Recent content by quidamschwarz

Sorry for not answering yet. There was a lot of work to do. My confusion was primarily that I posted the 'linear subspace solution' in my first answer and I couldn't see how the problem could be solved with the implicit function theorem (or the rank-version of it) without using this...

Sorry, but now I'm a little confused. From where to where is your map f(A):=A-A^\top going? From M(n) to M(n)? But then the diagonal funktions f_{ii}(A) = A_{ii}-A_{ii} = 0 vanish, so their 'gradients' are 0 and so not independent of the others. In general: Is a submanifold M defined as...

I don't think so. As far as I understand is the time parameter t in non-relativistic QM (what we are discussing here) something different as position and momentum. Well, physically this is trivial. What I mean is the following: A quantum state is given as a vector in a Hilbert space. Which...

Hi Eugene, you missed the t in your time evolution operator. It should be | \psi (t) \rangle = \exp(-\frac{i}{\hbar}\,t\,\hat{H}) | \psi (0) \rangle \;, where we assume, that the Hamiltonian is time-independent. In case of a time-dependent Hamiltonian, things get more complicated and...

Since this is a question about 'submanifolds', I guess, you're using the inherited topology. You should think about Sym(n) as a linear subspace of M(n). Then it's not hard to construct a (global!) parametrisation ...

It might help to view some of your expressions as binomials!? \frac{[2(k+m)]!}{(k+m)!^{2}} = {2(k+m) \choose k+m} and \frac{(k-j+m)!^{2}}{[2(k-j+m)]!} = {2(k-j+m) \choose k-j+m}^{-1}\;. So we have S=\sum_{k=0}^{m}\sum_{j=0}^{m+k-1}(-1)^{k}{m \choose k} {2(m+k) \choose m+k}...

Thanks for reminding me to think of L and K as operators instead just as tangentvectors. For a moment I thought only vectorfields can act on functions, so I got confused. Here you have to keep track of the position you’re taking the derivatives: (\phi_*X)_g(f) :=...

Hi Fredrik, I also tried to prove the equivalence of the left and right Lie bracket. But unfortunately it’s just not true – the right Lie bracket is the reverse of the left one. You can find a note on http://planetmath.org/encyclopedia/LieGroup.html The main idea is to consider the...