Show that $\mathrm{SO}_3(\mathbb{Q}_p) \simeq \mathrm{SL}_2(\mathbb{Q}_p) $ 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 $\mathbb{R}$?
 A: The thing is, as opposed to the case of $\mathbb{R}$, there are few anisotropic quadratic forms over $\mathbb{Q}_p$. In fact over $\mathbb{Q}_p$, the standard quadratic form $\langle 1,1,1\rangle$  is not anisotropic but it is isometric to $\langle 1,-1,-1\rangle$, so given your statement for $SO_{2,1}(\mathbb{R})$ this shouldn't be too surprising.
The situation is as follows (all quadratic forms are supposed to be non-degenrerate) : any quadratic form $q$ over $\mathbb{Q}_p$ can be written $\langle pa_1,\dots,pa_r, b_1,\dots,b_s\rangle$ with $a_i,b_i\in \mathbb{Z_p}^*$. Then $q$ is uniquely characterized by $q_1 = \langle \overline{a_1},\dots,\overline{a_r}\rangle$ and $q_2 = \langle \overline{b_1},\dots,\overline{b_s}\rangle$ which are quadratic forms over $\mathbb{F}_p$ ; and a quadratic form over $\mathbb{F}_p$ is uniquely characterized by its dimension and its discriminant.

Now an elegant way to see that $SO_3(\mathbb{Q}_p)\simeq SL_2(\mathbb{Q}_p)$ is the following : in general, if $Q$ is a quaternion algebra over a field $K$, then the quaternion norm $N$ of $Q$ is a quadratic form, and so is its restriction $q$ to pure quaternions $Q_0$. 
And it is known that the map $Q_1\to SO(Q_0,q)$ defined by $z\mapsto (x\mapsto zx\overline{z})$, where $Q_1 = \{ z\in Q\,|\, N(z)=1\}$, is an isomorphism.
EDIT : no, actually it has kernel $\{\pm 1\}$, and this is also a mistake in your question I think. What we will get is $SL_2(\mathbb{Q}_p)/\{\pm1\} \simeq SO_3(\mathbb{Q}_p)$.
Now over $\mathbb{R}$ there are two quaternion alegbras : $M_2(\mathbb{R})$ and the Hamilton quternions $\mathbb{H}$, and the corresponding $q$ are respectively $\langle 1,-1,-1\rangle$ and $\langle 1,1,1\rangle$, while the corresponding $Q_1$ are $SL_2(\mathbb{R})$ and $\mathbb{H}_1$ (which doesn't have an obvious simpler expression), which in particular gives us $SL_2(\mathbb{R})/\{\pm1\} \simeq SO_{2,1}(\mathbb{R})$.
But over $\mathbb{Q}_p$, the hamilton quaternions and the split quaternions $M_2(\mathbb{Q}_p)$ are isomorphic, precisely because $\langle 1,-1,-1\rangle$ and $\langle 1,1,1\rangle$ are isometric, so this gives $SL_2(\mathbb{Q}_p)/\{\pm1\}\simeq SO_3(\mathbb{Q}_p)\simeq SO_{2,1}(\mathbb{Q}_p)$.
A: 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.
