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For a complex semisimple Lie algebra and an integer $k\ge 0$, the Weyl character formula and Weyl denominator formula giveApply the dilation $e^{\mu }\mapsto e^{(k+1)\mu }$ to the denominator identity to obtain the numerator. Cancellation leaves a finite geometric series for each positive root of a root system. Evaluating the resulting polynomial at all formal exponentials equal to one gives the dimension, without taking an undefined quotient at the identity.

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 root of a root system of a complex semisimple Lie algebra, define $t_{\alpha }$ by $B(t_{\alpha },h)=\alpha (h)$ using the Killing form on its Cartan subalgebra. Then $\alpha (t_{\alpha })\ne 0$. Otherwise vectors $x\in {\mathfrak{g}}_{\alpha }$, $y\in {\mathfrak{g}}_{-\alpha }$ with $B(x,y)=1$ would generate a solvable Lie algebra with $[x,y]=t_{\alpha }$ central. The Lie theorem would make $\mathrm{ad}{t}_{\alpha }$ nilpotent, while the root-space decomposition makes it diagonalizable, forcing $t_{\alpha }$ into the zero center of a Lie algebra.

By irreducible traceless matrix Lie algebra is semisimple, let $R$ be the solvable radical. Lie's theorem triangularizes its action, with finitely many diagonal characters ${\lambda }_i:R\to \mathbb{C}$, and supplies a nonzero common eigenspace for one of them. For $x\in L$, conjugation by $e^{tx}$ preserves $R$ and gives an equivalent representation of it. Thus the multiset of characters ${\lambda }_i\circ e^{t\mathrm{ad}x}$ is constant. Each character varies continuously in $t$ and has values in that finite set of characters, so it is individually constant. Differentiating at zero gives ${\lambda }_i([x,r])=0$ for all $r\in R$.

Choose a common eigenspace $V_{\lambda }\ne 0$. For $v\in V_{\lambda }$,Thus $V_{\lambda }$ is $L$-invariant, so irreducibility makes it all of ${\mathbb{C}}^n$. Every element of $R$ acts as a scalar. Its trace is zero because $L\subseteq {\mathfrak{sl}}_n$, so that scalar is zero. Faithfulness of the matrix inclusion gives $R=0$. Hence $\boxed{L\text{ is semisimple}}$.

One sufficient diagonal-matrix condition for simplicity is that $L$ contain all traceless diagonal matrices. Indeed invariance under their adjoint actions then decomposes $L$ into that diagonal algebra and some matrix-unit spaces $\mathbb{C}{e}_{ij}$. Put a directed edge $j\to i$ when $e_{ij}\in L$. Irreducibility makes the directed graph strongly connected; otherwise a terminal component spans a proper invariant coordinate subspace. Bracketing matrix units along a simple directed path supplies the matrix unit between its endpoints. Strong connectivity therefore supplies every off-diagonal matrix unit, so $L={\mathfrak{sl}}_n$, which is simple. The mere existence of a diagonal matrix with distinct diagonal entries would not be a sufficient condition.

For the specified construction, $H$ has dimension two. Its diagonal adjoint action and the twelve distinct nonzero weights give a direct sum $L=H\oplus {⨁}_{i=1}^{12}\mathbb{C}{g}_i$, of dimension fourteen. The stated bracket condition and the weight-vector relations make it a Lie algebra. A subspace invariant under it is invariant under all the $g_i$ and hence their associative algebra, so it acts irreducibly on ${\mathbb{C}}^7$. The preceding result makes it semisimple.

Use prime-dimensional faithful irreducible module forces simplicity. To see this, decompose the algebra into simple ideals. Complete reducibility for one ideal and commutation with the others make its isotypic component invariant under the whole algebra, hence all of the representation. Writing that component as $W\otimes \mathrm{Hom}(W,V)$ and repeating produces a tensor product of irreducibles for the simple ideals. Faithfulness makes every factor nontrivial, of dimension at least two. Their dimensions multiply to seven, so only one factor is possible. Thus the constructed $L$ is simple.

The diagonal algebra $H$ is toral, and its centralizer inside $L$ is just $H$, since no $g_i$ has zero weight. A maximal toral subalgebra containing $H$ lies in that centralizer, so equals $H$; it is a Cartan subalgebra. Thus the rank is two and the twelve weights are the roots. Pair them as ${\alpha }_{i+6}=-{\alpha }_i$. The opposite-root brackets have weight zero and belong to $H$. For distinct $i,j\le 6$, the sum cannot be zero, and the supplied root-bracket result says their bracket is nonzero exactly when ${\alpha }_i+{\alpha }_j$ is a root. This is equivalent to $-{\alpha }_i-{\alpha }_j$ being a root, proving the requested equivalence for the paired brackets.

Among rank-two simple root systems, $A_2,B_2,G_2$ have respectively six, eight and twelve roots. ThereforeChoose simple roots $\alpha$ short and $\beta$ long. In $G_2$, $\beta +\alpha$ and $\beta +2\alpha$ are roots. Renumber with $g_1\in L_{\alpha }$, $g_3\in L_{\beta }$, $g_4\in L_{\beta +\alpha }$ and $g_5\in L_{\beta +2\alpha }$, retaining opposite partners in the other six positions. Root spaces are one dimensional, and the allowed bracket equality gives

Use the convention $A_{ij}=⟨{\alpha }_j,{\alpha }_i^{\vee }⟩=2({\alpha }_i,{\alpha }_j)/({\alpha }_i,{\alpha }_i)$. For a finite reduced crystallographic root system, diagonal entries are two and off-diagonal entries belong to $\{0,-1,-2,-3\}$, with $A_{ij}{A}_{ji}\in \{0,1,2,3\}$. Transposing this convention merely transposes the Cartan matrix.

The $\alpha$-string through $\beta$ has the consecutive roots $\beta -p\alpha ,\dots ,\beta +q\alpha$, where $p,q\ge 0$ and $p-q=⟨\beta ,{\alpha }^{\vee }⟩$. For distinct simple roots ${\alpha }_i,{\alpha }_j$, the negative step is absent, so $p=0$ and $q=-A_{ij}$.

The matrix determines each simple reflection on the simple-root basis:It therefore determines their generated Weyl group as linear transformations. Every root is in the Weyl orbit of a simple root. For a positive nonsimple root $\beta =\sum c_i{\alpha }_i$, some $(\beta ,{\alpha }_i)$ is positive because $\sum c_i(\beta ,{\alpha }_i)=(\beta ,\beta )>0$. Its reflection subtracts a positive integer multiple of ${\alpha }_i$ and lowers height. It stays positive unless the root is simple, since the other simple-root coefficients are unchanged and at least one is positive. Induction on height reaches a simple root; negative roots follow by sign. Hence $\Phi =W\Delta$, which is completely reconstructed by those reflection matrices. This determines the system up to root-system isomorphism, rather than fixing an absolute length scale; irreducible components may be rescaled.

For positive roots for a chain Cartan matrix, symmetry and adjacency imply all simple roots have equal length; normalize their squared lengths to two. Adjacent inner products are minus one and all others zero. Since every root is Weyl-conjugate to a simple root, every root also has squared length two. Write a positive root as $\beta ={\sum }_{i=1}^l{c}_i{\omega }_i$ with nonnegative integers $c_i$, and set $c_0=c_{l+1}=0$. ThenThere must be exactly two nonzero integer differences, one $+1$ and one $-1$. Nonnegativity and the two zero endpoint values imply that the coefficients are one on one consecutive interval and zero elsewhere.

Conversely a single simple root is a root. If ${\omega }_i+\cdots +{\omega }_j$ is a root and $j<l$, its pairing with ${\omega }_{j+1}^{\vee }$ is minus one. Reflecting in ${\omega }_{j+1}$ therefore adds that simple root, making the next longer interval a root. Induction constructs every interval. Thus

Identify the Cartan subalgebra with the coordinate space $(t_1,\dots ,t_n)$ used above. The root reflection in any root of a root system fixes the perpendicular hyperplane. For every root of the Cn root system, its complete coordinate action isEvery unlisted coordinate is fixed, and changing the sign of the root leaves its reflection unchanged. These formulas follow directly from $s_{\alpha }(t)=t-\alpha (t){\alpha }^{\vee }$, with the coroot regarded as a vector of the Cartan subalgebra.

The Cn Weyl group is the signed symmetric group: its elements permute the $n$ coordinates and independently change their signs. The Weyl reflections associated with the simple roots interchange consecutive coordinates for $i<n$ and negate the last coordinate for $i=n$. Thus

For a complex finite-dimensional simple Lie algebra, a Cartan subalgebra $\mathfrak{h}$ is a maximal commuting subalgebra of elements whose adjoint maps are semisimple. Equivalently in this setting it is a nilpotent self-normalizing subalgebra. Its dimension is the rank of a semisimple Lie algebra. Simultaneous diagonalization of its Adjoint representation gives the root-space decompositionA root of a root system is a nonzero linear functional $\alpha$ for which this root space is nonzero. For a complex semisimple algebra each root space is one-dimensional. A Cartan-Weyl basis consists of a basis $H_i$ of $\mathfrak{h}$ and one nonzero root vector $E_{\alpha }$ for every root.

The general Lie brackets have the formFor the opposite-root bracket, use the Killing form to define $h_{\alpha }$ by $\kappa (h_{\alpha },H)=\alpha (H)$. Its invariant bilinear form on a Lie algebra property givesOne may normalize the root vectors so that the pairing is one. If instead one uses a coroot as the opposite-root bracket, the root-vector normalization changes accordingly. In particular, root evaluation coordinates cannot simply be used as coefficients in a nonorthonormal Cartan basis.

For the matrix calculation take $N\ge 2$. The complexification of a Lie algebra of the special unitary group Lie algebra is the special linear Lie algebra ${\mathfrak{sl}}_N(\mathbb{C})$: traceless complex matrices. Its Cartan subalgebra consists of traceless diagonal matrices. Write $E_{jk}={\mathcal{T}}^{(j,k)}$ for the matrix units. The given Cartan basis is $H_i=E_{ii}-E_{i+1,i+1}$, $1\le i<N$, and the other basis elements are $E_{jk}$ with $j\ne k$.

The matrix-unit identity $E_{jk}{E}_{lm}={\delta }_{kl}{E}_{jm}$ givesThus all the roots, expressed as evaluation vectors in this precise Cartan basis, areThey are the functionals $e_j-e_k$ on traceless diagonal matrices; there are $N(N-1)$ of them. The corresponding root vector is $E_{jk}$. Together with $N-1$ Cartan generators, they give $N^2-1$ basis elements. The simple roots can be chosen as ${\alpha }_{i,i+1}$, whose evaluation vectors are the rows of the type-$A_{N-1}$ Cartan matrix, with $2$ on the diagonal and $-1$ on adjacent entries. These vectors are evaluations on $H_i$, not coordinates in an orthonormal realization of the root system.

To express every bracket strictly in the chosen basis, introduce the abbreviationThen all pairs are covered byHere both input root vectors have distinct row and column indices. The first case is the only one producing diagonal matrix units, and the displayed sum of $H_p$ resolves them completely into the chosen Cartan basis. Reversing the order gives the negative bracket. This also shows explicitly that the two nonzero non-Cartan cases have structure constants $+1$ and $-1$, and verifies the required root-addition rule.

Suppose the Cartan matrix has diagonal two, entries minus one between adjacent indices, and all other entries zero. Normalize the simple-root squared lengths to two. Every root is a Weyl group translate of a simple root, so all roots have squared length two. If a positive root has coefficients $c_1,\dots ,c_l$ in the simple-root basis, let $c_0=c_{l+1}=0$. ThenThe nonnegative integer coefficients force exactly one upward jump of one and one downward jump of one. Thus the coefficients are one on a consecutive interval and zero elsewhere. Conversely, an interval root can be extended by the next simple root: its Cartan pairing with that root is minus one, so the corresponding simple reflection adds it. Starting from a simple root constructs every interval and proves the displayed description. This also identifies the root system as type $A_l$ without needing to enumerate reflection words individually.

Let a semisimple complex Lie algebra $L={⨁}_i{L}_i$ act faithfully and irreducibly on a space $V$ of prime dimension. Complete reducibility for $L_1$ splits $V$ into isotypic components. The commuting other ideals preserve these components, so irreducibility leaves just one, of the form $W_1\otimes {\mathrm{Hom}}_{L_1}(W_1,V)$. Repeating with the other factors gives $V\cong {⨂}_i{W}_i$, with each $W_i$ an irreducible module for its simple ideal.

Faithfulness makes every factor nontrivial. A nonabelian simple Lie algebra has only the trivial one-dimensional module, so $\dim W_i\ge 2$. Their product is the prime $\dim V$, forcing just one simple ideal. Hence $L$ is simple. This is a representation-dimension obstruction, independent of the classification of root systems.

The hyperplane perpendicular to a root of a root system is its root hyperplane, interpreted as the kernel of the associated linear functional on the Cartan space. The corresponding reflection is $s_{\alpha }(v)=v-2⟨v,\alpha ⟩\alpha /⟨\alpha ,\alpha ⟩$. For the B2 root system, the four distinct hyperplanes are the two coordinate axes and the two diagonals; their reflections generate the signed-permutation Weyl group of order eight.