## Not signed in

Want to take part in these discussions? Sign in if you have an account, or apply for one below

## Site Tag Cloud

Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.

• CommentRowNumber1.
• CommentAuthorTim_Porter
• CommentTimeDec 6th 2011
• (edited Feb 25th 2012)

I have started various entries on parts of combinatorial group theory, including adding to group presentation, and started Tietze transformation.

• CommentRowNumber2.
• CommentAuthorTim_Porter
• CommentTimeFeb 25th 2012

I have added a skethc proof into Nielsen-Schreier theorem with a little discussion of the Schreier index formula. An interesting question is whether the topological proof using covering graphs has some sort of higher dimensional analogue. I have no feeling for it however. It would not generalise in any obvious way but perhaps some complicated analogue of the Schreier index formula might exist for (free?) 2-groups? Is there a categorification of those results?

• CommentRowNumber3.
• CommentAuthorTim_Porter
• CommentTimeFeb 25th 2012
• (edited Feb 25th 2012)

I had a further thought on this. The Reidemeister Schreier result and the Schreier index formula may interact with Abels and Holz’s ideas on ’higher generation by subgroups’. They have several subgroups in a group and from information on the syzygies of presentations of the subgroups they find information on a presentation of the big group by combining the ’local’ syzygies with homotopy information on the nerve of the cover by the cosets of groups in the family. (This construction is related to Volodin spaces as used in algebraic K-theory.)

(Added later: I have added some material into Vietoris complex relating to these ideas.)

• CommentRowNumber4.
• CommentAuthorMike Shulman
• CommentTimeFeb 26th 2012

A different question that occurs to me is to what extent the topological/groupoid argument could be phrased in homotopy type theory. Bouquets of circles are easy to describe as higher inductive types. It seems that the whole proof can’t possibly go through, since the theorem is not valid constructively, but it would be interesting to see where it breaks down. That might provide some insight into the higher-dimensional version as well.

• CommentRowNumber5.
• CommentAuthorTim_Porter
• CommentTimeFeb 26th 2012

@Mike: The Higgins proof is in the TAC reprint. I suspect the freeness proof does need a maximal tree and doesn’t that cause problems under certain circumstances. This could be rephrased as a collapsing argument reducing the graph to a bouquet of sphere, but I do not know how that could be done.

On a perhaps related topic, I have been wondering if the constructive homotopy type theory has looked at simple homotopy theory since naively simple homotopy equivalences are more ’constructible’ than general ones. In a few hours I leave to go to the LI2012 meeting in Marseille and there will be a lot about HTT in that, so perhaps I will have a better idea afterwards.

• CommentRowNumber6.
• CommentAuthorMike Shulman
• CommentTimeFeb 26th 2012
• (edited Feb 29th 2012)

Yes, I decided last night that it was probably the maximal tree where it would go wrong. But I think the rest of it should go through.

Here’s one way to think about categorifying the Schreier index formula. The ordinary Schreier index formula can be viewed as a statement about the multiplicativity of Euler characteristic. Namely, if $G$ is a free group, then its classifying space $B G$ (being a bouquet of circles) is a finite cell complex, hence Spanier-Whitehead dualizable. In particular, therefore, we can calculate its Euler characteristic as the trace of its identity map in the stable homotopy category. Of course if $G$ has rank $r$, then $\chi(B G) = 1-r$. Now if $H$ is a subgroup of $G$, we have a fibration sequence $G/H \to B H \to B G$, and Euler characteristics are multiplicative along fibrations:

$\chi(B H) = \chi(B G) \cdot \chi(G/H)$

Hence if $H$ has index $i$ and rank $s$, we have

$1-s = (1-r)\cdot i$

yielding the Schreier index formula. Now this is all stuff that can be said more generally about any dualizable spaces, not just those of the form $B G$. However, I kind of doubt that there is a notion of “free 2-group” whose classifying space would be dualizable, because $S^2$ is not a 2-type (unlike $S^1$, which is a 1-type). We could consider free $\infty$-groups that are generated only by 1- and 2-dimensional generators. Then their classifying spaces would be 2-dimensional CW complexes and hence dualizable, and the multiplicativity of Euler characteristic would apply to any fibration sequence we might construct.

By the way, it’s possible to prove in generality (and in fact, for abstract reasons) that if $E \to B$ is a fibration whose fibers are Spanier-Whitehead dualizable and whose base $B$ is Costenoble-Waner dualizable over itself (a priori a stronger condition than being S-W-dualizable, but all finite cell complexes are still C-W-dualizable), then the total space $E$ is again S-W-dualizable. This implies in particular that for any subgroup $H\lt G$ of a free group, the classifying space $B H$ is S-W-dualizable. I don’t know whether this proves the Nielsen-Schreier theorem; are there non-free groups whose classifying space is dualizable?

Finally, to your last thought: my initial reaction is that simple homotopy theory won’t naturally have anything to do with HoTT, since the latter is really about $\infty$-groupoids while the notion of simple homotopy depends on having actual topological spaces. But I could be wrong.

• CommentRowNumber7.
• CommentAuthorTim_Porter
• CommentTimeFeb 26th 2012

Simple homotopy uses either an explicit cylinder construction or a set of nice generating cofibrations and exists in abstract settings (Eckmann and Siebenmann) and so it may be much more applicable that at first it seems.

• CommentRowNumber8.
• CommentAuthorMike Shulman
• CommentTimeFeb 29th 2012

Mark Grant on MO has pointed me to Brown’s Cohomology of groups, where I’ve learned that many non-free groups are dualizable. Brown also defines the Euler characteristic of a dualizable group to be that of its classifying space, and includes the corresponding version of the multiplicativity formula as Theorem 6.3:

$\chi(\Gamma') = (\Gamma : \Gamma') \cdot\chi(\Gamma)$

He doesn’t remark that this includes the Schreier index formula, though.

• CommentRowNumber9.
• CommentAuthorTim_Porter
• CommentTimeFeb 29th 2012

Thanks, Mike. I had seen that formula but had forgotten it. :-(