Four, as shown in Figure 3, in a cube $A B C D-A_{1} B_{1} C_{1} D$ with edge length $a$, cut off its two corners along $A C B_{1}, A_{1} B I$, and then cut off two more corners along the faces $B D C_{1}, A C D_{1}$. Find the volume of the remaining geometric body.
The geometric body intercepted is shown in Figure 6. The base is a square, and there are 4 isosceles right triangles congruent to $\triangle M A_{1} B_{1}$, as well as 4 trapezoids congruent to $N M A_{1} G$.
See the original Figure 3, the cube has 4 tetrahedra of the form $B_{1}-A B C$ cut off, each with a volume of $V_{1}=\frac{1}{6} a^{3}$. The intersecting parts of two of these are small tetrahedra of the form $M-A B N$, with 4 in total, each having a volume of $V_{2}=\frac{1}{24} a^{3}$. The truncations of every 3 tetrahedra of the form $B_{1}-A B C$ do not intersect. Therefore, the volume of the remaining geometric body is
$$
V=a^{3}-4 \times \frac{1}{6} a^{3}+4 \times \frac{1}{24} a^{3}=\frac{1}{2} a^{3} .
$$