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This post is under construction. See the previous post in this series for context, or the bottom of the first in the series for a table of contents

We’ve explicitly defined \(\cW_1\) to be equivalent to \(\RRP^2\), and by connectedness, immediately concluded that there is exactly one weave with one ring. We now go to the second level:

Proposition 2: \(\cW_2 \simeq \prn{S^1 \coprod \RRP_2} \times \RRP_2\); in particular, there are exactly two weaves consisting of two rings, given by two circles which are either linked or not linked.

To prove this is a bit painful. First, I want to illustrate that this morally ought to be true: fix a component \(c\) of \(w \in \cW_2\), and call the other compoent \(c'\).. Then, \(c'\)’s disk is either punctured by \(c\) or not; if it’s not, then there is a contractible space of choices for the position of \(c'\) (given by the complement of the ball around \(c'\), and \(\RRP_2\)’s worth of choices for the directions of the components. If it is punchred, then there is an \(S^1 \times D^2 \simeq S^1\)’s worth of positions for \(c'\), and a contractible space of positions (we can deformation retract onto the space where the directions are orthogonal).

To actually prove this, we proceed in two cases given a weave \(w\) of components \(c,c' \hookrightarrow w\); first suppose that \(w\) is composite, so that \(w = c \oplus c'\). Picking the point on the line between the centers of \(c\) and \(c'\) which is equidistant between them, pushing centers linearly towards or away from that point, we may defomration retract the connected component \(\cW_w\) containing \(w\) onto one where the centers of \(c\) and \(c'\) are distance \(2\) from each other. We may further deformation retract to assume that \(c,c'\) are centered at two fixed such points. Note that the space we’ve arrived at is \(\RRP_2^2\).

Now suppose that \(w\) is prime. First, note that we may defomration retract to the subspace \(X \subset \cW_w\) where \(c\) is centered at the origin with direction the \(z\) axis and \(c\) is centered at a point of \(c'\). Further, note that this defomration retracts onto the subspace \(Y \subset X\) where the direction of \(c\) is tangent to \(c'\); to see this, note that \(X = Y \times (\RRP^2 - S^1)\), pulling out the possible directions of \(c\), and further note that \(\RRP^2 - S^1 \simeq *\).

Finally, note that \(Y \simeq S^1 \times \RRP^2\). This together with the composite computation finishes the proof.