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 $\mathbb{R}$?