If we remove $3$ lines from $\mathbb{R}^3,$ is this the $3$-torus? I think that I am thinking about this wrong.
So, I was trying to figure out the knot complement of the Borromean Rings embedded in $S^3.$ Because $S^3$ is the one-point compactification of $\mathbb{R}^3,$ I thought that the Borromean rings are equivalent to $3$ non-intersecting, perpendicular lines (similar to how the unknot in $S^3$ is equivalent to a single line in $\mathbb{R}^3$). Then, the complement of the Borromean Rings in $S^3$ is the same as removing $3$ lines from $\mathbb{R}^3,$ which is a $3$-torus (?).
From Wikipedia, the complement of the Borromean rings in $S^3$ is two regular octehedra in hyperbolic space.
I am unsure of how to come across this result. Any help would be appreciated.
I watched Not Knot which was really good for visualization, but I am confused about a more formal argument for the above result.
 A: In the case of the Borromean rings, unlike with the unknot, you'll need to treat the link components as copies of the circle, not lines. This is because under the one point compactification of $\mathbb{R}^3$, your perpendicular lines will all intersect at infinity, and thus will fail to even be a link.
As for understanding how the complement of the Borromean rings in $S^3$ is two octohedra glued together, there is no better place to learn than from the master of low dimensional topology, Bill Thurston:
http://library.msri.org/books/gt3m/PDF/7.pdf
On page 11 of the PDF linked above, he explicitly gives the volume of the Borromean rings' complement in $S^3$ under its complete hyperbolic metric. Thurston's entire collection of notes on 3 manifolds are freely available online thanks to MSRI. You should definitely take a look at them - they're chock-full of wonderful insight!
A: When you take the one-point compactification you only get to move at most one of the circles to infinity to make it a line. If you try to move a second circle to infinity the two circles will intersect at infinity and then you don't have a link anymore. So you can get one line and two circles and they're still linked.
A: The complement of three skew lines in $\Bbb R^3$ is the same as the complement
of a bouquet of three circles in $S^3$. Call this bouquet $A$. By Alexander duality,
$$\tilde{H}_k(S^3\setminus A)\cong\tilde{H}^{2-k}(A).$$
Therefore
$$\tilde{H}_2(S^3\setminus A)\cong\tilde{H}^0(A)=0.$$
Now let $B$ be the Borromean rings in $\Bbb R^3$. Similarly,
$$\tilde{H}_2(S^3\setminus B)\cong\tilde{H}^0(B)\cong\Bbb Z^2$$
since $B$ has three components. So $S^3\setminus A$ and $S^3\setminus B$ are not
homeomorphic.
