TITLE: Show that $\mathrm{SO}_3(\mathbb{Q}_p) \simeq \mathrm{SL}_2(\mathbb{Q}_p) $
QUESTION [13 upvotes]: I read in a paper that $\mathrm{SO}_3(\mathbb{Q}_p) \simeq \mathrm{SL}_2(\mathbb{Q}_p) $ this is counterintuitive / surprising since $\mathrm{SO}_3(\mathbb{R}) \not \simeq \mathrm{SL}_2(\mathbb{R}) $

I don't know about $p$-adic numbers, but for reals numbers small rotations are $1$ plus a skew-symmetric matrix:
$$\left[ \begin{array}{ccc} 1 &  0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]
+ \left[ \begin{array}{rrr} 0 &  a & b \\ -a & 0 & c \\ -b & -c & 0\end{array}\right]$$
Therefore a rotation of a sphere is defined by the three numbers $a,b,c \in \mathbb{R}$.  Next an element of $\mathrm{SL}_2(\mathbb{R})$:
$$ \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right]: ad-bc=1 $$
These also have three numbers, so at least the both spaces are $3$ dimensions, but $\mathrm{SO}_3(\mathbb{R})$ is compact, while $ \mathrm{SL}_2(\mathbb{R})$ is not compact.  For example a matrix like:
$$ \left[ \begin{array}{cc} 2 & 0 \\ 0 & \frac{1}{2} \end{array} \right] $$
and we can use  any number instead of $2$ (such as a million, $10^6$).  
However, I do recall reading that $\mathrm{SO}(2,1, \mathbb{R})\simeq \mathrm{SL}_2(\mathbb{R})$ to make matters more confusing.

Do we get any sort of identification when we try $\mathbb{Q}_p$ instead of $\mathbb{R}$?

REPLY [3 votes]: For a field $k$, the group $SL_2(k)$ acts on the space $S\subset \operatorname{End}(k^2)$ of symmetric matrices by the map $g.\alpha = g\alpha g^t$. This action preserves $\det \alpha$, which is a quadratic form in $\dim S = 3$ variables. Thus we get a map $SL_2(k) \to SO(V)$ for some quadratic form $V/k$. In nice cases (I'm not sure about the details offhand), we actually get an isomorphism, or at least a covering map. That gives a map $SL_2(k) \to SO(V)$ for the quadratic form
\begin{align*}
V &= \det \begin{pmatrix}X & Z \\ Z & Y\end{pmatrix} = XY - Z^2.
\end{align*}
For $k = \mathbb{Q}_p$, this form is actually isomorphic to the usual one $X^2 + Y^2 + Z^2$, and so we get a map (possibly an isomorphism?) $SL_2(\mathbb{Q}_p) \to SO_3(\mathbb{Q}_p)$, unlike the real case.