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Consider the two-dimensional vector subspaces $V_1, V_2 \subseteq \mathbb{R}^4$ given by \begin{align} V_1 & = \{(x_1, x_2, x_3, x_4) \in \mathbb{R}^4 : x_3 = x_4 = 0 \} \\ V_2 & = \{(x_1, x_2, x_3, x_4) \in \mathbb{R}^4 : x_1 = x_2 = 0 \}. \end{align}

Consider the space $X = \mathbb{R}^4 \backslash (V_1 \cup V_2)$. By continuously deforming every $x \in X$ to its corresponding unit vector $\frac{x}{||x||}$, it seems to me that the space $X$ is homotopic to $S^3 \backslash (C_1 \cup C_2)$ where $C_1$ and $C_2$ are disjoint copies of $S^1$.

My question is about the topology of the circles $C_1, C_2$. Are they linked? If so what does their "link" look like?

Alan Yan
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1 Answers1

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As Moishe Kohan pointed out in the comment, $C_1 \cup C_2$ forms the so called Hopf link. The definition via Hopf map is in order.

Recall that the homotopy classes of continuous (pointed) maps $S^3\to S^2$ are in one-to-one correspondence with $\pi_3[S^2]=\mathbb{Z}$. We define the Hopf map (a.k.a. Hopf fibration) as the class represented by $1\in\mathbb{Z}$, up to a prescribed convention on which is $1$ and which is $-1$.

A convenient representative of the Hopf map can be described by using quaternions:

Let $$ z\in S^3\subseteq \mathbb{H} = \{a+bi+cj+dk: a,b,c,d\in\mathbb{R}\} $$

denote a unit quaternion. Then $\overline{z}iz$ is a purely imaginary unit quaternion, which means

$$ \overline{z}iz\in S^2, $$

where we identified $S^2$ with $\{q\in\mathbb{H}: |q|=1,\;\operatorname{Re}q=0\}.$

As a result, we have a mapping $f: S^3\to S^2: z \mapsto \overline{z}iz$ which is well known to represent the unit element $\pm1\in\pi_3[S^2].$ In particular, the set $f^{-1}(pt_1)\cup f^{-1}(pt_2)\subset S^3$ has linking number $1$ whenever $pt_1, pt_2$ are two distinct points in $S^2$; Look up Hopf link.

Now we are in position to show that $C_1\cup C_2$ is such a set. Identifying $\mathbb{R}^4\simeq\mathbb{H}$, we have

$$ C_1=V_1\cap S^3=\{e^{i\theta}\in\mathbb{H}:\theta\in\mathbb{R}\}, \\ C_2=V_2\cap S^3=\{e^{i\theta}j\in\mathbb{H}:\theta\in\mathbb{R}\}.$$

Furthermore, it is an easy exercise to show that

$$ C_1 = f^{-1}(i),\; C_2 = f^{-1}(-i), $$

so that $C_1 \cup C_2$ is a typical example of Hopf link.

Hyeongmuk LIM
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