User:Lethe/orthogonal

The orthogonal group SO(g) is defined as those members O of Aut(V,g) such that
 * $$O^*g=g$$

or more concretely,
 * $$g(Ov,Ow)=g(v,w).$$

Assuming a time dependence of O, taking the derivative, and setting O(0) to 1, one finds
 * $$g(Xv,w)+g(v,Xw)=0.$$

Choosing a basis {en}n and switching to coordinate notation, we have
 * $$g_{cb}X^c{}_av^aw^b+g_{ac}v^aX^c{}_bw^b=0.$$

Since v and w are arbitrary, we conclude that
 * $$g_{cb}X^c{}_a=-g_{ac}X^c{}_b$$

or in more concise notation,
 * $$X_{ab}=-X_{ba}.$$

Commutation relations
First we must choose a basis for so(g). Denote by Eji the matrix that returns ej on ei and 0 on ek for k≠i, the matrix with a 1 in the j-th row and i-th column and 0 everywhere else. Thus the components of Eij are given by
 * $$(E_j{}^i{})^a{}_b=\delta^a_j\delta^i_b$$

and by construction we have
 * $$E_j{}^ie_k=\delta^i_k e_j.$$

Nota bene: In this notation, the labels which index the matrices have the first index (which indicates the row) lowered and the second index (which indicates the column) raised, while the indices which indicate a particular component of a matrix have the first index (which indicates the row) raised and the second index (which indicates the column) lowered. Thus (Eab)cd indicates the c-th row of the d-th column of the matrix which has a 1 in the a-th row and b-th column. Deviations from this notation will indicate the raising and lowering of indices by the metric.

Observe that
 * $$([E_i{}^j,E_k{}^\ell])^a{}_c= (E_i{}^j)^a{}_b(E_k{}^\ell)^b{}_c - (E_k{}^\ell)^a{}_b (E_i{}^j )^b{}_c = \delta^j_b\delta^a_i\delta^\ell_c\delta^b_k-\delta^\ell_b\delta^a_k\delta^j_c\delta^b_i = \delta^j_k\delta^a_i\delta^\ell_c-\delta^\ell_i\delta^a_k\delta^j_c=\delta^j_k(E_i{}^\ell)^a{}_c-\delta^\ell_i(E_k{}^j)^a{}_c,$$

or in short
 * $$[E_i{}^j,E_k{}^\ell] = \delta^j_kE_i{}^\ell-\delta^\ell_iE_k{}^j.$$

Members of so(g) are anti-Hermitian; they satisfy X†= –X. For any operator, X – X† is anti-Hermitian. Thus Eji – (Eji)† will always be anti-Hermitian. In fact, it is easily shown that these matrices span the n×n anti-Hermitian operators. First let's see what the Hermitian adjoint looks like in arbitrary inner product.

One defines the Hermitian adjoint by
 * $$g(A^\dagger v,w)=g(v,Aw).$$

Once again working in coordinates,
 * $$g_{ac}(A^\dagger)^a{}_bv^bw^c = g_{ba}v^bA^a{}_cw^c$$

or after canceling the vectors,
 * $$g_{ac}(A^\dagger)^a{}_b=g_{ba}A^a{}_c$$

then after inverting g, and re-indexing,
 * $$(A^\dagger)^a{}_b=g_{bc}g^{ad}A^c{}_d=A_b{}^a.$$

We note also that
 * $$[(E_i{}^j)^\dagger,(E_k{}^\ell)^\dagger] = - [E_i{}^j,E_k{}^\ell]^\dagger = \delta^\ell_i(E_k{}^j)^\dagger  - \delta^j_k(E_i{}^\ell)^\dagger$$

and
 * $$asb$$

It's also convenient to have these relations with lowered indices:
 * $$[E_{ij},E_{k\ell}] = g_{kj}E_{i\ell}-g_{i\ell} E_{kj}$$
 * $$(E_{ij})^\dagger = E_{ji}$$

So our basis for so(g) will be Lij= Eji – (Eji)† and it remains to calculate the commutation relations among them.


 * $$[L_i{}^j,L_k{}^\ell]=[$$

The commutation relations are
 * $$[L_{ij},L_{k\ell}] = [E_{ij}-E_{ji},E_{k\ell}-E_{\ell k}] = [E_{ij},E_{k\ell}] - [E_{ji},E_{k\ell}] - [E_{ij},E_{\ell k}] + [E_{ji},E_{\ell k}] = g_{kj}L_{i\ell}-g_{j\ell}L_{ik}-g_{ik}L_{j\ell}+g_{\ell i}L_{jk}$$