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Take the Verma module for the sl2 Lie algebra of highest weight $-2$, with basis $v_m$ and $f{v}_m=v_{m+1}$, $h{v}_m=(-2-2m)v_m$, $e{v}_m=-m(m+1)v_{m-1}$. It is simple. In ${\mathfrak{sl}}_2\otimes V$, the highest vectors of weight $-2$ are multiples of $w=e\otimes v_1-h\otimes v_0=f(e\otimes v_0)$. Every homomorphism to $V$ kills $e\otimes v_0$, whose weight is zero, and hence kills $w$. Every embedded copy of $V$ would have its highest vector proportional to $w$, so no projection onto such a copy can exist. Thus semisimplicity of the Lie algebra alone does not provide an action summand in infinite dimension.

A root string of length $\Lambda +1$ gives an irreducible sl2 Lie algebra module $R(\Lambda )$ when restricting the Adjoint representation to the sl2 subalgebra associated with a root. Include also the root's own $R(2)$ and one commuting Cartan direction $R(0)$. For $A_2$ either simple root gives $R(2)\oplus 2R(1)\oplus R(0)$. For $B_2$, the long root gives $R(2)\oplus 2R(1)\oplus 3R(0)$ and the short root gives $3R(2)\oplus R(0)$. For $G_2$, the long root gives $R(2)\oplus 4R(1)\oplus 3R(0)$ and the short root gives $R(2)\oplus 2R(3)\oplus 3R(0)$.

On the sl2 Lie algebra with basis $e=E_{12},h=E_{11}-E_{22},f=E_{21}$, the integral adjoint root polynomials areThese are conjugation by the upper and lower elementary matrices of the special linear group. Those matrices generate $S{L}_2(K)$, and a matrix commuting with both $E_{12}$ and $E_{21}$ is scalar. Thus the kernel is precisely $\{aI:a^2=1\}$ over every field, even in characteristic two, and the image is the displayed quotient.

Over a field of characteristic zero, a locally nilpotent operator has an algebraically defined exponential, because the sum is finite for each vector. Its inverse is $\exp (-T)$. The nilpotence bound may depend on the vector. In an sl2 Lie algebra Verma module, the raising operator is locally nilpotent but the lowering operator is not, so the latter's exponential does not define an endomorphism of the ordinary algebraic module.

// Target: lie-algebra.bigb

For the B2 root system with ${\alpha }_1={\epsilon }_1-{\epsilon }_2$ long and ${\alpha }_2={\epsilon }_2$ short, the spin fundamental weight is ${\Lambda }_2=({\epsilon }_1+{\epsilon }_2)/2$. Its four-dimensional highest-weight representation has the minuscule weight orbit $\frac{1}{2}(\pm {\epsilon }_1\pm {\epsilon }_2)$, each of multiplicity one. The crystal basis is the chain $++\stackrel{2}{\to }+-\stackrel{1}{\to }-+\stackrel{2}{\to }--$, where signs denote the coordinates. Each Kashiwara operator subtracts its indicated simple root; the sl2 Lie algebra strings have length one. This fixes every edge and gives a connected highest-weight graph with the correct character.

Over an algebraically closed field of characteristic zero, every three-dimensional perfect Lie algebra is isomorphic to the sl2 Lie algebra. Indeed the trace of a commutator gives $\mathrm{tr}(\mathrm{ad}x)=0$ for all $x$, because each element is a sum of Lie brackets. If every $\mathrm{ad}x$ were a nilpotent linear map, the Engel theorem would make the Lie algebra nilpotent, contradicting its constant nonzero lower central series of a Lie algebra. Some $\mathrm{ad}x$ therefore has a nonzero eigenvalue $\lambda$. Its other eigenvalues are $0,-\lambda$, since it kills $x$ and has zero trace. They are distinct, so its eigenvectors give a basis $h,e,f$ after scaling $x$ to $h=2x/\lambda$, with $[h,e]=2e$ and $[h,f]=-2f$. The Jacobi identity puts $[e,f]$ in the zero eigenspace $kh$. Its coefficient cannot vanish, since otherwise the derived algebra would lie in $ke\oplus kf$. Rescaling $f$ gives $[e,f]=h$. Thus this basis is an sl2 triple, proving the claimed Lie algebra isomorphism.

For the defining sl2 Lie algebra module with basis $x,y$, put $A_k=x\otimes x^{3-k}{y}^k$ and $B_k=y\otimes x^{3-k}{y}^k$. The raising operator has $e{A}_k=k{A}_{k-1}$ and $e{B}_k=A_k+k{B}_{k-1}$; the lowering operator has $f{A}_k=B_k+(3-k)A_{k+1}$ and $f{B}_k=(3-k)B_{k+1}$. The highest-weight vectors $A_0$ and $B_0-A_1$ generate irreducible submodules of highest weights four and two. Bases are respectively $(A_0,3A_1+B_0,A_2+B_1,A_3+3B_2,B_3)$ and $(B_0-A_1,B_1-A_2,B_2-A_3)$. Their vectors are independent in every weight space, proving the displayed direct sum without only comparing dimensions.

Choose simple root vectors with $[e_i,f_i]=h_i={\alpha }_i^{\vee }$, and let ${\rho }^{\vee }$ be the half-sum of positive coroots. It satisfies ${\alpha }_i({\rho }^{\vee })=1$. Expand $h=2{\rho }^{\vee }={\sum }_i{c}_i{h}_i$. For $i\ne j$, $[e_i,f_j]=0$, since the difference of distinct simple roots is not a root of a root system. Thus $[e,f]=h$, while $[h,e]=2e$ and $[h,f]=-2f$. These identities construct an embedded sl2 Lie algebra in any nonzero complex semisimple Lie algebra. Its raising vector is the principal nilpotent element with a nonzero component in each simple-root space.

For a finite-dimensional complex sl2 Lie algebra module $V$, the number of simple summands, counted with multiplicity, is $\dim V_0+\dim V_1$. Indeed, the simple module of highest weight $m\ge 0$ has weights $m,m-2,\dots ,-m$, each once. Exactly one of zero and one appears: zero when $m$ is even, one when $m$ is odd. The classification of finite-dimensional sl2 representations decomposes $V$ into these modules, and dimensions of weight spaces add over the direct sum.

Let $e_i=x^{4-i}{y}^i$ and $w_{ij}=e_i\wedge e_j$ for $i<j$. For the sl2 Lie algebra, the raising and lowering operators are $E=x{\partial }_y$, $F=y{\partial }_x$, and $w_{ij}$ has weight $8-2(i+j)$. A weight basis of the irreducible representation $L(6)$ isA weight basis of the complementary $L(2)$ isThe first vector in each list is a highest-weight vector. Applying $F$ along the first list gives successive coefficients $3,2,1,12,5,2$; along the second it gives $1,2$. The final vectors are killed by $F$. At the overlapping weights $2,0,-2$, the two listed vectors are independent, proving the direct sum.

// Target: semisimple-lie-algebra.bigb

For a complex vector space with a nondegenerate symmetric bilinear form $Q$, the map $u\wedge v\mapsto [w\mapsto Q(v,w)u-Q(u,w)v]$ identifies the exterior square of the defining representation with the Adjoint representation of a Lie algebra of $\mathfrak{so}(V)$. Equivariance follows from invariance of $Q$; an orthogonal basis verifies bijectivity. It is irreducible when $\mathfrak{so}(V)$ is simple. In dimension four it splits into the adjoint modules of the two commuting sl2 Lie algebras.

For a short root $\alpha$ in the G2 root system, restriction of the Adjoint representation to the sl2 subalgebra associated with a root givesHere $V(n)$ is the irreducible sl2 Lie algebra representation of highest weight $n$ and dimension $n+1$. Every nonzero root vector $x\in {\mathfrak{g}}_{\alpha }$ has $\dim Z_{\mathfrak{g}}(x)=6$: the raising operator has a one-dimensional kernel in each of these six irreducible 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.

For a finite-dimensional sl2 Lie algebra representation, write $\chi (q)={\sum }_j{c}_j{q}^j$. Each irreducible summand has character $q^m+q^{m-2}+\cdots +q^{-m}$, so $c_j=c_{-j}$ and $c_j\ge c_{j+2}$ for nonnegative $j$. Thus each parity subsequence is symmetric and unimodal. If all weights have one parity, removing the unused exponents gives an ordinarily unimodal coefficient sequence. The full sequence including every integer exponent need not be unimodal: $q+q^{-1}$ has coefficients $1,0,1$.

For a perfect Lie algebra $\mathfrak{s}$, every element is a sum of Lie brackets. The Adjoint representation satisfies $\mathrm{ad}[x,y]=[\mathrm{ad}x,\mathrm{ad}y]$, and the trace of a commutator is zero. ThereforeSome $\mathrm{ad}x$ must have a nonzero eigenvalue. Otherwise, over the algebraically closed field $k$, every $\mathrm{ad}x$ would be a nilpotent linear map, and the Engel theorem would imply that $\mathfrak{s}$ is a nilpotent Lie algebra. This contradicts perfection: the lower central series of a Lie algebra ${\gamma }_1=\mathfrak{s}$, ${\gamma }_{r+1}=[\mathfrak{s},{\gamma }_r]$ would equal $\mathfrak{s}\ne 0$ at every step instead of terminating.

Choose such an $x$. Since $[x,x]=0$, zero is an eigenvalue of $\mathrm{ad}x$. Its trace is zero and its dimension is three, so the remaining eigenvalues are $\lambda ,-\lambda$ with $\lambda \ne 0$. These three eigenvalues are distinct in characteristic zero, hence $\mathrm{ad}x$ is diagonalizable. Put $h=2x/\lambda$, and choose nonzero eigenvectors $e,f$ withThe zero eigenspace is $kh$, so $h,e,f$ form a basis. Applying the Jacobi identity givesThus $[e,f]=ch$ for some $c\in k$. If $c=0$, all Lie brackets of basis vectors lie in $⟨e,f⟩$, contradicting $[\mathfrak{s},\mathfrak{s}]=\mathfrak{s}$. Therefore $c\ne 0$, and replacing $f$ by $f/c$ gives $[e,f]=h$.

We have constructed an sl2 triple that is a basis. The linear map to the sl2 Lie algebra sendingpreserves all Lie brackets of basis vectors and is bijective. HenceThis is the classification of three-dimensional perfect Lie algebras without appealing to a classification of all simple Lie algebras.

Take the semidirect product of a Lie algebra and a modulewhere $k^2$ is the defining representation of the sl2 Lie algebra. To see directly that this is a Lie algebra, realize $(X,u)$ as the block matrix $\left(\begin{array}{cc}X& u\\ 0& 0\end{array}\right)$. These matrices are closed under the operator commutator, whose Jacobi identity holds in every associative algebra.

Using the standard sl2 triple $e,h,f$, its Lie brackets show $[{\mathfrak{sl}}_2,{\mathfrak{sl}}_2]={\mathfrak{sl}}_2$: indeed $h=[e,f]$, $e=[h,e]/2$, and $f=-[h,f]/2$. Also $h=\mathrm{diag}(1,-1)$ is invertible on $k^2$. For any $u\in k^2$, put $v=h^{-1}u$; thenIt follows that the derived algebra contains both ${\mathfrak{sl}}_2\oplus 0$ and $0\oplus k^2$, so $[L,L]=L$: this is a perfect Lie algebra.

However, $0\oplus k^2$ is a nonzero ideal of a Lie algebra with zero Lie bracket. It is therefore a solvable Lie algebra and is contained in the solvable radical of $L$. A semisimple Lie algebra has zero solvable radical, soThis illustrates the perfect semidirect product criterion: the Lie brackets internal to the first summand generate it, while its action generates the second summand.

There is a sign error in the printed assertion under the standard sl2 triple convention. Work first with the operators exactly as printed. Since the $v_i$ form a Hamel basis, their prescribed images extend to linear maps on $V$; each vector is a finite linear combination of basis vectors.

For $i\ge 1$, the operator commutators with $h$ areAt $i=0$, the first identity holds because both sides vanish; the second holds because $hf(v_0)=-2v_1$ and $fh(v_0)=0$. On the other hand, for $i\ge 1$,At $i=0$, both $ef(v_0)=e(v_1)=0$ and $fe(v_0)=0$, so this identity also holds. Thus the actual relations areIn particular, $[e,f]v_1=2v_1\ne -2v_1=h{v}_1$ in characteristic zero; the requested equality cannot hold as printed. Settingcorrects the sign and gives $[H,E]=2E$, $[H,F]=-2F$, $[E,F]=H$. These are the standard sl2 triple relations, so the standard basis of the sl2 Lie algebra acts on $V$ by $E,H,F$, giving a Lie algebra representation.

For completeness, these operators are linearly independent: applying $aE+bH+cF=0$ to $v_0$ gives $c=0$; applying it to $v_1$ then gives $b=0$; applying it to $v_2$ gives $a=0$. Hence the printed operators span a Lie subalgebra isomorphic to the sl2 Lie algebra, even though their printed normalization fails. Replacing the coefficient $i-1$ in the printed definition of $e(v_i)$ by $1-i$ is one equivalent way to repair the statement.

Use directed color-$i$ edges $u\stackrel{i}{\to }v$ for the Kashiwara operator ${\tilde{f}}_i$, so every edge subtracts ${\alpha }_i$ from the weight. The defining crystal of the defining odd-orthogonal representation isThe defining module has highest weight ${\Lambda }_1$ and precisely these five weights, all of multiplicity one. The color-two string through zero has length two, since $⟨{\epsilon }_2,{\alpha }_2^{\vee }⟩=2$. The other indicated strings have length one. Restriction to each simple-root sl2 Lie algebra and the root-string property of a crystal justify the edges; no other weight difference is the relevant simple root.

For the B2 spin crystal $B({\Lambda }_2)$, denote the four weights $\frac{1}{2}(\pm {\epsilon }_1\pm {\epsilon }_2)$ by their sign pairs. Its graph isThe fundamental weight ${\Lambda }_2$ is a minuscule weight: its pairings with all positive coroots are zero or one. Its Weyl orbit has four weights, and the dimension formula gives four, so each occurs once and there are no additional weights. All simple-root strings have length one. Subtracting ${\alpha }_1$ or ${\alpha }_2$ where another weight is present gives exactly this graph.

For the B2 adjoint crystal, write $x_{\beta }$ for the vertex of root weight $\beta$, and use two distinct zero-weight vertices $z_1,z_2$. There are eight root vertices and two zero vertices because the zero-weight space is the two-dimensional Cartan subalgebra. The complete lists of colored strings areEvery arrow in a given row has that row's color; the reverse arrows are the raising operators, and all unlisted lowering operations vanish. In particular the two zero vertices must not be merged.

For a constructive justification, let $V={\mathbb{C}}^5$ with the invariant symmetric form from part(a). The mapis an equivariant isomorphism ${\Lambda }^{2}V\cong {\mathfrak{so}}_5$, identifying the highest weight as ${\epsilon }_1+{\epsilon }_2$. Compute its crystal as the highest-weight component starting at $1\otimes 2$ in $B({\Lambda }_1)\otimes B({\Lambda }_1)$, with labels $1,2,0,\bar{2},\bar{1}$ for the defining chain. The crystal tensor-product rule iswhere ${\epsilon }_i$ counts available raising steps and ${\phi }_i$ counts lowering steps in the color-$i$ string. It generates the following ten vertices:Applying the displayed rule to these labels gives precisely the listed strings, proving the entire diagram rather than only its weight multiplicities. Tensor labels here are crystal vertices in the combinatorial limit, not an assertion that a bare ordinary tensor such as $v_0\otimes v_0$ lies in ${\Lambda }^{2}V$.

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 $h=\mathrm{diag}(1,-1)$, $e=E_{12}$ and $f=E_{21}$ in the sl2 Lie algebra, so $hx=x$, $hy=-y$, $ey=x$ and $fx=y$. On the symmetric power, a monomial $x^{3-k}{y}^k$ has weight $3-2k$. The tensor product of Lie algebra representations adds weights, because $h$ acts as $h\otimes I+I\otimes h$.

Put $A_k=x\otimes x^{3-k}{y}^k$ and $B_k=y\otimes x^{3-k}{y}^k$ for $0\le k\le 3$. These eight vectors are a basis of $U$; their weights are $4-2k$ and $2-2k$. Thus the weight-space decomposition and a weight basis for each summand areIn particular $\boxed{U=U_4\oplus U_2\oplus U_0\oplus U_{-2}\oplus U_{-4}}$, with dimensions summing to eight. Here the subscripts are the eigenvalues of the chosen Cartan element $h$.