Compute the homology of this configuration space.

Question

Let $U\subset V$ be finite labeling sets, and $K:\mathbb S^1\to\mathbb R^3$ be a knot. Consider the configuration space with points labeled $U$ lying on the knot, to make this space connected we fix an isotopic class $[i]$of injections $U\to K(\mathbb S^1)$: $$C_{V,U}(K)=\{\text{injective maps }c:V\to \mathbb R^3: c\in [i]\} $$ note that $C_{V,U}(K)$ is a submanifold of $(\mathbb R^3)^{V\setminus U}\times(\mathbb S^1)^{U}$. My questions are

(1) Does the toopology of $C_{V,U}(K)$ depends on the knot $K$.(For example, if $K$ is the unknot and trefoil)

(2) Is there a good way to compute the homology of $C_{V,U}(K)$?(again, compute for the unknot and trefoil)

Any comment or reference is welcome!

Answer 1

Since $K$ is injective, $C_{V,U}(K)$ is isomorphic to the $C_{V,U}(K)$ with $K$ the unknot. You can imagine, continuously sliding any or all of a given labeling of your knot around. That is exactly described by how those points move along just the circle. So forming a knot does not change the topology, up to isomorphism, whether a trivial map $\mathbb S^1 \to \mathbb R$ or a trefoil or other knot $\mathbb S^1 \to \mathbb R$ will induce isomorphic $C_{V,U}$, for fixed $U \subset V$.

As for computing the homology, it will be the same as the situation for doing it on just the circle, which has been treated in a variety of ways. For example,

• Configuration space of a circle

It's not exactly in the format you've described, but it should be relatively the same except with $\mathbb R$ labelings allowing each coordinate to differ in smaller and smaller increments..