Knot at Lunch, July 5, 2007

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Dear Knot at Lunch People,

We will have our next summer lunch on Thursday July 5, 2007, at the usual place, Bahen 6180, at 12 noon.

As always, please bring brown-bag lunch and fresh ideas. I'm not sure what we will be talking about; perhaps just continue with last week's topics.

As always, if you know anyone I should add to this mailing list or if you wish to be removed from this mailing list please let me know. To prevent junk accumulation in mailboxes, I will actively remove inactive people unless they request otherwise.



Some Content

Definition. Let \varphi:B\to S be a group homomorphism; denote its action by b\mapsto\bar b; i.e., let \bar b:=\varphi(b) for every b\in B. Let "the virtualization \operatorname{VB} of B", or more precisely, "the virtualization \operatorname{VB}_\varphi of B with respect to \varphi", be the following quotient of the free product B\star S of B and S:

\operatorname{VB}:=B\star S\left/\bar b^{-1}b_1\bar b=b_2\right. whenever b,\,b_{1,2}\in B and b^{-1}b_1b=b_2 in B.

In words, this is "if two element b_{1,2} of B are conjugate with conjugator b, in \operatorname{VB} they are conjugate also using the shadow of b".

Though note that under the same circumstances we do not mod out by b^{-1}\bar b_1b=\bar b_2.

It is clear that \varphi extends to a homomorphism \hat\varphi:\operatorname{VB}\to S. Let "the pure virtualization \operatorname{PVB} of B" be the kernel of that homomorphism:


Question. Is this definition at all interesting? More precisely:

  • If B is a braid group and S is the corresponding symmetric group, can \operatorname{VB} be reasonably identified with "virtual braids"?
  • Does the \operatorname{PVB} that we get here agree with \operatorname{PVB}_n of last time?
  • Is this definition encountered anywhere else in mathematics?
  • Are there other examples in which this definition is interesting?
  • Do we gain any new insight by using this definition?

Added July 20, 2007

Well, following an email from Jana Comstock and Scott Morrison it is clear to me that the answer to the first question above is NO, and hence the other questions above become moot.