# Proof of Scott’s Criterion for separability (hard math) (UPDATED)

13 Sep

UPDATED: Thanks to my dear friend Teddy (who hasn’t updated his website and is at Cornell now, not UCSB), I’ve made the converse direction of the proof more correct.  There’s definitely still a glaring defect, but that’s entirely my fault.

This is out of character for this blog- it’s not accessible for most people.  If you have taken a course in algebraic topology, you can read this post and I’ll explain everything.  Otherwise, I’m not offering enough background to understand it.  Sorry!  Blame pregnancy!

I’ve been spending the past few months slowly slogging through a big paper that my advisor recently cowrote, on an alternative proof of Wise’s Malnormal Special Quotient Theorem.  In the paper they spend a few paragraphs explaining Scott’s Criterion for separability, from Scott’s 1978 paper (need access for this).  I did not understand it when reading, but after meeting with my advisor and drawing some pictures it made a lot more sense!  So I’m going to draw some pictures for anyone trying to understand this- probably other graduate students.

Here’s the theorem as it appears in the MSQT paper.

Theorem (Scott, 1978) Suppose X is a connected complex and $H \leq \pi_1(X)$.  Then H is separable in $\pi_1(X)$ if and only if for every finite subcomplex $A \subset X^H$, there exists an intermediate finite degree cover $\hat{X}$ such that A embeds in $\hat{X}$.

Okay let’s unpack the theorem.  First we need to say what it means for a subgroup to be separableis separable in if, whenever you pick an element which is not in H, there exists some finite index subgroup of so that H is contained in and isn’t contained in K.  Intuitively, you can “separate” from via some finite index subgroup.  There are other equivalent definitions, but this is the one we’re going with.

Recall that a finite degree cover is a covering space where each point in has finitely many preimages. Left side is an infinite cover, the real numbers covering the circle. Middle is a happy finite cover, three circles triple covering the circle. Right is a happy finite cover, boundary of the Mobius strip double covering the circle.

Notation wise, $X^H$ just means the cover of corresponding to H, so that $\pi_1(X^H)=H$.  Also, recall that an embedding is an injective homeomorphism onto the image of the map.  So, for instance, a circle definitely embeds into the middle cover above, but not into the infinite one.  You can map a circle injectively to a subset of the real line, say to [0,1), but it’s not a homeomorphism where the two ends meet.  And by connected complex let’s say CW-complex.

Okay it’s proof time!  For notation we’re going to let $G = \pi_1(X)$.

Suppose that the condition is met, and we want to show that is separable.  Take an element not in H.  Since is the fundamental group of Xcorresponds to a (class of homotopy equivalent) loop(s) in X.  Since isn’t in H, it isn’t a loop in $latex X^H$- let’s say it’s a line segment.  By the condition, since this line segment is a compact subset of $X^H$, there exists some intermediate finite degree cover $\hat{X}$ so that the line segment embeds into it.  This finite degree cover corresponds to a finite degree subgroup K.  Since $\hat{X}$ is intermediate, is contained in K.  So is separable.

Here’s a schematic:

Okay let’s do from the other side now.  Suppose is a separable subgroup of G.  Pick a finite subcomplex of (the actual criterion just says compact, but we’re sticking with finite).  Look at all the elements $g_i$ of which have preimages in A- since is finite, we only have finitely many of these.  For any given $latex g_i$, since is separable we have a finite index subgroup $K_i$ which doesn’t include $g_i$ and which contains H.

I guess we still need to show embeds- do you believe me that it does?  I’m not sure I believe me.

Pick a compact subcomplex of H.  Since it’s compact, there are finitely many open sets that we need to consider, which cover A.  And since it’s a subcomplex of a CW-complex, this means we’re only looking at finitely many open cells in H.  These open cells project down to X, say in a set D.  Look at all the preimages of up in $X^H$– there’s infinitely many, since we assumed $X^H$ is an infinite cover.  And is one of the preimages of by construction.  (Also let’s make D, small enough so we have a “stack of pancakes” instead of batter all over the place).  Again, schematic: I know it looks like three, but there are actually Infinitely many preimages of the image of A

Now if we want an intermediate cover into which embeds, it can’t include any elements of that send to itself- that is, if $g.D\cap D \neq \emptyset$we don’t want g in $\pi_1(\hat{X})$.  Because then wouldn’t embed.  How many bad are there?  Well, since deck transformations act properly discontinuously, for any point in $X^H$ there’s an open neighborhood that never gets sent to itself (besides when is the identity, of course).  And we’re in CW-complexes, so we mean an open cell.  Look at the other cells of this particular component of D.  Again by proper discontinuity, there’s only finitely many g that’ll send this to some other copy of D.

Since is separable, for any one of these we have a finite index subgroup that doesn’t include and which does contain H.  Take the intersection of all these subgroups– since they all contain H, this intersection (call it K) does too.  And K doesn’t include any of the bad g.  Back in topology-land, K corresponds to a finite degree cover of X, since the intersection of finitely many finite-index subgroups has finite index.  And this cover is intermediate by construction, and embeds in it since there aren’t any bad g.

And that’s a proof of Scott’s criterion!  My next blog post will either be baking/cooking or a reasonable math post.

### 3 Responses to “Proof of Scott’s Criterion for separability (hard math) (UPDATED)”

1. Tom Loeza September 14, 2014 at 10:05 pm #

This is great. Really a big help to those who take so hard in math. A good reference. Really thanks.