# Is the unitary group of a $p$-adic anisotropic hermitian space commutative?

Let $$E/\mathbb Q_p$$ be a quadratic extension where $$p \not = 2$$. Let $$V$$ be an $$E$$-vector space equipped with a non-degenerate hermitian form, and assume that $$V$$ is anisotropic. In particular, $$\dim(V) \leq 2$$. Let $$G := \mathrm U(V)$$ be the associated unitary group.

Is $$G$$ always commutative? Alternatively, is $$G$$ an algebraic torus?

Of course, the question is trivial when $$\dim(V) \leq 1$$ so let us assume that $$V$$ has dimension $$2$$. For instance, if $$E/\mathbb Q_p$$ is unramified, then we can construct such a $$V$$ with two basis vectors $$v,w$$ with hermitian form $$(\cdot,\cdot)$$ given by $$(v,v) = 1, \quad (w,w) = p, \quad (v,w) = 0.$$ In general, I can not find a good presentation of $$G$$ which would make it obvious that all its elements commute with eachother. However, all the elements of the group that I could produce so far do commute.

• Is $U(V)$ supposed to be the subgroup of $GL_2(E)$ such that $(a,b)=(ga,gb)$ ? May 9 at 11:32
• There won't be many possible forms up to equivalence. May 9 at 11:40
• @reuns Absolutely, this is the group I have in mind! May 10 at 2:30

Assume that $$U(V)$$ is commutative. Since $$V$$ is anisotropic any line $$D$$ is a non-degenerate subspace of $$V$$, so that one may consider the orthogonal symetry $$s_D$$ relative to $$D$$ (this is an element of $$U(V)$$). Then for any $$f\in U(V)$$, we have $$fs_D f^{-1} =s_D$$, whence $$f(D)=D$$. This implies that any element $$f$$ of $$U(V)$$ stabilizes any line of $$V$$, so is an homothety. But if $${\rm dim}(V)\geqslant 2$$, it is easy to produce elements of $$U(V)$$ that are not homotheties !

• Do you have one concrete counter-example? May 10 at 13:31
• Take two lines $D$, $D'$ such that $D\not= D'$ and $D^{\perp}\not= D'$. Then $s_D$ and $s_{D'}$ do not commute. May 10 at 13:42
• Ok, I'm convinced, say $V=E^2$, for $v\ne 0\in V$ take $w\ne 0$ such that $(v,w)=0$, let $s_v(av+bw)=av-bw\in GL_2(E)$ May 10 at 13:50
• Thank you, this is a very nice argument! May 10 at 21:56

Not only is that group not commutative (as the other answer shows nicely), it is indeed largely analogous to the well-known unitary group $$U(2)$$ over the reals. In fact, if we take that basis $$(v,w)$$ you suggested for the anisotropic space $$V$$, your group $$U(V)$$ identifies with the matrices

$$\pmatrix{x \alpha & -p \bar \beta \\x \beta &\bar \alpha} = \pmatrix{ \alpha & -p \bar \beta \\ \beta &\bar \alpha} \cdot \pmatrix{x & 0 \\0 &1}$$

where $$\bar{()}$$ denotes the non-trivial automorphism of $$E$$, and $$\alpha, \beta, x \in E$$ are subject to the conditions $$N(x) = 1 = N(\alpha)+pN(\beta)$$, where $$N(e)= e \cdot \bar e$$ is the field norm $$E \rightarrow \mathbb Q_p$$. Note that the determinant of the above general matrix is $$x$$, and the splitting on the right amounts to $$U(V)$$ being the semidirect product of $$SU(V) := \{\pmatrix{ \alpha & -p \bar \beta \\ \beta &\bar \alpha} : \alpha, \beta \in E: N(\alpha)+pN(\beta)=1\}$$ with the torus given by the norm-$$1$$-group of your quadratic extension, $$\{x \in E: N(x)=1 \}$$.

And the analogy with the classical case does not end here, because just like over the reals, this $$SU(V)$$ identifies with the norm-$$1$$-group of "the" $$p$$-adic quaternions. In fact, if $$E= \mathbb Q_p(\sqrt a)$$ so that $$N(x+\sqrt a y)=x^2-ay^2$$, we can define the quaternion algebra (following K. Conrad's excellent notes)

$$Q:= (a, -p)_{\mathbb Q_p}$$

as the four-dimensional $$\mathbb Q_p$$-algebra with basis $$1,u,v, uv$$ and $$u^2=a, v^2=-p, vu=-uv$$, and its quaternion norm

$$N_Q (x_0 + x_1u +x_2v +x_3 uv) = x_0^2-ax_1^2+px_2^2-pax_3^2$$

which of course is the above $$N(\alpha) +p N(\beta)$$ for $$\alpha = x_0+x_1 \sqrt a$$ and $$\beta = x_2+x_3\sqrt a$$.

So $$SU(V) \simeq \{q \in Q: N_Q(q)=1\}$$ as $$p$$-adic Lie groups. (In fact, you'll see which of the matrices correspond to multiplication with $$\mathbb Q_p$$-multiples of $$1,u,v, uv$$ -- multiplication from the right that is, to make them $$E$$-linear from the left on the $$2$$-dimensional $$E$$-left-vector space $$Q$$. [I hope have not mixed up left and right here.])

To see that this group has ($$\mathbb Q_p$$-)dimension $$3$$, note e.g. that you can choose $$x_1,x_2,x_3$$ freely as long as they are "small" ($$p$$-adic absolute value less than $$p^{-2}$$ will do) and still find an $$x_0$$ to get to norm $$1$$. (Here is one thing that is a bit more cumbersome than in the classical case, where Hamilton's $$i,j,k$$, conveniently, already have norm $$1$$.) By the way, likewise you can see that that torus $$\{x \in E: N(x)=1\}$$ also has $$(\mathbb Q_p$$-)dimension $$1$$, i.e. $$U(V)$$ is a four-dimensional ($$p$$-adic) Lie group. And the Lie algebra of $$SU(V)$$ is given by the matrices

$$\{ \pmatrix{x_1\sqrt a& -p(x_2-x_3\sqrt a)\\x_2+x_3\sqrt a &-x_1\sqrt a} : x_1, x_2, x_3 \in \mathbb Q_p \}$$

a.k.a. the "pure" quaternions. Of course the group (or Lie algebra) is "the" compact $$\mathbb Q_p$$-form of the group $$SL_2$$ (or Lie algebra $$\mathfrak{sl}_2$$), and becomes split after scalar extension to $$E$$.

Further, all this generalizes in various ways: First of all you can replace the ground field $$\mathbb Q_p$$ by any finite extension $$K$$ thereof (now just take a uniformizer $$\pi_K$$ instead of $$p$$); then, you don't need the unramified quadratic extension as $$E$$, just take any quadratic extension, now you just make sure to choose instead of $$-\pi_K$$ any element in $$K$$ which is not a norm from $$E$$. In fact, by general facts about those quadratic extensions and/or quaternions and/or anisotropic quadratic forms over $$p$$-adic fields, although this seems like you have a lot of choices, for any given base field $$K$$ the results are isomorphic, you might as well call them $$SU_2 (K)$$.

Finally, I am also pretty sure everything important goes through for $$p=2$$ as well, but as always, some things in that case are odd.

• Wow, thank you very much for this excellent and enlightening answer! It really helps a lot May 18 at 1:14