Root Systems, Dynkin Diagrams, and the Cartan-Killing Classification — Examples

$\mathrm{\Phi }$ is finite, spans $E$, omits $0$, and $\mathbb{R}\alpha \cap \mathrm{\Phi }=\{\pm \alpha \}$; the reflection $s_{\alpha }$ sends $\alpha$ to $-\alpha$ and $-\alpha$ to $\alpha$, so $s_{\alpha }(\mathrm{\Phi })=\mathrm{\Phi }$; the only Cartan integers are $2(\pm \alpha ,\pm \alpha )\mathrm{/}(\alpha ,\alpha )=\pm 2$, which are integers. Hence $\mathrm{\Phi }$ is a reduced crystallographic root system.

Each positive system consists of one of the two roots, so the two bases are $\{\alpha \}$ and $\{-\alpha \}$. The Weyl group is $\{1,s_{\alpha }\}$ of order $2$, and relative to either one-element base the Cartan matrix is $[2]$ by [L2]. This is the rank-one system $A_1$.

For $A_2$ take ${\mathrm{\Phi }}_A=\{u_0,\dots ,u_5\}$. This finite set spans the plane, is reduced, and each root reflection is a symmetry of the regular hexagon. Its Cartan integers are $2\cos \theta \in \{0,\pm 1,\pm 2\}$. Put $\alpha =u_0$ and $\beta =u_2$; then ${\mathrm{\Phi }}_A^{+}=\{\alpha ,\beta ,\alpha +\beta \}$ is a positive system with base $(\alpha ,\beta )$, and $(\alpha ,\beta )=-1\mathrm{/}2$. Thus its Cartan matrix is $\left(\begin{array}{cc}2& -1\\ -1& 2\end{array}\right)$ and its off-diagonal Cartan product is $1$.

For $B_2$ take ${\mathrm{\Phi }}_B=\{\pm e_1,\pm e_2,\pm e_1\pm e_2\}$. This finite set spans the plane, omits zero, and is reduced. The reflections in the coordinate roots change one sign, while those in $e_1\pm e_2$ interchange the coordinates with possible sign changes, so every root reflection preserves ${\mathrm{\Phi }}_B$; direct pairings give Cartan integers in $\{0,\pm 1,\pm 2\}$. The roots $\alpha =e_1-e_2$ and $\beta =e_2$ form a base, since the positive roots are $\alpha ,\beta ,\alpha +\beta ,\alpha +2\beta$. Moreover $(\alpha ,\alpha )=2$, $(\beta ,\beta )=1$, and $(\alpha ,\beta )=-1$, so the Cartan matrix for $(\alpha ,\beta )$ is $\left(\begin{array}{cc}2& -1\\ -2& 2\end{array}\right)$ and its off-diagonal Cartan product is $2$.

For $G_2$, choose $\alpha ,\beta$ with $(\alpha ,\alpha )=6$, $(\beta ,\beta )=2$, $(\alpha ,\beta )=-3$ and put ${\mathrm{\Phi }}_G=\{\pm \alpha ,\pm \beta ,\pm (\alpha +\beta ),\pm (\alpha +2\beta ),\pm (\alpha +3\beta ),\pm (2\alpha +3\beta )\}\mathrm{.}$ The Gram determinant is positive, so $\alpha ,\beta$ form a basis; the displayed coefficient pairs then show that ${\mathrm{\Phi }}_G$ is finite, spans the plane, omits zero, and is reduced. The reflection $s_{\alpha }$ interchanges $\beta$ with $\alpha +\beta$ and $\alpha +3\beta$ with $2\alpha +3\beta$, and fixes $\alpha +2\beta$; the reflection $s_{\beta }$ interchanges $\alpha$ with $\alpha +3\beta$ and $\alpha +\beta$ with $\alpha +2\beta$, and fixes $2\alpha +3\beta$. Together with the images of $\alpha$ and $\beta$, these permutations show that the long and short roots are the two orbits of the simple reflections. Conjugating $s_{\alpha }$ or $s_{\beta }$ therefore proves reflection closure for every root. Direct pairings give integral Cartan integers in $\{0,\pm 1,\pm 2,\pm 3\}$. The six displayed unnegated roots are positive and have base $(\alpha ,\beta )$, whose Cartan matrix is $\left(\begin{array}{cc}2& -1\\ -3& 2\end{array}\right)$ and whose off-diagonal Cartan product is $3$.

By [L1] the three systems are exactly the irreducible rank-two reduced crystallographic systems, and the displayed Cartan products $1,2,3$ are those of $A_2,B_2,G_2$ respectively.

$({\alpha }_i,{\alpha }_i)=2$ and ${\alpha }_i^{\vee }={\alpha }_i$, since ${\alpha }_i$ has two nonzero coordinates equal to $\pm 1$.

For every $i,k$ one has $({\omega }_k,{\alpha }_i)=({\omega }_k,e_i)-({\omega }_k,e_{i+1})$; the vector ${\omega }_k$ has coordinates $1-k\mathrm{/}(n+1)$ in positions $1,\dots ,k$ and $-k\mathrm{/}(n+1)$ in positions $k+1,\dots ,n+1$, so the difference equals $1$ when $i=k$ and $0$ otherwise. Hence $({\omega }_k,{\alpha }_i^{\vee })={\delta }_{ki}$ and the displayed vectors are the fundamental weights of $A_n$; they form a basis of the weight lattice by [L2].

Let $\rho :S_{n+1}\to O(E)$ be the homomorphism obtained by restricting coordinate permutations to $E$. For $x\in E$, the reflection formula gives $s_{{\epsilon }_i-{\epsilon }_j}(x)=x-(x_i-x_j)({\epsilon }_i-{\epsilon }_j)=\rho ((ij))x$. Hence [L2] and the fact that the transpositions generate $S_{n+1}$ give $W(A_n)=\rho (S_{n+1})$.

The homomorphism $\rho$ is injective. Indeed, if $\rho (\sigma )$ is the identity on $E$, then for every $i\mathrm{\ne }j$ it fixes ${\epsilon }_i-{\epsilon }_j$, so ${\epsilon }_{\sigma (i)}-{\epsilon }_{\sigma (j)}={\epsilon }_i-{\epsilon }_j$; uniqueness of the positive and negative coordinate positions gives $\sigma (i)=i$ and $\sigma (j)=j$. Thus $\sigma =1$, and step 1.1 yields $W(A_n)\cong S_{n+1}$ with the asserted action.

For $B_n$: $\mathrm{\mid }{\alpha }_i{\mathrm{\mid }}^2=2$ for $i<n$ and $\mathrm{\mid }{\alpha }_n{\mathrm{\mid }}^2=1$; $({\alpha }_{n-1},{\alpha }_n)=-1$, so $a_{n-1,n}=2(-1)\mathrm{/}2=-1$ and $a_{n,n-1}=2(-1)\mathrm{/}1=-2$; all other off-diagonal entries of adjacent pairs are $-1$ and the rest vanish. Thus the diagram is a chain with a double edge at the end, the arrow being governed by $\mathrm{\mid }{a}_{n,n-1}\mathrm{\mid }=2>\mathrm{\mid }{a}_{n-1,n}\mathrm{\mid }=1$ and pointing toward the shorter root ${\alpha }_n$.

For $C_n$: $\mathrm{\mid }{\beta }_i{\mathrm{\mid }}^2=2$ for $i<n$ and $\mathrm{\mid }{\beta }_n{\mathrm{\mid }}^2=4$; $({\beta }_{n-1},{\beta }_n)=-2$, so $a_{n-1,n}=2(-2)\mathrm{/}2=-2$ and $a_{n,n-1}=2(-2)\mathrm{/}4=-1$; the diagram is again a chain with a double edge exactly at the end, and the arrow now points toward the shorter root ${\beta }_{n-1}$.

The matrices of steps 1.1 and 1.2 are transposes of one another, which is exactly coroot duality by [L2]; so transposing the Cartan matrix exchanges $B_n$ and $C_n$, reversing the arrow while keeping the chain and the double-edge position.

Extending the coordinate notation to rank one gives $B_1=\{\pm e_1\}$ and $C_1=\{\pm 2e_1\}$. The linear maps $e_1\mapsto \alpha$ and $2e_1\mapsto \alpha$ identify these systems with $A_1$ from [L1]. Thus $B_1=C_1=A_1$ up to root-system isomorphism.

The explicit similarity in [L2] identifies the eight roots of $B_2$ with those of $C_2$ and preserves every Cartan integer. Hence $B_2=C_2$ up to root-system isomorphism.

For $D_2$, [L3] gives $D_2=\{\pm (e_1+e_2),\pm (e_1-e_2)\}$, the orthogonal disjoint union of two rank-one systems, so $D_2=A_1\bigsqcup A_1$ by [L1].

For the simple roots of $D_3$ in [L3], all squared lengths are $2$, while $({\delta }_1,{\delta }_2)=({\delta }_1,{\delta }_3)=-1$ and $({\delta }_2,{\delta }_3)=0$. Their Dynkin graph therefore has the three-vertex path ${\delta }_2-{\delta }_1-{\delta }_3$, the $A_3$ diagram. Hence $D_3=A_3$ up to root-system isomorphism.

The images satisfy the Cartan and generator relations: $[h_i,h_j]=0$; $[h_1,E_{12}]=2E_{12}$, $[h_1,E_{23}]=-E_{23}$, $[h_2,E_{12}]=-E_{12}$, $[h_2,E_{23}]=2E_{23}$ and the negatives on the $f$'s; $[E_{12},E_{21}]=h_1$, $[E_{23},E_{32}]=h_2$, and the cross brackets $[E_{12},E_{32}]$ and $[E_{23},E_{21}]$ vanish.

Direct multiplication also gives $(\mathrm{ad}{E}_{12}{)}^2{E}_{23}=(\mathrm{ad}{E}_{23}{)}^2{E}_{12}=(\mathrm{ad}{E}_{21}{)}^2{E}_{32}=(\mathrm{ad}{E}_{32}{)}^2{E}_{21}=0$, so all four Serre relations hold. Hence the assignment induces a Lie-algebra homomorphism $\phi :\mathfrak{g}(A)\to {\mathfrak{s}\mathfrak{l}}_3(\mathbb{C})$.

The images generate ${\mathfrak{s}\mathfrak{l}}_3(\mathbb{C})$: $[E_{12},E_{23}]=E_{13}$, $[E_{23},E_{12}]=-E_{13}$, and $[E_{21},E_{32}]=-E_{31}$. Thus the image contains all six off-diagonal matrix units and the independent diagonal matrices $h_1,h_2$, which form a basis of the eight-dimensional space of traceless $3\times 3$ matrices. Hence $\phi$ is surjective.

Put $z=[e_1,e_2]$. The positive Serre relations give $[e_1,z]=[e_2,z]=0$, so $\mathrm{span}(e_1,e_2,z)$ is a Lie subalgebra containing the positive generators and contained in the subalgebra they generate; hence it equals ${\mathfrak{n}}_{+}$. With $w=[f_1,f_2]$, the negative Serre relations likewise give $[f_1,w]=[f_2,w]=0$, so ${\mathfrak{n}}_{-}=\mathrm{span}(f_1,f_2,w)$ and each half has dimension at most three. Since $\mathfrak{h}$ is spanned by $h_1,h_2$, the triangular decomposition in [L1] gives $\dim \mathfrak{g}(A)\le 8$. Surjectivity from step 3.1 onto the eight-dimensional algebra ${\mathfrak{s}\mathfrak{l}}_3(\mathbb{C})$ gives the reverse inequality, so $\dim \mathfrak{g}(A)=8$ and $\ker \phi =0$. Thus $\phi$ is the asserted isomorphism.

Write ${\alpha }_0$ for the long root and ${\beta }_0$ for the short root of the model, so that $\alpha ={\beta }_0$, $\beta ={\alpha }_0$. The twelve roots listed in the model become, in terms of $\alpha ,\beta$: $\pm \alpha ,\pm \beta ,\pm (\alpha +\beta ),\pm (2\alpha +\beta ),\pm (3\alpha +\beta ),\pm (3\alpha +2\beta )$. Hence the positive roots with respect to the base $\{\alpha ,\beta \}$ are exactly the six nonnegative combinations displayed, of heights $1,1,2,3,4,5$.

The root $3\alpha +2\beta$ has height $5$, the largest among the positive roots, and it is the unique highest root by [L2]. Directly, each of the other five coefficient pairs $(1,0),(0,1),(1,1),(2,1),(3,1)$ is coordinatewise at most $(3,2)$ and is not equal to it, so every other positive root is strictly below $3\alpha +2\beta$ in the root order.

The matrix $A=2I-\mathrm{Adj}(G)$ is symmetric and satisfies $a_{ii}=2$, $a_{ij}=-1$ for adjacent $i\mathrm{\ne }j$ and $a_{ij}=0$ otherwise; its diagram, as in [L2], is the cycle $G$ on $m\ge 3$ vertices.

The nonzero vector $x=(1,\dots ,1)$ satisfies $Ax=0$, because every row has diagonal entry $2$ and exactly two entries $-1$. Equivalently $x^{T}Ax=2m-2m=0$. Hence $A$ is not positive definite; moreover every diagonal conjugate $DA{D}^{-1}$ has the nonzero null vector $Dx$, so no symmetric diagonal conjugate can be positive definite.

By [L1] a finite-type Cartan matrix must be symmetrizable to a positive definite matrix; $A$ is not. Moreover, [L2] makes $A$ the only Cartan matrix with this unoriented simple-edge cycle: on each edge the nonpositive integral entries have product $1$, so both are $-1$. Thus the cycle is not a finite-type Dynkin diagram even though it is finite and connected. This explicit family suffices to refute the claimed statement; no broader tree assertion is needed.

The open set ${\mathrm{GL}}_2(\mathbb{C})\subset M_2(\mathbb{C})$ is a complex Lie group: multiplication is polynomial and inversion is the adjugate divided by the nonzero determinant. The determinant-one subset is a complex submanifold: on the open set where the entry $a\mathrm{\ne }0$, the equation $ad-bc=1$ solves $d=(1+bc)\mathrm{/}a$; at any other matrix at least one entry is nonzero and one solves for its opposite entry in the same way. These charts cover the subset, and the restricted group operations are holomorphic. Differentiating the determinant at $I$ gives $\mathrm{tr}X$, and the chart at $I$ shows that every traceless $X$ is tangent to this subset. For either matrix group the left-invariant extension of $X$ is $A\mapsto AX$; the field commutator with $A\mapsto AY$ is $A\mapsto A(XY-YX)$. Thus their tangent Lie algebras are respectively $M_2(\mathbb{C})$ and ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{C})$.

The algebra ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{C})$ is simple, hence semisimple. Indeed a nonzero ideal is invariant under $\mathrm{ad}h$, whose distinct eigenvalues on $e,f,h$ are $2,-2,0$. Polynomial spectral projections show that the ideal contains a nonzero multiple of at least one of these basis vectors. Bracketing with the others then puts all three in the ideal. Moreover $[{\mathfrak{s}\mathfrak{l}}_2,{\mathfrak{s}\mathfrak{l}}_2]={\mathfrak{s}\mathfrak{l}}_2$, so the whole algebra is not solvable; it therefore has no nonzero solvable ideal.

The group ${\mathrm{SL}}_2(\mathbb{C})$ is path connected. Indeed, if $g=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ has $a\mathrm{\ne }0$, then $g=\left(\begin{array}{cc}1& 0\\ c\mathrm{/}a& 1\end{array}\right)\left(\begin{array}{cc}a& 0\\ 0& a^{-1}\end{array}\right)\left(\begin{array}{cc}1& b\mathrm{/}a\\ 0& 1\end{array}\right)\mathrm{.}$ Each unipotent factor is joined to $I$ by multiplying its off-diagonal entry by $t\in [0,1]$, and the diagonal factor is joined to $I$ along $\mathrm{diag}(\gamma (t),\gamma (t{)}^{-1})$ for any path $\gamma$ in ${\mathbb{C}}^{\times }$ from $1$ to $a$. If $a=0$, then $c\mathrm{\ne }0$, and the path $\left(\begin{array}{cc}1& t\\ 0& 1\end{array}\right)g$ joins $g$ to a matrix whose upper-left entry is $c\mathrm{\ne }0$, reducing to the preceding case.

$Z({\mathrm{SL}}_2(\mathbb{C}))=\{\pm I\}$: a central matrix commutes in particular with the unipotent one-parameter subgroups generated by $E_{12}$ and $E_{21}$, hence with $E_{12}$ and $E_{21}$; it is therefore scalar, and determinant one leaves precisely $\pm I$.

$Z({\mathrm{PGL}}_2(\mathbb{C}))$ is trivial: a central projective transformation commutes with every dilation $z\mapsto az$, so it preserves their common fixed set $\{0,\mathrm{\infty }\}$; commuting also with the inversion $z\mapsto 1\mathrm{/}z$ and translations $z\mapsto z+b$ forces it to fix $0,1,\mathrm{\infty }$, hence it is the identity Möbius transformation.

The scalar subgroup is closed in ${\mathrm{GL}}_2(\mathbb{C})$, being defined there by zero off-diagonal entries and equal diagonal entries. It is normal, and its tangent algebra is $\mathbb{C}I$. Consequently [L3], applied to the underlying real groups, gives the quotient tangent algebra $M_2(\mathbb{C})\mathrm{/}\mathbb{C}I$. The quotient is also a complex Lie group: on the set of classes with a selected matrix entry nonzero, normalize that entry to $1$. The other three entries give an open subset of ${\mathbb{C}}^3$ with determinant nonzero. Transition functions and the locally expressed group operations are rational with nonzero denominators, hence holomorphic. These charts agree with the smooth quotient charts since normalization is a smooth local section. The tangent quotient map is complex linear. Finally $[X]\mapsto X-\frac{1}{2}\mathrm{tr}(X)I$ is a well-defined complex-linear bijection to ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{C})$ preserving commutators, since scalar matrices commute and commutators have trace zero.

The group ${\mathrm{GL}}_2(\mathbb{C})$ is path connected: for $g\in {\mathrm{GL}}_2(\mathbb{C})$ choose $z\in {\mathbb{C}}^{\times }$ with $z^2=\det g$; then $z^{-1}g\in {\mathrm{SL}}_2(\mathbb{C})$, and paths in ${\mathrm{SL}}_2(\mathbb{C})$ and ${\mathbb{C}}^{\times }$ join $g=z(z^{-1}g)$ to $I$. Its quotient ${\mathrm{PGL}}_2(\mathbb{C})$ is therefore connected.

An isomorphism of groups carries the center onto the center, so ${\mathrm{SL}}_2(\mathbb{C})$ and ${\mathrm{PGL}}_2(\mathbb{C})$ are not isomorphic, although steps 1.3 and 2.2 show both are connected and steps 1.1, 1.2, and 2.1 give both the complex semisimple Lie algebra ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{C})$. This witnesses the failure of the claim: the two groups are distinct global forms of the same Lie algebra.