Cartan subalgebras and root systems

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A Cartan matrix AA is a square matrix whose elements aija_{ij} satisfy the following conditions:

  1. aija_{ij} is an integer, one of {−3,−2,−1,0,2}\left\{ -3,-2,-1,0,2\right\}

  2. ajj=2a_{jj} = 2 for all diagonal elements of AA

  3. aij≤0a_{ij} \leq 0 off of the diagonal

  4. aij=0a_{ij} = 0 iff aji=0a_{ji} = 0

  5. There exists an invertible diagonal matrix DD such that DAD−1DAD^{-1} gives a symmetric and positive definite quadratic form. Give a 2×22 \times 2 non-diagonal Cartan matrix and its eigenvalues.

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The roots of a semisimple Lie algebra are the Lie algebra weights occurring in its adjoint representation. The set of roots forms the root system, and is completely determined by the semisimple Lie algebra. Consider the semisimple Lie algebra sl(2,R)\mathfrak {sl}(2,\mathbb {R}) with the generators

E=(0100),F=(0010),H=(100−1). E = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \qquad F = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, \qquad H = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.

Find the roots.

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A Chevalley basis for the semisimple Lie algebra sl(3,R)\mathfrak {sl}(3,\mathbb {R}) is given by

X1=(000001000),X2=(010000000),X3=(001000000), X_{1} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}, \qquad X_{2} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}, \qquad X_{3} = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}, Y1=(000000010),Y2=(000100000),Y3=(000000100), Y_{1} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}, \qquad Y_{2} = \begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}, \qquad Y_{3} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix}, H1=(00001000−1),H2=(1000−10000) H_{1} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix}, \qquad H_{2} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{pmatrix}

where Yj=XjTY_{j} = X_{j}^{T} for j=1,2,3j = 1,2,3. The Lie algebra has rank 2 owing to H1H_{1}, H2H_{2} and [H1,H2]=03[H_{1},H_{2}] = 0_{3}. Another basis could be formed by looking at the linear combinations

Uj=Xj+Yj,Vj=Xj−Vj. U_{j} = X_{j}+Y_{j}, \qquad V_{j} = X_{j}-V_{j}.
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(i)

Find the table of the commutator.

(ii)

Calculate the vectors of commutators

\begin{pmatrix} {[H_{1},X_{1}]} \\[H_{2},X_{1}] \end{pmatrix}, \qquad \begin{pmatrix} {[H_{1},X_{2}]} \\[H_{2},X_{2}] \end{pmatrix}, \qquad \begin{pmatrix} {[H_{1},X_{3}]} \\[H_{2},X_{3}] \end{pmatrix}

and thus find the roots.

Consider the semisimple Lie algebra sl(n+1,F)\mathfrak {sl}(n+1,\mathbb {F}). Let Ei,jE_{i,j} (i,j∈{1,2,…,n+1}i,j \in \{ 1,2,\ldots ,n+1\}) denote the standard basis, i.e. (n+1)×(n+1)(n+1) \times (n+1) matrices with all entries zero except for the entry in the ii-th row and jj-th column which is one. We can form a Cartan-Weyl basis with

Hj:=Ej,j−Ej+1,j+1,j∈{1,2,…,n}. H_{j} := E_{j,j} - E_{j+1,j+1}, \qquad j \in \{ 1,2,\ldots ,n\} .

Show that Ei,jE_{i,j} are root vectors for i≠ji \neq j, i.e. there exists λH,i,j∈F\lambda_{H,i,j} \in \mathbb {F} such that

[H,Ei,j]=λH,i,jEi,j [H,E_{i,j}] = \lambda _{H,i,j}E_{i,j}

for all H∈span⁡{ H1,…,Hn }H \in \operatorname {span}\{ \, H_{1}, \ldots , H_{n} \, \}.

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The Cartan matrix AA of a rank rr root system of a semi-simple Lie algebra is an r×rr \times r matrix whose entries are derived from the simple roots. The entries of the Cartan matrix are given by

ajk=2(αj,αk)αj,αj a_{jk} = 2\frac{(\alpha _{j},\alpha _{k})}{\alpha _{j},\alpha _{j}}

where ( , )(\, , \, ) is the Euclidean inner product and αj\alpha_{j} are the simple roots. The entries are independent of the choice of simple roots (up to ordering).

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(i)

The Cartan matrix of the Lie algebra e8e_{8} is given by the 8×88 \times 8 matrix

(2−1000000−12−1000000−12−1000−100−12−1000000−12−1000000−12−1000000−12000−100002). \begin{pmatrix} 2 & -1 & 0 & 0 & 0 & 0 & 0 & 0 \\ -1 & 2 & -1 & 0 & 0 & 0 & 0 & 0 \\ 0 & -1 & 2 & -1 & 0 & 0 & 0 & -1 \\ 0 & 0 & -1 & 2 & -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 & 2 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 & -1 & 2 & -1 & 0 \\ 0 & 0 & 0 & 0 & 0 & -1 & 2 & 0 \\ 0 & 0 & -1 & 0 & 0 & 0 & 0 & 2 \end{pmatrix}.

Show that the trace is equal to 1616. Show that the determinant is equal to 11. Find the eigenvalues and eigenvectors of the matrix.

(ii)

The Lie algebra sl(3,R)\mathfrak {sl}(3,\mathbb {R}) is the rank 22 Lie algebra with Cartan matrix

C=(2−112). C = \begin{pmatrix} 2 & -1 \\ 1 & 2 \end{pmatrix}.

Find the eigenvalues and normalized eigenvectors of CC.

(iii)

Let τ=2cos⁡(π/5)\tau =2\cos (\pi /5). Find the eigenvalues and eigenvectors of the three Cartan matrices

A2=(2−τ−τ2),A3=(2−10−12−τ0−τ2),A4=(2−100−12−100−12−τ00−τ2). A_{2} = \begin{pmatrix} 2 & -\tau \\ -\tau & 2 \end{pmatrix}, \qquad A_{3} = \begin{pmatrix} 2 & -1 & 0 \\ -1 & 2 & -\tau \\ 0 & -\tau & 2 \end{pmatrix}, \qquad A_{4} = \begin{pmatrix} 2 & -1 & 0 & 0 \\ -1 & 2 & -1 & 0 \\ 0 & -1 & 2 & -\tau \\ 0 & 0 & -\tau & 2 \end{pmatrix}.

The semisimple Lie algebra sl(3,R)\mathfrak {sl}(3,\mathbb {R}) has dimension 8. The standard basis is given by

h1=(1000−10000),h2=(00001000−1), h_{1} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{pmatrix}, \qquad h_{2} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix}, e1=(010000000),e2=(000001000), e_{1} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}, \qquad e_{2} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}, f1=(000100000),f2=(000000010), f_{1} = \begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}, \qquad f_{2} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}, e13=(001000000),f13=(000000100) . e_{13} = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}, \qquad f_{13} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix} \, .

Find the commutator table. Find the rank of the Lie algebra.

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There is only one simple complex Lie algebra of rank 1. If α=λe1\alpha = \lambda e_{1} is a nonzero root, −λe1-\lambda e_{1} is the other. The canonical choice for λ\lambda is 1/21/\sqrt{2}. The commutators for the basis

E−1/2,H1,E+1/2 E_{-1/\sqrt{2}}, \quad H_{1}, \quad E_{+1/\sqrt{2}}

are

[H1,E±1/2]=±12E±1/2,[E+1/2,E−1/2]=12H1 . [H_{1}, E_{\pm 1/\sqrt{2}}] = \pm \frac{1}{\sqrt{2}} E_{\pm 1/\sqrt{2}}, \qquad [E_{+1/\sqrt{2}}, E_{-1/\sqrt{2}}] = \frac{1}{\sqrt{2}} H_{1} \, .

Find a 2×22 \times 2 matrix representation for this semisimple Lie algebra.

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Consider the vectors in R2\mathbb {R}^{2}

α=123(1,3),β=123(1,−3) . \alpha = \frac{1}{2\sqrt{3}}(1,\sqrt{3}), \qquad \beta = \frac{1}{2\sqrt{3}}(1,-\sqrt{3}) \, .

Then

α+β=13(1,0) . \alpha + \beta = \frac{1}{\sqrt{3}}(1,0) \, .

Find eight 3×33 \times 3 matrices

H1,  H2,  Eα,  E−α,  Eβ,  E−β,  Eα+β,  E−(α+β) H_{1}, \; H_{2}, \; E_{\alpha }, \; E_{-\alpha }, \; E_{\beta }, \; E_{-\beta }, \; E_{\alpha +\beta }, \; E_{-(\alpha +\beta )}

such that the following commutation relations are satisfied

[H1,E±α]=±123E±α,[H2,E±α]=±12E±α [H_{1}, E_{\pm \alpha }] = \pm \frac{1}{2\sqrt{3}}E_{\pm \alpha }, \qquad [H_{2}, E_{\pm \alpha }] = \pm \frac{1}{2}E_{\pm \alpha } [H1,E±β]=±123E±β,[H2,E±β]=±12E±β [H_{1}, E_{\pm \beta }] = \pm \frac{1}{2\sqrt{3}}E_{\pm \beta }, \qquad [H_{2}, E_{\pm \beta }] = \pm \frac{1}{2}E_{\pm \beta } [H1,E±(α+β)]=13E±(α+β),[H2,E±(α+β)]=0 [H_{1}, E_{\pm (\alpha +\beta )}] = \frac{1}{\sqrt{3}}E_{\pm (\alpha +\beta )}, \qquad [H_{2}, E_{\pm (\alpha +\beta )}] = 0 [Eα,E−α]=H123+H22,[Eβ,E−β]=H123−H22, [E_{\alpha }, E_{-\alpha }] = \frac{H_{1}}{2\sqrt{3}} + \frac{H_{2}}{2}, \qquad [E_{\beta }, E_{-\beta }] = \frac{H_{1}}{2\sqrt{3}} - \frac{H_{2}}{2}, [Eα+β,E−(α+β)]=H13,[Eα,Eβ]=16Eα+β, [E_{\alpha +\beta }, E_{-(\alpha +\beta )}] = \frac{H_{1}}{\sqrt{3}}, \qquad [E_{\alpha }, E_{\beta }] = \frac{1}{\sqrt{6}}E_{\alpha +\beta }, [Eα,Eα+β]=0,[Eβ,Eα+β]=0 [E_{\alpha }, E_{\alpha +\beta }] = 0, \qquad [E_{\beta }, E_{\alpha +\beta }] = 0 [Eα,E−(α+β)]=−16E−β,[Eβ,E−(α+β)]=16E−α [E_{\alpha }, E_{-(\alpha +\beta )}] = -\frac{1}{\sqrt{6}}E_{-\beta }, \qquad [E_{\beta }, E_{-(\alpha +\beta )}] = \frac{1}{\sqrt{6}}E_{-\alpha }

and [H1,H2]=0[H_{1},H_{2}] = 0. We can assume that H1,H2H_{1}, H_{2} are diagonal matrices.

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Show that the dimension of a Cartan sub-Lie algebra is not in general the maximal dimension of an abelian sub-Lie algebra. Consider the Lie algebra sl(2n,C)\mathfrak {sl}(2n,\mathbb {C}) of 2n×2n2n \times 2n matrices of trace 0.

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The semisimple Lie algebra sl(3,C)\mathfrak {sl}(3,\mathbb {C}) has dimension 8 and rank 2. The standard basis is given by

h1=(1000−10000),h2=(00001000−1), h_{1} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{pmatrix}, \qquad h_{2} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix}, e1=(010000000),e2=(000001000), e_{1} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}, \qquad e_{2} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}, f1=(000100000),f2=(000000010), f_{1} = \begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}, \qquad f_{2} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}, e13=(001000000),f13=(000000100) . e_{13} = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}, \qquad f_{13} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix} \, .

Note that e1=f1Te_{1} = f_{1}^{T}, e2=f2Te_{2} = f_{2}^{T}, e13=f13Te_{13} = f_{13}^{T}. Find the commutation relation table.

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Consider the Lie algebra sl(3,C)\mathfrak {sl}(3,\mathbb {C}). Given the roots find the structure constants. The semisimple Lie algebra sl(3,C)\mathfrak {sl}(3,\mathbb {C}) has dimension 8 and rank 2. The root system of sl(3,C)\mathfrak {sl}(3,\mathbb {C}) consists of 6 vectors which go from the origin to vertices of regular hexagon. Check the Jacobi identities for the resulting structure constants.

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Consider the semisimple Lie algebra sl(n+1,F)\mathfrak {sl}(n+1,\mathbb {F}). Let Ei,jE_{i,j} (i,j∈{1,2,…,n+1}i,j \in \{ 1,2,\ldots ,n+1\}) denote the standard basis, i.e. (n+1)×(n+1)(n+1) \times (n+1) matrices with all entries zero except for the entry in the iith row and jjth column which is one. We can form a Cartan-Weyl basis with

Hj:=Ej,j−Ej+1,j+1,j∈{1,2,…,n} . H_{j} := E_{j,j} - E_{j+1,j+1}, \qquad j \in \{ 1,2,\ldots ,n\} \, .

Show that Ei,jE_{i,j} are root vectors for i≠ji \neq j, i.e. there exists λH,i,j∈F\lambda_{H,i,j} \in \mathbb {F} such that

[H,Ei,j]=λH,i,jEi,j [H,E_{i,j}] = \lambda _{H,i,j}E_{i,j}

for all H∈⟨H1,…,Hn⟩H \in \left\langle H_{1},\ldots ,H_{n}\right\rangle.

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The exceptional Lie algebra g2\mathfrak {g}_{2} has the simple roots

(0,1,−1)(1,−2,1) . (0,1,-1) \qquad (1,-2,1) \, .

Find the Cartan matrix. Verify the properties of the Cartan matrix.

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The exceptional Lie algebra g2\mathfrak {g}_2 has rank 2 and dimension 14. A basis is given by

H1,H2,X1,X2,X3,X4,X5,X6,Y1,Y2,Y3,Y4,Y5,Y6 H_1, H_2, X_1, X_2, X_3, X_4, X_5, X_6, Y_1, Y_2, Y_3, Y_4, Y_5, Y_6

with the commutation relations [H1,H2]=0[H_1,H_2] = 0 and

H1=[X1,Y1],H2=[X2,Y2],[H1,X1]=2X1,[H2,X2]=2X2. H_1 = [X_1,Y_1], \quad H_2 = [X_2,Y_2], \quad [H_1,X_1] = 2X_1, \quad [H_2,X_2] = 2X_2 .

Thus

[H1,Y1]=−2Y1,[H2,Y2]=−2Y2. [H_1,Y_1] = -2Y_1, \quad [H_2,Y_2] = -2Y_2 .

Thus H1,X1,Y1H_1, X_1, Y_1 and H2,X2,Y2H_2, X_2, Y_2 each span the sub Lie algebra sl(2,C)\mathfrak {sl}(2,\mathbb {C}) of g2\mathfrak {g}_2. The commutator table for g2\mathfrak {g}_2 (Table 4.1) is, where entry (i,j)(i,j) gives [rowi,columnj][\text{row}_i, \text{column}_j] for i<ji < j and is left blank (not part of the table) otherwise:

H2X1Y1X2Y2X3Y3X4Y4X5Y5X6Y6H102X1−2Y1−3X23Y2−X3Y3X4−Y43X5−3Y500H2−X1Y12X2−2Y2X3−Y300−X5Y5X6−Y6X1H1X302X4−3Y2−3X5−2Y30Y400Y10−Y33X2−2Y42X33Y5−X4000X2H20Y100−X600Y5Y2−X10000Y6−X50X3H1+3H2−3X62Y1000Y4Y3−2X13Y600−X40X42H1+3H20−Y10Y3Y4X10X30X5H1+H20−Y2Y5X20X6H1+2H2 \begin{array}{c|ccccccccccccc}& H_2 & X_1 & Y_1 & X_2 & Y_2 & X_3 & Y_3 & X_4 & Y_4 & X_5 & Y_5 & X_6 & Y_6 \\ \hline H_1 & 0 & 2X_1 & -2Y_1 & -3X_2 & 3Y_2 & -X_3 & Y_3 & X_4 & -Y_4 & 3X_5 & -3Y_5 & 0 & 0 \\ H_2 & & -X_1 & Y_1 & 2X_2 & -2Y_2 & X_3 & -Y_3 & 0 & 0 & -X_5 & Y_5 & X_6 & -Y_6 \\ X_1 & & & H_1 & X_3 & 0 & 2X_4 & -3Y_2 & -3X_5 & -2Y_3 & 0 & Y_4 & 0 & 0 \\ Y_1 & & & & 0 & -Y_3 & 3X_2 & -2Y_4 & 2X_3 & 3Y_5 & -X_4 & 0 & 0 & 0 \\ X_2 & & & & & H_2 & 0 & Y_1 & 0 & 0 & -X_6 & 0 & 0 & Y_5 \\ Y_2 & & & & & & -X_1 & 0 & 0 & 0 & 0 & Y_6 & -X_5 & 0 \\ X_3 & & & & & & & H_1+3H_2 & -3X_6 & 2Y_1 & 0 & 0 & 0 & Y_4 \\ Y_3 & & & & & & & & -2X_1 & 3Y_6 & 0 & 0 & -X_4 & 0 \\ X_4 & & & & & & & & & 2H_1+3H_2 & 0 & -Y_1 & 0 & Y_3 \\ Y_4 & & & & & & & & & & X_1 & 0 & X_3 & 0 \\ X_5 & & & & & & & & & & & H_1+H_2 & 0 & -Y_2 \\ Y_5 & & & & & & & & & & & & X_2 & 0 \\ X_6 & & & & & & & & & & & & & H_1+2H_2 \end{array}

Note that sl(3,C)\mathfrak {sl}(3,\mathbb {C}) is a sub Lie algebra of g2\mathfrak {g}_2. Write a SymbolicC++ program that finds the adjoint representation.

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References

[StHa17] Steeb W., Hardy Y. Problems and Solutions in Introductory and Advanced Matrix Calculus. 2nd ed., World Scientific Publishing, 2017.
[STH12] Steeb W., Tanski I., Hardy Y. Problems and Solutions for Groups, Lie Groups, Lie Algebras with Applications. World Scientific Publishing, 2012.