How to check two circles are linked or not? (without using topology) In $\mathbb{R}^6$, three loops
$$C_1:=\{(0,x,-x;0,y,-y)\mid x^2+y^2=1\}\\
C_2:=\{(x,0,-x;y,0,-y)\mid x^2+y^2=1\}\\
C_3:=\{(x,-x,0;y,-y,0)\mid x^2+y^2=1\}$$
are embedded. Is there a pair of circles that are linked?
 A: No, the subspaces are not linked. Consider that if I can move $C_1$ arbitrarily far away from $C_2$ without them ever intersecting, then they are unlinked. Define
$$ C_{1,h} = \{(h,x,-x,0,y,-x) \mid x^2+y^2=1 \}, \qquad h \in \mathbf{R}$$
Now let us move $C_1$ far away from $C_2$. Suppose these two subspaces intersected one another as $h$ in $C_{1,h}$ got large, so we had
$$ (h,x_1,-x_1,0,y_1,-y_1) = (x_2,0,-x_2,y_2,0,-y_2), \qquad
   x_1^2+y_1^2=1, \quad x_2^2+y_2^2=1, \quad h > 0 $$
That is,
$$ (h-x_2,x_1,-x_1+x_2,-y_2,y_1,y_2-y_1) = 0 $$
so $x_1=0$ and $y_1=0$, violating the requirement $x_1^2+y_1^2=1$. Then the above equality never holds, so the spaces never intersect as we vary $h$. So we can move $C_1$ as far away from $C_2$ as we want without intersections; thus the spaces are not linked.
Now what's the point behind this problem? Probably to show you why two one-dimensional subspaces of a high-dimensional space are never linked. It is always possible to move through the larger space to move the subspaces far from one another, without the subspaces ever intersecting.
