{\mathbf{P}_{\mathrm{mech}}}}{\mathop{}\!\mathrm{d}{t}} = \int_D (\rho \mathbf{E} + \mathbf{J} \times \mathbf{B}) \mathop{}\!\mathrm{d}{x} \] We would like to derive an equation similar to the analogous statement for total energy: \[ \frac{\mathop{}\!\mathrm{d}{E_{\mathrm{mech}}}}{\mathop{}\!\mathrm{d}{t}} + \frac{\mathop{}\!\mathrm{d}{}}{\mathop{}\!\mathrm{d}{t}} \int_D \frac{\epsilon_0}{2} \left( \|\mathbf{E}\|^2 + c^2 \|\mathbf{B}\|^2 \right) \mathop{}\!\mathrm{d}{\mathbf{x}} = - \oint_S \mathbf{S} \cdot d \mathbf{a}, \] where \(\mathbf{S}\) is the Poynting vector, \(S\) is the surface bounding \(D\), and the integral on the left hand side can be identified with total field energy. Even if these quantities mean nothing to the reader, we emphasize that is a particularly desirable sort of equation: it is a conservation law, reducing rates of change across \(D\) to a surface integral.

Using Maxwell’s equations to eliminate charge and current densities, \[ \rho = \epsilon_0 \operatorname{div}\mathbf{E} \quad \text{ and } \quad \mathbf{J} = \frac{1}{\mu_0} \operatorname{curl}\mathbf{B} - \epsilon_0 \frac{\mathop{}\!\partial {\mathbf{E}}}{\mathop{}\!\partial {t}}, \] along with Maxwell’s \(\operatorname{div}\mathbf{B} = 0\) and the product rule \[ \frac{\mathop{}\!\partial {}}{\mathop{}\!\partial {t}} (\mathbf{E} \times \mathbf{B}) = \frac{\mathop{}\!\partial {\mathbf{E}}}{\mathop{}\!\partial {t}} \times \mathbf{B} + \mathbf{E} \times \frac{\mathop{}\!\partial {\mathbf{B}}}{\mathop{}\!\partial {t}}, \] we get \[ \begin{split} \frac{\mathop{}\!\mathrm{d}{\mathbf{P}_{\mathrm{mech}}}}{\mathop{}\!\mathrm{d}{t}} & + \frac{\mathop{}\!\mathrm{d}{}}{\mathop{}\!\mathrm{d}{t}} \int_D \epsilon_0 (\mathbf{E} \times \mathbf{B}) \mathop{}\!\mathrm{d}{\mathbf{x}} \\ & = \int_D \epsilon_0 \left( \mathbf{E} \operatorname{div}\mathbf{E} - \mathbf{E} \times \operatorname{curl}\mathbf{E} + c^2 \mathbf{B} \operatorname{div}\mathbf{B} - c^2 \mathbf{B} \times \operatorname{curl}\mathbf{B} \right) \mathop{}\!\mathrm{d}{\mathbf{x}}. \end{split} \] The integral on the left hand side is identified as the total electromagnetic momentum; the integral on the right has a peculiar sort of symmetry to it, but is ungainly. We would like to use the divergence theorem to reach another conversation law, but cannot apply it outright because our integrals involve vectors rather than scalars!

We proceed by using the standard basis to express, say, \(\mathbf{E} = E_1 \mathbf{e}_1 + E_2 \mathbf{e}_2 + E_3 \mathbf{e}_3\). Then \[ \mathbf{E} \operatorname{div}\mathbf{E} - \mathbf{E} \times \operatorname{curl}\mathbf{E} = \sum_{i=1}^3 \left( E_i \operatorname{div}\mathbf{E} - \mathbf{e}_i \cdot (\mathbf{E} \times \operatorname{curl}\mathbf{E}) \right) \cdot \mathbf{e}_i \] We can rearrange such expressions with vector calculus identities such as \[ \begin{aligned} \operatorname{div}(E_i \mathbf{E}) & = E_i \operatorname{div}\mathbf{E} + \mathbf{E} \cdot \operatorname{grad}E_i, \text{ and } \\ \tfrac{1}{2} \operatorname{grad}\|\mathbf{E}\|^2 & = \mathbf{E} \times \operatorname{curl}\mathbf{E} + (\mathbf{E} \cdot \nabla) \mathbf{E}, \end{aligned} \] arriving ultimately at the equation \[ \begin{split} \int_D \left( \mathbf{E} \operatorname{div}\mathbf{E} - \mathbf{E} \times \operatorname{curl}\mathbf{E} \right) \mathop{}\!\mathrm{d}{\mathbf{x}} & = \sum_{i=1}^3 \mathbf{e}_i \int_D \operatorname{div}\left( E_i \mathbf{E} - \tfrac{1}{2} \mathbf{e}_i \|\mathbf{E}\|^2 \right) \mathop{}\!\mathrm{d}{\mathbf{x}} \\ & = \sum_{i=1}^3 \mathbf{e}_i \oint_S \left( E_i \mathbf{E} - \tfrac{1}{2} \mathbf{e}_i \|\mathbf{E}\|^2 \right) \cdot \mathop{}\!\mathrm{d}{\mathbf{a}}, \end{split} \] with similar results for the \(\mathbf{B}\) terms. We have reduced the change in momentum to a surface integral, but its current form is unsatisfactory for a number of reasons: not only did we have to choose a basis, but the integrand looks suspiciously like products of \(\mathbf{E}\) with itself.

Ultimately, we might hope for a way of multiplying vectors, \(\otimes\), satisfying \[ \sum_{i=1}^3 \mathbf{e}_i \otimes E_i \mathbf{E} = \mathbf{E} \otimes \mathbf{E}. \] In our wildest dreams, we might even hope that the seemingly basis-dependent combination that would come up in the second term, \(\sum_i \mathbf{e}_i \otimes \mathbf{e}_i\), turns out to have a basis-independent expression. That is, if \(\{\mathbf{u}_1,\mathbf{u}_2,\mathbf{u}_3\}\) is another orthonormal basis, then we want to have \[ \sum_{i=1}^3 \mathbf{e}_i \otimes \mathbf{e}_i = \sum_{i=1}^3 \mathbf{u}_i \otimes \mathbf{u}_i. \] We are doing the work of dreaming up a new sort of algebraic object! The question is: what should this product \(\otimes\) do on an arbitrary vector space \(V\)? An important observation is that we have not assigned physical meaning to quantities like \(\mathbf{E} \otimes \mathbf{E}\). Instead, we only interact with this expression while integrating, i.e., we want \(\mathbf{E} \otimes \mathbf{E} \cdot \mathop{}\!\mathrm{d}{\mathbf{a}}\) to act like the vector obtained by dotting one of the \(\mathbf{E}\)s with the surface area element \(\mathop{}\!\mathrm{d}{\mathbf{a}}\). Instead of phrasing this in the language of dot products, which depends unnecessarily on our Euclidean structure, we can ask for a way of associating a functional \(f: V \to \mathbb{R}\) and product \(x \otimes y\) to \(f(x) y \in V\).

All told, we seem to be describing a mapping \(\otimes: V \times V \to V \otimes V\), where \(V \otimes V\) is the place (in particular, a vector space) we will keep our fancy new objects in. Let’s emphasize a couple of properties we should require of this pairing.

• Combinations of the form \(x \otimes y\) should be linear in \(x\) and \(y\), reflecting the algebra in manipulating the integrands. That is, for all \(x,y,z \in V\) and \(\alpha, \beta \in F\), we want \[ \begin{aligned} (\alpha x + \beta y) \otimes z & = \alpha x \otimes z + \beta y \otimes z, \\ x \otimes (\alpha y + \beta z) & = \alpha x \otimes y + \beta x \otimes z. \end{aligned} \]

• For all \(f \in V^*\), we have an induced map \[ \begin{aligned} V \otimes V & \to V \\ x \otimes y & \mapsto f(x) y. \end{aligned} \]

We could go on, but at this point we can appeal to the modern framework: tensor products exist, they are characterized (up to canonical isomorphism) by some universal properties, etc. For many a working mathematician, physicist, engineer, etc., we can takeaway that writing expressions like \(\mathbf{E} \otimes \mathbf{E}\) won’t cause trouble so long as we keep to some rules.

As far as the electromagnetism story goes, we can write the conservation law as \[ \frac{\mathop{}\!\mathrm{d}{\mathbf{P}_{\mathrm{mech}}}}{\mathop{}\!\mathrm{d}{t}} + \frac{\mathop{}\!\mathrm{d}{}}{\mathop{}\!\mathrm{d}{t}} \int_D \epsilon_0 (\mathbf{E} \times \mathbf{B}) \mathop{}\!\mathrm{d}{\mathbf{x}} = \oint_S \mathcal{T} \cdot \mathop{}\!\mathrm{d}{\mathbf{a}}, \] where the rightmost integrand is the Maxwell stress tensor given by \[ \mathcal{T} = \epsilon_0 \left( \mathbf{E} \otimes \mathbf{E} + c^2 \mathbf{B} \otimes \mathbf{B} - \frac{1}{2} (\|\mathbf{E}\|^2 + c^2\|\mathbf{B}\|^2) \sum_{i=1}^3 \mathbf{e}_i \otimes \mathbf{e}_i \right). \]

Formal construction

For a full-fledged formal definition of the tensor product, say of the spaces \(V\) and \(W\), we start with a formal vector space spanned by the Cartesian product \(V \times W\). In other words, basis elements of this new space are of the form \((x,y) \in V \times W\)⁴, so it consists of finite formal sums \[ (x_1,y_1)+\dots+(x_k,y_k). \] We then obtain the desired tensor product by quotienting via a subspace cutting out by the relations we want \(\otimes\) to satisfy.

Definition 6.1 (As a quotient) Given vector spaces \(V\) and \(W\) over the same field \(F\), we construct the tensor product \(V \otimes W\) as follows. Consider \(Z\) a vector space with the Cartesian product \(V \times W\) as a basis, together with the subspace \(R\) spanned by the vectors \[ \begin{aligned} (x+x',y) - (x,y) - (x',y),\\ (x,y+y') - (x,y) - (x,y'), \\ (\alpha x,y) - \alpha(x,y), \\ (x,\alpha y) - \alpha(x,y). \end{aligned} \] for all \(x,x' \in V, y,y' \in W,\) and \(\alpha \in F\). We define \[ V \otimes W \coloneqq Z/R, \] where the equivalence class of \((x,y)\) is written \(x \otimes y\).

Remark 6.1. It is not obvious from this definition that the tensor product of two finite-dimensional \(\mathbb{R}\) (or \(\mathbb{C}\)) vector spaces will again be finite-dimensional, since we are taking the quotient of a vector space with uncountable dimension, though this turns out to be true.

Instead of relying on this construction, we mention a more practical perspective and also a universal property that give, respectively, concrete and useful ways of realizing tensors.

Definition 6.2 (From bases) Given vector spaces \(V\) and \(W\) over the same field \(F\), with respective bases \(\{v_1,\dots,v_n\}\) and \(\{w_1,\dots,w_n\}\), the tensor product \(V \otimes W\) is a vector space which has as a basis all formal products \(v_i \otimes w_j\). In particular, \(\dim V \otimes W = \dim V \dim W\). The tensor product of \(x \in V\) and \(y \in W\) is defined from their decomposition on the bases by introducing a distribution property of \(\otimes\). In other words, if \[ x = \alpha_1 v_1 + \cdots + \alpha_n v_n \quad \text{ and } \quad y = \beta_1 w_1 + \cdots + \beta_m w_m, \] then the tensor product of \(x\) and \(y\) is \[ x \otimes y = (\alpha_1 v_1 + \cdots + \alpha_n v_n) \otimes (\beta_1 w_1 + \cdots + \beta_m w_m) = \sum_{i,j} (\alpha_i \beta_j) v_i \otimes w_j. \]

A noteworthy difficulty here is that different choices of bases give rise to different (though canonically isomorphic) tensor products, which is why mathematicians don’t usually take this as the definition of the tensor product. Verifying that this construction is well-defined and reconciles with the first is outside our immediate focus—see, e.g., (Dummit and Foote 2003, 359; Strang 2005, 461).

Another avenue towards understanding tensors is to think of them directly as objects which allow us to translate the theory of multi-linear maps \(V_1 \times \cdots \times V_n \to Z\) into that of linear maps. In this sense, tensors are exactly what we need in order to make such a translation possible. For example, if \(T: V \times W \to Z\) is a bi-linear map, then we always have \[ \begin{split} T(x + x', y) & = T(x,y) + T(x',y), \\ T(x,y + y') & = T(x,y) + T(x,y'), \\ T(\alpha x, y) & = \alpha T(x,y) = T(x,\alpha y), \\ \end{split} \] for any \(x, x' \in V, y, y' \in W,\) and \(\alpha, \beta \in F\). To translate \(T\) into a linear transformation, we need to design a product \((x,y) \mapsto x \otimes y\) to encompass the bi-linearity, i.e., it should satisfy \[ \begin{split} (x + x') \otimes y & = x \otimes y + x' \otimes y, \\ x \otimes ( y + y') & = x \otimes y + x \otimes y', \\ \quad (\alpha x) \otimes y & = \alpha x \otimes y = x \otimes (\alpha y ). \end{split} \tag{6.5}\] These allow us to define a linear map \[ \begin{split} \tilde{T}: V \otimes W & \to Z \\ x \otimes y & \mapsto T(x,y), \end{split} \] extended linearly (not every vector in \(V \otimes W\) is of the form \(x \otimes y\), but rather a sum of such terms).

In summary, tensor products are the unique (up to canonical isomorphism, which is an abstract way of quantifying just how unique we mean) object satisfying the following:

Theorem 6.1 (Universal property of tensors) The tensor product of two vector spaces \(V\) and \(W\) is a vector space denoted \(V \otimes W\) together with a bi-linear map \(\otimes: V \times W \to V \otimes W\) such that, for every bi-linear \(T: V \times W \to Z\), there is a unique linear map \(\tilde{T}: V \otimes W \to Z\) such that \(T = \tilde{T} \circ \otimes\). In other words, the following diagram commutes:

\[ \begin{CD} V \times W @>T>> Z \\ @V\otimes VV @VV\operatorname{\mathbb{I}} V \\ V \otimes W @>>\tilde{T}> Z \end{CD} \tag{6.6}\]

Corollary 6.1 Given a pair of linear transformations \(A: U \to W\) and \(B: V \to Z\), there is a unique linear map \(A \otimes B: U \otimes V \to W \otimes Z\) satisfying \[ (A \otimes B)(x \otimes y) = Ax \otimes By. \]

Proof. The map \(U \times V \to W \times Z\) given by \((x,y) \mapsto (Ax, By)\) satisfies \[ \begin{split} (\alpha x+\beta x',y) & \mapsto (\alpha Ax + \beta Ax',y), \\ (x,\alpha y+\beta y') & \mapsto (x,\alpha By + \beta By'), \end{split} \] from which it follows that the composition \(U \times V \to W \otimes Z\) given by \((x,y) \mapsto Ax \otimes By\) is bi-linear. We conclude by applying the universal property of tensors.

Proposition 6.2 (Composition law) If \(A_1, A_2: V \to V\) and \(B_1, B_2: W \to W\), then \[ (A_1 \otimes B_1) \circ (A_2 \otimes B_2) = (A_1 A_2) \otimes (B_1 B_2). \] In particular, \(\mathbb{I}_V \otimes \mathbb{I}_W\) is the identity on \(\operatorname{End}(V) \otimes \operatorname{End}(W) \cong \operatorname{End}(V \otimes W)\).

Indeed, if \(\rho: G \to \operatorname{GL}(V)\) and \(\sigma: G \to \operatorname{GL}(W)\) are representations, then \[ \begin{split} V \times W & \to V \otimes W \\ (x,y) & \mapsto \rho(g) x \otimes \sigma(g) y \end{split} \] is bi-linear. Thus, we can define \((\rho \otimes \sigma)(g)\) as the unique map \[ \rho(g) \otimes \sigma(g): V \otimes W \to V \otimes W \] sending \(x \otimes y\) to \(\rho(g) x \otimes \sigma(g) y\). More fashionably, we can write \[ g \cdot (x \otimes y) \coloneqq (g \cdot x) \otimes (g \cdot y) \tag{6.7}\] for the desired induced action \(G \to \operatorname{GL}(V \otimes W)\).

Proposition 6.3 If \(A: V \to V\) and \(B: W \to W\) are linear maps, then \[ \operatorname{Tr}(A \otimes B) = \operatorname{Tr}(A) \operatorname{Tr}(B). \]

Proof. We can compute this directly by fixing bases \(\{ v_1, \dots, v_n \} \subset V\) and \(\{ w_1, \dots, w_n \} \subset W\), so that the \(v_i \otimes w_j\) constitute a basis for \(V \otimes W\). If \(A: V \to V\) and \(B: W \to W\), we can write these as matrices \((a_{i,j})\) and \((b_{i,j})\) with respect to our bases. The matrix for \(A \otimes B\) in our basis for \(V \otimes W\) is called the Kronecker product of \(A\) and \(B\), given by \[ \begin{aligned} (A \otimes B)(v_i \otimes w_j) & = (A v_i \otimes B w_j) \\ & = \left( \sum_{k=1}^n a_{k, i} v_k \right) \otimes \left( \sum_{\ell=1}^n b_{\ell, j} w_\ell \right) \\ & = \sum_{k,\ell} (a_{k, i} b_{\ell, j}) v_k \otimes w_\ell. \end{aligned} \] We compute the trace by summing over the \((i,j)\)-coefficient for each \(v_i \otimes w_j\): \[ \operatorname{Tr}(A \otimes B) = \sum_{i, j} (a_{i, i} b_{j, j}) = \left( \sum_{i=1}^n a_{i,i} \right)\left( \sum_{j=1}^m b_{j,j} \right) = \operatorname{Tr}(A) \operatorname{Tr}(B). \]

We have established the necessary machinery for a neat alternative argument, without relying on bases:

Proof (Proof by universal property). Note that the assignment \[ \begin{split} \operatorname{End}(V) \times \operatorname{End}(W) & \to F \\ (A,B) & \mapsto \operatorname{Tr}(A)\operatorname{Tr}(B) \end{split} \] is bi-linear. Hence we have a linear map \[ \begin{split} \operatorname{End}(V) \otimes \operatorname{End}(W) & \stackrel{f}{\longrightarrow} F \\ A \otimes B & \mapsto f(A \otimes B) = \operatorname{Tr}(A) \operatorname{Tr}(B). \end{split} \] Furthermore, we can compute \[ \begin{split} f((A_1 \otimes B_1) \circ (A_2 \otimes B_2)) & = f(A_1A_2 \otimes B_1B_2) \\ & = \operatorname{Tr}(A_1A_2) \operatorname{Tr}(B_1B_2) \\ & = \operatorname{Tr}(A_2A_1) \operatorname{Tr}(B_2B_1) \\ & = f(A_2A_1 \otimes B_2B_1) \\ & = f((A_2 \otimes B_2) \circ (A_1 \otimes B_1)). \end{split} \] Moreover, since \[ f(\mathbb{I}_V \otimes \mathbb{I}_W) = \operatorname{Tr}(\mathbb{I}_V) \operatorname{Tr}(\mathbb{I}_W) = \dim V \dim W = \dim V \otimes W, \] we can apply Exercise A.4 to conclude that \(f = \operatorname{Tr}\).

6.4 Hom Spaces

If \(V\) and \(W\) are vector spaces, the set \(\operatorname{Hom}(V,W)\) of linear maps \(V \to W\) is itself a vector space. Given bases \(\mathscr{A} = \{v_1,\dots,v_n\} \subset V\) and \(\mathscr{B} = \{w_1,\dots,w_n\} \subset W\), encoding a linear map \(L\) as the \(m \times n\) matrix \({}_{\mathscr{B}}[L]_{\mathscr{A}}\) gives an isomorphism \(\operatorname{Hom}(V,W) \cong \operatorname{Mat}_{m \times n}(F)\). In other words, there is a basis \(\{ E_{i,j}: 1 \leq i \leq m, 1 \leq j \leq n \}\) of \(\operatorname{Hom}(V,W)\) given by \[ E_{i,j}(\alpha_1 v_1 + \cdots + \alpha_n v_n) = \alpha_j w_i. \] Notice that \({}_{\mathscr{B}}[E_{i,j}]_{\mathscr{A}}\) is simply the matrix that is \(1\) in the \((i,j)\) entry but zero elsewhere. In particular, we see that \(\dim \operatorname{Hom}(V,W) = mn\).

If \(V\) and \(W\) are equipped with \(G\)-actions, what action can we equip \(\operatorname{Hom}(V,W)\) with? There are several choices, but one in particular is desirable in light of yet another way that one can understand tensor products, namely the natural isomorphism \[ \begin{split} V^* \otimes W & \stackrel{\Phi}{\to} \operatorname{Hom}(V,W) \\ f \otimes y & \mapsto \Phi(f \otimes y)(x) = f(x) y. \end{split} \tag{6.8}\] While this array of symbols looks intimidating on first glance, the word “natural” is applied for good reason—somehow, there is only really one way we could try to write down such a map, and it turns out to be the isomorphism we are after!

Let’s think about this correspondence in the context of bases, so we can think of vectors as columns, functionals as rows, and maps as matrices. Take \(\{v_1,\dots,v_n\} \subset V\) and \(\{w_1,\dots,w_n\} \subset W\) as bases, then fix the dual basis \(\{f_1,\dots,f_n\} \subset V^*\). We compute \[ \Phi(f_j \otimes w_i)(\alpha_1 v_1 + \cdots + \alpha_n v_n) = f_j(\alpha_1 v_1 + \cdots + \alpha_n v_n) w_i = \alpha_j w_i \] In other words, \(\Phi(f_i \otimes w_j) = E_{i,j}\), the \((i,j)\) basis element of \(\operatorname{Hom}(V,W)\)! Therefore our correspondence \(\Phi\) sends a basis of \(V^* \otimes W\) to a basis of \(\operatorname{Hom}(V,W)\), i.e., it is an isomorphism. Informally, the isomorphism \(V^* \otimes W \cong \operatorname{Hom}(V,W)\) is a way of matching row vectors in \(F^n\) and column vectors in \(F^m\) into matrices of \(\operatorname{Mat}_{m \times n}(F)\).

The upshot of all this is we already have a way for \(G\) to act on dual spaces and tensor products, so we can think of the isomorphism \(\Phi\) as teaching us how to act on \(\operatorname{Hom}(V,W)\). In other words, we want to define our action of \(G\) so that \[ \Phi(g \cdot (f \otimes y)) = g \cdot \Phi(f \otimes y). \] Writing \(\rho: G \to \operatorname{GL}(V)\) and \(\sigma: G \to \operatorname{GL}(W)\), we know that \[ \begin{split} \Phi(g \cdot (f \otimes y))(x) & = \left( (g \cdot f) \otimes (g \cdot y) \right)(x) = \overbrace{(g \cdot f)(x)}^{\in F} \overbrace{(g \cdot y)}^{\in W} \\ & = f( g^{-1} \cdot x ) \; (g \cdot y), \end{split} \] i.e., \(g\) acting on the vector \(\Phi(f \otimes y)(g^{-1} \cdot x) \in W\). So, given \(L \in \operatorname{Hom}(V,W)\), we should define \[ (g \cdot L)(x) \coloneqq g \cdot L( g^{-1} \cdot x ). \]

We can rephrase this condition as requiring the following diagram to commute:

\[ \begin{CD} V @>{\rho(g)}>> V \\ @VV{L}V @VV{g \cdot L}V \\ W @>{\sigma(g)}>> W \end{CD} \tag{6.9}\]

which should be compared with Equation 5.1. Hence \(\Phi\) is an isomorphism of \(G\)-representations.

Theorem 6.2 If \(V\) and \(W\) are \(G\)-representations and \(\varphi \in \operatorname{Hom}(V,W)\), then \(\varphi\) is \(G\)-linear (i.e., an intertwiner) if and only if \(\varphi \in \operatorname{Hom}(V,W)^G\) (i.e., \(\varphi\) is fixed by \(G\)). That is, \[ \operatorname{Hom}_G(V,W) = \operatorname{Hom}(V,W)^G. \tag{6.10}\]

Proof. If \(\varphi\) is an intertwiner and \(g \in G\), then \[ (g \cdot \varphi)(x) = g \cdot \varphi(g^{-1} \cdot x) = g \cdot \left( g^{-1} \cdot \varphi(x) \right) = \varphi(x), \] so \(\varphi\) is fixed by \(G\). On the other hand, if \(\varphi\) is fixed by \(g \in G\) and \(x \in V\), take \(x' = g \cdot x\). Thus \[ \varphi(x') = (g \cdot \varphi)(x') = g \cdot \varphi(g^{-1} \cdot x'), \] i.e., \(\varphi(g \cdot x) = g \cdot \varphi(x)\). If this holds for all \(g \in G\), then \(\varphi\) is \(G\)-linear.

Dummit, D. S., and R. M. Foote. 2003. Abstract Algebra. Wiley.

Jackson, John David. 1998. Classical Electrodynamics. 3rd ed. John Wiley & Sons.

Strang, Gilbert. 2005. Linear Algebra and Its Applications. 4th ed. Brooks/Cole, Cengage Learning.

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1. If you’ve never heard of generating functions, they are handy objects for keeping track of sequences of numbers. The idea is we package them all up into a single object—namely an infinite series that might even converge to some tidy rational function—rather than thinking of them as an unwieldy list.

2. This might not seem impressive, since we need to know \(A^k\) for all \(k \in \mathbb{Z}^+\). Remember that, for an element of finite order—in particular, any \(A\) in the image of a representation of a finite group—this list is finite!

3. The great miracle of representation theory is that we will be able to do this!

4. Compare this Definition 2.8 and Theorem 2.3. In particular, the direct sum \(V \oplus W\) gives a way of making \(V \times W\) itself into a vector space, but this formulation of the tensor product makes \(V \times W\) into a basis for some very large space that we proceed to quotient into \(V \otimes W\).