Representations of Lie algebras

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Define on VnV_{n} the structure of an su(2)\mathfrak {su}(2)-module, using formulas (2), returning to the basis elements

k1=12(e−1+e2),k2=i2(e−1−e1),k3=−i2e0 \mathbf{k}_{\mathbf{1}}=\frac{1}{2}\left(e_{-1}+e_{2}\right), \quad \mathbf{k}_{2}=\frac{i}{2}\left(e_{-1}-e_{1}\right), \quad \mathbf{k}_{3}=-\frac{i}{2} e_{0}

and associating with them the linear operators K1,K2,K3K_{1}, K_{2}, K_{3}.

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Let L=⟨e−1,e0,e1⟩L= \left\langle e_{-1}, e_{0}, e_{1}\right\rangle be a simple Lie algebra with multiplication table (1) over an algebraically closed field FF of characteristic p>2p>2. Consider the square matrices of order pp

E−1=∥0γ10…0000γ2…00………………000…γp−20000…0γp−1γ000…00∥,E1=∥00…00β10…00001…000………………00…10000…010∥,E0=[E−1,E1]=diag⁡(λ,λ+2,…,λ+2(p−2),λ+2(p−1)). \begin{aligned} E_{-1}= & \left\Vert \begin{smallmatrix} 0 & \gamma _{1} & 0 & \ldots & 0 & 0 \\ 0 & 0 & \gamma _{2} & \ldots & 0 & 0 \\ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\ 0 & 0 & 0 & \ldots & \gamma _{p-2} & 0 \\ 0 & 0 & 0 & \ldots & 0 & \gamma _{p-1} \\ \gamma _{0} & 0 & 0 & \ldots & 0 & 0 \end{smallmatrix}\right\Vert , E_{1}=\left\Vert \begin{smallmatrix} 0 & 0 & \ldots & 0 & 0 & \beta \\ 1 & 0 & \ldots & 0 & 0 & 0 \\ 0 & 1 & \ldots & 0 & 0 & 0 \\ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\ 0 & 0 & \ldots & 1 & 0 & 0 \\ 0 & 0 & \ldots & 0 & 1 & 0 \end{smallmatrix}\right\Vert , \\ & E_{0}=\left[E_{-1}, E_{1}\right]=\operatorname {diag}(\lambda , \lambda +2, \ldots , \lambda +2(p-2), \lambda +2(p-1)). \end{aligned}

Verify that the correspondence ei↦Ei,i=−1,0,1e_{i} \mapsto E_{i}, \quad i=-1,0,1, establishes an irreducible matrix representation of the Lie algebra LL, depending on three parameters (λ,β,γ0)\left(\lambda , \beta , \gamma_{0}\right). Here

γk=βγ0+kλ+k(k−1),k=1,2,…,p−1 \gamma _{k}=\beta \gamma _{0}+k \lambda +k(k-1), \quad k=1,2, \ldots , p-1

(This entirely new situation is typical for representations over a field of finite characteristic.)

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Consider the Lie algebra with basis {e1,e2,e3}\{ e_{1}, e_{2}, e_{3}\} and the commutators

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

Find the adjoint representation.

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Give a Lie algebra which is semisimple, but not simple. Provide a matrix representation.

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(i)

Classify all real two-dimensional Lie algebras, where F=R\mathbb {F} = \mathbb {R}. Denote the basis elements by XX and YY.

(ii)

Find a representation of the non-abelian two-dimensional Lie algebra using differential operators.

(iii)

Find a representation of the non-abelian two-dimensional Lie algebra using 2×22 \times 2 matrices.

The simple Lie algebra sl(2,R)\mathfrak {sl}(2,\mathbb {R}) is spanned by the 2×22 \times 2 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 [e,f]=h[e,f] = h, [h,e]=2e[h,e] = 2e, [h,f]=−2f[h,f] = -2f. Show that the bracket relations are also satisfied by the matrices

ρ2(h)=(20000000−2), \rho _{2}(h) = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -2 \end{pmatrix}, ρ2(e)=(020001000),ρ2(f)=(000100020) \rho _{2}(e) = \begin{pmatrix} 0 & 2 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}, \qquad \rho _{2}(f) = \begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 2 & 0 \end{pmatrix}

and

ρ3(h)=(3000010000−10000−3), \rho _{3}(h) = \begin{pmatrix} 3 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -3 \end{pmatrix}, ρ3(e)=(0300002000010000),ρ3(f)=(0000100002000030) . \rho _{3}(e) = \begin{pmatrix} 0 & 3 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix}, \qquad \rho _{3}(f) = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 3 & 0 \end{pmatrix} \, .
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Consider the Lie algebra gl(2,R)\mathfrak {gl}(2,\mathbb {R}). The 2×22 \times 2 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}).

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(i)

Find the adjoint representation.

(ii)

Find the Killing form and the metric tensor for the Lie algebra. From these results decide whether the Lie algebra is semisimple.

Let e1,e2,e3e_{1}, e_{2}, e_{3} be the generators of a three-dimensional Lie algebra with the commutation relation

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

Find the adjoint representation. Is the Lie algebra semisimple?

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Let InI_{n} be the n×nn \times n unit matrix. Consider the 2n×2n2n \times 2n matrices

E1=12(In0n0n−In),E2=12(0nInIn0n),E3=12(0nIn−In0n) . E_{1} = \frac{1}{2} \begin{pmatrix} I_{n} & 0_{n} \\ 0_{n} & -I_{n} \end{pmatrix}, \qquad E_{2} = \frac{1}{2} \begin{pmatrix} 0_{n} & I_{n} \\ I_{n} & 0_{n} \end{pmatrix}, \qquad E_{3} = \frac{1}{2} \begin{pmatrix} 0_{n} & I_{n} \\ -I_{n} & 0_{n} \end{pmatrix} \, .

Find the commutators [E1,E2],[E1,E3],[E3,E2][E_{1},E_{2}], [E_{1},E_{3}], [E_{3},E_{2}]. Discuss.

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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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(i)

Is the three dimensional Heisenberg algebra with the basis {x,y,z}\{ x,y,z\} and the commutators [x,y]=z[x,y] = z, [x,z]=0[x,z] = 0, [y,z]=0[y,z] = 0 nilpotent?

(ii)

Find a matrix representation with 3×33 \times 3 matrices.

Let z∈Cz \in \mathbb {C}. Consider the vector fields

V1=−∂∂z−∂∂zˉ,V2=i∂∂zˉ−i∂∂z,V3=2i(z∂∂zˉ−zˉ∂∂z) . V_{1} = -\frac{\partial }{\partial z} - \frac{\partial }{\partial \bar{z}}, \qquad V_{2} = i\frac{\partial }{\partial \bar{z}} - i\frac{\partial }{\partial z}, \qquad V_{3} = \frac{2}{i}\left( z\frac{\partial }{\partial \bar{z}} - \bar{z}\frac{\partial }{\partial z} \right) \, .

Find the commutators and show that we have a basis of a Lie algebra.

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Consider the vector fields

V12=x1∂∂x2−x2∂∂x1,V31=x3∂∂x1−x1∂∂x3,V23=x2∂∂x3−x3∂∂x2 . V_{12} = x_{1}\frac{\partial }{\partial x_{2}} - x_{2}\frac{\partial }{\partial x_{1}}, \qquad V_{31} = x_{3}\frac{\partial }{\partial x_{1}} - x_{1}\frac{\partial }{\partial x_{3}}, \qquad V_{23} = x_{2}\frac{\partial }{\partial x_{3}} - x_{3}\frac{\partial }{\partial x_{2}} \, .

Find the commutators [V12,V31],[V31,V23],[V23,V12][V_{12},V_{31}], [V_{31},V_{23}], [V_{23},V_{12}] and show that V12,V31,V23V_{12}, V_{31}, V_{23} form a basis of a Lie algebra. Is the Lie algebra semisimple?

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(i)

Consider the non-commutative two dimensional Lie algebra with [A,B]=A[A,B] = A and

A=(0100),B=(0001) . A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \qquad B = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \, .

Show that the matrices

{A⊗I2+I2⊗A,  B⊗I2+I2⊗B} \{ A \otimes I_{2} + I_{2} \otimes A, \; B \otimes I_{2} + I_{2} \otimes B \}

also form a non-commutative Lie algebra under the commutator, where ⊗\otimes denotes the Kronecker product. Discuss.

(ii)

We denote by sl(2,R)\mathfrak {sl}(2,\mathbb {R}) the semisimple Lie algebra spanned by a basis {E,F,H}\{ E,F,H\} with the Lie bracket defined by

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

Using 2×22 \times 2 matrices we have the representation

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

Let ⊗\otimes be the Kronecker product. Do the three 4×44 \times 4 matrices

H⊗I2+I2⊗H,F⊗I2+I2⊗F,E⊗I2+I2⊗E H \otimes I_{2} + I_{2} \otimes H, \qquad F \otimes I_{2} + I_{2} \otimes F, \qquad E \otimes I_{2} + I_{2} \otimes E

form a basis of a Lie algebra? Calculate the commutators.

Consider the Lie algebra sl(2,R)\mathfrak {sl}(2,\mathbb {R}) with

x=(0100),y=(0010),h=(100−1) x = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \qquad y = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, \qquad h = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}

and the commutators [x,h]=−2x[x,h] = -2x, [x,y]=h[x,y] = h, [y,h]=2y[y,h] = 2y. Consider

h↦h⊗I2+I2⊗h,x↦x⊗h+h⊗x,y↦y⊗h+h⊗y . h \mapsto h \otimes I_{2} + I_{2} \otimes h, \qquad x \mapsto x \otimes h + h \otimes x, \qquad y \mapsto y \otimes h + h \otimes y \, .

Find the three commutators

[h⊗I2+I2⊗h,  x⊗h+h⊗x],[h⊗I2+I2⊗h,  y⊗h+h⊗y],[x⊗h+h⊗x,  y⊗h+h⊗y] [h\otimes I_{2}+I_{2}\otimes h, \; x\otimes h+h\otimes x], \quad [h\otimes I_{2}+I_{2}\otimes h, \; y\otimes h+h\otimes y], \quad [x\otimes h+h\otimes x, \; y\otimes h+h\otimes y]

and thus show that we have a Lie algebra. Is this Lie algebra isomorphic to sl(2,R)\mathfrak {sl}(2,\mathbb {R})?

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The four matrices

X=(0100),Y=(0010),H=(100−1),I2=(1001) X = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \qquad Y = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, \qquad H = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, \qquad I_{2} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}

form a basis of the Lie algebra gl(2,R)\mathfrak {gl}(2,\mathbb {R}). We have

[X,Y]=H,[H,X]=2X,[H,Y]=−2Y [X,Y] = H, \qquad [H,X] = 2X, \qquad [H,Y] = -2Y

and [I2,X]=[I2,Y]=[I2,H]=02[I_{2},X] = [I_{2},Y] = [I_{2},H] = 0_{2}. Can one find an invertible matrix SS over R\mathbb {R} such that

SXS−1=Y,SYS−1=X,SHS−1=−H  ? SXS^{-1} = Y, \qquad SYS^{-1} = X, \qquad SHS^{-1} = -H \; ?
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Consider the semisimple Lie algebra so(2,1)\mathfrak {so}(2,1) whose elements X1,X2,X3X_{1}, X_{2}, X_{3} satisfy the commutation relations

[X1,X2]=X3,[X2,X3]=−X1,[X3,X1]=X2 . [X_{1},X_{2}] = X_{3}, \qquad [X_{2},X_{3}] = -X_{1}, \qquad [X_{3},X_{1}] = X_{2} \, .
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(i)

Find the adjoint representation.

(ii)

Let YjY_{j} (j=1,…,nj = 1,\ldots ,n) be a basis for a Lie algebra. Then gjk=κ(Yj,Yk)g_{jk} = \kappa (Y_{j},Y_{k}) is called the metric tensor. From the adjoint representation find the metric tensor.

Consider the Lie algebra sl(2,R)\mathfrak {sl}(2,\mathbb {R}) with the basis H,X,YH, X, Y and the commutators

[H,X]=2X,[H,Y]=−2Y,[X,Y]=H [H,X] = 2X, \qquad [H,Y] = -2Y, \qquad [X,Y] = H

with

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

Assume that v→∈R\overrightarrow {v} \in \mathbb {R} is an eigenvector of HH, i.e. Hv→=αv→H\overrightarrow {v} = \alpha \overrightarrow {v}, where α\alpha is the eigenvalue. Calculate H(Xv→)H(X\overrightarrow {v}). Discuss.

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(i)

Consider the differential operators

J3=i2(p∂∂x−x∂∂p) J_{3} = \frac{i}{2}\left( p\frac{\partial }{\partial x} - x\frac{\partial }{\partial p} \right)

and

K1=i2(p∂∂p−x∂∂x),K2=−i2(x∂∂p+p∂∂x) . K_{1} = \frac{i}{2}\left( p\frac{\partial }{\partial p} - x\frac{\partial }{\partial x} \right), \qquad K_{2} = -\frac{i}{2}\left( x\frac{\partial }{\partial p} + p\frac{\partial }{\partial x} \right) \, .

Find the commutators. Show that we have a basis for a semisimple Lie algebra.

(ii)

Consider the differential operators

J~3=14(x2−d2dx2) \tilde{J}_{3} = \frac{1}{4}\left( x^{2} - \frac{d^{2}}{dx^{2}} \right)

and

K~1=i4(xddx+ddxx),K~2=14(x2+d2dx2) . \tilde{K}_{1} = \frac{i}{4}\left( x\frac{d}{dx} + \frac{d}{dx}x \right), \qquad \tilde{K}_{2} = \frac{1}{4}\left( x^{2} + \frac{d^{2}}{dx^{2}} \right) \, .

Find the commutators. Show that we have a basis for a semisimple Lie algebra. Are the two Lie algebras isomorphic?

Consider the spin-matrices (spin-1)

S+=(020002000),S−=(000200020),Sz=(10000000−1) . S_{+} = \begin{pmatrix} 0 & \sqrt{2} & 0 \\ 0 & 0 & \sqrt{2} \\ 0 & 0 & 0 \end{pmatrix}, \qquad S_{-} = \begin{pmatrix} 0 & 0 & 0 \\ \sqrt{2} & 0 & 0 \\ 0 & \sqrt{2} & 0 \end{pmatrix}, \qquad S_{z} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{pmatrix} \, .

Find the commutators and thus show that we have a basis of a Lie algebra. Show that we have a faithful representation using differential operators by

S+→z∂∂ζ,S−→ζ∂∂z,Sz→12(z∂∂z−ζ∂∂ζ) . S_{+} \to z\frac{\partial }{\partial \zeta }, \qquad S_{-} \to \zeta \frac{\partial }{\partial z}, \qquad S_{z} \to \frac{1}{2}\left( z\frac{\partial }{\partial z} - \zeta \frac{\partial }{\partial \zeta } \right) \, .
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Consider the Lie algebra sl(2,R)\mathfrak {sl}(2,\mathbb {R}) with the basis e1,e2,e3e_{1}, e_{2}, e_{3} and the commutators

[e2,e1]=2e1,[e2,e3]=−2e3,[e1,e3]=e2 . [e_{2},e_{1}] = 2e_{1}, \qquad [e_{2},e_{3}] = -2e_{3}, \qquad [e_{1},e_{3}] = e_{2} \, .
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(i)

Find the adjoint representation.

(ii)

Calculate exp(ad⁡e1),exp(ad⁡e2),exp(ad⁡e3)\mathrm{exp}\left(\operatorname {ad}e_{1}\right), \mathrm{exp}\left(\operatorname {ad}e_{2}\right), \mathrm{exp}\left(\operatorname {ad}e_{3}\right).

The 2×22 \times 2 matrices

E=(0100),F=(0010),H=(1001) 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 for the Lie algebra sl(2,R)\mathfrak {sl}(2,\mathbb {R}) with the commutators [H,E]=2E[H,E] = 2E, [H,F]=−2F[H,F] = -2F, [E,F]=H[E,F] = H. Let k≥1k \geq 1. Do the matrices (mj=j(k−j+1)m_{j} = j(k-j+1))

ρk(H)=(k00…00k−20…000k−4…0⋮⋮⋮⋱⋮000…−k) \rho _{k}(H) = \begin{pmatrix} k & 0 & 0 & \ldots & 0 \\ 0 & k-2 & 0 & \ldots & 0 \\ 0 & 0 & k-4 & \ldots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \ldots & -k \end{pmatrix} ρk(F)=(000…⋅0100…⋅0010…⋅0⋮⋮⋮…⋅⋮000…10),ρk(E)=(0m10…000m2…0000…0⋮⋮⋮…mk000…0) \rho _{k}(F) = \begin{pmatrix} 0 & 0 & 0 & \ldots & \cdot & 0 \\ 1 & 0 & 0 & \ldots & \cdot & 0 \\ 0 & 1 & 0 & \ldots & \cdot & 0 \\ \vdots & \vdots & \vdots & \ldots & \cdot & \vdots \\ 0 & 0 & 0 & \ldots & 1 & 0 \end{pmatrix}, \qquad \rho _{k}(E) = \begin{pmatrix} 0 & m_{1} & 0 & \ldots & 0 \\ 0 & 0 & m_{2} & \ldots & 0 \\ 0 & 0 & 0 & \ldots & 0 \\ \vdots & \vdots & \vdots & \ldots & m_{k} \\ 0 & 0 & 0 & \ldots & 0 \end{pmatrix}

satisfy the same commutation relations?

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Let q>0q > 0. The algebra suq(2)\mathfrak {su}_{q}(2) is generated by X+,X−=(X+)∗X_{+}, X_{-} = (X_{+})^{*} and H=H∗H = H^{*} with deformed commutation relations

[H,X+]=2X+,[H,X−]=−2X−,[X+,X−]=qH−q−Hq−q−1 . [H,X_{+}] = 2X_{+}, \qquad [H,X_{-}] = -2X_{-}, \qquad [X_{+},X_{-}] = \frac{q^{H}-q^{-H}}{q-q^{-1}} \, .

Show by direct computation of the commutators that this algebra has the same matrix representation as the undeformed Lie algebra su(2)\mathfrak {su}(2), i.e.

X+↦(0100),X−↦(0010),H↦(100−1) . X_{+} \mapsto \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \qquad X_{-} \mapsto \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, \qquad H \mapsto \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \, .
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(i)

Consider the Lie algebra so(2,3)\mathfrak {so}(2,3). The ten generators L1,L2,L3,D1,D2,D3,F1,F2,F3L_1, L_2, L_3, D_1, D_2, D_3, F_1, F_2, F_3 and VV satisfy the commutation relations

\begin{aligned} [L_j, L_k] & = i \epsilon _{jk\ell } L_\ell \\[L_j, V] & = 0 \\[L_j, D_k] & = i \epsilon _{jk\ell } D_\ell \\[L_j, F_k] & = i \epsilon _{jk\ell } F_\ell \\[D_j, F_k] & = -i \delta _{jk} V \\[D_j, D_k] & = -i \epsilon _{jk\ell } L_\ell \\[F_j, F_k] & = -i \epsilon _{jk\ell } L_\ell \\[D_j, V] & = -i F_j \\[F_j, V] & = i D_j \end{aligned}

where j,k=1,2,3j,k = 1,2,3 and ϵ123=ϵ312=ϵ231=+1\epsilon_{123} = \epsilon_{312} = \epsilon_{231} = +1. Find the adjoint representation of so(2,3)\mathfrak {so}(2,3).

(ii)

Find the Killing form κ(x,y):=\tr[ad⁡(x)ad⁡(y)]\kappa (x,y) := \tr [\operatorname {ad}(x)\operatorname {ad}(y)]. Then show that the Lie algebra is semisimple.

(iii)

Find the two quadratic Casimir operators of so(2,3)\mathfrak {so}(2,3). The Casimir operators commute with all elements of the Lie algebra.

Let z1,z2z_1, z_2 be complex variables. Show that differential operators (vector fields)

J1=12(z1∂∂z2+z2∂∂z1),J2=12i(z1∂∂z2−z2∂∂z1), J_1 = \frac{1}{2}\left( z_1 \frac{\partial }{\partial z_2} + z_2 \frac{\partial }{\partial z_1} \right), \quad J_2 = \frac{1}{2}i\left( z_1 \frac{\partial }{\partial z_2} - z_2 \frac{\partial }{\partial z_1} \right), J3=12(z2∂∂z2−z1∂∂z1) J_3 = \frac{1}{2}\left( z_2 \frac{\partial }{\partial z_2} - z_1 \frac{\partial }{\partial z_1} \right)

form a basis of a Lie algebra under the commutator.

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Let j=0,1/2,1,3/2,…j = 0, 1/2, 1, 3/2, \ldots. Show that the differential operators

J+=2jz−z2∂∂z,J−=∂∂z,Jz=z∂∂z−j J_+ = 2jz - z^2 \frac{\partial }{\partial z}, \quad J_- = \frac{\partial }{\partial z}, \quad J_z = z\frac{\partial }{\partial z} - j

form a basis of a Lie algebra under the commutator.

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Let j=0,1/2,1,3/2,…j = 0, 1/2, 1, 3/2, \ldots. Show that the differential operators

J+=z(j−z∂∂z),J−=1z(j+z∂∂z),Jz=z∂∂z J_+ = z\left( j - z\frac{\partial }{\partial z} \right), \quad J_- = \frac{1}{z}\left( j + z\frac{\partial }{\partial z} \right), \quad J_z = z\frac{\partial }{\partial z}

form a basis of a Lie algebra under the commutator.

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Consider the differential operators

J+=−x(y∂∂y−x∂∂x+n),J−=−1x((y−1)∂∂y+x∂∂x+n) J_+ = -x\left( y\frac{\partial }{\partial y} - x\frac{\partial }{\partial x} + n \right), \quad J_- = -\frac{1}{x}\left( (y-1)\frac{\partial }{\partial y} + x\frac{\partial }{\partial x} + n \right) J3=x∂∂x J_3 = x\frac{\partial }{\partial x}

where nn is an integer.

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(i)

Calculate the commutators [J+,J−][J_+,J_-], [J+,J3][J_+,J_3], [J−,J3][J_-,J_3] and thus show that we have a basis of the Lie algebra sl(2,R)\mathfrak {sl}(2,\mathbb {R}).

(ii)

Find

exp⁡(γJ3)J+exp⁡(−γJ3),exp⁡(γJ3)J−exp⁡(−γJ3) \exp (\gamma J_3) J_+ \exp (-\gamma J_3), \quad \exp (\gamma J_3) J_- \exp (-\gamma J_3) exp⁡(αJ+)J−exp⁡(−αJ+),exp⁡(αJ+)J3exp⁡(−αJ+) \exp (\alpha J_+) J_- \exp (-\alpha J_+), \quad \exp (\alpha J_+) J_3 \exp (-\alpha J_+) exp⁡(βJ−)J+exp⁡(−βJ−),exp⁡(βJ−)J3exp⁡(−βJ−). \exp (\beta J_-) J_+ \exp (-\beta J_-), \quad \exp (\beta J_-) J_3 \exp (-\beta J_-) .
(iii)

Use the result form (ii) to calculate

eαJ+eβJ−(λJ3−m)e−βJ−e−αJ+. e^{\alpha J_+} e^{\beta J_-} (\lambda J_3 - m) e^{-\beta J_-} e^{-\alpha J_+} .

Consider the vector fields

V1=ddz,V2=zddz,V3=z2ddz. V_1 = \frac{d}{dz}, \quad V_2 = z\frac{d}{dz}, \quad V_3 = z^2\frac{d}{dz} .

They satisfy the commutation relations

[V2,V1]=−V1,[V2,V3]=V3,[V3,V1]=−2V2. [V_2,V_1] = -V_1, \quad [V_2,V_3] = V_3, \quad [V_3,V_1] = -2V_2 .

Thus we have a realization of the Lie algebra sl(2,C)\mathfrak {sl}(2,\mathbb {C}). Extend the vector fields to the operators

W1=ddz+f1(z),W2=zddz+f2(z),W3=z2ddz+f3(z). W_1 = \frac{d}{dz} + f_1(z), \quad W_2 = z\frac{d}{dz} + f_2(z), \quad W_3 = z^2\frac{d}{dz} + f_3(z) .

Find the condition on the analytic functions f1,f2,f3f_1, f_2, f_3 such that W1,W2,W3W_1, W_2, W_3 also satisfy the commutation relations

[W2,W1]=−W1,[W2,W3]=W3,[W3,W1]=−2W2. [W_2,W_1] = -W_1, \quad [W_2,W_3] = W_3, \quad [W_3,W_1] = -2W_2 .
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(i)

Let b†,bb^\dagger , b be Bose creation and annihilation operators with the commutation relations

[b,b†]=I,[b,b]=0,[b†,b†]=0 [b,b^\dagger ] = I, \quad [b,b] = 0, \quad [b^\dagger ,b^\dagger ] = 0

where II is the identity operator. Consider the operators

b†b,b†,b,I. b^\dagger b, \quad b^\dagger , \quad b, \quad I .

Find the commutators.

(ii)

Show that there is a non-hermitian faithful representation by 3×33 \times 3 matrices

b†b→M22=(000010000) b^\dagger b \to M_{22} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix} b†→M23=(000001000),b→M12=(010000000) b^\dagger \to M_{23} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}, \quad b \to M_{12} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}

and

I→M13=(001000000). I \to M_{13} = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} .

Note that the identity operator II is not mapped into the 3×33 \times 3 identity matrix.

Consider the Lie algebra sl(2,R)\mathfrak {sl}(2,\mathbb {R}) with the basis H,X,YH, X, Y and the commutators

[H,X]=2X,[H,Y]=−2Y,[X,Y]=H. [H,X] = 2X, \quad [H,Y] = -2Y, \quad [X,Y] = H .

Give faithful representations of sl(2,R)\mathfrak {sl}(2,\mathbb {R}) using (i) 2×22\times 2 matrices, (ii) 3×33\times 3 matrices, (iii) vector fields, (iv) second order differential operators, (v) Bose operators, (vi) Fermi operators.

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(i)
(ii)
(iii)
(iv)
(v)
(vi)

Let b1†,b2†b_1^\dagger , b_2^\dagger be Bose creation operator and let II be the identity operator. The semi-simple Lie algebra su(1,1)\mathfrak {su}(1,1) is generated by

K+:=b1†b2†,K−:=b1b2,K0:=12(b1†b1+b2†b2+I) K_+ := b_1^\dagger b_2^\dagger , \quad K_- := b_1b_2, \quad K_0 := \frac{1}{2}(b_1^\dagger b_1+b_2^\dagger b_2+I)

with the commutation relations

[K0,K+]=K+,[K0,K−]=−K−,[K−,K+]=2K0. [K_0,K_+] = K_+, \quad [K_0,K_-] = -K_-, \quad [K_-,K_+] = 2K_0 .

We use the ordering K+,K−,K0K_+, K_-, K_0 for the basis. Write a computer algebra program that finds the ajoint representation for the Lie algebra.

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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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Let b1,b2,b3b_1, b_2, b_3 be Bose annihilation operators. Show that

H1=b1†b1−b2†b2,H2=b2†b2−b3†b3 H_1 = b_1^\dagger b_1 - b_2^\dagger b_2, \quad H_2 = b_2^\dagger b_2 - b_3^\dagger b_3 E12=b1†b2,E23=b2†b3,E13=b1†b3 E_{12} = b_1^\dagger b_2, \quad E_{23} = b_2^\dagger b_3, \quad E_{13} = b_1^\dagger b_3 E21=b2†b1,E32=b3†b2,E31=b3†b1 E_{21} = b_2^\dagger b_1, \quad E_{32} = b_3^\dagger b_2, \quad E_{31} = b_3^\dagger b_1

are a representation of the Lie algebra su(3)\mathfrak {su}(3). Write a SymbolicC++ program that implements this representation.

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Let A,BA, B be 2×22 \times 2 matrices over C\mathbb {C}. We define

A⋆B:=(a1100a120b11b1200b21b220a2100a22) . A \star B := \begin{pmatrix} a_{11} & 0 & 0 & a_{12} \\ 0 & b_{11} & b_{12} & 0 \\ 0 & b_{21} & b_{22} & 0 \\ a_{21} & 0 & 0 & a_{22} \end{pmatrix} \, .

Consider the Lie algebra sl(2,R)\mathfrak {sl}(2,\mathbb {R}) with the basis

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} \, .

Find h⋆hh \star h, e⋆ee \star e, f⋆ff \star f. Calculate the commutators

[e⋆e,f⋆f],[h⋆h,e⋆e],[h⋆h,f⋆f] . [e\star e, f\star f], \qquad [h\star h, e\star e], \qquad [h\star h, f\star f] \, .

Discuss.

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Let s=0,1/2,1,3/2,…s = 0, 1/2, 1, 3/2, \ldots. The spin ss generators of the Lie algebra sl(2,C)\mathfrak {sl}(2,\mathbb {C}) at site kk are given by

S+,k=zk2∂∂zk−2szk,S−,k=−∂∂zk,S3,k=zk∂∂zk−s . S_{+,k} = z_{k}^{2}\frac{\partial }{\partial z_{k}} - 2sz_{k}, \qquad S_{-,k} = -\frac{\partial }{\partial z_{k}}, \qquad S_{3,k} = z_{k}\frac{\partial }{\partial z_{k}} - s \, .

Find the commutators.

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References

[KiGv88] Kirillov A.A., Gvishiani A.D. Теоремы и задачи функционального анализа. 2-е изд., Наука, 1988.
[Кос04] Кострикин А.И. Основные структуры алгебры. ФИЗМАТЛИТ, 2004.
[STH12] Steeb W., Tanski I., Hardy Y. Problems and Solutions for Groups, Lie Groups, Lie Algebras with Applications. World Scientific Publishing, 2012.