4

Let $n \geq 0$ be an integer, let $A = (a_{ij})$ be the $(2n+1) \times (2n+1)$ matrix defined by $a_{ij} = 0$ unless $i + j = 2n+2$, in which case $a_{ij} = 1$. Let $G$ be the group scheme over $\mathbb{Z}$ consisting of $(2n+1) \times (2n+1)$ matrices of determinant equal to $1$ that are orthogonal for the bilinear form defined by $A$. (So $G$ is a split special orthogonal group scheme.)

Question: How does one compute the volume of $G(\mathbb{Z}_2)$ (with respect to the Haar measure)?

What I know: Let $p$ be an odd prime. Then $G$ is smooth over $\mathbb{Z}_p$, and so the map $\pi_p \colon G(\mathbb{Z}_p) \to G(\mathbb{Z}/p\mathbb{Z})$ is surjective. Let $G'_p = \ker \pi_p$. Then we have that $G(\mathbb{Z}_p) = \bigcup_{g \in G(\mathbb{Z}/p\mathbb{Z})} g \cdot G'_p$, so if we normalize the Haar measure so that $\operatorname{Vol}(G'_p) = p^{-\dim G}$, we find that $$\operatorname{Vol}(G(\mathbb{Z}_p)) = \frac{\#G(\mathbb{Z}/p\mathbb{Z})}{p^{\dim G}}.$$ Now suppose $p = 2$. The above argument fails because $G$ is not smooth over $\mathbb{Z}_2$. For example, for each $m \geq 1$, the involution $$\left[\begin{array}{ccc} 1 & 0 & 2^{m-1} \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \in G(\mathbb{Z}/2^m\mathbb{Z})$$ does not admit a lift to $G(\mathbb{Z}/2^{m+1}\mathbb{Z})$. I'm not at all familiar with it, but if I understand correctly, the theory of Bruhat and Tits gives a model $\mathcal{G}$ that is smooth over $\mathbb{Z}_2$ and such that for any etale $\mathbb{Z}_2$-algebra $R$, we have that $\mathcal{G}(R) = G(R)$. In particular, this means that $\mathcal{G}(\mathbb{Z}_2) = G(\mathbb{Z}_2)$, and the image of the mod-$2$ reduction map $\pi_2\colon G(\mathbb{Z}_2) \to G(\mathbb{Z}/2\mathbb{Z})$ is equal to $\mathcal{G}(\mathbb{Z}/2\mathbb{Z})$. Am I right to then conclude that $$\operatorname{Vol}(G(\mathbb{Z}_2)) = \operatorname{Vol}(\mathcal{G}(\mathbb{Z}_2)) = \frac{\#\mathcal{G}(\mathbb{Z}/2\mathbb{Z})}{2^{\dim G}} = \frac{\#\operatorname{image}(\pi_2)}{2^{\dim G}}?$$ Is there a reference that computes $\#\operatorname{image}(\pi_2)$? According to the post https://mathoverflow.net/a/19327/76440, in the context of $2n \times 2n$ orthogonal matrices, the subgroup of $\operatorname{O}_{2n}(\mathbb{Z}/2\mathbb{Z})$ generated by the elementary orthogonal matrices "presumably" has index $2$. Since the image of $\operatorname{O}_{2n}(\mathbb{Z}_2) \to \operatorname{O}_{2n}(\mathbb{Z}/2\mathbb{Z})$ contains the elementary orthogonal matrices, I think this image also has index at most $2$.

Ashvin Swaminathan
  • 655
  • 4
  • 13
  • 2
    What's the "usual $p$-adic Haar measure"? I can think only of a well-defined element of $\mathbf{Z}/(p-1)\mathbf{Z}$. – YCor Jul 28 '19 at 21:53
  • @YCor I believe what I mean is this: there is a Haar measure on $\mathbb{Q}_p$, and that measure gives rise to a measure on $\mathbb{Q}_p^N$ for any integer $N \geq 1$, and pulling this measure back via local charts to $G(\mathbb{Q}_p)$ determines the usual $p$-adic Haar measure (which does not depend on the choice of the charts). Is this reasonable to you? – Ashvin Swaminathan Jul 28 '19 at 22:03
  • @YCor: I think he meant the normalized Haar measure so that $\mathbb{Z}{p}$ has measure 1, and $p\mathbb{Z}{p}$ has measure $\frac{1}{p}$. This is quite standard. But I have never heard of measure on $G(\mathbb{Q}_{p})$. I also do not understand why smoothness is an issue at here. – Bombyx mori Jul 28 '19 at 22:15
  • 4
    I understand none of these two answers. Yes, the measure will depend on the charts. Maybe for a semisimple $\mathbf{Q}_p$-group $G$ some subtlety ensures that there's a canonical normalization depending only on the structure of algebraic group over $\mathbf{Q}_p$ (certainly we shouldn't expect any such canonical normalization for a unipotent group). – YCor Jul 28 '19 at 22:31
  • 3
    As @YCor says, the measure depends on the charts; there's no reason randomly chosen measures on local charts should even agree on the overlap. Certainly there is a Haar measure, but it's unique only up to normalisation. One reasonably common choice is to declare that the measure of a hyperspecial maximal compact, such as yours, is 1, in which case the answer is easy. Waldspurger considers a measure whose definition is less natural but behaves better; according to it the measure of a hyperspecial maximal compact is the square root of the order of the associated finite-field Lie algebra. – LSpice Jul 28 '19 at 22:42
  • 2
    Echoing other comments: on adele groups the ambiguities in the aggregate local Haar measures admit normalization ("Tamagawa measure"). At p-adic places, as @LSpice comments, some variant of the uniqueness-up-to-conjugation of Iwahori subgroups (manifest, for example, in somewhat subtle facts about hyperspecial maximal compacts) can give a canonical normalization. The minor hazard of giving local Iwahoris measure $1$ (which is incompatible with global/adelic measure considerations) is overcome by making a canonical choice of maximal compacts (e.g., hyperspecial), to be compatible with global. – paul garrett Jul 28 '19 at 23:47
  • @Bombyxmori The reason why smoothness is an issue is that while I know how to compute $#G(\mathbb{Z}/2\mathbb{Z})$, I do not know how to compute $#\operatorname{image}(\pi_2)$. I have modified the question to reflect this. – Ashvin Swaminathan Jul 31 '19 at 03:49
  • @LSpice Shouldn't the measure be less than 1? If the measure is the square root of the size of the associated finite-field Lie algebra, then I would think that the measure is at least $2^n$. Also, do you know if the measure considered by Waldspurger gives something like $#\operatorname{image}(\pi_2)/2^{\dim G}$? – Ashvin Swaminathan Jul 31 '19 at 05:49
  • I misremembered the normalisation; it's the size of what you call $\mathcal G(\mathbb Z/2\mathbb Z)$, divided by the square root of the finite-field orthogonal Lie algebra. (I'm not sure about why the measure should be less than 1; it's a choice of normalisation and can be anything you want, although there are good Fourier-analytic reasons for Waldspurger's choice.) – LSpice Jul 31 '19 at 11:37
  • Your question now asks about the normalised Haar measure, but there is no unique such thing. You've chosen to set the measure of $G'_p$ to $p^{-\dim G}$, which is fine but not at all required; Waldspurger sets the measure to $q^{-\dim G/2}$. Anyway, I now see that your question is really about the special fibre in characteristic 2, and unfortunately I don't know anything worthwhile there. – LSpice Jul 31 '19 at 11:38
  • 1
    Oh, one more thing. For any split, (connected) reductive group, the reductive quotient at a hyperspecial vertex, in this case $\mathcal G(\mathbb Z/2\mathbb Z)$, is the split, (connected) reductive group of the same type (i.e., with the same root system). In particular, in this case, it is the split orthogonal group of the same rank over $\mathbb Z/2\mathbb Z$. I don't know the size of that group, but it can be computed without any reference to the $p$-adic group. – LSpice Jul 31 '19 at 12:23
  • 3
    I see that you have edited "the normalised Haar measure" to "a normalised Haar" measure—which, really, is just a longer way of saying "Haar meaasure"—but I just want to be repetitive to make sure it's clear: your statement "since $\operatorname{Vol}(G'_p) = p^{-\dim G}$" in the odd-$p$ case is a choice; it does not, and cannot, follow from any of what has been said previously (since one can normalise the Haar measure to assign any fixed open, compact subset any fixed positive, finite measure). So the question is really "what is vol. of hyperspecial parahoric wrt this choice of measure?" – LSpice Jul 31 '19 at 17:52
  • @LSpice Thanks for these very helpful comments; I've tried to correct the phrasing of the question accordingly! – Ashvin Swaminathan Jul 31 '19 at 18:14

0 Answers0