A symmetric bilinear form corresponds to a quadratic formQ(v)=⟨v,v⟩=∑xixj⟨ei,ej⟩.We do base change so that Q(x,x)=∑(−1)kxk2. If n is even, we write them ask=1∑mzkzk+mk=1∑mzkzk+m+z2m+12The matrix is(ImIm)and ⎝⎛ImIm1⎠⎞
Let V×V→C be a skew-symmetric nondegenerate bilinear pairing. Then dimV must be even. There exists a basis with matrix(−ImIm)
sp(2m) consists of the matrix(ACB−At)such that B,C are symmetric.
so(2m) consists of the matrix(ACB−At)such that B,C are skew-symmetric.
so(2m+1) consists of the matrix⎝⎛AC−FtB−At−EtEF0⎠⎞such that B,C are skew-symmetric.
Let Hi=Eii−Em+i,m+i, and h the subspace generated by Hi (the Cartan subalgebra).
For any g→gl(V) and any β∈h∗,Vβ={x∈V∣Hx=β(H)x,∀H∈h}We haveV=β∈h⨁Vβ.
Denote Li the dual basis of Hi in h∗. These constructions apply to both so and sp.
For 1≤i,j≤m, let Xij=Eij−Em+j,m+i,i=jYij=Ei,m+j+Ej,m+i,i=jZij=Em+i,j+Em+j,i,i=jUi=Ei,m+iVi=Em+i,i[H,Xij]=(Li−Lj)(H)Xij[H,Yij]=(Li+Lj)(H)Yij[H,Zij]=−(Li+Lj)(H)Zij[H,Ui]=2Li(H)Ui[H,Vi]=−2Li(H)ViRecallSp(2m)={(ACBD)t(−ImIm)+(−ImIm)(ACBD)=0.}
For so, we letXij=Eij−Em+j,m+iYij=Ei,m+j−Ej,m+iZij=Em+1,j−Em+j,iFor so(2n+1), we letUi=Ei,2m+1−E2m+1,m+iVi=Em+i,2m+1−E2m+1,i