Talk:Table of Lie groups

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[edit] Table layout

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)

I updated the table layout per above. Wow, was that ever tedious!

[edit] A template?

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)

[edit] SE(3)

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)

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)

[edit] S(n) is not like the others

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)

It's indeed not like the others since the group multiplication given here is not the matrix multiplication. It is not an interesting construction in the following sense: You can take any manifold diffeomorphic to Rn and give it a multiplication just by pulling forward (back) the multiplication (=addition) on Rn. What you get is only a Lie group isomorphic to Rn, so nothing new.

The symmetric positive definite matrices are a symmetric space of non-compact type diffeomorph to GL(n,R) / O(n), but not a Lie group.

129.187.111.52 10:03, 21 September 2007 (UTC)