李代数和表示理论导学-Definitions and first examples

Definitions and first examples

1. Let L L L be the real vector space R 3 R^3 R3. Define [ x y ] = x × y [xy]=x\times y [xy]=x×y (cross product of vectors) for x x x, y y y ∈ L \in L ∈L, and verify that L L L is a Lie algebra. Write down the structure constants relative to the usual basis of R 3 R^3 R3.Verify L L L is a Lie algebra
   ( L 1 ) (L1) (L1) Cross product is bilinear
   ( L 2 ) (L2) (L2) [ x x ] = x × x = 0 [xx]=x\times x=0 [xx]=x×x=0
   ( L 3 ) (L3) (L3) [ x [ y z ] ] = [ x , y × z ] = x × ( y × z ) = ( x , z ) y − ( x , y ) z [x[yz]]=[x,y\times z]=x\times (y\times z)=(x,z)y-(x,y)z [x[yz]]=[x,y×z]=x×(y×z)=(x,z)y−(x,y)z
   [ y [ z x ] ] = y × ( z × x ) = ( y , x ) z − ( y , z ) x [y[zx]]=y\times (z\times x)=(y,x)z-(y,z)x [y[zx]]=y×(z×x)=(y,x)z−(y,z)x
   [ z [ x y ] ] = z × ( x × y ) = ( z , y ) x − ( z , x ) y [z[xy]]=z\times(x\times y)=(z,y)x-(z,x)y [z[xy]]=z×(x×y)=(z,y)x−(z,x)y
   [ x [ y z ] ] + [ y [ z x ] ] + [ z [ x y ] ] = ( x , z ) y − ( x , y ) z + ( y , x ) z − ( y , z ) x + ( z , y ) x − ( z , x ) y = [ ( x , z ) − ( z , x ) ] y + [ ( y , x ) − ( x , y ) ] z + [ ( z , y ) − ( y , z ) ] x = 0 [x[yz]]+[y[zx]]+[z[xy]]=(x,z)y-(x,y)z+(y,x)z-(y,z)x+(z,y)x-(z,x)y=[(x,z)-(z,x)]y+[(y,x)-(x,y)]z+[(z,y)-(y,z)]x=0 [x[yz]]+[y[zx]]+[z[xy]]=(x,z)y−(x,y)z+(y,x)z−(y,z)x+(z,y)x−(z,x)y=[(x,z)−(z,x)]y+[(y,x)−(x,y)]z+[(z,y)−(y,z)]x=0
   So L L L is a Lie algebra and the multiplication table as follows
   

2. Verify that the following equations and those implied by ( L 1 ) ( L 2 ) (L1) (L2) (L1)(L2) define a Lie algebra structure on a three dimensional vector space with basis ( x , y , z ) : [ x y ] = z , [ x z ] = y , [ y z ] = 0. (x,y,z):[xy]=z,[xz]=y,[yz]=0. (x,y,z):[xy]=z,[xz]=y,[yz]=0.
   We only to verify Jacobi identity
   [ x [ y z ] ] + [ y [ z x ] ] + [ z [ x y ] ] = [ x 0 ] + [ y , − y ] + [ z z ] = 0 [x[yz]]+[y[zx]]+[z[xy]]=[x0]+[y,-y]+[zz]=0 [x[yz]]+[y[zx]]+[z[xy]]=[x0]+[y,−y]+[zz]=0

3. Let x = ( 0 1 0 0 ) x=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} x=(00​10​), h = ( 1 0 0 − 1 ) h=\begin{pmatrix} 1&0 \\ 0&-1 \end{pmatrix} h=(10​0−1​), y = ( 0 0 1 0 ) y=\begin{pmatrix} 0&0 \\ 1&0 \end{pmatrix} y=(01​00​) be an ordered basis for s l ( 2 , F ) {sl}(2,F) sl(2,F). Computer the matrices of a d x , a d h , a d y ad x, ad h, ad y adx,adh,ady relative to this basis.
   ∵ \because ∵ [ x , h ] = x h − h x = − 2 x [x,h]=xh-hx=-2x [x,h]=xh−hx=−2x
   [ x , y ] = x y − y x = h [x,y]=xy-yx=h [x,y]=xy−yx=h
   [ y , h ] = y h − h y = 2 y [y,h]=yh-hy=2y [y,h]=yh−hy=2y
   ∴ \therefore ∴ a d x ( x , h , y ) = ( a d x ( x ) , a d x ( h ) , a d x ( y ) ) = ( [ x x ] , [ x h ] , [ x y ] ) = ( 0 , − 2 x , h ) = ( x , h , y ) ( 0 − 2 0 0 0 1 0 0 0 ) adx(x,h,y)=(adx(x),adx(h),adx(y))=([xx],[xh],[xy])=(0,-2x,h)=(x,h,y)\begin{pmatrix} 0&-2&0\\0&0&1\\0&0&0\end{pmatrix} adx(x,h,y)=(adx(x),adx(h),adx(y))=([xx],[xh],[xy])=(0,−2x,h)=(x,h,y)⎝⎛​000​−200​010​⎠⎞​
   a d y ( x , h , y ) = ( a d y ( x ) , a d y ( h ) , a d y ( y ) ) = ( [ y x ] , [ y h ] , [ y y ] ) = ( − h , 2 y , 0 ) = ( x , h , y ) ( 0 0 0 − 1 0 0 0 2 0 ) ady(x,h,y)=(ady(x),ady(h),ady(y))=([yx],[yh],[yy])=(-h,2y,0)=(x,h,y)\begin{pmatrix} 0&0&0 \\ -1&0&0 \\ 0&2&0 \end{pmatrix} ady(x,h,y)=(ady(x),ady(h),ady(y))=([yx],[yh],[yy])=(−h,2y,0)=(x,h,y)⎝⎛​0−10​002​000​⎠⎞​
   a d h ( x , h , y ) = ( a d h ( x ) , a d h ( h ) , a d h ( y ) ) = ( [ h x ] , [ h h ] , [ h y ] ) = ( 2 x , 0 , − 2 y ) = ( x , h , y ) ( 2 0 0 0 0 0 0 0 − 2 ) adh(x,h,y)=(adh(x),adh(h),adh(y))=([hx],[hh],[hy])=(2x,0,-2y)=(x,h,y)\begin{pmatrix}2&0&0\\0&0&0\\0&0&-2\end{pmatrix} adh(x,h,y)=(adh(x),adh(h),adh(y))=([hx],[hh],[hy])=(2x,0,−2y)=(x,h,y)⎝⎛​200​000​00−2​⎠⎞​

4. Find a linear Lie algebra isomorphic to the nonabelian two dimensional algebra constructed in (1.4).
   Define a map φ : L → a d L \varphi : L\rightarrow adL φ:L→adL by x ↦ a d x , y ↦ a d y , [ x y ] ↦ a d [ x y ] x\mapsto adx, y\mapsto ady, [xy]\mapsto ad[xy] x↦adx,y↦ady,[xy]↦ad[xy] give an operation a d [ x y ] = [ a d x a d y ] ad[xy]=[adx ady] ad[xy]=[adxady]in a d L adL adL, then a d L ≅ L adL\cong L adL≅L because of [ a d x a d y ] = a d [ x y ] = a d x [adxady]=ad[xy]=adx [adxady]=ad[xy]=adx

5. Verify the assertions made in (1.2) about t ( n , F ) , δ ( n , F ) , n ( n , F ) t(n,F),\delta(n,F),n(n,F) t(n,F),δ(n,F),n(n,F), and compute the dimension of each algebra, by exhibiting bases.
   (i) t ( n , F ) ， n ( n , F ) t(n,F)，n(n,F) t(n,F)，n(n,F)and δ ( n , F ) \delta(n,F) δ(n,F) are closed under the bracket.
   If A ∈ t ( n , F ) A\in t(n,F) A∈t(n,F), B ∈ t ( n , F ) B\in t(n,F) B∈t(n,F), then [ A , B ] = A B − B A [A,B]=AB-BA [A,B]=AB−BA also lies in t ( n , F ) t(n,F) t(n,F)
   If A ∈ n ( n , F ) A\in n(n,F) A∈n(n,F), B ∈ n ( n , F ) B\in n(n,F) B∈n(n,F), then [ A , B ] = A B − B A [A,B]=AB-BA [A,B]=AB−BA also lies in n ( n , F ) n(n,F) n(n,F)
   If A ∈ δ ( n , F ) A\in \delta(n,F) A∈δ(n,F), B ∈ δ ( n , F ) B\in \delta(n,F) B∈δ(n,F), then [ A , B ] = A B − B A [A,B]=AB-BA [A,B]=AB−BA also lies in δ ( n , F ) \delta(n,F) δ(n,F)
   (ii) A ∈ δ ( n , F ) A \in \delta(n,F) A∈δ(n,F), B ∈ n ( n , F ) B \in n(n,F) B∈n(n,F), [ A B ] = A B − B A [AB]=AB-BA [AB]=AB−BA is also a strictly upper matrix
   (iii) If A i ∈ t ( n , F ) A_{i} \in t(n,F) Ai​∈t(n,F), i=1,2 A i = B i + C i A_{i}=B_{i}+C_{i} Ai​=Bi​+Ci​, where B i ∈ δ ( n , F ) B_{i} \in \delta(n,F) Bi​∈δ(n,F), C i ∈ n ( n , F ) C_{i}\in n(n,F) Ci​∈n(n,F) [ A 1 , A 2 ] = [ B 1 + C 1 , B 2 + C 2 ] = [ B 1 B 2 ] + [ B 1 C 2 ] + [ C 1 B 2 ] + [ C 1 C 2 ] = B 1 B 2 − B 2 B 1 + B 1 C 2 − C 2 B 1 + C 1 B 2 − B 2 C 1 + C 1 C 2 − C 2 C 1 ∈ n ( n , F ) [A_{1},A_{2}]=[B_{1}+C_{1},B_{2}+C_{2}]=[B_{1}B_{2}]+[B_{1}C_{2}]+[C_{1}B_{2}]+[C_{1}C_{2}]=B_{1}B_{2}-B_{2}B_{1}+B_{1}C_{2}-C_{2}B_{1}+C_{1}B_{2}-B_{2}C_{1}+C_{1}C_{2}-C_{2}C_{1} \in n(n,F) [A1​,A2​]=[B1​+C1​,B2​+C2​]=[B1​B2​]+[B1​C2​]+[C1​B2​]+[C1​C2​]=B1​B2​−B2​B1​+B1​C2​−C2​B1​+C1​B2​−B2​C1​+C1​C2​−C2​C1​∈n(n,F)
   for δ ( n , F ) \delta(n,F) δ(n,F), we take all the diagonal matrices e i i e_{ii} eii​, 1 ⩽ i ⩽ n 1 \leqslant i \leqslant n 1⩽i⩽n n in number
   for n ( n , F ) n(n,F) n(n,F), we take all the matrices e i j e_{ij} eij​, $ 1 \leqslant i \leq j \leqslant n$, n ( n − 1 ) 2 \frac{n(n-1)}{2} 2n(n−1)​ in number\
   and t ( n , F ) t(n,F) t(n,F) is the direct sum of δ ( n , F ) \delta(n,F) δ(n,F) and n ( n , F ) n(n,F) n(n,F) as space, so d i m ( t ( n , F ) ) = d i m ( δ ( n , F ) ) + d i m ( n ( n , F ) ) = n + n ( n − 1 ) 2 dim(t(n,F))=dim(\delta(n,F))+dim(n(n,F))=n+\frac{n(n-1)}{2} dim(t(n,F))=dim(δ(n,F))+dim(n(n,F))=n+2n(n−1)​

6. Let x ∈ g l ( n , F ) x\in gl(n,F) x∈gl(n,F) have n n n distinct eigenvalues a 1 , … a n a_{1},\dots a_{n} a1​,…an​ in F F F. Prove that the eigenvalues of a d x ad x adx are precisely the n 2 n^2 n2 scalars a i − a j ( 1 ⩽ i , j ⩽ n ) a_{i}-a_{j} (1 \leqslant i,j\leqslant n) ai​−aj​(1⩽i,j⩽n), which of course need not be distinct.
   proof
   We can find a suitable basis such that the matrix of x x x on this basis is diag ( a 1 , … a n ) (a_{1},\dots a_{n}) (a1​,…an​), and let e i j , 1 ⩽ i , j ⩽ n e_{ij}, 1 \leqslant i,j \leqslant n eij​,1⩽i,j⩽n be the standard basis of g l ( n , F ) gl(n,F) gl(n,F), and a d x ( e i j ) = [ x , e i j ] = x e i j − e i j x = ( a 1 ⋱ a n ) ( 1 ) − ( 1 ) ( a 1 ⋱ a n ) = ( ⋱ a i ⋱ ) − ( ⋱ a j ⋱ ) = ( ⋱ a i − a j ⋱ ) = ( a i − a j ) e i j adx(e_{ij})=[x,e_{ij}]=xe_{ij}-e_{ij}x=\begin{pmatrix} a_{1}& & \\ &\ddots& \\ & & a_{n} \end{pmatrix} \begin{pmatrix} & & &\\& & 1\\& & & \end{pmatrix}-\begin{pmatrix} & & &\\& & 1\\& & & \end{pmatrix}\begin{pmatrix} a_{1}& & \\ &\ddots& \\ & & a_{n} \end{pmatrix}=\begin{pmatrix} \ddots & &\\& & a_{i}\\& & \ddots \end{pmatrix} -\begin{pmatrix} \ddots & &\\& & a_{j}\\& & \ddots \end{pmatrix}=\begin{pmatrix} \ddots & &\\& & a_{i}-a_{j}\\& & \ddots \end{pmatrix}=(a_{i}-a_{j})e_{ij} adx(eij​)=[x,eij​]=xeij​−eij​x=⎝⎛​a1​​⋱​an​​⎠⎞​⎝⎛​​​1​​⎠⎞​−⎝⎛​​​1​​⎠⎞​⎝⎛​a1​​⋱​an​​⎠⎞​=⎝⎛​⋱​​ai​⋱​⎠⎞​−⎝⎛​⋱​​aj​⋱​⎠⎞​=⎝⎛​⋱​​ai​−aj​⋱​⎠⎞​=(ai​−aj​)eij​

7. Let s ( n , F ) s(n,F) s(n,F) denote the scalar matrices(=scalar multiples of the identity) in g l ( n , F ) gl(n,F) gl(n,F). If char F is 0 0 0 or else a prime not diving n n n, prove that g l ( n , F ) = s l ( n , F ) + s ( n , F ) gl(n,F)=sl(n,F)+s(n,F) gl(n,F)=sl(n,F)+s(n,F) (direct sum of vector spaces), with [ s ( n , F ) , g l ( n , F ) ] = 0 [s(n,F),gl(n,F)]=0 [s(n,F),gl(n,F)]=0.
   proof
   A ∈ s l ( n , F ) , B = a E ∈ s ( n , F ) , [ A , B ] = A B − B A = a A − a A = 0 , a ∈ F A\in sl(n,F), B=aE\in s(n,F), [A,B]=AB-BA=aA-aA=0, a\in F A∈sl(n,F),B=aE∈s(n,F),[A,B]=AB−BA=aA−aA=0,a∈F, what`s more d i m ( s l ( n , F ) ) + d i m ( s ( n , F ) ) = n 2 − 1 + 1 = n 2 = d i m ( g l ( n , F ) ) dim(sl(n,F))+dim(s(n,F))=n^2-1+1=n^2=dim(gl(n,F)) dim(sl(n,F))+dim(s(n,F))=n2−1+1=n2=dim(gl(n,F)), so g l ( n , F ) = s l ( n , F ) + s ( n , F ) gl(n,F)=sl(n,F)+s(n,F) gl(n,F)=sl(n,F)+s(n,F)

8. Verify the stated deimension of D l D_{l} Dl​
   Since s x = − x t s sx=-x^ts sx=−xts, where x = ( m n p q ) x=\begin{pmatrix} m &n\\p&q \end{pmatrix} x=(mp​nq​), s = ( 0 I l I l 0 ) s=\begin{pmatrix} 0 &I_{l}\\ I_{l} &0 \end{pmatrix} s=(0Il​​Il​0​), we have ( 0 I l I l 0 ) \begin{pmatrix} 0 &I_{l}\\ I_{l} &0 \end{pmatrix} (0Il​​Il​0​) ( m n p q ) \begin{pmatrix} m &n\\ p &q \end{pmatrix} (mp​nq​)= − ( m t p t n t q t ) -\begin{pmatrix} m^t &p^t\\ n^t &q^t\end{pmatrix} −(mtnt​ptqt​) ( 0 I l I l 0 ) \begin{pmatrix} 0 &I_{l}\\I_{l} &0 \end{pmatrix} (0Il​​Il​0​), which induces p = − p t , q = − m t , n = − n t p=-p^t, q=-m^t, n=-n^t p=−pt,q=−mt,n=−nt, since m = − q t m=-q^t m=−qt, we take e i i − e l + i   l + i e_{ii}-e_{l+i\ l+i} eii​−el+i l+i​, l l l in number; e i   j − e i + l   j + l t e_{i\ j}-e_{i+l\ j+l}^t ei j​−ei+l j+lt​, l 2 − l l^2-l l2−l in number; since n = − n t n=-n^t n=−nt , we take e i   j + l − e j   i + l , 1 ⩽ i ≠ j ⩽ l , 1 / 2 l ( l − 1 ) e_{i\ j+l}-e_{j\ i+l}, 1\leqslant i\neq j\leqslant l, 1/2l(l-1) ei j+l​−ej i+l​,1⩽i​=j⩽l,1/2l(l−1) in number, similar p = − p t p=-p^t p=−pt, 1 / 2 l ( l − 1 ) 1/2l(l-1) 1/2l(l−1) in number, so the total number of basis elements is l + l 2 − l + 1 / 2 l ( l − 1 ) + 1 / 2 l ( l − 1 ) = 2 l 2 − l l+l^2-l+1/2l(l-1)+1/2l(l-1)=2l^2-l l+l2−l+1/2l(l−1)+1/2l(l−1)=2l2−l.

9. When char F = 0 F=0 F=0, show that each classical algebra L = A l , B l , C l , o r D l L=A_{l}, B_{l}, C_{l}, or D_{l} L=Al​,Bl​,Cl​,orDl​ is equal to [ L L ] [LL] [LL].
   proof
   Let x 1 , … x l x_{1},\dots x_{l} x1​,…xl​ be the basis of L ( A l , B l , C l , o r D l ) L(A_{l}, B_{l}, C_{l}, or D_{l}) L(Al​,Bl​,Cl​,orDl​), [ L L ] = { [ x i x j ] = ∑ k = 1 l a i j x k , x i , x j ∈ L } [LL]=\{[x_{i}x_{j}]=\sum\limits_{k=1}^{l}a_{ij}x_{k}, x_{i},x_{j}\in L\} [LL]={[xi​xj​]=k=1∑l​aij​xk​,xi​,xj​∈L}, dim [ L L ] = l [LL]=l [LL]=l

10. For small values of ℓ \ell ℓ, isomorphisms occur among certain of the classical algebra. Show that $ A_{l}, B_{l}, C_{l}$ are isomorphic, while D 1 D_{1} D1​ is the one dimensional Lie algebra. Show that B 2 B_{2} B2​ is isomorphic to C 2 , D 3 C_{2}, D_{3} C2​,D3​ to A 3 A_{3} A3​, What can you say about D 2 D_{2} D2​?
    Since dim A l = ( l + 1 ) 2 − 1 A_{l}=(l+1)^2-1 Al​=(l+1)2−1, dim B l = 2 l 2 + l B_{l}=2l^2+l Bl​=2l2+l, dim C l = 2 l 2 − l C_{l}=2l^2-l Cl​=2l2−l, dim D l = 2 l 2 − l D_{l}=2l^2-l Dl​=2l2−l, dim A 1 A_{1} A1​=dim B 1 B_{1} B1​=dim C 1 = 3 C_{1}=3 C1​=3, dim D 1 = 1 D_{1}=1 D1​=1, dim B 2 B_{2} B2​=dim C 2 = 10 C_{2}=10 C2​=10, dim D 3 D_{3} D3​=dim A 3 = 15 A_{3}=15 A3​=15, while Dim D 2 = 6 D_{2}=6 D2​=6

11. Verify that the commutator of two derivations of an F F F-algebra is again a derivation, whereas the ordinary product need not be.
    proof
    If δ \delta δ, δ ′ \mathop{{\delta}'} δ′ ∈ D e r U \in Der\mathscr{U} ∈DerU, a , b ∈ U a,b\in \mathscr{U} a,b∈U,
      [ δ , δ ′ ] ( a b ) \ [\delta,\delta'](ab)  [δ,δ′](ab)= ( δ δ ′ − δ ′ δ ) ( a b ) = δ δ ′ ( a b ) − δ ′ δ ( a b ) = a δ δ ′ ( b ) + δ δ ′ ( a ) b − a δ ′ δ ( b ) − δ ′ δ ( a ) b = a ( δ δ ′ ( b ) − δ ′ δ ( b ) ) + ( δ δ ′ ( a ) − δ ′ δ ( a ) ) b = a [ δ δ ′ ] ( b ) + [ δ δ ′ ] ( a ) b (\delta\mathop{{\delta}'}-\mathop{{\delta}'}\delta)(ab)= \delta\mathop{{\delta}'}(ab)-\mathop{{\delta}'}\delta(ab)= a\delta\mathop{{\delta}'}(b)+\delta\mathop{{\delta}'}(a)b-a\mathop{{\delta}'}\delta(b)-\mathop{{\delta}'}\delta(a)b= a(\delta\delta'(b)-\delta'\delta(b))+(\delta\delta'(a)-\delta'\delta(a))b=a[\delta\delta'](b)+[\delta\delta'](a)b (δδ′−δ′δ)(ab)=δδ′(ab)−δ′δ(ab)=aδδ′(b)+δδ′(a)b−aδ′δ(b)−δ′δ(a)b=a(δδ′(b)−δ′δ(b))+(δδ′(a)−δ′δ(a))b=a[δδ′](b)+[δδ′](a)b
    while δ δ ′ ( a b ) = δ ( a δ ′ ( b ) + δ ′ ( a ) b ) = δ ( a δ ′ ( b ) ) + δ ( δ ′ ( a ) b ) = a δ δ ′ ( b ) + δ ( a ) δ ′ ( b ) + δ ′ ( a ) δ ( b ) + δ δ ′ ( a ) b ≠ a δ δ ′ ( b ) + δ δ ′ ( a ) b \delta\mathop{{\delta}'}(ab)=\delta(a\mathop{{\delta}'}(b)+\mathop{{\delta}'}(a)b)= \delta(a\mathop{{\delta}'}(b))+\delta(\mathop{{\delta}'}(a)b)= a\delta\mathop{{\delta}'}(b)+\delta(a)\mathop{{\delta}'}(b)+\mathop{{\delta}'}(a)\delta(b)+\delta\mathop{{\delta}'}(a)b\neq a\delta\mathop{{\delta}'}(b)+\delta\mathop{{\delta}'}(a)b δδ′(ab)=δ(aδ′(b)+δ′(a)b)=δ(aδ′(b))+δ(δ′(a)b)=aδδ′(b)+δ(a)δ′(b)+δ′(a)δ(b)+δδ′(a)b​=aδδ′(b)+δδ′(a)b

12. Let L L L be a Lie algebra and let x ∈ L x\in L x∈L. Prove that the subspace of L L L spanned by the eigenvectors of a d x ad x adx is a subalgebra.
    proof
    Let v 1 , … v l v_{1},\dots v_{l} v1​,…vl​ be the eigenvectors of a d x ad x adx belong to eigenvalues a i , … a l a_{i},\dots a_{l} ai​,…al​, i.e. a d x ( v i ) = a i v i adx(v_{i})=a_{i}v_{i} adx(vi​)=ai​vi​, 1 ⩽ i ⩽ l 1\leqslant i \leqslant l 1⩽i⩽l, a d x ( [ v i v j ] ) = [ x , [ v i v j ] ] = [ v i [ x v j ] ] − [ v j [ x v i ] ] = [ v i , a j v j ] − [ v j , a i v i ] = a j [ v i v j ] + a i [ v i v j ] = ( a i + a j ) [ v i v j ] adx([v_{i} v_{j}])=[x,[v_{i}v_{j}]]=[v_{i}[xv_{j}]]-[v_{j}[xv_{i}]]=[v_{i},a_{j}v_{j}]-[v_{j},a_{i}v_{i}]=a_{j}[v_{i}v_{j}]+a_{i}[v_{i}v_{j}]=(a_{i}+a_{j})[v_{i}v_{j}] adx([vi​vj​])=[x,[vi​vj​]]=[vi​[xvj​]]−[vj​[xvi​]]=[vi​,aj​vj​]−[vj​,ai​vi​]=aj​[vi​vj​]+ai​[vi​vj​]=(ai​+aj​)[vi​vj​]

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来源： https://blog.51cto.com/u_15255081/2870566