On a monomial $X^a{Y}^b$, direct differentiation givesFor example $(EF-FE)(X^a{Y}^b)=[a(b+1)-b(a+1)]X^a{Y}^b=(a-b)X^a{Y}^b$. The same calculation with $H$ givesThese are exactly the brackets of $h,e,f$, so the linear map $\rho$ respects every bracket and is a Lie algebra representation.

For $m\ge 0$, let $W_m$ be the space of homogeneous polynomials of total degree $m$. Each operator preserves total degree. Every polynomial is a finite sum of homogeneous polynomials, soIn $W_m$, the monomials $v_j=X^{m-j}{Y}^j$ have distinct weights $m-2j$ and exactly the actions described in part (i). To see simplicity directly, a nonzero invariant subspace is invariant under $H$ and hence contains one of its weight components: polynomial interpolation in $H$ projects onto each of its distinct eigenspaces. Starting with a nonzero multiple of $v_j$, repeated application of $E$ reaches $v_0$, and repeated application of $F$ then reaches all $v_j$. All coefficients used are nonzero in characteristic zero. Therefore $W_m$ is simple and isomorphic to $L_m$.

ThusThis is an algebraic direct sum, not an infinite product or completion. Its simple submodules are precisely the $W_m$: they are pairwise nonisomorphic and each occurs once, so projection onto homogeneous degrees and Schur's lemma exclude diagonal copies between distinct degrees. Complete reducibility here follows from the explicit decomposition, rather than applying a finite-dimensional theorem directly to the infinite-dimensional $W$.

Every $2\times 2$ matrix has the unique decompositionBoth summands define invariant subspaces for the Adjoint representation of a Lie algebra. Scalar matrices commute with everything, so $\mathbb{C}I$ is $L_0$. The traceless matrices form ${\mathfrak{sl}}_2$ with basis $e,h,f$ of weights $2,0,-2$. Their brackets satisfy $[e,f]=h$, $[h,e]=2e$, $[h,f]=-2f$; the raising and lowering operations connect all three weight lines. This is the simple module $L_2$. ThereforeAs a check, its zero-weight space has dimension two and its one-weight space is zero, agreeing with part (ii).

In the simple module $L_m$, weight $0$ occurs once if $m$ is even and never if $m$ is odd. Weight $1$ occurs once if $m$ is odd and never if $m$ is even. Thus each simple summand contributes exactly one to $\dim V_0+\dim V_1$.

Complete reducibility of semisimple Lie algebra representations writes $V={⨁}_m{L}_m^{\oplus a_m}$. Taking the two weight spaces commutes with this finite direct sum, soThe right side counts irreducible summands with multiplicity, independently of a chosen decomposition.

By the classification of finite-dimensional sl2 representations over $\mathbb{C}$, every finite-dimensional module is a direct sum of simple modules $L_m$, indexed by integers $m\ge 0$. The module $L_m$ has dimension $m+1$ and a basis $v_0,\dots ,v_m$ withwith out-of-range vectors interpreted as zero. In particular $v_0$ is a highest weight vector of weight $m$. The weights are $m,m-2,\dots ,-m$, each with multiplicity one. The multiplicities of the $L_m$ in a finite module determine that module up to isomorphism.

Suppose $\mathrm{ad}(x)$ has a nonzero eigenvalue $\lambda$ and corresponding vector $y$. Then $[x,y]=\lambda y$, soFor a diagonalizable endomorphism $T$, its eigenspace decomposition shows $\ker T^2=\ker T$. The displayed equations therefore contradict diagonalizability of $\mathrm{ad}(y)$.

Thus no $\mathrm{ad}(x)$ has a nonzero eigenvalue. Each is diagonalizable over the algebraically closed field, so each is the zero endomorphism. Hence $[x,z]=0$ for every $x,z\in \mathfrak{g}$, proving

If $\mathfrak{g}$ is nilpotent, every Lie subalgebra $\mathfrak{a}$ is nilpotent, since ${\mathfrak{a}}^j\subseteq {\mathfrak{g}}^j$ in the respective lower central series of a Lie algebra. A two-dimensional nilpotent algebra has nonzero centre by part (i). If $0\ne z$ is central, complete it to a basis $(z,u)$; every bracket of these basis elements is zero. Thus $\mathfrak{a}$ is abelian.

Conversely, suppose $\mathfrak{g}$ is not nilpotent. By Engel's theorem, some $\mathrm{ad}(x)$ is not nilpotent. Since the field is algebraically closed, it has an eigenvalue $\lambda \ne 0$ with eigenvector $y\ne 0$. ThusThe vectors $x,y$ are independent: otherwise their bracket would be zero. Their span is therefore a two-dimensional Lie subalgebra and it is nonabelian. This contradicts the hypothesis. Hence

Fix $x$ with $x^N=0$ as a matrix. On the full matrix algebra define $L_x(y)=xy$ and $R_x(y)=yx$. These endomorphisms commute and each has $N$th power zero. The Adjoint representation of a Lie algebra is $\mathrm{ad}(x)=L_x-R_x$. The binomial theorem givesFor $m=2N-1$, each summand has either $j\ge N$ or $m-j\ge N$, and hence vanishes. Therefore $\mathrm{ad}(x)$ is nilpotent on the full matrix algebra and on its invariant subspace $\mathfrak{g}$.

Engel's theorem, in its adjoint form, states that a finite-dimensional Lie algebra is nilpotent if every $\mathrm{ad}(x)$ is a nilpotent endomorphism. Applying it to all $x\in \mathfrak{g}$ provesThis is distinct from assuming that an arbitrary representation of a nilpotent algebra must act by nilpotent matrices, which is false.

Write the lower central series of a Lie algebra as ${\mathfrak{g}}^1=\mathfrak{g}$ and ${\mathfrak{g}}^{j+1}=[\mathfrak{g},{\mathfrak{g}}^j]$. Since the nilpotent Lie algebra is nonzero, there is a last nonzero term ${\mathfrak{g}}^c$, with ${\mathfrak{g}}^{c+1}=0$. Thus $[\mathfrak{g},{\mathfrak{g}}^c]=0$, soConsequently the centre is nonzero. This argument works over every field and includes an abelian algebra, for which $c=1$.

A tabloid is the row-equivalence class $\{t\}$ of a tableau filled bijectively with $1,\dots ,n$: order inside each row is forgotten, but the rows remain distinguished. The permutation module $M^{\lambda }$ has these tabloids as an $F$-basis, with ${\Sigma }_n$ permuting their entries. Let $C_t$ be the column stabilizer of $t$ and defineThe Specht module $S^{\lambda }$ is the submodule spanned by all polytabloids $e_t$.

If a module homomorphism $\varphi :M^{\lambda }\to M^{\mu }$ is nonzero on $S^{\lambda }$, it is nonzero on some spanning $e_t$. Equivariance gives $0\ne \varphi (e_t)={\kappa }_t\varphi (\{t\})$, so ${\kappa }_t{M}^{\mu }\ne 0$. The allowed criterion yields $\lambda \ge \mu$.

In characteristic zero, Maschke's theorem gives an equivariant projection $M^{\lambda }\to S^{\lambda }$. If $S^{\lambda }\cong S^{\mu }$, compose that projection with the isomorphism and the inclusion into $M^{\mu }$. The preceding argument gives $\lambda \ge \mu$. Reversing the roles gives $\mu \ge \lambda$, so antisymmetry of dominance givesFor a map defined only on a Specht module in arbitrary characteristic, dominance need not hold. Take $F={\mathbb{F}}_2$, $n=2$, $\lambda =(1,1)$ and $\mu =(2)$. The first Specht module is the one-dimensional span ofThe transposition fixes $e$, so this is the trivial module. The one-row permutation module $M^{(2)}$ is also trivial and one-dimensional. Sending $e$ to its single tabloid is a nonzero equivariant map, yet $(1,1)\ngeqq (2)$. In this characteristic such a map need not extend to the surrounding permutation module; that is exactly where the earlier proof ceases to apply.

For a Young diagram cell $(i,j)$, its hook length is one plus the number of cells to its right and below. The hook-length formula isApplying it to the two shapes givesThe shapes here require $n\ge 4$.

A hook tabloid remembers its first-row set and the ordered singleton entries below it. The proposed map keeps that first-row set and merges the remaining entries into one unordered row. Thus it is well-defined on tabloids. Permuting entries before merging gives the same tabloid as merging before permuting, so it is a module homomorphism. Every target tabloid is reached by putting its second-row entries, in any order, into the singleton rows. Therefore it is surjective.

For $r=2$, the strictly dominating partitions of $\lambda =(n-2,1,1)$ are exactly $(n),(n-1,1),(n-2,2)$. Indeed dominance forces the first part to be at least $n-2$, and the possibilities for the remaining sum give precisely those shapes or $\lambda$ itself.

In characteristic zero, Maschke's theorem makes the permutation module semisimple, and Young's rule gives one copy of $S^{\lambda }$ and only strictly dominating additional shapes. The trivial multiplicity is one, since a transitive permutation module has a one-dimensional invariant space. For the standard representation $S^{(n-1,1)}$, Frobenius reciprocity identifies its multiplicity with the dimension of fixed vectors under the row stabilizer $S_{n-2}$. In its realization $\{(z_1,\dots ,z_n):\sum z_i=0\}$, those fixed vectors have the form $(c,\dots ,c,a,b)$ with $(n-2)c+a+b=0$, a two-dimensional space. Thus its multiplicity is two.

Finally $\dim M^{\lambda }=n(n-1)$. Subtracting the dimensions of one $S^{\lambda }$, one trivial module and two standard modules leaves $n(n-3)/2$, exactly one $S^{(n-2,2)}$. ThereforeThe additional multiplicities are $1,2,1$ respectively. If the printed weak summation index is read literally alongside its separate $S^{\lambda }$, then $m_{\lambda ,\lambda }=0$; the total multiplicity of $S^{\lambda }$ is one.

A reduced crystallographic root system $\Phi$ is a finite spanning subset of a real Euclidean vector space, excluding zero, such that $\Phi \cap \mathbb{R}\alpha =\{\pm \alpha \}$, each reflection $s_{\alpha }(v)=v-2(v,\alpha )\alpha /(\alpha ,\alpha )$ preserves $\Phi$, and all Cartan integers $⟨\beta ,\alpha ⟩=2(\beta ,\alpha )/(\alpha ,\alpha )$ are integers. A base is a vector-space basis $\Delta \subset \Phi$ in which each root has integer coordinates either all nonnegative or all nonpositive. The Weyl group is generated by the root reflections.

For nonparallel roots put $p=⟨\beta ,\alpha ⟩$ and $q=⟨\alpha ,\beta ⟩$. Then $pq=4{\cos }^2\theta$ is an integer strictly below $4$ and at least zero. ThereforeIf $(\alpha ,\beta )>0$, the positive integers $p,q$ have product $1,2$ or $3$, so at least one is one. If $p=1$, reflection gives $\beta -\alpha \in \Phi$ and hence its negative. If $q=1$, reflection directly gives $\alpha -\beta \in \Phi$. This proves the positive-inner-product assertion.

To prove that an $\alpha$-string has no gaps, suppose consecutive occurring integer positions $a<b$ in $\beta +t\alpha$ had $b-a\ge 2$. If $(\beta +a\alpha ,\alpha )<0$, apply the previous result to $\beta +a\alpha$ and $-\alpha$ to obtain the missing root at $a+1$. Thus this inner product must be nonnegative. Similarly the absence of a root at $b-1$ forces $(\beta +b\alpha ,\alpha )\le 0$. But the latter inner product exceeds the former by $(b-a)(\alpha ,\alpha )>0$, impossible. Thus every intermediate integer position occurs.

Write the complete finite string as $\beta -r\alpha ,\dots ,\beta +s\alpha$. Reflection maps position $t$ to $-p-t$, so it interchanges the two endpoints. HenceHere $r,s$ are the maximal endpoint distances, not arbitrary shorter distances mentioned in the existence assertion.

Use the Cartan-matrix convention $C_{ij}=⟨{\alpha }_i,{\alpha }_j⟩$. The displayed matrix is realized bywith the standard orthonormal coordinates. Closure under the two simple reflections givesThe positive roots are ${\alpha }_1,{\alpha }_2,{\alpha }_1+{\alpha }_2,2{\alpha }_1+{\alpha }_2$. Conversely the standard fact that every root is Weyl-conjugate to a simple root, or successive height reduction by simple reflections, shows these two reflection orbits exhaust the roots. This is the rank-two system $B_2$, also called $C_2$ after exchanging the length convention. With a transposed Cartan-matrix convention, interchange which simple-root label is short; the geometric system and Weyl group are unchanged.

The reflection in $e_2$ flips the second coordinate, while the reflection in $e_1-e_2$ swaps the two coordinates. These generate all eight signed permutation matrices in two coordinates. Thuswhere $D_8$ means the symmetry group of a square.