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.

  • 1
    $\begingroup$ Is $U(V)$ supposed to be the subgroup of $GL_2(E)$ such that $(a,b)=(ga,gb)$ ? $\endgroup$
    – reuns
    May 9 at 11:32
  • $\begingroup$ There won't be many possible forms up to equivalence. $\endgroup$
    – reuns
    May 9 at 11:40
  • $\begingroup$ @reuns Absolutely, this is the group I have in mind! $\endgroup$
    – Suzet
    May 10 at 2:30

2 Answers 2


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 !

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

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

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