Knot at Lunch, July 5, 2007: Difference between revisions
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'''Definition.''' Let <math>\varphi:B\to S</math> be a group homomorphism; denote its action by <math>b\mapsto\bar b</math>; i.e., let <math>\bar b:=\varphi(b)</math> for every <math>b\in B</math>. Let "the virtualization <math>\operatorname{VB}</math> of <math>B</math>", or more precisely, "the virtualization <math>\operatorname{VB}_\varphi</math> of <math>B</math> with respect to <math>\varphi</math>", be the following quotient of the free product <math>B\star S</math> of <math>B</math> and <math>S</math>: |
'''Definition.''' Let <math>\varphi:B\to S</math> be a group homomorphism; denote its action by <math>b\mapsto\bar b</math>; i.e., let <math>\bar b:=\varphi(b)</math> for every <math>b\in B</math>. Let "the virtualization <math>\operatorname{VB}</math> of <math>B</math>", or more precisely, "the virtualization <math>\operatorname{VB}_\varphi</math> of <math>B</math> with respect to <math>\varphi</math>", be the following quotient of the free product <math>B\star S</math> of <math>B</math> and <math>S</math>: |
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{{Equation*|<math>\operatorname{VB}:=B\star S\left/ |
{{Equation*|<math>\operatorname{VB}:=B\star S\left/\bar b^{-1}b_1\bar b=b_2\right.</math> whenever <math>b,\,b_{1,2}\in B</math> and <math>b^{-1}b_1b=b_2</math> in <math>B</math>.}} |
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In words, this is "if |
In words, this is "if two element <math>b_{1,2}</math> of <math>B</math> are conjugate with conjugator <math>b</math>, in <math>\operatorname{VB}</math> they are conjugate also using the shadow of <math>b</math>". |
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Though note that we do not mod out by <math>b^{-1} |
Though note that under the same circumstances we do not mod out by <math>b^{-1}\bar b_1b=\bar b_2</math>. |
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It is clear that <math>\varphi</math> extends to a homomorphism <math>\hat\varphi:\operatorname{VB}\to S</math>. Let "the pure virtualization <math>\operatorname{PVB}</math> of <math>B</math>" be the kernel of that homomorphism: |
It is clear that <math>\varphi</math> extends to a homomorphism <math>\hat\varphi:\operatorname{VB}\to S</math>. Let "the pure virtualization <math>\operatorname{PVB}</math> of <math>B</math>" be the kernel of that homomorphism: |
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* Are there other examples in which this definition is interesting? |
* Are there other examples in which this definition is interesting? |
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* Do we gain any new insight by using this definition? |
* Do we gain any new insight by using this definition? |
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===Added July 20, 2007=== |
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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. |
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Latest revision as of 20:36, 20 July 2007
Invitation
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.
Best,
Dror.
Some Content
Definition. Let be a group homomorphism; denote its action by ; i.e., let for every . Let "the virtualization of ", or more precisely, "the virtualization of with respect to ", be the following quotient of the free product of and :
In words, this is "if two element of are conjugate with conjugator , in they are conjugate also using the shadow of ".
Though note that under the same circumstances we do not mod out by .
It is clear that extends to a homomorphism . Let "the pure virtualization of " be the kernel of that homomorphism:
Question. Is this definition at all interesting? More precisely:
- If is a braid group and is the corresponding symmetric group, can be reasonably identified with "virtual braids"?
- Does the that we get here agree with 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.
