8
$\begingroup$

Studying stability of certain non-autonomous dynamical systems on Lie groups I have come across the following question: Exactly which finite-dimensional, real Lie groups have adjoint representations that are bounded away from zero?

Edit: by "bounded away from zero" I mean that the image of the adjoint representation avoids an open neighborhood of zero in End(g), where g is the Lie algebra. Equivalently, the closure of the image does not contain zero, or, the norm (pick your favorite one) of every element of the adjoint representation is bounded from below by one and the same positive number. By Hadamard's inequality, a determinant bound will do as well. [end edit]

This should include compact Lie groups since for those there exists an inner product on the Lie algebra with respect to which all inner automorphisms are orthogonal, i.e. the elements of the adjoint representation have norm 1. Correct?

Also, for abelian Lie groups the adjoint representation is trivial, hence again bounded away from zero.

I believe that semisimple Lie groups should also be included but can not think of a valid argument.

Is there actually a counter example? I tried the general linear group GL(2) but the elements of the adjoint representation that I tried always have (some) unit eigenvalues. Is this an accident? I would have thought that GL(n) itself occurs as an adjoint representation somehow which would then not be bounded away from zero. But evidently I am not quite understanding the different dimensions here (the adjoint representation of GL(2) is a subgroup of GL(4)).

My apologies if this is trivial but I could not find anything that looked relevant in several books on Lie groups.

Improve this question
$\endgroup$
2
  • 3
    $\begingroup$ It would help to say more precisely what "bounded away from zero" means in this context. Also, note that the adjoint representation of a reductive group such as a general linear group sends the center to the identity element. In particular, GL(n) itself won't occur as (the image of) an adjoint representation. $\endgroup$
    – Jim Humphreys
    Commented Aug 19, 2013 at 14:08
  • $\begingroup$ Thanks, I'll edit the question. Appreciate the comment about GL(n). $\endgroup$
    – Jochen Trumpf
    Commented Aug 20, 2013 at 2:12

2 Answers 2

Reset to default
6
$\begingroup$

The adjoint rep is always bounded away from $0$. Let $\mathfrak{g}_0$ be a simple quotient of $\mathfrak{g}$. (I consider the $1$-dimensional Lie algebra to be simple, so there is always a simple quotient.) Let $\mathfrak{h}$ be the kernel of $\mathfrak{g} \to \mathfrak{g}_0$ and let $H = \exp(\mathfrak{h})$.

The adjoint action preserves $\mathfrak{h}$, so the adjoint representation is block upper triangular. The upper left block is the adjoint action of $G/H$ on its Lie algebra, which is $\mathfrak{g}_0$. Since $\mathfrak{g}_0$ is simple, it is unimodular, meaning that the adjoint rep has determinant $1$. This idea is taken from anton's answer.

In summary, the adjoint rep of $G$ can be put in block upper triangular form with the upper left block a matrix of determinant $1$ (and not a $0 \times 0$ matrix).

Improve this answer
$\endgroup$
0
3
$\begingroup$

I am not sure what you mean by "bounded away from zero", but if you mean that the closure of the image of the adjoint representation does not contain zero, that is correct. A proof might proceed by showing that the image lies in the adjoint group of the lie algebra and the latter lies in the group of elements of determinant one.

Improve this answer
$\endgroup$
6
  • 1
    $\begingroup$ Is this not only the case for the Lie groups with unimodular Lie algebras? $1 = \det \operatorname{Ad}(\exp X) = \det \exp \operatorname{ad}(X) = e^{\operatorname{tr} \operatorname{ad}(X)}$, hence $\operatorname{ad}(X)$ must be traceless for all $X$. Unless I'm missing something, this is not the case for every Lie algebra. $\endgroup$
    – José Figueroa-O'Farrill
    Commented Aug 19, 2013 at 18:41
  • $\begingroup$ Which orbit of the adjoint representation are we taking? Are we asking that some orbit doesn't have zero in its closure? $\endgroup$
    – Ben McKay
    Commented Aug 19, 2013 at 19:24
  • $\begingroup$ @Jose: Oh yes right, I was thinking of reductive groups. Sorry. $\endgroup$
    – user1688
    Commented Aug 19, 2013 at 20:07
  • $\begingroup$ Edited the question to spell out what I mean by "bounded away from zero". Thanks already for the reductive idea. Still evaluating David's answer above. $\endgroup$
    – Jochen Trumpf
    Commented Aug 20, 2013 at 2:22
  • $\begingroup$ Does this not require the Lie group to be connected (and have a unimodular Lie algebra)? The proof idea given by @Jose seems to require that a group element can be written as a finite product of exponentials. $\endgroup$
    – Jochen Trumpf
    Commented Aug 21, 2013 at 1:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .