Talk:Table of Lie groups

message: I'm having trouble parsing the following:
:connected, not compact, for n≥2: not simply connected
I think that means that:
* For all n, its connected and not compact
* For n=1 it is simply connected
* For n≥2 it is not simply connected
Is that right?
I'm thinking of creating multiple lines for different values of n so that the if-then-else confusion can be eliminated.  Am also thinking of replacing above with five columns:
* connected: y/n
* simply connected: y/n
* compact: y/n
* simple: y/n
* semisimple: y/n
This would make the table narrower and easier to read. 
Is this a good idea?
I'm thinking of converting from html tables to wiki tables too.
I'm wondering if there should be a column that says
* nilpotent  (refering to algebra)
* solvable (ditto)
* center 
Everything listed so far seems to not be nilpotent.  e.g. Heisenberg group is not listed. 
Good or bad idea?
linas 04:30, 7 September 2005 (UTC)
replies: text: I updated the table layout per above. Wow, was that ever tedious!
Listing S and SS as Y/N is a good idea, but I only see the 'Y'. I don't see the 'N'. Sure the reader can figure out that if left blank, it is equivalent to 'N'. but since you said Y/N, you really should do that instead of leaving it blank. It is not that tedious. And why are all the ones listed not nilpotent? How complete do you want this list to be? The impression I had on reading it was that it was meant to be pretty complete. If not, please say what the limits to completeness are intended to be.
For Real Lie Groups that have blank in the column for universal cover, does this mean they have none, or that the cover is uninteresting? From the Wikipedia page on universal covering groups, it would seem that they do have covers, since one would expect "locally simply connected" to imply "simple connected".
Finally, I don't think you really need centers: Cartan subalgebras would be more interesting. Or Dynkin diagrams, which you would probably want to represent by links, since they are covered elsewhere in Wikipedia.
64.105.137.254 (talk) 08:30, 16 July 2013 (UTC)
thread_title: Table layout

message: I wonder if it would be useful if we made a little template that can be placed at the bottom of each Lie group related page for navigational purposes? --HappyCamper 16:53, 18 August 2006 (UTC)
thread_title:  A template? 

message: I am missing info on SE(3) = Lie group of rigid body transformations.
It's a semi-direct product of SO(3) and R^3.
--Benjamin.friedrich 13:42, 15 November 2006 (UTC)
replies: text: Yes, I was looking for SE(3) too. Shouldn't that be included in the list of Lie groups? gnusbiz (talk) 20:10, 5 May 2008 (UTC)
replies: text:  Ditto. Why no SE(3)? —Ben FrantzDale (talk) 14:53, 17 May 2011 (UTC)
thread_title:  SE(3) 

message: Do symmetric matrices belong in this table?
Every other group on this table comes from a Lie algebra of matrices with the Lie bracket given by commutator.
This one comes from take a vector subspace of matrices, decreeing that it has trivial Lie bracket, and then taking the Exponential map (Lie theory), and you can do this with any vector subspace of matrices; it's not an interesting construction, as far as I can tell.
The reason this concerns me is that symmetric matrices do not form a Lie algebra under the commutator; consider
:A=\begin{bmatrix}0 & 1\\1& 0\end{bmatrix}\qquad
B=\begin{bmatrix}1 & 0\\0& 0\end{bmatrix}
Then AB-BA=\begin{bmatrix}0 & 1\\-1& 0\end{bmatrix} is not symmetric.
Symmetric matrices naturally form a Jordan algebra, which is a nice object but complementary to Lie algebras.
Nbarth 00:23, 21 June 2007 (UTC)
thread_title:  ''S(n)'' is not like the others 

message: Is the surface of a 2-sphere a Lie group? It seems like this would be a subgroup of SO(3), but I'm not sure... If it is, what's it called; if it isn't can someone explain why not? That is, isn't it the group generated by the matrix exponential like so:
:M=\exp\begin{bmatrix}0 & -z & y\\ z & 0 & 0\\ -y & 0 &0\end{bmatrix}
Thanks. —Ben FrantzDale (talk) 01:40, 19 November 2009 (UTC)
replies: text: No, this is not right. The only spheres that are Liegroups are the 1 and 3-dimesional (and the zerodimensional if you like). Unfortunately i can not explain exactly why this is the case.  —Preceding unsigned comment added by 80.143.234.243 (talk) 13:16, 25 November 2009 (UTC)
replies: text:  OK. I'd love an explanation of why a 2-sphere isn't a group. I think that would help me understand exactly what constitutes a group versus a topological space. —Ben FrantzDale (talk) 17:12, 25 November 2009 (UTC)
replies: text: A quick answer is that all lie groups have euler characteristic equal to zero. But an easy calculation shows that the euler characteristic of S2  =  2. which excludes it from being a lie group. (The euler chracteristic is zero precisely for manifolds having a nowhere zero vector field. And its easy to create one of these on a lie group.) th2601:200:C082:2EA0:B9AD:9DB1:DF58:CB13 (talk) 21:36, 24 May 2023 (UTC)
thread_title:  2-sphere?