4. In tetrahedron $ABCD$, faces $ABC$ and $BCI$ are both equilateral triangles with side length $2a$, and $AD = 2\sqrt{2}a$. Points $M$ and $N$ are the midpoints of edges $AB$ and $CD$, respectively. The shortest distance between $M$ and $N$ on the tetrahedron is ( ).
(A) $2a$
(B) $\frac{3}{2}a$
(C) $a$
(D) $\frac{5}{2}a$
4. A.

On the surface of the tetrahedron, the possible paths from point $M$ to $N$ are as follows: (1) passing through edge $A C$; (2) passing through edge $A D$; (3) passing through edge $B C$; (4) passing through edge $B D$. When the distances from $M$ to $N$ in the above four cases are not all equal, take the smallest value among them. For this purpose, let's first unfold according to case (1) as shown in the figure, then $M N_{\min }=M N$. Connect $M C$, by $M C^{2}=B C^{2}-B M^{2}$ we get $M C^{2}=3 a^{2}$. Also, $A C=C D=2 a, A D=2 \sqrt{2} a$, so $A C \perp C D$. Therefore,
$$
\begin{aligned}
M N^{2} & =M C^{2}+C N^{2}+2 M C \cdot C N \cdot \cos \angle M C N \\
& =(4+\sqrt{3}) a^{2}, \\
M N & =\sqrt{4+\sqrt{3}} a .
\end{aligned}
$$

Similarly, for cases (2) and (3), we get $M N=2 a$.
Case (4) is the same as case (1).
Therefore, the minimum value of $M N$ is $2 a$.