# Is every subgroup of a connected unimodular (matrix) Lie group also unimodular?

My intuition is that the answer is yes: Let $G$ be the original group, and let $H$ be a subgroup of $G$. Let $\mu$ be a Haar measure on $G$ that is both right- and left-invariant. I think that if we restrict $\mu$ to $H$ and restrict the translation to translations by elements of $H$, then invariance must be preserved. Thus, by hand-waving, I guess that $\mu$ must be both right- and left-invariant on $H$, and not just on $G$. The reason that I'm not sure of it is the following: The 2x2 affine matrix group is not unimodular. It is, however, a subgroup of $GL(2)$.

Now, this may be just fine: Both the affine group and GL(2) are not connected. And, while $GL(2)_+$ is unimodular, I think that $GL(2)$ is not unimodular. Thus, the fact that the 2x2 affine group is not unimodular does not cause any problem. Can someone please verify I'm right on this?

If I'm right, this brings me to another question: Is every connected matrix group unimodular? I think that, pending on a positive answer to my original question, this must be the case since they are all subgroups of $GL(n)_+$ for some $n$.

Any help is much appreciated.

Thanks!

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While the left-invariance of a left Haar measure is perhaps the property that is used most often, a Haar measure has to satisfy other properties as well, see e.g. wikipedia. In this case, the restriction of $\mu$ to $H$ is of course still gives rise to a left-invariant measure, but this measure is trivial, so it's not a Haar measure. –  Guntram Nov 27 '12 at 21:58
A connected Lie group $G$ with Lie algebra $g$ is unimodular iff $\operatorname{ad}x:g\to g$ has trace zero for all $x\in g$. One can easily see that this implies that connected normal subgroups of an unimodular Lie group are themselves unimodular and find examples of non-normal subgroups which are not; indeed, this reduces the verifications required to linear algebra. –  Mariano Suárez-Alvarez Nov 27 '12 at 22:07
Thank you. Guntram, why would the restricted measure be trivial? For example, let the measure of $GL(2)_+$ be defined by $\mu(S)=\int_S 1/(det(X)^2)dX$ where dX is the Lebesgue measure. This is both right and left invariant. And suppose I restrict it to the group given as an example by Yemon Choi (with a being positive). Why would this result in a trivial measure? –  John Nov 27 '12 at 23:10
Your integral is zero as one would see by considering the case of ${\mathbb R}$ inside ${\mathbb R}^2$. –  Yemon Choi Nov 27 '12 at 23:46
Ah, ok. Thanks. –  John Nov 28 '12 at 0:00

Take the so-called ax+b group, i.e. the connected component of the affine group of the real line. Or, even more concretely, $\left\{ \left( \matrix{ a & b \\ 0 & 1 } \right) \colon a>0, b\in{\mathbb R} \right\}$ This is not unimodular.
So I think the proof by handwaving'' has a mistake somewhere. Probably your conception of restricting a measure from a space to a closed subset may need rethinking ...
Thank you all. Are you sure about your example? That is, are you sure it is not unimodular? If $a$ can have negative values, then yes, I know this is non-unimodular (this is what I referred to in the original post). But when $a$ is positive, this is a connected group, and when computing the trace of $ad_x(y)$ the result is zero (where x and y are 2x2 matrices with zeros in the last row). Doesn't this contradict that "g is unimodular iff ad_x:g→g has trace zero for all x∈g"? Or maybe I'm missing something? –  John Nov 27 '12 at 22:57
Clarification: I computed $ad_x(y)$ by $xy-yx$. –  John Nov 27 '12 at 22:58
This group is most definitely not unimodular - just try actually working out left and right Haar measures. I seem to recall that $a\,da$ and $a^2\, da$ will do the job. –  Yemon Choi Nov 27 '12 at 23:39
Thanks. How about $\mu(S)=\int_S 1/(\det(X)^2)dX$? –  John Nov 27 '12 at 23:42
Also, isn't the result you want that $ad_x$ has trace zero when regarded as a map from the Lie algebra to itself, i.e. as a linear map on gothic-g? That is different from saying that commutators have trace zero, which is what you seem to be observing above. –  Yemon Choi Nov 27 '12 at 23:43