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Give $\mathfrak{g}$ its Adjoint representation. The representation identity gives $a(y\cdot (x\otimes v))=[y,x]v+xyv=y(xv)$, so the action map is a Lie algebra representation homomorphism. For a nontrivial simple module it is surjective. If the acting algebra is semisimple and the module is finite dimensional, Weyl complete reducibility theorem splits the surjection and makes $V$ a direct summand.

The derivations of a Lie algebra form a Lie subalgebra of $\mathrm{End}(\mathfrak{g})$ with the commutator bracket. The identity $[D,\mathrm{ad}x]=\mathrm{ad}(Dx)$ makes the inner derivations an ideal of a Lie algebra. For a complex semisimple Lie algebra, the Weyl complete reducibility theorem and the vanishing center of a Lie algebra imply $\mathrm{Der}(\mathfrak{g})=\mathrm{ad}\mathfrak{g}$.

For $V=L({\omega }_1)$ with the G2 root system numbered short-first,In the chain basis of the crystal of the seven-dimensional G2 representation, normalize $v_1=f_1{v}_0$, $v_2=f_2{v}_1$, $v_3=f_1{v}_2$. Then $v_0\wedge v_1$ and $v_0\wedge v_3-2v_1\wedge v_2$ are nonzero highest-weight vectors of weights ${\omega }_2$ and ${\omega }_1$; the identities $e_1{v}_1=v_0$, $e_1{v}_3=2v_2$, $e_2{v}_2=v_1$ verify both raising conditions. The Weyl complete reducibility theorem and G2 dimension polynomial exhaust the twenty-one dimensions of the exterior square. In the symmetric square, $v_0^2$ generates the twenty-seven-dimensional summand. Self-duality supplies a Lie-invariant bilinear form, which is symmetric because a nondegenerate alternating bilinear form cannot have odd dimension; its inverse gives the remaining invariant line.

For a finite-dimensional continuous complex representation $R$ of a compact group, normalize Haar measure to mass one and average any positive-definite Hermitian inner product:The resulting inner product is positive-definite and invariant. If $W$ is an invariant subspace, then $W^{\perp }$ is invariant as well, since $⟨R(g)v,w{⟩}_K=⟨v,R(g^{-1})w{⟩}_K$. Thus $V=W\oplus W^{\perp }$. Restriction to a compact real form, followed by integration of a Lie-algebra representation, proves the Weyl complete reducibility theorem.

For a finite-dimensional representation of a complex semisimple Lie algebra, select a maximal remaining dominant weight, subtract that multiplicity of the corresponding irreducible formal character, and repeat. The Weyl complete reducibility theorem guarantees nonnegative residual multiplicities and termination. Checking the full character or the exhausted total dimension confirms that no summands remain.

For a complex semisimple Lie algebra and a choice of positive roots, every finite-dimensional Irreducible Lie algebra representation has a unique highest weight, which is a dominant integral weight. Conversely each dominant integral weight labels exactly one such irreducible module up to isomorphism. Its highest-weight vector spans the highest-weight space and generates the module. This classification and Weyl complete reducibility theorem reduce decomposing any finite-dimensional module to its irreducible highest-weight summands.

In the irreducible sl2 Lie algebra module $V(m)$, use $v_j=f^j{v}_0$, with $0\le j\le m$. The Lie-invariant bilinear form $B(v_j,v_k)=(-1{)}^j{1}_{\{j+k=m\}}$ is nondegenerate: its anti-diagonal is nonzero. Invariance follows from $h{v}_j=(m-2j)v_j$, $f{v}_j=v_{j+1}$ and $e{v}_j=j(m-j+1)v_{j-1}$. Its transpose equals $(-1{)}^{m}B$, so it is a symmetric bilinear form when $m$ is an even number, and an alternating bilinear form when $m$ is an odd integer. The Weyl complete reducibility theorem gives a nondegenerate invariant form on any finite-dimensional module by taking the orthogonal direct sum of these forms.

The Verma module $M(0)$ has basis $v_n=f^n{v}_0$, with $h{v}_n=-2n{v}_n$, $f{v}_n=v_{n+1}$ and $e{v}_n=n(1-n)v_{n-1}$. Its submodule spanned by $v_n$ for $n\ge 1$ is $M(-2)$, and the quotient is the trivial one-dimensional module. A complementary submodule would be trivial and hence lie in the zero-weight space $\mathbb{C}{v}_0$, but $f{v}_0=v_1\ne 0$. Therefore the displayed extension does not split. This explains why the finite-dimensional hypothesis in the Weyl complete reducibility theorem is essential.

Use the corrected Lie algebra representation $E=-e$, $H=h$, $F=f$ from part (i). A vector subspace is invariant under $E,H,F$ exactly when it is invariant under the printed $e,h,f$, since changing the sign of a linear map does not change its invariant subspaces. DefineThis is a nonzero proper submodule: $H$ preserves each $k{v}_i$, $F$ increases the index, $E(v_1)=0$, and $E(v_i)=-(i-1)v_{i-1}\in W$ for $i\ge 2$.

We must rule out submodules consisting only of mixtures of weight vectors. Let $U\ne 0$ be any submodule, and choose $u={\sum }_{j\in S}{a}_j{v}_j\in U$ with $S$ finite and every $a_j\ne 0$. The eigenvalues $-2j$ of $H$ are distinct. For $i\in S$, the Lagrange interpolation polynomialsatisfies $p_i(H)u=a_i{v}_i$. Since $U$ is invariant under $H$, it is invariant under every polynomial in $H$, so $v_i\in U$.

If $i=0$, repeated applications of $F$ give $F^m{v}_0=m!v_m$, and $U=V$. If $i\ge 1$, repeated applications of $E$ give a nonzero multiple of $v_1$, because the successive coefficients $-(i-1),\dots ,-1$ are nonzero in characteristic zero. Repeated applications of $F$ to $v_1$ then give every $v_j$ with $j\ge 1$, so $W\subseteq U$.

Finally, if such a $U$ contains any vector outside $W$, that vector has a nonzero coefficient of $v_0$; the same polynomial interpolation extracts $v_0$, forcing $U=V$. Hence the complete list of submodules is $0,W,V$. In particular, $W$ is a simple module. The quotient module $V/W$ has basis ${\overline{v}}_0$; $E{\overline{v}}_0=H{\overline{v}}_0=0$ and $F{\overline{v}}_0={\overline{v}}_1=0$. It is the one-dimensional trivial Lie algebra representation.

The module extension does not split. If $V=W\oplus U$ for a submodule $U$, then $U$ would be a one-dimensional trivial Lie algebra representation, since it maps isomorphically to $V/W$. But $\ker H=k{v}_0$ and $F{v}_0=v_1\ne 0$, so no such $U$ exists. ThusHere semisimple module means a direct sum of modules that are simple modules, equivalently a module in which every submodule has a complementary submodule. This is the nonsplit zero-weight Verma extension for sl2: rescaling to $w_i=i!v_i$ gives $F{w}_i=w_{i+1}$ and $E{w}_i=i(1-i)w_{i-1}$. Its infinite-dimensional vector space is outside the hypotheses of the Weyl complete reducibility theorem.

Assume $0\le k\le n$. Let $V=L_{n-1}$ be the $n$-dimensional irreducible sl2 Lie algebra representation. Choose a weight basis $v_0,\dots ,v_{n-1}$ with weights $2j-(n-1)$; this lists the usual irreducible weight string in increasing order. Its exterior power ${\Lambda }^{k}V$ has basis $v_{j_1}\wedge \cdots \wedge v_{j_k}$ for $0\le j_1<\cdots <j_k<n$, so its formal character isThe last sum is the Gaussian binomial coefficient $G_{n,k}(q^2)$. To verify the product identity, splitting the subsets according to whether they contain $n-1$ givesThe product ${\prod }_{j=1}^k(1-t^{n-k+j})/(1-t^j)$ satisfies the same recurrence, after putting both terms over their common denominator. Thus induction identifies it with the subset sum, proving it is a polynomial with nonnegative integer coefficients.

The quantum integer convention in the PDF hasMultiplying these factors in the factorial quotient givesThis establishes the required representation-theoretic interpretation.

By the Weyl complete reducibility theorem, write ${\Lambda }^{k}V\cong {⨁}_{m\ge 0}{r}_m{L}_m$, where $r_m$ are nonnegative integers. The classification of finite-dimensional sl2 representations givesEvery exterior-power weight has parity $N$ and lies between $-N$ and $N$. Therefore only $m\le N$ with $m\equiv N(\mathrm{mod}2)$ occur. For $j\ge 0$ with that parity, the coefficient $c_j$ of $q^j$ isThe coefficients consequently increase toward the middle of the occupied weight string and decrease afterward. If $G_{n,k}(t)={\sum }_{s=0}^N{a}_s{t}^s$, then $a_s=c_{-N+2s}$. HenceThis is the unimodality of Gaussian binomial coefficients, proved through parity unimodality of an sl2 character. It includes $k=0,n$, when the character is one.

There is a normalization qualification in the printed wording. The symmetric quantum-binomial quotient is a Laurent polynomial in $q$, with exponents spaced by two; the unimodality just proved is for that occupied parity, or equivalently for the ordinary polynomial $G_{n,k}(t)$ after setting $t=q^2$ and removing the monomial shift. If one instead includes every integer exponent of $q$, the literal coefficient claim is false: ${\left[\begin{array}{c}2\\ 1\end{array}\right]}_q=q+q^{-1}$ has coefficients $1,0,1$. The representation-theoretic conclusion is unimodality on the parity string, and ordinary unimodality of the Gaussian polynomial.

Use the finite-dimensional complex-module setting of the course. We prove the Weyl complete reducibility theorem by Haar averaging produces invariant complements, defining its constructions rather than invoking complete reducibility itself.

Choose a compact real form $\mathfrak{k}\subset \mathfrak{g}$: it is a real Lie subalgebra with $\mathfrak{g}=\mathfrak{k}\oplus i\mathfrak{k}$, and its simply connected Lie group $K$ is compact. Existence of this form and compact group is a standard fact for complex semisimple Lie algebras. Restrict the representation to $\mathfrak{k}$; integration of a Lie-algebra representation gives a group representation $R:K\to GL(V)$. Since $K$ is connected and generated by exponentials, the $\mathfrak{g}$-invariant subspace $W$ is also $K$-invariant.

Choose any positive-definite Hermitian inner product on $V$. With the Haar measure of $K$ normalized to mass one, defineThis is a Hermitian inner product, and it is positive-definite: for $v\ne 0$ every $R(k)v$ is nonzero, so the continuous positive integrand has positive integral. Right invariance of Haar measure gives $⟨R(a)v,R(a)w{⟩}_K=⟨v,w{⟩}_K$ for $a\in K$.

Let $W^{\prime }=W^{\perp }$ for this averaged inner product. Finite-dimensional inner-product linear algebra gives $V=W\oplus W^{\prime }$. If $v\in W^{\prime }$ and $w\in W$, thenbecause $R(a^{-1})w\in W$. Thus $W^{\prime }$ is $K$-invariant. Differentiating makes it $\mathfrak{k}$-invariant, and since it is a complex vector subspace, $\mathfrak{g}=\mathfrak{k}+i\mathfrak{k}$ also preserves it. ThereforeThe zero or whole-space submodule cases are included. This proves the required complement theorem without circularly assuming it.

The finite-dimensional hypothesis is essential to the statement. The nonsplit zero-weight Verma extension for sl2 is a counterexample for unrestricted modules: $M(0)$ has basis $v_n$ with $h{v}_n=-2n{v}_n$, $f{v}_n=v_{n+1}$ and $e{v}_n=n(1-n)v_{n-1}$. Its submodule spanned by $v_n$ for $n\ge 1$ has trivial one-dimensional quotient. A complement would be trivial and contained in the zero-weight line $\mathbb{C}{v}_0$, yet $f{v}_0=v_1\ne 0$. Hence no such complement exists in that infinite-dimensional case. The paper's finite-dimensional representation setting supplies the necessary hypothesis.

In the defining representation $V={\mathbb{C}}^{2n}$, the basis vectors $v_i$ and $v_{\bar{i}}$ have weights ${\epsilon }_i$ and $-{\epsilon }_i$, respectively, each with multiplicity one. Thus its formal character isThe vector $v_1$ is killed by every positive root space, since all positive-root matrices are upper triangular, and is a highest-weight vector. Any nonzero invariant subspace splits into weight spaces and therefore contains one of the natural weight vectors, since those weight spaces are one-dimensional. The root matrices connect that vector to all the others, so this representation is irreducible for $n\ge 2$. In terms of the fundamental weights of Dn, its highest weight is ${\omega }_1$ for $n\ge 3$ and ${\omega }_1+{\omega }_2$ for $n=2$.

The flip of the two tensor factors is equivariant, henceUse the invariant symmetric bilinear form $Q(u,v)=u^{T}Jv$. Contraction $C(u\otimes v)=Q(u,v)$ is an equivariant map to the trivial Lie algebra representation. Its inverse-form tensoris invariant and satisfies $C(\tau )=2n$. ConsequentlyThe traceless summand contains the highest-weight vector $v_1\otimes v_1$ of weight $2{\epsilon }_1$. The Weyl complete reducibility theorem states that every finite-dimensional complex semisimple Lie algebra representation is a direct sum of irreducible representations, so the submodule generated by this highest vector is $L(2{\epsilon }_1)$. The Weyl dimension formula verifies that it is the entire traceless space: only positive roots involving ${\epsilon }_1$ change their factors, andThis equals $\dim {\mathrm{Sym}}^{2}V-1=n(2n+1)-1$. Thus the traceless symmetric square of the defining even orthogonal representation is irreducible, including $n=2$.

For the other summand, the exterior square of the defining orthogonal representation has the equivariant identificationThe image is skew relative to $Q$, and in a basis making $Q$ the identity the images are the independent elementary skew matrices. The map is therefore an isomorphism onto the Adjoint representation of a Lie algebra. For $n\ge 3$, the connected Dn Dynkin diagram gives a simple Lie algebra (with $D_3=A_3$); its adjoint module is irreducible because its invariant subspaces are exactly its ideals. Its highest weight is ${\epsilon }_1+{\epsilon }_2$, and $v_1\wedge v_2$ is a highest vector. The tensor-square decomposition of the defining even orthogonal representation is consequentlyThe three dimensions are $(n+1)(2n-1)$, $n(2n-1)$ and one, adding to $4n^2$. The highest weights are $2{\omega }_1$, ${\omega }_2$ and zero for $n\ge 4$; for $n=3$ the middle weight is ${\omega }_2+{\omega }_3$.

For $n=2$, the Chiral decomposition of the complexified so4 Lie algebra splits the adjoint module into its two three-dimensional ideals. ThusThe weights are $2{\omega }_1+2{\omega }_2$, $2{\omega }_1$, $2{\omega }_2$ and zero. Equivalently, labelling highest weights by the two sl2 Lie algebra factors, this is $(2,2)\oplus (2,0)\oplus (0,2)\oplus (0,0)$. If $n=1$ is allowed, the abelian algebra has $V=L({\epsilon }_1)\oplus L(-{\epsilon }_1)$ and $V\otimes V=L(2{\epsilon }_1)\oplus 2L(0)\oplus L(-2{\epsilon }_1)$; there is then no single highest weight for an irreducible natural module.

The weight diagram has weights $6,4,2,0,-2,-4,-6$ with multiplicities $1,1,2,2,2,1,1$. An Irreducible Lie algebra representation of the sl2 Lie algebra with highest weight $n$ has weights $n,n-2,\dots ,-n$, all of multiplicity one. The top weight therefore forces one $L(6)$ summand. Subtracting its formal character leaves weights $2,0,-2$, each once, which is $L(2)$. By Weyl complete reducibility theorem,Their dimensions $7+3=10$ account for the entire module.

Use the contraction splitting of a defining module tensor a dual symmetric square. For a quadratic polynomial $p$ in the dual variables, defineBoth maps are intertwiners. The first is the natural tensor contraction; the second uses the invariant tensor ${\sum }_i{e}_i\otimes y_i$. Directly, $C(Q_j)=4y_j$, so $C\iota =4I$ andThe vector $T_{1;33}=e_1\otimes y_3^2$ lies in $\ker C$ and is killed by both positive simple-root operators $E_{12},E_{23}$. It is a highest-weight vector of weight $L_1-2L_3={\omega }_1+2{\omega }_2$. By Weyl complete reducibility theorem, it generates the corresponding irreducible summand. The Weyl dimension formula for ${\mathfrak{sl}}_3$ gives $\dim {\Gamma }_{a,b}=(a+1)(b+1)(a+b+2)/2$, hence $\dim {\Gamma }_{1,2}=15$, exhausting the kernel. ThereforeThe kernel has the nine outer weights of multiplicity one and each inner weight $-L_j$ of multiplicity two; the dual summand has just the three weights $-L_j$, each once.

Let $f_1,f_2,f_3$ be the dual basis of $W^{\ast }$, and use $L_1+L_2+L_3=0$. The elementary tensors $e_i\otimes e_j\otimes f_k$ have weight $L_i+L_j-L_k$. Grouping them gives every weight space explicitly:The counts are $6+3\cdot 2+3\cdot 5=27$, so this table exhausts the tensor product. The diagrams below use equilateral coordinates $L_1=(1,0)$, $L_2=(-1/2,\sqrt{3}/2)$, $L_3=(-1/2,-\sqrt{3}/2)$; numbers at points are weight multiplicities.

To decompose the module, first split the first two factors into their symmetric and alternating parts. Put $t_{ij}=(e_i\otimes e_j+e_j\otimes e_i)/2$, including $t_{ii}=e_i\otimes e_i$. The equivariant contractionis onto, since $C(t_{ii}\otimes f_i)=2e_i$. Its kernel $K$ has dimension fifteen and contains the highest-weight vector $t_{11}\otimes f_3$ of weight $2L_1-L_3$.

Here is an explicit weight basis of the trace-free symmetric-square dual tensor module, and thus the requested fifteen-vector basis:Every displayed vector contracts to zero. The first two families occupy distinct one-dimensional weights; at each weight $L_i$ the two vectors in the last family are independent because their respective $t_{ij}\otimes f_j$ terms are distinct. Thus all fifteen are independent and span $K$.

For completeness, $K$ is irreducible, not just an invariant kernel of the right dimension. On elementary tensors the raising action isAt the six weights $2L_i-L_k$, both $E_{12}$ and $E_{23}$ kill the vector only for $(i,k)=(1,3)$. At each weight $-2L_k$, at least one of those operators is nonzero. At weight $L_1$, write a vector as $a(t_{12}\otimes f_2-t_{11}\otimes f_1/2)+b(t_{13}\otimes f_3-t_{11}\otimes f_1/2)$. The two raising equations are $3a+b=0$ and $-a+b=0$, forcing zero. On the two-dimensional weight-$L_2$ space $E_{12}$ is injective, and on the weight-$L_3$ space $E_{23}$ is injective, as substitution in the same action formula shows. Therefore the common raising kernel in $K$ is the single line $\mathbb{C}(t_{11}\otimes f_3)$. By Weyl complete reducibility theorem, every irreducible summand supplies a highest-weight line, so $K$ has only one summand and is irreducible.

The symmetric-factor contraction splits off a defining copy $W$; explicitly $e_i\mapsto {\sum }_j{t}_{ij}\otimes f_j$ is equivariant and its contraction is $4e_i$. For the alternating factor, the invariant volume form identifies ${\Lambda }^{2}W\cong W^{\ast }$, sending $e_1\wedge e_2$ to $f_3$, $e_2\wedge e_3$ to $f_1$ and $e_3\wedge e_1$ to $f_2$. ThereforeThe symmetric dual module is irreducible of dimension six and highest weight $-2L_3$; its symmetric-monomial basis has weights $-2L_i$ and $-L_i-L_j=L_k$, all of multiplicity one. In the original tensor product its highest-weight vector isThe raising action above annihilates it, and its weight is $L_1+L_2-L_3=-2L_3$.

Altogether the Triple tensor decomposition for the defining sl3 representation isThe fifteen-dimensional module has multiplicity one at each $2L_i-L_k$ and $-2L_i$, and multiplicity two at each $L_i$. The six-dimensional module has multiplicity one at $-2L_i$ and $L_i$. Each defining module has the three weights $L_i$, once each. These add to the original table of multiplicities.

The two defining summands are not intrinsically distinguished: other diagonal copies inside their isotypic component have the same three-point weight diagram. Thus the displayed diagrams account for every irreducible submodule type, as well as showing both copies in this decomposition.

Give $\mathfrak{g}$ its Adjoint representation and use the tensor product of Lie algebra representations. Define the action map of a Lie algebra representationFor $y\in \mathfrak{g}$, the action on the source is $y(x\otimes v)=[y,x]\otimes v+x\otimes yv$. The defining representation identity then givesThus $a$ is a Lie algebra representation homomorphism.

If $V$ is nontrivial and simple, the action is not identically zero, so the image of $a$ is a nonzero submodule. Simplicity forces $a$ to be surjective. For a finite-dimensional $V$ and a semisimple Lie algebra, the Weyl complete reducibility theorem provides an invariant complement $U$ to $\ker a$ in $\mathfrak{g}\otimes V$. The restriction $a{|}_U$ is then an isomorphism onto $V$, givingThis proves the requested summand assertion in the finite-dimensional setting.

If finite dimensionality is not implicit, the final assertion needs that hypothesis: the action summand can fail for an infinite-dimensional simple module. Here is an explicit counterexample. Take $\mathfrak{g}={\mathfrak{sl}}_2$ with $[h,e]=2e$, $[h,f]=-2f$, $[e,f]=h$, and the Verma module of highest weight $-2$:The lowering coefficient never vanishes for $m>0$, so an invariant subspace, using the distinct $h$-weights to isolate a basis vector and then raising, contains $v_0$ and hence all of $V$. Thus $V$ is simple.

In ${\mathfrak{sl}}_2\otimes V$, a vector of weight $-2$ is $Ae\otimes v_1+Bh\otimes v_0$. Applying $e$ gives $-2(A+B)e\otimes v_0$, so its highest vectors are precisely the multiples of $w=e\otimes v_1-h\otimes v_0$. But $w=f(e\otimes v_0)$, and $e\otimes v_0$ has weight zero, absent from $V$. Every homomorphism $r:\mathfrak{g}\otimes V\to V$ therefore kills this vector and then kills $w$. Any embedded copy of $V$ would send its highest vector to a nonzero multiple of $w$, so it cannot admit a projection back to $V$. This rules out a direct summand, not merely a splitting of the particular action map.

Let $h$ be the standard Cartan element of the sl2 Lie algebra, with $[h,e]=2e$ and $[h,f]=-2f$. For a finite-dimensional representation, its formal character records its weight multiplicities:The classification of finite-dimensional sl2 representations and the Weyl complete reducibility theorem give $V={⨁}_{m\ge 0}{a}_m{L}_m$, with finitely many nonzero $a_m$. Each $L_m$ has weights $m,m-2,\dots ,-m$, each of multiplicity one. ThereforeIn particular $c_j=c_{-j}$, and for $j\ge 0$ one has $c_j-c_{j+2}=a_j\ge 0$. This proves parity unimodality of an sl2 character: within each parity, the coefficients increase towards zero and decrease away from zero. For a representation whose weights have a single parity, deleting the intervening zero coefficients gives a symmetric unimodal sequence.

The parity convention is necessary for the printed assertion. If unimodality means the full coefficient sequence at every consecutive integer exponent, it is false: the standard two-dimensional representation has character $q+q^{-1}$, whose coefficients at exponents $-1,0,1$ are $1,0,1$. The statement established above is unimodality on cosets of the root lattice for ${\mathfrak{sl}}_2$, and is the version needed in part ii. Infinite-dimensional representations need not even have Laurent-polynomial characters.

Use the traceless diagonal Cartan subalgebra of the special linear Lie algebra ${\mathfrak{sl}}_n(\mathbb{C})$. Write ${\epsilon }_i(\mathrm{diag}(h_1,\dots ,h_n))=h_i$, so ${\sum }_i{\epsilon }_i=0$. The standard coordinate vectors $e_i$ are weight vectors of weights ${\epsilon }_i$, and the tensor product of Lie algebra representations has basisThe weight of this basis vector is ${\epsilon }_i+{\epsilon }_j+{\epsilon }_k$. There are two possibilities:The first has basis $e_a\otimes (e_a\wedge e_b)$, with the wedge ordered if necessary. In the second, the tensor's first factor can be any one of the three indicated coordinate vectors. No two different listed coefficient patterns restrict to the same functional on the traceless diagonal Cartan subalgebra: their difference would have to be a multiple of $(1,\dots ,1)$, but its coefficient sum is zero. All unlisted weight spaces are zero. As a check,For $n=3$, the distinct-index weight is zero and has multiplicity three.

To determine the irreducible summands, take the upper triangular raising operators. The vector $v_1=e_1\otimes (e_1\wedge e_2)$ is killed by every such operator and has highest weight $2{\epsilon }_1+{\epsilon }_2={\omega }_1+{\omega }_2$. The exterior product mapis a Lie algebra representation homomorphism. It has an equivariant right inverseIndeed $A\iota =I$. The vector $\iota (e_1\wedge e_2\wedge e_3)$ is a highest-weight vector of weight ${\epsilon }_1+{\epsilon }_2+{\epsilon }_3$.

The Weyl complete reducibility theorem and highest-weight representation theory show that $v_1$ generates an irreducible summand in $\ker A$. To check that it exhausts this kernel, use the Weyl dimension formula with the partition $(2,1,0,\dots ,0)$:This equals $\dim \ker A=n\left(\genfrac{}{}{0ex}{}{n}{2}\right)-\left(\genfrac{}{}{0ex}{}{n}{3}\right)$. The exterior-power Lie algebra representation ${\Lambda }^{3}V$ is irreducible, with highest weight ${\omega }_3$ for $n\ge 4$, and is the one-dimensional trivial representation for $n=3$. Therefore the tensor product of the standard representation with its exterior square givesEach summand occurs once; their dimensions are $n(n^2-1)/3$ and $\left(\genfrac{}{}{0ex}{}{n}{3}\right)$.

The Weyl complete reducibility theorem states that every finite-dimensional Lie algebra representation of a finite-dimensional complex semisimple Lie algebra is a direct sum of Irreducible Lie algebra representations. Equivalently, every invariant vector subspace has an invariant complement.

We use the permitted Casimir operator properties in the following precise form. There is a central quadratic operator $C$ commuting with the action on every module; it is zero on the trivial Lie algebra representation, and on every nontrivial finite-dimensional Irreducible Lie algebra representation it is a nonzero scalar $c$. This follows from the Schur lemma and the Casimir eigenvalue $c_{\lambda }=(\lambda ,\lambda +2\rho )$, using the Killing form normalization. We also use the permitted one-dimensional-representation fact: a complex semisimple Lie algebra has only trivial one-dimensional representations. Equivalently, it is a perfect Lie algebra, $[\mathfrak{g},\mathfrak{g}]=\mathfrak{g}$, so a character $\mathfrak{g}\to \mathbb{C}$ annihilating brackets must vanish. Neither fact assumes complete reducibility of the module being proved reducible.

First prove that every finite-dimensional short exact sequencewith trivial quotient splits. If $W$ is nontrivial irreducible, the Casimir operator $C_E$ has image in $W$ and restricts to $c{I}_W$ there. Consequently $\ker C_E$ is a one-dimensional invariant complement to $W$. If $W$ is trivial irreducible, $E$ has a basis in which every action is $\left(\begin{array}{cc}0& a(x)\\ 0& 0\end{array}\right)$. The Lie algebra representation identity makes $a([x,y])=0$, so the one-dimensional-representation fact gives $a=0$ and again the sequence splits.

For general $W$, induct on $\dim W$. Choose an irreducible submodule $W_1\subseteq W$. The induced sequence with kernel $W/W_1$ and middle term $E/W_1$ splits by induction. The inverse image $E_1\subseteq E$ of its invariant complement is a submodule fitting into $0\to W_1\to E_1\to \mathbb{C}\to 0$. The irreducible-kernel case gives an invariant line in $E_1$ mapping isomorphically to the quotient. It is also an invariant complement to $W$ in $E$. The case $W=0$ starts this induction. This proves splitting of a trivial quotient for a semisimple Lie algebra, including kernels that are not assumed completely reducible.

Now let $U\subseteq V$ be any invariant subspace. On the Hom representation $\mathrm{Hom}(V,U)$ the action isLet $E$ consist of the maps whose restriction to $U$ is a scalar multiple of $I_U$. This is a submodule, and restriction givesThe right-hand map is surjective because an ordinary linear projection $V\to U$ exists; its quotient action is trivial because a commutator with $I_U$ is zero. The splitting just proved supplies an invariant $T\in E$ with $T{|}_U=I_U$. Thus $T$ intertwines the actions, $T^2=T$, andThe cases $U=0$ and $U=V$ are immediate. Choosing an irreducible submodule and repeating this complement construction proves the Weyl complete reducibility theorem. This last step is invariant complement from an equivariant projection.

Dropping finite dimensionality gives an example with the complex simple Lie algebra ${\mathfrak{sl}}_2$. Its Verma module of highest weight zero has a basis $v_0,v_1,\dots$ withwhere $e{v}_0=0$. These actions obey $[h,e]=2e$, $[h,f]=-2f$, $[e,f]=h$. The span of $v_1,v_2,\dots$ is a proper submodule and the quotient is trivial. It has no invariant complement: such a complement would be a trivial line, while $f$ is injective on the entire module. Thus a simple Lie algebra can have an infinite-dimensional representation that is not completely reducible, even over $\mathbb{C}$.

Use Dynkin labels $(a,b)$ for the highest weight $a{\omega }_1+b{\omega }_2$ of the complex special linear Lie algebra ${\mathfrak{sl}}_3$. The A2 root system has ${\alpha }_1=(2,-1)$ and ${\alpha }_2=(-1,2)$ in these coordinates. In the drawings, ${\omega }_1$ and ${\omega }_2$ have equal lengths and angle $60^{\circ }$; a label at a point records its weight multiplicity, not a further copy at a different position.