Timeline for Does a finite-dimensional Lie algebra always exponentiate into a universal covering group
Current License: CC BY-SA 2.5
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 13, 2017 at 12:57 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
|
|
| Oct 10, 2010 at 4:44 | comment | added | algori | Victor -- I think the answer to that is no. Take e.g. $g=sl_2(\mathbf{C})$ and the defining representation. There are matrices in $SL_2(\mathbf{C})$ not in the image of the exponential map of $g$. Take e.g. $\begin{pmatrix}-1&1\\0&-1\end{pmatrix}$. But this is the exponential of some matrix in $gl_2(\mathbf{C})$, which, consequently, can't be written as a linear combination of the generators of $g$. | |
| Oct 10, 2010 at 4:11 | comment | added | Victor Galitski | Hi Algori, Thank you for your answer. I want to start with a Lie algebra, A, as a primary structure w/o reference to a particular Lie group(s). Then, take a faithful matrix representation, T[A], of A: T[B] are matrices corresponding to the abstract elements B =B^k X_k from A. Form all possible products exp(i T[B]) exp( i T[C]) = exp(i M) and calculate the result (i.e., a matrix M) via standard matrix multiplication. For finite-d representation the existence and convergence is perhaps a non-issue. Now I rephrase my main question: Can M always be written as a linear combination of T[X_k]? | |
| Oct 10, 2010 at 3:54 | history | answered | algori | CC BY-SA 2.5 |