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For the defining representation, the standard coordinate vectors $e_1,e_2,e_3,e_4,e_5$ are weight vectors of weights ${\epsilon }_1,{\epsilon }_2,-{\epsilon }_1,-{\epsilon }_2,0$, respectively. Thus the weight diagram has four axial points and the origin, each of multiplicity one.

For the Adjoint representation, the eight nonzero weights of a representation are the roots from part (ii), each of multiplicity one, with the displayed $X_{\alpha }$ as basis. The zero weight space is the Cartan subalgebra, with basisThe Adjoint representation diagram therefore has the four axial points, four diagonal points, and the origin of multiplicity two. The multiplicities sum to ten, as they must.

Write $\bar{i}=2n+1-i$, and let $E_{ab}$ denote a matrix unit. The diagonal Cartan subalgebra consists ofFor a diagonal matrix, $[t,E_{ab}]=(t_a-t_b)E_{ab}$. Substituting into the symplectic relation shows that the following are root vectors:The zero weight space is spanned by $H_i=E_{ii}-E_{\bar{i}\bar{i}}$. The symplectic equation pairs the remaining entries exactly as in this table; the displayed vectors and the $H_i$ are independent and span the algebra. This proves the root-space decompositionThere are $2n^2$ roots of a root system, each with one-dimensional root space, giving $\dim \mathfrak{g}=n(2n+1)$.

The upper triangular root vectors select preciselyThe simple roots of this Cn root system areIndeed, the positive roots of a root system have expansionsThese expansions also identify the highest root:Normalize the coordinate inner product by $({\epsilon }_i,{\epsilon }_j)={\delta }_{ij}$. The simple coroots are ${\alpha }_i^{\vee }={\epsilon }_i-{\epsilon }_{i+1}$ for $i<n$ and ${\alpha }_n^{\vee }={\epsilon }_n$. Solving $⟨{\omega }_j,{\alpha }_i^{\vee }⟩={\delta }_{ij}$ gives the fundamental weightsIn the sum of the positive roots of a root system, the pair $({\epsilon }_i-{\epsilon }_j)+({\epsilon }_i+{\epsilon }_j)$ contributes $2{\epsilon }_i$, and the long root contributes another $2{\epsilon }_i$. Therefore the Weyl vector isThese are the Positive-root data for Cn.

For $n\ge 2$, the finite Dynkin diagram is a chain of $n$ vertices. The bonds between ${\alpha }_i$ and ${\alpha }_{i+1}$ for $i<n-1$ are single. The last bond is double, with its arrow pointing from the long root ${\alpha }_n$ toward the short root ${\alpha }_{n-1}$. For the Extended Dynkin diagram, add ${\alpha }_0=-\theta =-2{\epsilon }_1$ at the other end, doubly bonded to ${\alpha }_1$, with arrow toward ${\alpha }_1$. The following Cn Dynkin diagrams include both ends and the low-rank cases.

At $n=1$, there is just the root ${\alpha }_1=2{\epsilon }_1$ and one finite vertex. The affine Cartan matrix is $\left(\begin{array}{cc}2& -2\\ -2& 2\end{array}\right)$; its two vertices have equal root length, so the double bond has no arrow.

Choose a Cartan subalgebra, simple roots ${\alpha }_1,\dots ,{\alpha }_r$, and root vectors $e_i,f_i$ normalized by $[e_i,f_i]=h_i={\alpha }_i^{\vee }$. LetHere the coroots lie in the Cartan subalgebra, and ${\alpha }_i({\rho }^{\vee })=1$. This last identity is the Weyl vector identity applied to the dual root system. For distinct simple roots, ${\alpha }_i-{\alpha }_j$ is not a root, hence $[e_i,f_j]=0$. ConsequentlyThese are exactly the sl2 Lie algebra relations. For nonzero $\mathfrak{g}$ the resulting homomorphism from ${\mathfrak{sl}}_2$ is nonzero and therefore injective, since ${\mathfrak{sl}}_2$ is simple. Its image is a Principal sl2 subalgebra. Equivalently its raising element is a principal nilpotent element, a nilpotent element with Lie algebra centralizer of dimension equal to the rank. The construction works for a semisimple Lie algebra with several simple factors by taking the triple in every factor, giving one diagonally embedded ${\mathfrak{sl}}_2$. This is the construction of a principal sl2 triple.

For a representation $V$, restrict to this subalgebra and take its formal character. If $V_{\mu }$ is the weight space for the original Cartan subalgebra, the resulting q-character of a highest-weight representation, or the analogous q-character for any finite-dimensional representation, isIt is independent of conjugating the chosen principal triple. In particular a root $\beta ={\sum }_i{k}_i{\alpha }_i$ has grading $\beta (h)=2{\sum }_i{k}_i$. Some conventions use $q^{h/2}$ instead; that convention divides all exponents here by two. The normalization above agrees with the ordinary sl2 Lie algebra character in the preceding part.

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 $\left(\genfrac{}{}{0ex}{}{n}{3}\right)$.

Write an element of the symplectic Lie algebra in $n\times n$ blocks. Direct multiplication in $AJ+J{A}^T=0$ givesThe diagonal Cartan subalgebra consists of $H=\mathrm{diag}(h_1,\dots ,h_n,-h_1,\dots ,-h_n)$. Define ${\epsilon }_i(H)=h_i$. Under the Adjoint representation, the off-diagonal matrix units in $P$ have weights ${\epsilon }_i-{\epsilon }_j$; the symmetric entries in $Q$ have weights ${\epsilon }_i+{\epsilon }_j$, including $2{\epsilon }_i$ on the diagonal; those in $R$ have the negatives of these weights. All nonzero root spaces are one-dimensional, while the zero weight space is the diagonal Cartan subalgebra. Thus the Cn root system isChoose the positive roots and simple roots asTheir highest root and Weyl vector areFor the last identity, each pair $i<j$ contributes $2{\epsilon }_i$ before division by two, and the long positive root contributes another ${\epsilon }_i$ afterward. Successive simple roots have equal length except for ${\alpha }_n$, whose squared length is twice that of ${\alpha }_{n-1}$. The Cn Dynkin diagram is a chain with a double last bond, whose arrow points toward ${\alpha }_{n-1}$, the short root:

The displayed schematic is for $n\ge 4$; omit intermediate nodes for $n=2,3$. For $n=1$ the diagram has just the single node ${\alpha }_1=2{\epsilon }_1$.

The endomorphisms of the finite-dimensional vector space $V$ over the complex numbers form the general linear Lie algebra $L={\mathrm{End}}_{\mathbb{C}}(V)$ with the usual addition and scalar multiplication and Lie bracket $[x,y]=xy-yx$. This commutator is bilinear and antisymmetric, and expanding the six products verifies the Jacobi identity.

For a Lie subalgebra $\mathfrak{g}=L_1$, being an abelian Lie algebra means $[\mathfrak{g},\mathfrak{g}]=0$. A nilpotent Lie algebra has ${\gamma }_1=\mathfrak{g}$, ${\gamma }_{j+1}=[\mathfrak{g},{\gamma }_j]$ eventually zero; a solvable Lie algebra has ${\mathfrak{g}}^{(0)}=\mathfrak{g}$, ${\mathfrak{g}}^{(j+1)}=[{\mathfrak{g}}^{(j)},{\mathfrak{g}}^{(j)}]$ eventually zero. These are the lower central series of a Lie algebra and derived series of a Lie algebra, respectively. Nilpotence is a condition on the Lie bracket, and does not require every member to be a nilpotent endomorphism: a nonzero scalar multiple of the identity spans an abelian Lie algebra.

A flag of a vector space is an increasing chain of vector subspaces. The flag we construct is a complete flag, $0=V_0\subset V_1\subset \cdots \subset V_d=V$, where $\dim V_i=i$, and each $V_i$ is an invariant subspace for $\mathfrak{g}$. We first prove the common-eigenvector assertion in the Lie theorem, by induction on $\dim \mathfrak{g}$; the zero algebra is immediate. For nonzero solvable $\mathfrak{g}$, its derived algebra is proper, so there is a codimension-one ideal of a Lie algebra $\mathfrak{k}$ containing $[\mathfrak{g},\mathfrak{g}]$. Write $\mathfrak{g}=\mathfrak{k}\oplus \mathbb{C}x$. By induction there are $v\ne 0$ and a linear functional $\lambda$ on $\mathfrak{k}$ such that $kv=\lambda (k)v$ for every $k\in \mathfrak{k}$.

Let $W$ be the cyclic subspace spanned by $v,xv,x^{2}v,\dots$. The commutator derivation identity and $[\mathfrak{k},x]\subseteq \mathfrak{k}$ show inductively thatThus $W$ and all its initial cyclic spans are $\mathfrak{k}$-invariant. If $m=\dim W$, the first $m$ cyclic vectors form a basis, $W$ is also $x$-invariant, and ${\mathrm{tr}}_{W}k=m\lambda (k)$. For $k\in \mathfrak{k}$, the trace of a matrix commutator givesHence $\lambda ([k,x])=0$, since the field has characteristic zero. The nonzero common weight spaceis $x$-invariant: $k(xw)=x(kw)+[k,x]w=\lambda (k)xw$. The restriction of $x$ to $V_{\lambda }$ has an eigenvector, because $\mathbb{C}$ is an algebraically closed field. This is a common eigenvector for $\mathfrak{g}$. Its line is invariant, and repeating the argument on the quotient vector space gives the complete invariant flag. Equivalently, this proves simultaneous triangularization of a Lie algebra representation.

In a basis adapted to this complete flag, every member of $\mathfrak{g}$ is upper triangular, so every member of its derived algebra is strictly upper triangular. Products of $d$ strictly upper triangular matrices vanish, and each iterated Lie bracket of $d$ such matrices is a sum of these products. Consequently the derived algebra is a nilpotent Lie algebra. We may therefore take $L_2=[L_1,L_1]$: it is an ideal, and the quotient Lie algebra $L_1/L_2$ is abelian. This also covers $\dim V=0$.

A finite-dimensional Lie algebra over the complex numbers $L$ is a semisimple Lie algebra when its solvable radical is zero, equivalently when it has no nonzero solvable ideals. Its Killing form isThe cyclic property of the trace makes this bilinear form symmetric and gives its invariance of a bilinear form on a Lie algebra:It follows that $R=\{x:B_L(x,L)=0\}$ is an ideal of a Lie algebra. For $x\in R$, $\mathrm{ad}x$ induces the zero map on $L/R$. Therefore, for $x,y\in R$, the matrix trace splits over the invariant subspace $R$ and the quotient to give $B_R(x,y)=B_L(x,y)=0$. In particular $B_R([R,R],R)=0$. The Cartan solvability criterion implies that $R$ is solvable. Since $L$ is semisimple, $R=0$: the Killing form is nondegenerate.

For completeness, the trace step in the Cartan solvability criterion is precisely the mechanism of the previous solution. For a complex matrix Lie algebra $\mathfrak{a}$ with $\mathrm{tr}(xy)=0$ for $x\in [\mathfrak{a},\mathfrak{a}]$, $y\in \mathfrak{a}$, set $U=[\mathfrak{a},\mathfrak{a}]$ and $W=\mathfrak{a}$. If $\beta \in M$ and $x=[a,b]$, then $\mathrm{tr}(x\beta )=\mathrm{tr}(a[b,\beta ])=0$, since $[b,\beta ]\in U$. Linearity and the trace orthogonality nilpotence lemma show that every member of $U$ is nilpotent. The Engel theorem makes $U$ nilpotent and hence $\mathfrak{a}$ solvable. Apply this to $\mathfrak{a}=\mathrm{ad}R$; the kernel of this Adjoint representation of a Lie algebra is the abelian center of $R$, so $R$ is solvable as claimed.

For an arbitrary complex Lie algebra, a Cartan subalgebra $H$ means a nilpotent Lie algebra that is self-normalizing: $N_L(H)=\{x:[x,H]\subseteq H\}=H$. This definition does not assume that $H$ is abelian. We prove that it is abelian when $L$ is semisimple.

Use the generalized-weight decomposition for a nilpotent Lie algebra for the action of $H$ on $L$. Its zero generalized weight space isWe have $H\subseteq L^0$, since $H$ is nilpotent. If $L^0/H\ne 0$, the Engel theorem gives a nonzero coset annihilated by every $h\in H$. Its representative $x$ satisfies $[H,x]\subseteq H$, contradicting $N_L(H)=H$. Thus $L^0=H$.

For a nonzero generalized weight $\lambda$, choose $h_0\in H$ with $\lambda (h_0)\ne 0$. The operator $\mathrm{ad}{h}_0$ is invertible on $L^{\lambda }$ and nilpotent on $H$. For $h\in H$, $z\in L^{\lambda }$, write $z=(\mathrm{ad}{h}_0{)}^{m}w$ with $m$ large enough that $(\mathrm{ad}{h}_0{)}^{m}h=0$. Invariance of the Killing form givesOn the other hand, $H$ is solvable, so the Lie theorem triangularizes its action on $L$. For $a,b,h\in H$, the matrix $\mathrm{ad}[a,b]$ is strictly upper triangular, while $\mathrm{ad}h$ is upper triangular. Thus $B_L([H,H],H)=0$. Together with $L=H\oplus {⨁}_{\lambda \ne 0}{L}^{\lambda }$, this yields $B_L([H,H],L)=0$. Nondegeneracy gives $[H,H]=0$.

Finally, if $x$ commutes with $H$, it normalizes $H$, hence lies in $H$. Any abelian subalgebra containing $H$ consists of such elements. Thus $H$ is a maximal abelian subalgebra, indeed $C_L(H)=H$.

Choose a positive system of a root system and the corresponding triangular decomposition of a Lie algebra $L={\mathfrak{n}}^{-}\oplus H\oplus {\mathfrak{n}}^{+}$. A primitive element of a Lie algebra representation of weight $\omega \in H^{\ast }$ is a nonzero vector $v$ with $hv=\omega (h)v$ for every $h\in H$ and ${\mathfrak{n}}^{+}v=0$. Thus it is a highest-weight vector; the choice of positive roots is part of this definition. A highest-weight representation is generated by such a vector.

Let $\mathfrak{b}=H\oplus {\mathfrak{n}}^{+}$ be the corresponding Borel subalgebra. Define its one-dimensional Lie algebra representation ${\mathbb{C}}_{\omega }$ by the scalar $\omega (h)$ on $H$ and zero on ${\mathfrak{n}}^{+}$. This respects the Lie bracket, since $[\mathfrak{b},\mathfrak{b}]\subseteq {\mathfrak{n}}^{+}$. The induced Verma moduleis nonzero: the Poincaré-Birkhoff-Witt theorem identifies its underlying vector space with $U({\mathfrak{n}}^{-})$, and $v=1\otimes 1$ is a primitive element of weight $\omega$. Its weight spaces are finite dimensional, its top weight space is the line $\mathbb{C}v$, and all other weights are $\omega -\sum n_i{\alpha }_i$ with $n_i\ge 0$ and at least one positive coefficient.

A proper submodule cannot contain $v$, because $v$ generates $M(\omega )$. More strongly it has no component of weight $\omega$: any finite sum of distinct weight vectors can be projected onto its individual components by a polynomial in a generic element of $H$. Therefore the sum $N$ of all proper submodules still misses the top weight line and is proper. It contains every proper submodule, so the irreducible quotient of a Verma moduleis irreducible and retains the nonzero primitive element $v+N$. This constructs the requested representation for every $\omega \in H^{\ast }$. It is not asserted to be finite dimensional for arbitrary $\omega$.

For the sl2 Lie algebra, use $h=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$, $e=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$ and $f=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)$, with $[h,e]=2e$, $[h,f]=-2f$, $[e,f]=h$. The classification of finite-dimensional sl2 representations gives one irreducible $V_n$ for every integer $n\ge 0$. In a basis $v_0,\dots ,v_n$ its action iswhere $v_{-1}=v_{n+1}=0$. These formulas satisfy the three Lie brackets. Any nonzero invariant subspace contains a weight vector by polynomial projection using $h$; repeated application of $e$ gives $v_0$, and applications of $f$ then give the entire basis. Thus $V_n$ is irreducible.

Conversely, in any finite-dimensional irreducible module, start with an eigenvector of $h$ and apply $e$ until reaching a nonzero vector $v$ with $ev=0$. This process terminates because $e$ increases the eigenvalue by two and only finitely many eigenvalues occur. Write $hv=av$. The sl2 highest-weight lowering formula givesLet $f^{n}v\ne 0$ and $f^{n+1}v=0$, which again follows from the finite set of weights. Applying $e$ to the latter identity yields $(n+1)(a-n)f^{n}v=0$, so $a=n$. The resulting $n+1$ distinct weight vectors span an invariant subspace, hence the entire irreducible module. Therefore $\dim V_n=n+1$ and its primitive weight is ${\omega }_n(h)=n$, with $n=0,1,2,\dots$; its weights are $n,n-2,\dots ,-n$.

Use the sl2 Lie algebra relations $[H,X]=2X$, $[H,Y]=-2Y$ and $[X,Y]=H$. A highest-weight vector satisfies $Xv=0$ and $Hv=mv$, and $H{Y}^{j}v=(m-2j)Y^{j}v$. For $k=0$, $XYv=YXv+Hv=mv$. Assuming the formula at $k-1$, one getsThus the sl2 highest-weight lowering formula isIn particular a finite-dimensional Irreducible Lie algebra representation has highest weight $m\in {\mathbb{Z}}_{\ge 0}$ and weight vectors $v,Yv,\dots ,Y^{m}v$, as in the classification of finite-dimensional sl2 representations.

Let $V_{\lambda }$ be a finite-dimensional Irreducible Lie algebra representation of a complex semisimple Lie algebra, with highest weight $\lambda$. Write $m_{\lambda }(\mu )=\dim (V_{\lambda }{)}_{\mu }$, taking it to be zero when $\mu$ is not a weight, and let $\rho$ be the half-sum of positive roots. The Killing form induces an inner product on the real span of weights. Freudenthal multiplicity formula statesThe sums are finite. For a weight $\mu \ne \lambda$, the coefficient on the left is positive: move $\mu$ into the dominant Weyl chamber, use that a weight is below $\lambda$ in dominance order, and note that $(\mu ,\rho )$ does not increase on moving back out of that chamber. The resulting recursion starts from $m_{\lambda }(\lambda )=1$.

Here is a proof using the allowed Casimir operator. Choose root vectors $E_{\alpha },F_{\alpha }$ for positive and negative roots, normalized by $B(E_{\alpha },F_{\alpha })=1$, and let $t_{\alpha }\in \mathfrak{h}$ satisfy $B(t_{\alpha },h)=\alpha (h)$. Invariance of a bilinear form on a Lie algebra gives $[E_{\alpha },F_{\alpha }]=t_{\alpha }$. If $h_i,h^i$ are dual bases of the Cartan subalgebra under $B$, the Casimir operator isOn the weight space of $\mu$, the first sum acts by $(\mu ,\mu )$. Replacing $E_{\alpha }{F}_{\alpha }$ with $F_{\alpha }{E}_{\alpha }+t_{\alpha }$ givesDefine $T_{\alpha }(\mu )=\mathrm{tr}(F_{\alpha }{E}_{\alpha }{|}_{(V_{\lambda }{)}_{\mu }})$. The cyclic trace identity between adjacent weight spaces and the commutator relation yieldIterate upwards until the weight spaces vanish to obtain $T_{\alpha }(\mu )={\sum }_{j\ge 1}(\mu +j\alpha ,\alpha )m_{\lambda }(\mu +j\alpha )$. Finally take the trace of the displayed restriction of $C$. Its Casimir eigenvalue is $(\lambda ,\lambda +2\rho )$, so subtracting $(\mu ,\mu +2\rho )m_{\lambda }(\mu )$ proves the formula. This trace argument handles weight multiplicities greater than one without choosing a separate sl2 Lie algebra string through each vector.

Over the complex numbers, every finite-dimensional representation of a solvable Lie algebra has a basis in which every representing matrix is upper triangular. The Lie theorem is often stated first as the existence of a common eigenvector in every nonzero finite-dimensional Lie algebra representation of a solvable Lie algebra. Applying that assertion successively to quotient representations gives an invariant complete flag, and hence the upper triangular form. The same proof works over any algebraically closed field of characteristic zero.

We prove the common eigenvector assertion by induction on $\dim \mathfrak{g}$, writing the action as $av$. The zero Lie algebra is immediate. If $\mathfrak{g}\ne 0$ is solvable, its derived series of a Lie algebra shows that $[\mathfrak{g},\mathfrak{g}]\ne \mathfrak{g}$. Choose a codimension-one ideal of a Lie algebra $\mathfrak{a}$ containing $[\mathfrak{g},\mathfrak{g}]$, and choose $x\notin \mathfrak{a}$. By induction there are $v\ne 0$ and a linear functional $\lambda :\mathfrak{a}\to \mathbb{C}$ such that $av=\lambda (a)v$ for all $a\in \mathfrak{a}$.

Let $W$ be the span of $v,xv,x^{2}v,\dots$. If $m=\dim W$, the first $m$ of these vectors form a basis, and $W$ is $x$-invariant. We claim that for each $a\in \mathfrak{a}$,For $j=0$ this is the definition of $v$. For the induction step, use $ax=xa+[a,x]$ and $[a,x]\in \mathfrak{a}$. Applying the induction hypothesis to both $a$ and $[a,x]$ proves the claim. Consequently $W$ is $\mathfrak{a}$-invariant, and every $a\in \mathfrak{a}$ acts on $W$ by an upper triangular matrix with all diagonal entries $\lambda (a)$.

Since both $x$ and $a$ preserve $W$, the matrix trace of their commutator on $W$ is zero. The claim applied to $[x,a]\in \mathfrak{a}$ givesHere characteristic zero is essential: $m\ne 0$ in the field, so $\lambda ([x,a])=0$. Now the common weight spaceis nonzero and $x$-invariant. Indeed, for $w\in E_{\lambda }$,An endomorphism of a nonzero finite-dimensional complex vector space has an eigenvector, so choose an eigenvector of $x$ in $E_{\lambda }$. It is a common eigenvector for $\mathfrak{g}=\mathfrak{a}+\mathbb{C}x$. This proves the Lie theorem.

For the printed matrices in characteristic $p$, $[x,y]=x$, and there is no common eigenvector. Index the standard basis by $e_0,\dots ,e_{p-1}$. The cyclic entry in the PDF givesThe $p$ diagonal eigenvalues of $y$ are distinct in $k$. Thus every eigenvector of $y$ is a scalar multiple of a single $e_j$. Since $p\ge 2$, $x{e}_j$ is never a scalar multiple of $e_j$, proving the assertion even when $k$ is not an algebraically closed field.

For $1\le j\le p-1$,At the cyclic boundary,Hence $[x,y]=x$. The two-dimensional Lie subalgebra $kx+ky$ has derived algebra $kx$, whose own derived algebra is zero, so it is solvable. It nevertheless has no common eigenvector, including after extending $k$ to its algebraic closure. This is a failure of Lie theorem in positive characteristic. In the proof above, the obstruction is precisely that $\dim W=p$ can vanish as a scalar in $k$.

The derived algebra of a complex solvable Lie algebra is nilpotent. First suppose $\mathfrak{g}\subseteq \mathfrak{gl}(V)$. By the Lie theorem, put every element of $\mathfrak{g}$ in upper triangular form. The diagonal of a commutator of upper triangular matrices is zero, so $\mathfrak{d}=[\mathfrak{g},\mathfrak{g}]$ consists of strictly upper triangular matrices. The Lie algebra of all such matrices is a nilpotent Lie algebra: if $F_r$ consists of matrices whose entries vanish whenever $j-i<r$, thenTherefore the lower central series of a Lie algebra of $\mathfrak{d}$ reaches zero. Alternatively, every element of $\mathfrak{d}$ is a nilpotent linear map, and the Engel theorem states that a finite-dimensional Lie subalgebra consisting of nilpotent linear maps is a nilpotent Lie algebra.

For an abstract complex solvable Lie algebra $\mathfrak{g}$, apply the preceding result to its Adjoint representation. The Lie algebra $\mathrm{ad}\mathfrak{d}=[\mathrm{ad}\mathfrak{g},\mathrm{ad}\mathfrak{g}]$ is nilpotent. Since $\ker (\mathrm{ad}{|}_{\mathfrak{d}})=\mathfrak{d}\cap Z(\mathfrak{g})$ is central in $\mathfrak{d}$, some term of the lower central series of a Lie algebra of $\mathfrak{d}$ lies in that central ideal; the next term is zero. Thus $\mathfrak{d}$ itself is nilpotent.

Conversely, every nilpotent Lie algebra is solvable, since its derived series of a Lie algebra is contained term by term in its lower central series of a Lie algebra. If $\mathfrak{d}$ is nilpotent, it is therefore solvable, and $\mathfrak{g}/\mathfrak{d}$ is Abelian. More directly, the derived series of a Lie algebra of $\mathfrak{g}$, after its first term, is the derived series of a Lie algebra of $\mathfrak{d}$. Consequently the derived algebra nilpotence criterion is

Give $V_n$ the action $(g\cdot P)(x,y)=P((x,y)g)$, where $(x,y)$ is a row vector. Substitution preserves homogeneous degree, and composition agrees with matrix multiplication, so this is a complex representation of SU(2) group. The diagonal torus $\mathrm{diag}(z,z^{-1})$ acts on $x^{n-j}{y}^j$ with weight $z^{n-2j}$; these weights are distinct.

A nonzero invariant subspace is invariant under the torus, so finite-dimensional Fourier projection onto its weight spaces supplies a nonzero monomial. Differentiating the group action and complexifying supplies the operators $E=x{\partial }_y$ and $F=y{\partial }_x$. Their repeated application moves between every successive monomial, with nonzero coefficients until an endpoint is reached:The invariant subspace therefore contains the whole monomial basis. ThusFor $n=0$ this is the one-dimensional trivial representation. Torus weight projections are legitimate because an invariant complex subspace is closed in finite dimension.

For $G={\mathbb{G}}_m$, write $k[G]=k[t,t^{-1}]$. Sincethe weight-space decomposition isas a $G\times G$-representation.

If $V={⨁}_r{V}_r$ is the weight decomposition of a finite-dimensional $G$-module, the matrix-coefficient argument again gives ${\mathrm{Hom}}_G(V,k[G])\cong V^{\ast }$. A map from $k[G]$ to $V$ is independently determined by the image in $V_r$ of the basis vector $t^r$; only finitely many $V_r$ are nonzero. Thusas vector spaces.

The weights of $M_{\lambda }$ lie below $\lambda$ in the positive-root order, and every weight space is finite-dimensional. A nonzero submodule $V$ is stable under the Cartan subalgebra, so it is a direct sum of its weight spaces. Choose a maximal weight $\mu$ occurring in $V$. Every positive-root operator would raise its weight; maximality therefore makes it kill any nonzero $v\in V_{\mu }$. Thus $v$ is a singular vector.

The Casimir element is central and acts throughout $M_{\lambda }$ byThe same element acts on the highest-weight vector $v$ of weight $\mu$ byBoth are the action of one operator on the same module, so the scalars agree and

Choose the Borel subalgebra $\mathfrak{b}=\mathfrak{t}\oplus {\mathfrak{n}}^{+}$ determined by the positive roots. Regard the one-dimensional space ${\mathbb{C}}_{\lambda }$ as a $\mathfrak{b}$-module on which ${\mathfrak{n}}^{+}$ acts by zero and $h\in \mathfrak{t}$ acts by $\lambda (h)$. The Verma module isThe Poincaré-Birkhoff-Witt theorem identifies it as a vector space with $U({\mathfrak{n}}^{-})$ acting on a highest-weight vector $v_{\lambda }$.

For each positive root $\alpha$, arbitrary powers of a negative-root vector contribute the geometric series $1+e^{-\alpha }+e^{-2\alpha }+\cdots$. Consequently the formal character of a weight module isThis product is interpreted in the completion of the group algebra in the negative-root direction; the PBW basis proves that every coefficient is the correct finite weight space dimension.

Use ${\mathrm{ch}}_{q}V={\mathrm{tr}}_V{q}^{2{\rho }^{\vee }}$. The positive roots of a root system of the G2 root system have simple-root heights $1,1,2,3,4,5$. Each root space has dimension one, and the zero weight space has dimension two. ConsequentlyBy the classification of finite-dimensional sl2 representations, restriction to the Principal sl2 subalgebra is $L_{10}\oplus L_2$, of dimensions eleven and three. If the grading uses ${\rho }^{\vee }$ instead, divide every exponent by two.

For a Cartan subalgebra $H$ of a finite-dimensional complex semisimple Lie algebra $L$, $H$ is abelian and $C_L(H)=H$, so it is maximal among abelian subalgebras. Starting from the nilpotent self-normalizing definition, the generalized-weight decomposition for a nilpotent Lie algebra and the Engel theorem show that the zero generalized weight space of its adjoint action is exactly $H$. Invariance of the Killing form makes $H$ orthogonal to every nonzero generalized weight space. The Lie theorem gives $B([H,H],H)=0$, and nondegeneracy then forces $[H,H]=0$. An element commuting with $H$ normalizes it, hence belongs to $H$.

For the unitary group $U(n)$ with $n\ge 3$, the third tensor power of its defining representation decomposes as displayed. The first two summands have highest weights $(3,0,\dots )$ and $(1,1,1,0,\dots )$ and dimensions $\left(\genfrac{}{}{0ex}{}{n+2}{3}\right)$ and $\left(\genfrac{}{}{0ex}{}{n}{3}\right)$. The mixed summand has highest weight $(2,1,0,\dots )$ and dimension $n(n^2-1)/3$. In the three-dimensional weight space spanned by permutations of $e_1\otimes e_1\otimes e_2$, the vectors killed by all positive root vectors have coefficient sum zero, giving multiplicity two. For $n=2$, the exterior cube vanishes and $V_{(2,1)}=\det \otimes {\mathbb{C}}^2$.