5
$\begingroup$

The question may be a little naive (or even appear as a duplicate) as I guess the result is well known. I saw on the other thread that

" c) A solvable Lie group G is linear iff its commutator subgroup G′ is closed, and G′ has no non-trivial compact subgroup (Thm 3.2 in Chap. XVIII) "

But I cannot find the reference book anywhere. May I ask someone to give a hint why this is true?

Background:

Try to find a non-trivial example of infranil manifold which is not a matrix Lie group. Then the speaker visiting my university claiming every nilpotent Lie group must be a matrix Lie group. Since nilpotent is defined entirely algebraically, I feel there should be a proof based on algebraic techniques. But I do not know how to prove the above statement.

$\endgroup$
  • 5
    $\begingroup$ Simply connected nilpotent Lie groups are linear; $\begin{pmatrix}1&\mathbf R&\mathbf R\\0&1&\mathbf R\\0&0&1\end{pmatrix}/\begin{pmatrix}1&0&\mathbf Z\\0&1&0\\0&0&1\end{pmatrix}$ isn't. See Ado's theorem. $\endgroup$ – Francois Ziegler Oct 3 '14 at 2:24
  • $\begingroup$ @FrancoisZiegler: I see. I always thought Ado's theorem only tell us the Lie algebra is linear, and the Lie group could have covers which are not linear. I did not thought about this example. $\endgroup$ – Bombyx mori Oct 3 '14 at 2:27
  • $\begingroup$ I still need to think carefully why your example would not be linear (is it a variation of Heisenberg group?) But thanks! $\endgroup$ – Bombyx mori Oct 3 '14 at 2:28
  • 2
    $\begingroup$ Yes, a variation due to Garrett Birkhoff, Lie groups simply isomorphic with no linear group (1936). $\endgroup$ – Francois Ziegler Oct 3 '14 at 2:31
  • $\begingroup$ @FrancoisZiegler: Thanks! I guess this is enough food for thought. I am glad I asked it at here and did not took the statement blindly. $\endgroup$ – Bombyx mori Oct 3 '14 at 2:34
5
$\begingroup$

Another reference which was not yet mentioned, I think, is the article of M. Moskowitz, "Faithful Representations and a local property of Lie groups", Math. Z. $143$, 1975. There the question is discussed when all analytic groups with a given Lie algebra $\mathfrak{g}$ have a faithful linear representation. He proves among other things the following result:

Theorem All connected Lie groups $G$ with Lie algebra $\mathfrak{g}$ have a faithful linear representation if and only if $Z(\mathfrak{g})\cap [\mathfrak{g},\mathfrak{g}]=0$ and $\tilde{S}$ has a faithful representation.

Here $\tilde{G}=rad(\tilde{G})\cdot \tilde{S}$ is the Levi decomposition for the simply connected group $\tilde{G}$ with Lie algebra $\mathfrak{g}$.

Corollary Let $G$ be a connected Lie group with Lie algebra $\mathfrak{g}$. If $ad(\mathfrak{g})$ has nontrivial center then there is a locally isomorphic group without any faithful representation.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.