Let $G$ be a connected, reductive group defined over a local field $F$. Let $P$ be a parabolic subgroup of $G$ which is defined over $F$, and let $N = \mathscr R_u(P)$ be the unipotent radical of $P$. Then $N(F)$ is a locally compact group which I understand should be unimodular. Why is this?

As an example, consider a "split solvable group," a connected, solvable algebraic group $N$ defined over $F$, for which there is a composition series of $F$-subgroups

$$1 = N_0 \triangleleft N_1 \triangleleft \cdots \triangleleft N_r = N $$

where each quotient $N_i/N_{i-1}$ is $F$-isomorphic to the additive group $\mathbb{A}^1$. Each group $(N_i/N_{i-1})(F) \cong \mathbb{A}^1(F) = F$ is unimodular. I have heard there is a way to construct the Haar measure on $N(F)$ from the Haar measures on the quotients $(N_i/N_{i-1})(F)$, and that this idea generalizes in some way to showing that $\mathscr R_u(P)$ is unimodular, but I am not familiar with this type of argument.

  • $\begingroup$ as mentioned below, this easy fact was already answered (in a broader generality) on math.stackexchange.com/questions/323017/… $\endgroup$
    – YCor
    Dec 30, 2016 at 6:44
  • $\begingroup$ @D_S: Can you clarify your remarks about "split solvable" (such a group might be an algebraic torus, for example)? See also my comment on Peter McNamara's answer. Your question needs some editing. $\endgroup$ Dec 30, 2016 at 16:23
  • $\begingroup$ As @JimHumphreys says, you have defined a split unipotent subgroup; a split soluble subgroup might have $\operatorname{GL}_1$ composition factors. $\endgroup$
    – LSpice
    Apr 24, 2019 at 17:26

3 Answers 3


The slickest way to handle all such matters is to exploit the underlying algebraic structure so as to avoid any need to compute anything (and to argue in a manner which works uniformly for groups of points over all local compact fields or adele rings).

Let $G$ be any smooth group scheme of relative dimension $d> 0$ over any commutative ring $k$ at all. Then $G$ has its natural adjoint representation on the vector bundle ${\rm{Lie}}(G)$, and so consequently (by Yoneda considerations, since the formation of ${\rm{Lie}}(G)$ commutes with any base change, by smoothness) we get a homomorphism $\chi:G \rightarrow {\rm{GL}}_1$ of $k$-group schemes that computes the induced action of $G$ on the line bundle $\wedge^d({\rm{Lie}}(G))$ over $k$.

I claim that (in a sense to be made precise) $\chi$ is an algebraic incarnation of the "modulus character" and over suitable locally compact $k$-algebras it will instantly yield unimodularity results when $G$ has no nontrivial $k$-homomorphism to ${\rm{GL}}_1$ at all (such as - for fields $k$ -- either unipotent $G$ or connected reductive $G$ whose maximal central $k$-torus is $k$-anisotropic). The significance of this goes further, because it also explains conceptually (i.e., without any calculations) what may otherwise seem quite mysterious and astonishing from a purely analytic approach, namely that modulus characters in such setting with algebraic groups always wind up being "absolute values" (in the sense of normalized absolute values for local fields or idelic norms) of algebraic characters.

To start cooking with gas, we use a marvelous algebro-geometric argument that in some sense does "all left-translations at once": for any group scheme $H$ whatsoever over any scheme $S$ at all one can rigorously construct (and prove the uniqueness of) a left-invariant global section $\omega \in \Omega^1_{H/S}(H)$ extending a given global section $\omega_0$ over $S$ of the relative cotangent sheaf $e^{\ast}(\Omega^1_{H/S})$: see Proposition 1 in section 4.2 of the awesome book N\'eron Models (for an argument that doesn't even use smoothness if one focuses on the viewpoint of relative 1-forms and abandons trying to relate it to a dual of a Lie algebra). That proof may look a bit notationally heavy, but in the setting of affine groups over rings it simplifies.

Applying that to the smooth $f:G \rightarrow {\rm{Spec}}(k)$, for which $e^{\ast}(\Omega^1_{G/k}) = {\rm{Lie}}(G)^{\ast}$, by globalizing from Zariski-local arguments over ${\rm{Spec}}(k)$ (where ${\rm{Lie}}(G)$ becomes free) we get a unique isomorphism of $O_G$-modules $$O_G \otimes_k e^{\ast}(\Omega^1_{G/k}) \simeq \Omega^1_{G/k}$$ that lifts the identity along $e$ and carries $e^{\ast}(\Omega^1_{G/k})$ isomorphically onto the $k$-module $f_{\ast}(\Omega^1_{G/k})^{\ell}$ of left-invariant sections of $f_{\ast}(\Omega^1_{G/k})$. Passing to top exterior powers, we get canonically $\Omega^d_{G/k} = O_G \otimes \Omega^d_{G/k}(e)$ so that the $k$-line $\Omega^d_{G/k}(e) = \wedge^d({\rm{Lie}}(G))^{\ast}$ is identified with the $k$-module $\Omega^d_{G/k}(G)^{\ell}$ of left-invariant top-degree differential forms.

But by identifying the dual representation $\wedge^d({\rm{Lie}}(G))^{\ast}$ in this manner with $\Omega^d_{G/k}(G)^{\ell}$ (via $e$-pullback), how does the adjoint action on the former get interpreted on the latter? Well, tautologically the effect of "global conjugation" on a left-invariant global differential form is the same as the effect of a global right-translation (since the left-multiplication part of the global conjugation has no effect!). So we conclude that for any $k$-algebra $A$ and any $g \in G(A)$, $\chi^{-1}(g) \in A^{\times}$ is the scaling effect on $\Omega^d_{G_A/A}(G_A)$ of pullback along right-multiplication by $g^{-1}$; the reason for $\chi^{-1}$ is because of the intervention of the dual on the top exterior power of the Lie algebra (better appreciated when we want to have a left action even at the $\Omega^1$-level, where the difference between left and right is more manifest than after passing to action on a line bundle). Since $\chi^{-1}(g^{-1}) = \chi(g)$, voila: we see that $\chi$ is an algebraic version of a modulus character for left-invariant top-degree differential forms.

Now we come to the original focus of interest, namely a smooth affine group scheme $G$ over either a locally compact field $k$ (archimedean or not) or a global field $k$. Consider the first case, for which we will use the standard normalized Haar measure on $k$ to define Lebesgue integration on $k^d$, and we define the normalized absolute value $|\cdot|_k$ to compute the scaling effect of $k^{\times}$-multiplication on Haar measures on $k$ in the sense that $m(cU) = |c|_k m(U)$ for $c \in k^{\times}$ and Borel sets $U \subset k$ (so for $k = \mathbf{C}$ this is the square of the usual absolute value). We associate a left Haar measure $\mu_{\omega}$ on $G(k)$ to a choice of nonzero left-invariant top-degree differential form $\omega$ on $G$ by integrating $|\omega|_k$ in the usual manner (with the help of partitions of unity). The role of the normalized $|\cdot|_k$ is due to its appearance in the $k$-analytic Change of Variables Formula; orientation are irrelevant here when $k$ is archimedean due to the $|\cdot|_k$ on the local coefficient functions for $\omega$ in etale-local coordinates on $G$. (This association of a non-negative regular Borel measure to any global top-degree differential form works purely analytically for $k$-analytic manifolds, having nothing to do with groups; in the group case with a non-zero left-invariant form we get a left Haar measure).

By design, for $c \in k^{\times}$ we have $\mu_{c\omega} = |c|_k \mu_{\omega}$. But $$(r_g^{\ast}(\mu_{\omega}))(U) = \mu_{\omega}(r_g^{-1}(U)) = \mu_{\omega}(Ug^{-1}) = \mu_{r_g^{\ast}\omega}(U)$$ and $r_g^{\ast}\omega = \chi(g)\omega$ (gee, I hope I didn't lose any inversions here...), so putting it all together gives $r_g^{\ast}\mu_{\omega} = |\chi(g)|_k \mu_{\omega}$. Hence, as promised, we see that the modulus character for $G(k)$ against all left Haar measures is $|\chi|_k$ (since it is enough to test against one left Haar measure, such as $\mu_{\omega}$ as above).

So boom, there you have it by completely conceptual arguments: if $k$ is a local field then the modulus character for $G(k)$ acting on left Haar measures is $|\chi_G|_k$ where $|\cdot|_k$ is the standard "normalized" absolute value on $k$ (square of usual absolute value for $k = \mathbf{C}$) and $\chi_G:G \rightarrow {\rm{GL}}_1$ is $g \mapsto \det({\rm{Ad}}_G(g))$. In particular, if $G$ has no nontrivial $k$-homomorphism to ${\rm{GL}}_1$ (a purely algebraic-groups assertion, no analytic content at all!) then $\chi_G=1$ and hence $G(k)$ is unimodular. This applies to all unipotent $k$-groups and connected reductive $k$-groups whose center does not contain ${\rm{GL}}_1$. And it even applies to reductive $G$ without any anisotropicity condition on the maximal central $k$-torus because by algebraicity it suffices to observe that the maximal central torus maps onto $G/\mathscr{D}(G)$ and that central torus obviously leaves invariant any Haar measure. Likewise, for parabolic $k$-subgroups $P$ of connected reductive $G$ with $U:= \mathscr{R}_{u,k}(P)$, so $M := P/U$ satisfies $M(k) = P(k)/U(k)$ by Borel-Tits relative structure theory, we get as the modulus character for $P$ the pullback to $P(k)$ of the determinant of the $M(k)$-action on ${\rm{Lie}}(U)$ (since $M$ is itself unimodular; no anisotropicity conditions on $M$!).

If $k$ is a global field and we work with $G(\mathbf{A}_k)$ then by limiting considerations with $S$-integral models (avoiding consideration of infinite direct product measures, because that yields some convergence headaches) one finds that $|\!|\chi_G|\!|_k$ is the corresponding modulus character by pure-thought methods.

The preceding disposes of pretty much all questions about computing modulus characters for smooth groups over local and global fields.


The group is unimodular because it is nilpotent. The why has been answered on math.SE.

  • $\begingroup$ As you point out, the question is already well answered for a nilpotent Lie group. Here the setting is less clear, since "unipotent" typically comes up only for algebraic groups and the context is a local field (presumably not just the real field). Indeed, the notion of "reductive" was first explored systematically by Borel and Tits, though the name is perhaps misleading in prime characteristic. Can you say a little more? $\endgroup$ Dec 30, 2016 at 16:19
  • 2
    $\begingroup$ @JimHumphreys Every nillpotent locally compact group is unimodular, and the linked answer does explain that (though the question which it answers only asks about the Lie case). There is nothing to add here. $\endgroup$
    – Uri Bader
    Dec 30, 2016 at 20:20

You are making it harder than it has to be, by perhaps needlessly combining a construction of the Haar measure with its properties. Ask yourself about continuous group homomorphisms from a unipotent group to the multiplicative group of positive reals, which is what the modular function must be, whatever else it is. Such a group hom must be trivial on the commutator subgroup, to begin. And once one starts thinking this way, the whole game may be short-circuited by noting that (often, anyway), the unipotent radical is the commutator subgroup of the Borel subgroup... Not always, sure, but this gets us a long way down that road. :)

  • $\begingroup$ There's perhaps a tiny delicacy here in that the question is about the unipotent radical of any parabolic, not just of the Borel (and it's not obvious to me how to reduce questions about the former to those about the latter). Nonetheless, in any case, we can use the fact that unipotent radicals of parabolics are split in the sense of the question, and that there are no non-trivial algebraic homomorphisms $\mathfrak{gl}_1 \to \operatorname{GL}_1$. $\endgroup$
    – LSpice
    Apr 24, 2019 at 17:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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