《表示理论导引》第二章习题选做

\(\newcommand{\CC}{\mathbb C} \newcommand{\ZZ}{\mathbb Z} \newcommand{\gg}{\mathfrak g} \newcommand{\UU}{\mathcal U} \DeclareMathOperator{\Id}{Id} \DeclareMathOperator{\End}{End} \DeclareMathOperator{\Hom}{Hom} \DeclareMathOperator{\ch}{char} \DeclareMathOperator{\Tr}{Tr} \DeclareMathOperator{\sl}{\mathfrak{sl}} \DeclareMathOperator{\Span}{span}\)

Chapter 2 Basic notions of representation theory

2.3 Representations

Problem 2.3.15. Let \(V\) be a nonzero finite dimensional representation of an algebra \(A\). Show that it has an irreducible subrepresentation. Then show by example that this does not always hold for infinite dimensional representations.

Proof. Let \(W\) be one nonzero subrepresentation of \(V\) with smallest dimension, then clearly \(W\) is irreducible.
But when \(V\) is infinite dimensional, for example, \(A=k[x]\) and \(V=A\), then each nonzero subrepresentation of \(V=A\) is a nonzero ideal, which has the form \((f)\). Since \(0 \subsetneq (xf)\subsetneq (f)\), \((f)\) cannot be irreducible. \(\square\)

Problem 2.3.16. Let \(A\) be an algebra over a field \(k\). The center \(Z(A)\) of \(A\) is the set of all elements \(z\in A\) which commute with all elements of \(A\). For example, if \(A\) is commutative then \(Z(A) = A\).
(a) Show that if \(V\) is an irreducible finite dimensional representation of \(A\) then any element \(z \in Z(A)\) acts in \(V\) by multiplication by some scalar \(\chi_V(Z)\). Show that \(\chi_V\colon Z(A)\to k\) is a homomorphism. It is called the central character of \(V\).

Proof. Since \(z\in Z(A)\), consider the linear transform \(\rho(z) \colon V\to V\), since it commutes with \(A\), it gives a homomorphism of representation. Since \(V\) is irreducible, by Schur's lemma, \(\rho(z)\) has the form \(\chi_V(z) \cdot I\).

Take some nonzero \(v\), then \(\chi_V\) can be identified by \(zv = \chi_V(z)v\). Therefore, for all \(x, y\in Z(A)\) we have \((x+y)v = xv + yv = (\chi_V(x) + \chi_V(y))\) and \(xyv = x \cdot \chi_V(y)v = \chi_V(x) \chi_V(y)v\), thus we have \(\chi_V\) is a homomorphism. \(\square\)

(b) Show that if \(V\) is an indecomposable finite dimensional representation of \(A\) then for any \(z \in Z(A)\), the operator \(\rho(z)\) by which \(z\) acts in \(V\) has only one eigenvalue \(\chi_V (z)\), equal to the scalar by which \(z\) acts on some irreducible subrepresentation of \(V\). Thus \(\chi_V \colon Z(A) \to k\) is a homomorphism, which is again called the central character of \(V\).

Proof. For \(z\in Z(A)\), let \(\lambda_1,\dots, \lambda_n\) be the different eigenvalues of \(\rho(z)\), let their generalized eigenspace be \(W_1,\dots,W_n\), we know that \(V = W_1\oplus \cdots \oplus W_n\), and each \(W_i\) is a subrepresentation, thus we must have \(n=1\). So each \(\rho(z)\) only has one eigenvalue.

For \(x, y\in Z(A)\), since \(\rho(x)\) and \(\rho(y)\) commutes, let \(W\) be the eigenspace of \(\rho(x)\), then \(W\) is also \(\rho(y)\)-invariant, thus \(\rho(y)|_W\) has an eigenvector \(v\), with \(xv = \chi_V(x)v\) and \(yv = \chi_V(y)V\). We can conclude that \(\chi_V\) is a homomorphism. \(\square\)

(c) Does \(\rho(z)\) in (b) have to be a scalar operator?

No. For the indecomposable representation of \(k[x]\), for example, \(k[x] / (x - \lambda)^n\), for \(n > 1\), clearly we can see that \(\rho(x)\) is not a scalar operator.

Problem 2.3.18 Prove the following "Infinite dimensional Schur's lemma" (due to Dixmier): Let \(A\) be an algebra over \(\mathbb C\) and \(V\) be an irreducible representation of \(A\) with at most countable basis. Then any homomorphism of representations \(\phi\colon V\to V\) is a scalar operator.

Proof. Consider \(\CC(\phi) \subset \End_A(V)\) as a subalgebra. Let \(v\) be a nonzero vector of \(V\), then \(W = Av\) is a subrepresentation of \(V\), thus \(W=V\). Let \(a_1,a_2,\dots\) be a set of representatives such that \(a_1v,a_2v,\dots\) forms a basis of \(V\), then each \(\psi \in \End_A(V)\) can be determined by \(\psi(a_iv) = a_i \psi(v)\), thus \(\End_A(V)\) has countably dimension. Therefore \(C(\phi)\) has countably dimension.
Suppose \(\phi\) is not a scalar, we show that \(\phi\) is a transcendental extension. Suppose \(\phi\) is annihilated by some polynomial \(f(x) = (x-\alpha_1)\cdots(x-\alpha_n)\). Since the subspace annihilated by \(\phi - \alpha_i\) is a subrepresentation, we must have \(\phi - \alpha_i\) is an isomorphism, then \(f(\phi)\) is an isomorphism, contraidiction.
Therefore, we have \(\phi\) is transcendental over \(\CC\). We know that the sets \(\{1/(x-\alpha)\} \subset \CC(x)\) is linearly independent, thus \(\CC(\phi)\) has uncountable dimension. \(\square\)

2.5 Quotients

Problem 2.5.1. Let \(A = k[x_1,\dots,x_n]\) and let \(I\neq A\) be any ideal in \(A\) containing all homogeneous polynomials of degree \(\geq N\). Show that \(A/I\) is an indecomposable representation of \(A\).

Proof. Suppose we have the decomposition \(A/I = V_1 \oplus V_2\), let \(f\in A\) be the element that \(f\cdot 1 = (1, 0)\) and \(g = 1 - f\). One of them satisfies the constant term is nonzero, we can assume \(f = f_0 + O(x)\). Therefore, we can find a polynomial \(h\) such that \(fh = 1 + O(x^N)\), thus \(f\cdot h = 1\) in \(A/I\), but \((1, 0)\) is not invertible, contraidiction. So such decomposition does not exist. \(\square\)

2.7 Examples of algebras

Problem 2.7.4 Let \(A\) be the Weyl algebra.
(a) If \(\ch k = 0\), what are the finite dimensional representations of \(A\)? What are the two-sided ideals in \(A\)?

For the finite dimensional space \(V\), we have \(\Tr(\rho(x)\rho(y) - \rho(y)\rho(x)) = \Tr(I) = \dim V\), but on another hand, \(\Tr(\rho(x)\rho(y) - \rho(y)\rho(x)) = \Tr(\rho(x)\rho(y)) - \Tr(\rho(y)\rho(x)) = 0\). This gives a contraidiction, so we cannot have a representation on \(V\).
Consier a quotient \(A/I\), suppose \(I\) is nonzero, then we have some nontrivial relation \(f = \sum c_{ij} x^i y^j = 0\). Let \(t\) be the degree of \(y\), one can verify that \(y^tf - fy^t\) is a nonzero polynomial of \(x\) in \(A\). Similarly, \(I\) contains a polynomial of \(y\). In conclusion, \(A/I\) has finite dimension, then we must have \(A/I=0\). Thus the ideals of \(A\) can only be \(A\) or \(0\).

Suppose for the rest of the problem that \(\ch k = p\).
(b) What is the center of \(A\)?

For \(z\in Z(A)\), we only need to verify that \(z\) commutes with \(x\) and \(y\). Note that \(y x^n - x^n y = n x^{n-1}\), and \(y^nx - xy^n = nx^{n-1}\). Clearly we have \(x^p, y^p\in Z(A)\). At last we show that \(Z(A) = \langle x^p, y^p \rangle\). Suppose \(f = \sum c_{ij}x^iy^j\) does not lie in \(\langle x^p, y^p \rangle\), then if some \(p\nmid i\) has nonzero term, we can verify that \(yf - fy\neq 0\), if \(p\nmid j\), then \(xf - fx \neq 0\).

(c) Find all irreducible finite dimensional representations of \(A\).

Let \(V\) be a finite dimensional representation of \(A\). Let \(v\) be an eigenvalue of \(y\), since \(V\) is irreducible, we have \(Av = V\). Since \(v\) is the eigenvector of \(y\), we have \(\{ v, xv, \dots, x^iv, \dots\}\) spans \(V\). Therefore, we have \(n=\dim V\) such that \(\{v, \dots, x^{n-1}v\}\) forms a basis of \(V\). Since \(x^p\in Z(A)\), we know that \(x^p\) acts constantly on \(V\), thus \(n \leq p\). The trace argument shows that \(p \mid n\), so we must have \(n = p\).

Now we have \(x^p v = \lambda v\) and \(yv = \mu v\) for some eigenvalues \(\lambda, \mu\). Furthermore, we have \(y x^i v = \mu x^{i}v + i x^{i-1} v\). This is the representation that \(x, (\partial+\mu)\) acts on \(k[x]/(x^p-\lambda)\).

2.8 Quivers

Problem 2.8.11. Let \(A\) be a \(\ZZ_+\)-graded algebra, i.e., \(A = \bigoplus_{n\geq 0}A[n]\) and \(A[n] \cdot A[m] \subset A[n + m]\). If \(A[n]\) is finite dimensional, it is useful to consider the Hilbert series \(h_A(t) = \sum \dim A[n] t^n\) (the generating function of dimensions of \(A[n]\)). Often this series converges to a rational function, and the answer is written in the form of such a function. For example, if \(A = k[x]\) and \(\deg(x^n) = n\), then

\[h_A(t) = 1 + t + t^2 + \cdots + t^n + \cdots = \frac 1{1-t}. \]

Find the Hilbert series of the following graded algebras:
(a) \(A = k[x_1, \dots , x_m]\) (where the grading is by degree of polynomials).

Since \(\dim A[n]\) is the number of solutions for the equation \(d_1+\cdots+d_m = n\), we have

\[h_A(t) = \sum_n \binom{n+m-1}{n} t^n = \frac 1{(1-t)^m}. \]

(b) \(A = k \langle x_1,\dots,x_m \rangle\) (the grading is by length of words).

\[h_A(t) = \sum_{n} m^n t^n = \frac 1{1-mt}. \]

(c) \(A\) is the exterior (= Grassmann) algebra \(\bigwedge k[x_1,\dots,x_m]\) generated over some field \(k\) by \(x_1,\dots,x_m\) with the defining relations \(x_ix_j + x_jx_i = 0\) and \(x^2_i = 0\) for all \(i,j\) (the grading is by degree).

\[h_A(t) = \sum_n \binom m n t^n = (1+t)^m. \]

(d) \(A\) is the path algebra \(P_Q\) of a quiver \(Q\) (the grading is defined by \(\deg(p_i) = 0\), \(\deg(a_h) = 1\)).

Let \(M_Q\) be the adjacency matrix of \(Q\), we have

\[h_A(t) = \sum_n j^{\mathsf T} M_Q^n j t^n, \]

where \(j\) denotes the all one vector, thus we have

\[h_A(t) = j^{\mathsf T} (1 - M_Q t)^{-1} j. \]

2.14 Tensor products and duals of representations of Lie algebras

Problem 2.14.3. Let \(V, W, U\) be finite dimensional representations of a Lie algebra \(\gg\). Show that the space \(\Hom_\gg(V \otimes W, U)\) is isomorphic to \(\Hom_\gg(V,U \otimes W^*)\). (Here \(\Hom_\gg := \Hom_{\UU(\gg)}\).)

Proof. We first show that \(\Hom_g(V, W)\) are the linear maps \(\varphi\colon V\to W\) such that for all \(x\in \gg\), \(\varphi\circ \rho_V(x) = \rho_W(x) \circ \varphi\). Suppose \(\varphi\in \Hom_\gg(V, W)\), then for each \(x\in \gg\), it can be identified as an element in \(\UU_\gg\), as an algebra representation, we should have \(\varphi(x v) = x \varphi(v)\), which is \(\varphi(\rho_V(x)) = \rho_W(x)\varphi(v)\). Conversely, suppose \(\varphi(x v) = x \varphi(v)\) holds for all \(x\in \gg\), then the equation holds for all \(x\in \UU_\gg\).
We first give the mapping \(\Hom_\gg(V, U\otimes W^*) \to \Hom_\gg(V\otimes W, U)\) by

\[\Phi\colon f \mapsto ((v\otimes w) \mapsto f(v) w), \]

clearly this maps each \(\Hom(V,U\otimes W^*)\) to \(\Hom(V\otimes W, U)\) linearly, then we verify that this is in \(\Hom_\gg\). Since \(f\in \Hom_\gg(V,U\otimes W^*)\), we have

\[\begin{align*} f\circ \rho_V(x) &= \rho_{U\otimes W^*}(x)\circ f\\ &= (\rho_U(x)\otimes I + I \otimes \rho_{W^*}(x)) \circ f\\ &= (\rho_U(x)\otimes I - I \otimes \rho_{W}(x)^*) \circ f. \end{align*}\]

Then we have

\[\begin{align*} \Phi(f) \circ \rho_{V\otimes W}(x) &= ((v\otimes w) \mapsto f(v) w) \circ (\rho_V(x)\otimes I + I \otimes \rho_W(x))\\ &= (u\otimes w) \mapsto (f(\rho_V(x)v)w + f(v) \rho_W(x) w)\\ &= (u\otimes w) \mapsto ((f\circ \rho_V(x))(v)w + [(I\otimes \rho_W(x)^*)\circ f](v) w)\\ &= (u\otimes w) \mapsto ([(\rho_U(x)\otimes I)\circ f](v)w)\\ &= (u\otimes w) \mapsto (\rho_U(x)[f(v)w])\\ &= \rho_U(x) \circ \Phi(f). \end{align*}\]

Thus we have \(\Phi(f) \in \Hom_\gg(U\otimes V, W)\).
In another direction, we give the mapping \(\Hom(V\otimes W, U) \to \Hom(V, U\otimes W^*)\) by

\[\Psi(g) = v\mapsto (w\mapsto g(v, w)), \]

we can verify that \(\Phi\) and \(\Psi\) are inverse to each other, and \(\Psi\) also preserves \(\Hom_\gg\).

2.15 Representations of \(\sl(2)\)

Problem 2.15.1. According to the above, a representation of \(\sl(2)\) is just a vector space \(V\) with a triple of operators \(E, F,H\) such that \(HE -EH = 2E, HF -F H = -2F, EF -F E = H\) (the corresponding map \(\rho\) is given by \(\rho(e)=E,\rho(f)=F,\rho(h)=H\)).
Let \(V\) be a finite dimensional representation of \(\sl(2)\) (the ground field in this problem is \(\CC\)).
(a) Take eigenvalues of \(H\) and pick one with the biggest real part. Call it \(\lambda\). Let \(\overline V(\lambda)\) be the generalized eigenspace corresponding to \(\lambda\). Show that \(E|_{\overline V(\lambda)} = 0\).

Proof. We prove by induction. Let \(V_k = \ker (H-\lambda)^k\), we prove that \(E|_{V_k} = 0\). Clearly that \(k=0\) holds. Now suppose \(E|_{V_{k-1}}=0\), for each \(v\in V_{k}\), we have \((H-\lambda)v \in V_{k-1}\), we have \(2Ev = ((H-\lambda)E - E(H-\lambda))v = (H-\lambda)E v\), thus we have \(HEv = (\lambda + 2)Ev\), if \(Ev\) is nonzero, then it is an eigenvector of \(H\) with eigenvalue \(\lambda+2\), contradicting to our choice of \(\lambda\), so we must have \(Ev = 0\), thus \(E|_{V_k}=0\).

Since \(E|_{V_k}=0\) holds for all \(k\), we have \(E|_{\overline V(\lambda)} = 0\). \(\square\)

(b) Let \(W\) be any representation of \(\sl(2)\) and let \(w\in W\) be a nonzero vector such that \(Ew = 0\). For any \(k > 0\) find a polynomial \(P_k(x)\) of degree \(k\) such that \(E^kF^kw = P_k(H)w\).

Proof. For \(k=0\) clearly we have \(P_0(x)=1\) holds. Then suppose for smaller \(k\) holds, we have

\[\begin{align*} E^k F^k w &= E^{k-1} (EF) F^{k-1}w \\ &= E^{k-1} (H + FE) F^{k-1} w\\ &= E^{k-1} H F^{k-1} w + E^{k-2} (EF) EF^{k-1} w \\ &= \cdots \\ &= FE^kF^{k-1}w + \sum_{i=0}^{k-1} E^{k-1-i} H E^i F^{k-1}w\\ &= FE P_{k-1}(H)w + \sum_{i=0}^{k-1} E^{k-1-i} H E^i F^{k-1}w, \end{align*}\]

and since

\[E^i H = E^{i-1}(H-2)E = \cdots = (H-2i)E^i, \]

we further have

\[\begin{align*} &= FE P_{k-1}(H)w + \sum_{i=0}^{k-1} (H - 2(k-1-i))E^{k-1}F^{k-1}w\\ &= (FE + kH - k(k-1)) P_{k-1}(H)w, \end{align*}\]

note that \(E P_{k-1}(H) w = 0\), we have

\[E^kF^k w = (kH - k(k-1))P_{k-1}(H) w. \]

Then \(P_k(x) = k(x - (k-1))P_{k-1}(x)\) holds, we have \(P_k(x)\) has degree \(k\). \(\square\)

(c) Let \(v\in \overline{V}(\lambda)\) be a generalized eigenvector of \(H\) with eigenvalue \(\lambda\). Show that there exists \(N > 0\) such that \(F^Nv = 0\).

Proof. Noticed that \((H-\mu + 2)^i F = F (H-\mu)^i\), we have \(F \left(\overline V(\mu)\right) \subset \overline{V}(\mu - 2)\). Suppose \(F^N v \neq 0\) for all \(N>0\), we have the subspaces \(\overline{V}(\lambda - 2N)\) are nonempty for all \(N >0\), but they are direct summands of \(V\), contradicting to the fact that \(V\) is finite dimensional. \(\square\)

(d) Show that \(H\) is diagonalizable on \(\overline{V} (\lambda)\).

Since we have \(F^N v = 0\) for each \(v\in \overline{V}(\lambda)\) and \(V\) is finite dimensional, we can conclude that there exists some \(N\) such that \(F^N|_{\overline{V}(\lambda)} = 0\), thus \(E^NF^N|_{\overline{V}(\lambda)} = 0\). By (b) we know that \(E^N F^N = P_N(H)\), and \(P_N(x) = N! x(x-1)\cdots(x-(N-1))\) has no multiple roots, we know that \(\overline V(\lambda)\) decomposes into the eigenspaces of eigenvalues \(k\) for \(0\leq k\leq N-1\).

(e) Let \(N_v\) be the smallest \(N\) satisfying (c). Show that \(\lambda = N_v - 1\).

Proof. \(N_v\) should be the smallest \(N\) such that \(P_N(\lambda)=0\) but \(P_{N-1}(\lambda)\neq 0\), clearly we have \(\lambda = N - 1\).

(f) Show that for each \(N > 0\), there exists a unique up to isomorphism irreducible representation of \(\sl(2)\) of dimension \(N\). Compute the matrices \(E,F,H\) in this representation using a convenient basis.

Proof. Suppose \(V\) is irreducible and \(N = \dim V\). Pick an eigenvector \(v\) of \(\lambda\), by the preceding argument, we know that \(v, Fv, \dots, F^{\lambda}v\) are linearly independent, since they are nonzero and are \(H\)-eigenvectors with eigenvalue \(\lambda, \lambda-2,\dots, -\lambda\). Moreover, we have \(E F^i v = i(\lambda-i+1)F^{i-1} v\), so \(\langle F \rangle v\) is a \(\sl(2)\)-subrepresentation, since \(V\) is irreducible, we must have \(\lambda + 1 = N\), we can write \(E,F,H\) as follows:

\[ E = \begin{pmatrix} & N - 1\\ & & 2(N - 2)\\ & & & \ddots \\ & & & & N-1\\ & \end{pmatrix}, \]

\[ F = \begin{pmatrix} \\ 1\\ & 1\\ & & \ddots\\ & & & 1 & \end{pmatrix}, \]

\[ H = \begin{pmatrix} N - 1 \\ & N - 3 \\ & & \dots \\ & & & -(N-1) \end{pmatrix}. \square \]

Denote the \((\lambda+1)\)-dimensional irreducible representation from (f) by \(V_\lambda\). Below you will show that any finite dimensional representation is a direct sum of \(V_\lambda\).
(g) Show that the operator \(C = EF + F E + H^2/2\) (the so-called \textbf{Casimir operator}) commutes with \(E, F, H\) and equals \(\frac{\lambda(\lambda+2)}2 \Id\) on \(V_\lambda\).

Proof. Can be proved by computation. \(\square\)

Now it is easy to prove the direct sum decomposition. Namely, assume the contrary, and let \(V\) be a reducible representation of the smallest dimension, which is not a direct sum of smaller representations.
(h) Show that \(C\) has only one eigenvalue on \(V\), namely \(\lambda(\lambda+2)/2\) for some nonnegative integer \(\lambda\).

Proof. Since \(C\) always lies in the center of \(\End V\), suppose \(C\) has several eigenvalues, then their generalized eigenspace decomposes to subrepresentation, which contradicts to the assumption that \(V\) is indecomposable. So \(C\) has only one eigenvalue, since the eigenvalue should be equal to the eigenvalue of some irreducible component, it must be equal to some \(\lambda(\lambda+2)/2\). \(\square\)

(i) Show that \(V\) has a subrepresentation \(W = V_\lambda\) such that \(V/W = nV_\lambda\) for some \(n\).

Proof. By the assumption on minimality of \(V\), we have \(V/W\) decomposes to irreducible components, which must be all \(V_\lambda\).

(j) Deduce from (i) that the eigenspace \(V(\lambda)\) of \(H\) is \((n + 1)\)-dimensional. If \(v_1,\dots,v_{n+1}\) is its basis, show that \(F^jv_i\), \(1 \leq i \leq n+1\), \(0 \leq j \leq \lambda\), are linearly independent and therefore form a basis of \(V\).

Proof. Now the \(\lambda\)-eigenspace of \(H|_W\) has dimension \(1\), and \(H|_{V/W}\) has dimension \(n\), thus \(H|_V\) has dimension \(n+1\).

Note that \(F^j v_i\) lies in the \((\lambda-2j)\)-eigenspace of \(H\), so if \(F^jv_i\) are linearly dependent, say, \(\sum c_{ij}F^jv_i = 0\), we have a relation such that \(c_{ij}\neq 0\) only for a minimal \(j\), thus \(F^j\left(\sum_{c_i} c_i v_i\right) = 0\). Since \(v_i\) are linearly independent, \(v = \sum_{c_i} c_i v_i\) is nonzero, but as argued in (e), we have \(v, Fv, \dots, F^\lambda v\) are nonzero. So we must have these \(F^j v_i\) are linearly independent. \(\square\)

(k) Define \(W_i = \Span(v_i,Fv_i,\dots,F^\lambda v_i)\). Show that \(W_i\) are subrepresentations of \(V\) and derive a contradiction to the fact that \(V\) cannot be decomposed.

Proof. Already derived in (f), that \(W_i\) is \(E,F,H\)-invariant, thus is a subrepresentation. This gives a decomposition \(V = W_1 \oplus \cdots \oplus W_{n+1}\).

(l) (Jacobson–Morozov lemma) Let \(V\) be a finite dimensional complex vector space and \(A \colon V \to V\) a nilpotent operator. Show that there exists a unique, up to an isomorphism, representation of \(\sl(2)\) on \(V\) such that \(E = A\).

Proof. By the Jordan normal form theorem, \(A\) can be decomposed into nilpotent Jordan blocks, each block we can treat \(A\) as \(E\) and construct \(F,H\) as derived in (f). Conversely, suppose there is a representation satisfying \(E=A\), we can identify the Jordan blocks through the decomposition of \(V\) into irreducible components, thus such representation is unique. \(\square\)

(m) (Clebsch–Gordan decomposition) Find the decomposition of the representation \(V_\lambda\otimes V_\mu\) of \(\sl(2)\) into irreducibles components.

Introduce the character \(\chi_V(x) = \Tr(e^{xH})\), then we have

\[\begin{align*} \chi_{V\oplus W}(x) &= \Tr(e^{xH})\\ &= \Tr(e^{xH}|_V + e^{xH}|_W)\\ &= \chi_V(x) + \chi_W(x),\\ \chi_{V\otimes W}(x) &= \Tr(e^{x(H_V \otimes I + I\otimes H_W)})\\ &= \Tr(e^{x H_V} \otimes e^{xH_W})\\ &= \Tr(e^{x H_V}) \Tr(e^{xH_W}) \\ &= \chi_V(x) \chi_W(x). \end{align*}\]

Now we compute the character of \(V_\lambda\), we have

\[\begin{align*} \chi_{V_\lambda}(x) &= \Tr(e^{xH})\\ &= \sum_{i=0}^\lambda e^{x(\lambda - 2i)}, \end{align*}\]

It's not hard to see that different decompositions gives different characters, so this can uniquely determine the decomposition. Then we compute

\[\begin{align*} \chi_{V_\lambda \oplus V_\mu}(x) &= \left(\sum_{i=0}^\lambda e^{x(\lambda - 2i)}\right) \left(\sum_{i=0}^\mu e^{x(\mu - 2i)}\right)\\ &= \sum_{i,j} e^{x(\lambda+ \mu - 2(i+j))}\\ &= \sum_{i=0}^{\lfloor (\lambda + \mu)/2\rfloor}\chi_{V_{\lambda + \mu - 2i}}(x). \end{align*}\]

This gives the decomposition

\[ V_\lambda \otimes V_\mu = \bigoplus_{i=0}^{\lfloor (\lambda + \mu)/2\rfloor} V_{\lambda + \mu - 2i}. \]