How to prove that two linked circles in $\mathbb{R^3}$ are not ambient isotopic to two circles showing no crossing

Question

Let $C_1$ and $C_2$ be a pair of linked circles in $\mathbb{R^3}$ showing exactly two crossings.

I want to know how to prove that no ambient isotopy takes them so that they will show no crossing, not using the theorem of Reidemeister moves.

If some point is unclear let me correct it.

Thank you in advance.

Answer 1

The linking number of two curves can be defined by an integral (see the Wikipedia entry for Linking number, and look for Gauss's integral formula).

Now suppose you had an isotopy, $H$. You could compute $$ u(s) = Link(H(\gamma, s), H(\zeta, s)) $$ for each $s$, where $\gamma$ and $\zeta$ are your two (parameterized) curves, and $H(\gamma, s)$ is shorthand for the curve $$ t \mapsto H(\gamma(t), s) $$ that represents how $\gamma$ looks after being deformed by stage $s$ of the isoptopy.

Now $u$ is evidently a smooth function of $s$, but it takes on integer values, hence must be constant. Since it's "2" at the start of the isotopy (i.e., $u(0) = 2$, it must be $2$ at the end. But two uncrossing circles have linking number 0.

Answer 2

Just for an alternative approach, you can use the fundamental group of the complement. That is, $\pi_1(\mathbb{R}^3-$unlink with 2 components$)=F_2$, the free group on two generators. The link you describe is the Hopf link, the simplest non-trivial link, and the only one with 2 crossings. This has fundamental group $\mathbb{Z}^2$. This can easily be proven using the Wirtinger presentation. See also wikipedia's page on the Hopf link.