Verifying bases and commutator tables of small concrete Lie algebras

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Consider the n×nn \times n matrices EijE_{ij} having 1 in the (i,j)(i,j) position and 0 elsewhere, where i,j=1,…,ni, j = 1, \ldots , n (elementary matrices). Calculate the commutator. Discuss.

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Consider the vector space of the 2×22 \times 2 matrices over R\mathbb {R} and the three 2×22 \times 2 matrices with trace equal to 0

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}.

Show that we have a basis of a Lie algebra.

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Consider the matrices

h1=(100000001),h2=(000010000),h3=(001000100), h_{1} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \qquad h_{2} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}, \qquad h_{3} = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix}, e=(010000010),f=(000101000). e = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}, \qquad f = \begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}.

Show that the matrices form a basis of a Lie algebra.

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An n×nn \times n matrix XX over C\mathbb {C} is skew-hermitian if X∗=−XX^{*} = -X. Show that the commutator of two skew-hermitian matrices is again skew-hermitian. Discuss.

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The Lie algebra su(m)\mathfrak {su}(m) consists of all m×mm \times m matrices XX over C\mathbb {C} with the conditions X∗=−XX^{*} = -X (i.e. XX is skew-hermitian) and tr⁡(X)=0\operatorname {tr}(X) = 0. Note that exp(X)\mathrm{exp}\left(X\right) is a unitary matrix. Find a basis for su(3)\mathfrak {su}(3).

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Any fixed element XX of a Lie algebra LL defines a linear transformation

ad⁡(X):Z→[X,Z]for anyZ∈L. \operatorname {ad}(X): Z \to [X,Z] \quad \text{for any} \quad Z \in L.

Show that for any K∈LK \in L we have

[ad⁡(Y),ad⁡(Z)]K=ad⁡([Y,Z])K. [\operatorname {ad}(Y), \operatorname {ad}(Z)]K = \operatorname {ad}([Y,Z])K.

The linear mapping ad⁡\operatorname {ad} gives a representation of the Lie algebra known as adjoint representation.

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There is only one non-commutative Lie algebra LL of dimension 2. If xx, yy are the generators (basis in LL), then

[x,y]=x. [x,y] = x.
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(i)

Find the adjoint representation of this Lie algebra. Let vv, ww be two elements of a Lie algebra. Then we define

ad⁡v(w):=[v,w] \operatorname {ad}v(w) := [v,w]

and wad⁡v:=[v,w]w\operatorname {ad}v := [v,w].

(ii)

The Killing form is defined by

κ(x,y):=tr⁡(ad⁡xad⁡y) \kappa (x,y) := \operatorname {tr}(\operatorname {ad}x \operatorname {ad}y)

for all x,y∈Lx, y \in L. Find the Killing form.

Consider the Lie algebra L=sl(2,F)L = \mathfrak {sl}(2,\mathbb {F}) with char⁡(F)≠2\operatorname {char}(\mathbb {F}) \neq 2. Take as the standard basis for LL the three matrices

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}.
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(i)

Find the commutators.

(ii)

Find the adjoint representation of LL with the ordered basis { E  H  F }\{ \, E \; H \; F \, \}.

(iii)

Show that LL is simple. If LL has no ideals except itself and 0, and if moreover [L,L]≠0[L,L] \neq 0, we call LL simple. A subspace II of a Lie algebra LL is called an ideal of LL if x∈Lx \in L, y∈Iy \in I together imply [x,y]∈I[x,y] \in I. In other words show that the Lie algebra sl(2,R)\mathfrak {sl}(2,\mathbb {R}) has no proper nontrivial ideals.

Consider the four-dimensional Lie algebra gl(2,R)\mathfrak {gl}(2,\mathbb {R}). The elementary matrices

e1=(1000),e2=(0100),e3=(0010),e4=(0001) e_{1} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \qquad e_{2} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \qquad e_{3} = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, \qquad e_{4} = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}

form a basis of gl(2,R)\mathfrak {gl}(2,\mathbb {R}). Find the adjoint representation.

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Show that the matrices

J1=(00000−1010),J2=(001000−100),J3=(0−10100000) J_{1} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{pmatrix}, \qquad J_{2} = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0 \end{pmatrix}, \qquad J_{3} = \begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}

form generators of a Lie algebra. Is the Lie algebra simple? A Lie algebra is simple if it contains no ideals other than LL and 00.

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The simple Lie algebra sl(2,R)\mathfrak {sl}(2,\mathbb {R}) is spanned by the matrices

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

with the commutators [H,E]=2E[H,E] = 2E, [H,F]=−2F[H,F] = -2F and [E,F]=H[E,F] = H.

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(i)

Define

C:=12H2+EF+FE. C := \frac{1}{2}H^{2} + EF + FE.

Find CC. Calculate the commutators [C,H][C,H], [C,E][C,E], [C,F][C,F]. Show that CC can be written in the form

C=12H2+H+2FE. C = \frac{1}{2}H^{2} + H + 2FE.
(ii)

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

v→=(10). \overrightarrow {v} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}.

Calculate the vectors Hv→H\overrightarrow {v}, Ev→E\overrightarrow {v}, Fv→F\overrightarrow {v} and Cv→C\overrightarrow {v}. Give an interpretation.

(iii)

The universal enveloping algebra U(sl(2,R))U(\mathfrak {sl}(2,\mathbb {R})) of the Lie algebra sl(2,R)\mathfrak {sl}(2,\mathbb {R}) is the associative algebra with generators HH, EE, FF and the relations

HE−EH=2E,HF−FH=−2F,EF−FE=H. HE-EH = 2E, \qquad HF-FH = -2F, \qquad EF-FE = H.

Find a basis of the Lie algebra sl(2,R)\mathfrak {sl}(2,\mathbb {R}) so that all matrices are invertible. Find the inverse matrices of these matrices. Give the commutation relations.

(iv)

Which of the EE, FF, HH matrices are nonnormal? Use linear combinations to find a basis where all elements are normal matrices.

We know that

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

is an ordered basis of the simple Lie algebra sl(2,R)\mathfrak {sl}(2,\mathbb {R}) with

[E,H]=−2E,[E,F]=H,[F,H]=2F. [E,H] = -2E, \qquad [E,F] = H, \qquad [F,H] = 2F.

Consider

E~=(0001001000000000),H~=(1000010000−10000−1),F~=(0000000001001000). \widetilde{E} = \begin{pmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}, \qquad \widetilde{H} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix}, \qquad \widetilde{F} = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{pmatrix}.

Find the commutators [E~,H~][\widetilde{E},\widetilde{H}], [E~,F~][\widetilde{E},\widetilde{F}], [F~,H~][\widetilde{F},\widetilde{H}].

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The 2×22 \times 2 matrices

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}

form a basis of the Lie algebra sl(2,R)\mathfrak {sl}(2,\mathbb {R}) with the commutation relations [E,H]=−2E[E,H] = -2E, [E,F]=H[E,F] = H, [F,H]=2F[F,H] = 2F.

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(i)

Let ⋆\star be the star product. Do the nine 4×44 \times 4 matrices

E⋆E,E⋆F,E⋆H,F⋆E,F⋆F,F⋆H,H⋆E,H⋆F,H⋆H E \star E, \quad E \star F, \quad E \star H, \quad F \star E, \quad F \star F, \quad F \star H, \quad H \star E, \quad H \star F, \quad H \star H

form a basis of a Lie algebra?

(ii)

Define the 4×44 \times 4 matrices

H~:=H⊗I2+I2⊗H,E~:=E⊗H−1+H⊗E,F~:=F⊗H−1+H⊗F. \widetilde{H} := H \otimes I_{2} + I_{2} \otimes H, \qquad \widetilde{E} := E \otimes H^{-1} + H \otimes E, \qquad \widetilde{F} := F \otimes H^{-1} + H \otimes F.

Find the commutators [H~,E~][\widetilde{H},\widetilde{E}], [H~,F~][\widetilde{H},\widetilde{F}], [E~,F~][\widetilde{E},\widetilde{F}].

(iii)

Consider

H~=H⊗I2+I2⊗H,E~=E⊗H+H⊗E,F~=F⊗H+H⊗F. \widetilde{H} = H \otimes I_{2} + I_{2} \otimes H, \qquad \widetilde{E} = E \otimes H + H \otimes E, \qquad \widetilde{F} = F \otimes H + H \otimes F.

Find the commutators [H~,E~][\widetilde{H},\widetilde{E}], [H~,F~][\widetilde{H},\widetilde{F}], [E~,F~][\widetilde{E},\widetilde{F}] and thus show that we have a basis of a Lie algebra. Is this Lie algebra isomorphic to sl(2,R)\mathfrak {sl}(2,\mathbb {R})?

A basis for the Lie algebra su(N)\mathfrak {su}(N), for odd NN, may be built from two unitary unimodular N×NN \times N matrices

g=(10000ω0000ω20⋮⋮⋮⋱⋮000…ωN−1),h=(010…0001…0⋮⋮⋮⋱⋮000…1100…0) g = \begin{pmatrix} 1 & 0 & 0 & & 0 \\ 0 & \omega & 0 & & 0 \\ 0 & 0 & \omega ^{2} & & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \ldots & \omega ^{N-1} \end{pmatrix}, \qquad h = \begin{pmatrix} 0 & 1 & 0 & \ldots & 0 \\ 0 & 0 & 1 & \ldots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \ldots & 1 \\ 1 & 0 & 0 & \ldots & 0 \end{pmatrix}

where ω\omega is a primitive NN-th root of unity, i.e. with period not smaller than NN, here taken to be exp(4πi/N)\mathrm{exp}\left(4\pi i/N\right). We obviously have

hg=ωgh.(1) hg = \omega gh. \tag {1}
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(i)

Find gNg^{N} and hNh^{N}.

(ii)

Find tr⁡(g)\operatorname {tr}(g).

(iii)

Let m→=(m1,m2)\overrightarrow {m} = (m_{1},m_{2}), n→=(n1,n2)\overrightarrow {n} = (n_{1},n_{2}) and define

m→×n→:=m1n2−m2n1 \overrightarrow {m} \times \overrightarrow {n} := m_{1}n_{2} - m_{2}n_{1}

where m1=0,1,…,N−1m_{1} = 0,1,\ldots ,N-1 and m2=0,1,…,N−1m_{2} = 0,1,\ldots ,N-1. The complete set of unitary unimodular N×NN \times N matrices

Jm1,m2:=ωm1m2/2gm1hm2 J_{m_{1},m_{2}} := \omega ^{m_{1}m_{2}/2}g^{m_{1}}h^{m_{2}}

suffice to span the Lie algebra su(N)\mathfrak {su}(N), where J0,0=INJ_{0,0} = I_{N}. Find J∗J^{*}.

(iv)

Calculate Jm→Jn→J_{\overrightarrow {m}}J_{\overrightarrow {n}}.

(v)

Find the commutator [Jm→,Jn→][J_{\overrightarrow {m}}, J_{\overrightarrow {n}}].

Consider the 2×22 \times 2 matrices over R\mathbb {R}

A=(−1010),B=(010−1). A = \begin{pmatrix} -1 & 0 \\ 1 & 0 \end{pmatrix}, \qquad B = \begin{pmatrix} 0 & 1 \\ 0 & -1 \end{pmatrix}.

Calculate the commutator C=[A,B]C = [A,B] and check whether CC can be written as a linear combination of AA and BB. If so we have a basis of a Lie algebra.

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Consider the 3×33 \times 3 matrices

h1=(100000001),h2=(000010000),h3=(001000100), h_{1} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \qquad h_{2} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}, \qquad h_{3} = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix}, e=(010000010),f=(000101000). e = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}, \qquad f = \begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}.

Show that the matrices form a basis of a Lie algebra.

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A basis of the simple Lie algebra sl(2,R)\mathfrak {sl}(2,\mathbb {R}) is given by the traceless 2×22 \times 2 matrices

X1=12(0110),X2=12(−1001),X3=12(01−10). X_{1} = \frac{1}{2}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \qquad X_{2} = \frac{1}{2}\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}, \qquad X_{3} = \frac{1}{2}\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}.
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(i)

Find the commutators [X1,X2][X_{1},X_{2}], [X2,X3][X_{2},X_{3}], [X3,X1][X_{3},X_{1}].

(ii)

Let z∈Cz \in \mathbb {C}. Find exp(zX1)\mathrm{exp}\left(zX_{1}\right), exp(zX2)\mathrm{exp}\left(zX_{2}\right), exp(zX3)\mathrm{exp}\left(zX_{3}\right).

(iii)

Let u,v,r∈Ru, v, r \in \mathbb {R}. Show that

g(u,v,r)=euX3erX2evX3 g(u,v,r) = e^{uX_{3}}e^{rX_{2}}e^{vX_{3}} =er/2(−sin⁡(u/2)sin⁡(v/2)cos⁡(v/2)sin⁡(u/2)−cos⁡(u/2)sin⁡(v/2)cos⁡(u/2)cos⁡(v/2)) = e^{r/2}\begin{pmatrix} -\sin (u/2)\sin (v/2) & \cos (v/2)\sin (u/2) \\ -\cos (u/2)\sin (v/2) & \cos (u/2)\cos (v/2) \end{pmatrix} +e−r/2(cos⁡(u/2)cos⁡(v/2)cos⁡(u/2)sin⁡(v/2)−cos⁡(v/2)sin⁡(u/2)−sin⁡(u/2)sin⁡(v/2)). + e^{-r/2}\begin{pmatrix} \cos (u/2)\cos (v/2) & \cos (u/2)\sin (v/2) \\ -\cos (v/2)\sin (u/2) & -\sin (u/2)\sin (v/2) \end{pmatrix}.
(iv)

Find g(u,v,r)−1g(u,v,r)^{-1}.

Consider vectors in the vector space R3\mathbb {R}^{3} and the vector product ×\times. Consider the mapping of the vectors in R3\mathbb {R}^{3} into 3×33 \times 3 skew-symmetric matrices

(abc)↔(0c−b−c0ab−a0). \begin{pmatrix} a \\ b \\ c \end{pmatrix} \leftrightarrow \begin{pmatrix} 0 & c & -b \\ -c & 0 & a \\ b & -a & 0 \end{pmatrix}.

Calculate the vector product

(a2b2c2)×(a1b1c1) \begin{pmatrix} a_{2} \\ b_{2} \\ c_{2} \end{pmatrix} \times \begin{pmatrix} a_{1} \\ b_{1} \\ c_{1} \end{pmatrix}

and the commutator [M1,M2][M_{1},M_{2}], where

M1=(0c1−b1−c10a1b1−a10),M2=(0c2−b2−c20a2b2−a20). M_{1} = \begin{pmatrix} 0 & c_{1} & -b_{1} \\ -c_{1} & 0 & a_{1} \\ b_{1} & -a_{1} & 0 \end{pmatrix}, \qquad M_{2} = \begin{pmatrix} 0 & c_{2} & -b_{2} \\ -c_{2} & 0 & a_{2} \\ b_{2} & -a_{2} & 0 \end{pmatrix}.

Discuss.

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A basis for the Lie algebra su(3)\mathfrak {su}(3) is given by the eight 3×33 \times 3 skew-hermitian matrices

X1=(0i0i00000),X2=(010−100000), X_{1} = \begin{pmatrix} 0 & i & 0 \\ i & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}, \qquad X_{2} = \begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}, X3=(i000−i0000),X4=(00i000i00), X_{3} = \begin{pmatrix} i & 0 & 0 \\ 0 & -i & 0 \\ 0 & 0 & 0 \end{pmatrix}, \qquad X_{4} = \begin{pmatrix} 0 & 0 & i \\ 0 & 0 & 0 \\ i & 0 & 0 \end{pmatrix}, X5=(001000−100),X6=(00000i0i0), X_{5} = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0 \end{pmatrix}, \qquad X_{6} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & i \\ 0 & i & 0 \end{pmatrix}, X7=(0000010−10),X8=13(i000i000−2i). X_{7} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{pmatrix}, \qquad X_{8} = \frac{1}{\sqrt{3}}\begin{pmatrix} i & 0 & 0 \\ 0 & i & 0 \\ 0 & 0 & -2i \end{pmatrix}.

Find

∑j=18(Xj⊗Xj),∑j=18((XjXk)⊗Xj). \sum _{j=1}^{8}(X_{j} \otimes X_{j}), \qquad \sum _{j=1}^{8}((X_{j}X_{k}) \otimes X_{j}).

Express the results as Kronecker products of two 3×33 \times 3 matrices.

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Consider the three 3×33 \times 3 matrices

A=(010−100000),B=(00−1000100),C=(0000010−10) A = \begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}, \qquad B = \begin{pmatrix} 0 & 0 & -1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix}, \qquad C = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{pmatrix}

which satisfy the commutation relations [A,B]=C[A,B] = C, [B,C]=A[B,C] = A, [C,A]=B[C,A] = B. Thus we have a basis of the simple Lie algebra so(3)\mathfrak {so}(3). Calculate the commutators

[A⊗I3,B⊗I3],[B⊗I3,C⊗I3],[C⊗I3,A⊗I3]. [A \otimes I_{3}, B \otimes I_{3}], \qquad [B \otimes I_{3}, C \otimes I_{3}], \qquad [C \otimes I_{3}, A \otimes I_{3}].
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Let LL be a finite dimensional Lie algebra. Let C∞(S1)C^{\infty }(S^{1}) be the set of all infinitely differentiable functions, where S1S^{1} is the unit circle manifold. In the product space L⊗C∞(S1)L \otimes C^{\infty }(S^{1}) we define the Lie bracket (g1,g2∈Lg_{1}, g_{2} \in L and f1,f2∈C∞(S1)f_{1}, f_{2} \in C^{\infty }(S^{1}))

[g1⊗f1,g2⊗f2]:=[g1,g2]⊗(f1f2). [g_{1} \otimes f_{1}, g_{2} \otimes f_{2}] := [g_{1},g_{2}] \otimes (f_{1}f_{2}).

Calculate

[g1⊗f1,[g2⊗f2,g3⊗f3]]+[g3⊗f3,[g1⊗f1,g2⊗f2]]+[g2⊗f2,[g3⊗f3,g1⊗f1]]. [g_{1} \otimes f_{1}, [g_{2} \otimes f_{2}, g_{3} \otimes f_{3}]] + [g_{3} \otimes f_{3}, [g_{1} \otimes f_{1}, g_{2} \otimes f_{2}]] + [g_{2} \otimes f_{2}, [g_{3} \otimes f_{3}, g_{1} \otimes f_{1}]].
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The elements (generators) Z1,Z2,…,ZrZ_{1}, Z_{2}, \ldots , Z_{r} of an rr-dimensional Lie algebra satisfy the conditions

[Zμ,Zν]=∑τ=1rcμντZτ [Z_{\mu }, Z_{\nu }] = \sum _{\tau =1}^{r} c^{\tau }_{\mu \nu }Z_{\tau }

with cμντ=−cνμτc^{\tau }_{\mu \nu } = -c^{\tau }_{\nu \mu }, where the cμντc^{\tau }_{\mu \nu }'s are called the structure constants. Let AA be an arbitrary linear combination of the elements

A=∑μ=1raμZμ. A = \sum _{\mu =1}^{r} a^{\mu }Z_{\mu }.

Suppose that XX is some other linear combination such that

X=∑ν=1rbνZνand[A,X]=ρX. X = \sum _{\nu =1}^{r} b^{\nu }Z_{\nu } \qquad \text{and} \qquad [A,X] = \rho X.

This equation has the form of an eigenvalue equation, where ρ\rho is the corresponding eigenvalue and XX the corresponding eigenvector. Assume that the Lie algebra is represented by matrices. Find the secular equation for the eigenvalues ρ\rho.

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Given the basis of the Lie algebra sl(2,R)\mathfrak {sl}(2,\mathbb {R})

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}

with the commutation relations [E,F]=H[E,F] = H, [E,H]=−2E[E,H] = -2E, [F,H]=−2F[F,H] = -2F.

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(i)

Let z∈Cz \in \mathbb {C}. We define

E~=ezH/2⊗E+E⊗e−z‾H/2,F~=ez‾H/2⊗F+F⊗e−zH/2,H~=H⊗I2+I2⊗H. \widetilde{E} = e^{zH/2} \otimes E + E \otimes e^{-\overline{z}H/2}, \qquad \widetilde{F} = e^{\overline{z}H/2} \otimes F + F \otimes e^{-zH/2}, \qquad \widetilde{H} = H \otimes I_{2} + I_{2} \otimes H.

Find the commutators [E~,F~][\widetilde{E},\widetilde{F}], [E~,H~][\widetilde{E},\widetilde{H}], [F~,H~][\widetilde{F},\widetilde{H}].

(ii)

Let U(sl(2,R))U(\mathfrak {sl}(2,\mathbb {R})) be the universal enveloping algebra. Then any element of U(sl(2,R))U(\mathfrak {sl}(2,\mathbb {R})) can be expressed as a sum of product of the form FjHkEℓF^{j}H^{k}E^{\ell } where j,k,ℓ≥0j,k,\ell \geq 0. Show that

EF2=−2F+2FH+F2E. EF^{2} = -2F+2FH+F^{2}E.
Remark.
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Utilize that EF2≡[E,F2]+F2EEF^{2} \equiv [E,F^{2}]+F^{2}E.

Study the Lie algebra sl(2,F)\mathfrak {sl}(2,\mathbb {F}), where char⁡(F)=2\operatorname {char}(\mathbb {F}) = 2.

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Do the matrices

E+=(010001000),E−=(000100010),H=(10000000−1) E_{+} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}, \qquad E_{-} = \begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}, \qquad H = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{pmatrix}

form a basis for the Lie algebra sl(2,R)\mathfrak {sl}(2,\mathbb {R})?

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Show that a basis of the Lie algebra sl(2,C)\mathfrak {sl}(2,\mathbb {C}) is given by

12(0ii0),12(01−10),12(−i00i), \frac{1}{2}\begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix}, \qquad \frac{1}{2}\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}, \qquad \frac{1}{2}\begin{pmatrix} -i & 0 \\ 0 & i \end{pmatrix}, 12(0110),12(0−ii0),12(−1001). \frac{1}{2}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \qquad \frac{1}{2}\begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \qquad \frac{1}{2}\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}.
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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.

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Let EjkE_{jk} (j,k=1,2,3,4j,k = 1,2,3,4) be the elementary matrices in the vector space of 4×44 \times 4 matrices. Show that the 15 matrices

X1=E12,X2=E23,X3=E13,X4=E34,X5=E24,X6=E14 X_{1} = E_{12}, \qquad X_{2} = E_{23}, \qquad X_{3} = E_{13}, \qquad X_{4} = E_{34}, \qquad X_{5} = E_{24}, \qquad X_{6} = E_{14} Y1=E21,Y2=E32,Y3=E31,Y4=E43,Y5=E42,Y6=E41 Y_{1} = E_{21}, \qquad Y_{2} = E_{32}, \qquad Y_{3} = E_{31}, \qquad Y_{4} = E_{43}, \qquad Y_{5} = E_{42}, \qquad Y_{6} = E_{41} H1=14diag⁡(3,−1,−1,−1,−1),H2=12diag⁡(1,1,−1,−1), H_{1} = \frac{1}{4}\operatorname {diag}(3,-1,-1,-1,-1), \qquad H_{2} = \frac{1}{2}\operatorname {diag}(1,1,-1,-1), H3=14diag⁡(1,1,1,−3) H_{3} = \frac{1}{4}\operatorname {diag}(1,1,1,-3)

form a basis (Cartan-Weyl basis) of the Lie algebra sl(4,C)\mathfrak {sl}(4,\mathbb {C}).

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Consider a four-dimensional vector space with basis e1e_{1}, e2e_{2}, e3e_{3}, e4e_{4}. Assume that the non-zero commutators are given

[e2,e3]=e1,[e1,e4]=2e1,[e2,e4]=e2,[e3,e4]=e2+e3. [e_{2},e_{3}] = e_{1}, \qquad [e_{1},e_{4}] = 2e_{1}, \qquad [e_{2},e_{4}] = e_{2}, \qquad [e_{3},e_{4}] = e_{2}+e_{3}.

Do these relations define a Lie algebra? For example we have

[e1,[e2,e3]]+[e3,[e1,e2]]+[e2,[e3,e1]]=[e1,e1]+[e3,0]+[e2,0]=0. [e_{1},[e_{2},e_{3}]]+[e_{3},[e_{1},e_{2}]]+[e_{2},[e_{3},e_{1}]] = [e_{1},e_{1}]+[e_{3},0]+[e_{2},0] = 0.

If so find the adjoint representation.

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(i)

Find the Lie algebra generated by the 2×22 \times 2 matrices

A=(0001),B=(00−10). A = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}, \qquad B = \begin{pmatrix} 0 & 0 \\ -1 & 0 \end{pmatrix}.

Note that [A,B]=A+B[A,B] = A+B.

(ii)

Find the Lie algebra generated by the 3×33 \times 3 matrices

A=(00001000−1),B=(000−100000),C=(000000100). A = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix}, \qquad B = \begin{pmatrix} 0 & 0 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}, \qquad C = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix}.

Note that [A,B]=B[A,B] = B, [B,C]=03[B,C] = 0_{3}, [C,A]=C[C,A] = C.

In the decomposition of the simple Lie algebra sl(3,R)\mathfrak {sl}(3,\mathbb {R}) one finds the 3×33 \times 3 matrices

A=(a11a120a21a22000−a11−a22),B=(00b1300b23b31b320) A = \begin{pmatrix} a_{11} & a_{12} & 0 \\ a_{21} & a_{22} & 0 \\ 0 & 0 & -a_{11}-a_{22} \end{pmatrix}, \qquad B = \begin{pmatrix} 0 & 0 & b_{13} \\ 0 & 0 & b_{23} \\ b_{31} & b_{32} & 0 \end{pmatrix}

where ajk,bjk∈Ra_{jk}, b_{jk} \in \mathbb {R}. Find the commutators [A,A′][A,A'], [B,B′][B,B'] and [A,B][A,B]. Discuss. Note that

[A,B]=(00a12b23+a22b1300a21b13+a11b23−a22b31−a21b32−a11b32−a12b310). [A,B] = \begin{pmatrix} 0 & 0 & a_{12}b_{23}+a_{22}b_{13} \\ 0 & 0 & a_{21}b_{13}+a_{11}b_{23} \\ -a_{22}b_{31}-a_{21}b_{32} & -a_{11}b_{32}-a_{12}b_{31} & 0 \end{pmatrix}.

Thus the matrix of the commutator [A,B][A,B] is of the form of matrix BB.

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Let α∈R\alpha \in \mathbb {R}. Show that the matrices

L1=(α+i00α),L2=(0100),L3=(0i00) L_{1} = \begin{pmatrix} \alpha +i & 0 \\ 0 & \alpha \end{pmatrix}, \qquad L_{2} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \qquad L_{3} = \begin{pmatrix} 0 & i \\ 0 & 0 \end{pmatrix}

satisfy the commutation relations

[L1,L2]=L3,[L2,L3]=02,[L3,L1]=L2. [L_{1},L_{2}] = L_{3}, \qquad [L_{2},L_{3}] = 0_{2}, \qquad [L_{3},L_{1}] = L_{2}.
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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.