AffineSymmetricGroup
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From: GeometricGroup on 20 May 2010 01:45 Let \tilde{S_n} be affine symmetric group. Elements x \in \tilde{S_n} are bijections from Z to itself satisfying: 1. x(i+n)=x(i)+n for all i \in Z. 2. \sum_{i=1}^n x(i)=(n+2)C2, where nCr denotes "n choose r". If the notation is not straightfoward, plz check it directly at source: http://www.math.ucdavis.edu/~anne/WQ2009/MAT280-Lecture20.pdf (first paragraph) I can't understand why 1 and 2 hold. Any explanation or simple example? Thanks.
From: Derek Holt on 20 May 2010 09:26 On 20 May, 10:45, GeometricGroup <ggx...(a)gmail.com> wrote: > Let \tilde{S_n} be affine symmetric group. Elements x \in \tilde{S_n} are bijections from Z to itself satisfying: > > 1. x(i+n)=x(i)+n for all i \in Z. > 2. \sum_{i=1}^n x(i)=(n+2)C2, where nCr denotes "n choose r". > > If the notation is not straightfoward, plz check it directly at source: > > http://www.math.ucdavis.edu/~anne/WQ2009/MAT280-Lecture20.pdf > > (first paragraph) > > I can't understand why 1 and 2 hold. Any explanation or simple example? It should be (n+1)C2 in 2., not (n+2)C2. This is the definition of the group, so it doesn't make sense to ask why 1 and 2 hold. Of course you do have to prove that bijections satisfying 1 and 2 form a group. Derek Holt.
From: GeometricGroup on 20 May 2010 05:41 > On 20 May, 10:45, GeometricGroup <ggx...(a)gmail.com> > wrote: > > Let \tilde{S_n} be affine symmetric group. Elements > x \in \tilde{S_n} are bijections from Z to itself > satisfying: > > > > 1. x(i+n)=x(i)+n for all i \in Z. > > 2. \sum_{i=1}^n x(i)=(n+2)C2, where nCr denotes "n > choose r". > > > > If the notation is not straightfoward, plz check it > directly at source: > > > > > http://www.math.ucdavis.edu/~anne/WQ2009/MAT280-Lectur > e20.pdf > > > > (first paragraph) > > > > I can't understand why 1 and 2 hold. Any > explanation or simple example? > > It should be (n+1)C2 in 2., not (n+2)C2. > > This is the definition of the group, so it doesn't > make sense to ask > why 1 and 2 hold. Of course you do have to prove that > bijections > satisfying 1 and 2 form a group. > > Derek Holt. Can we derive that definition from the Coxeter graph of Affine symmetric group shown in http://en.wikipedia.org/wiki/Coxeter_group#Affine_Weyl_groups Another thing is that 2 does not seem to be definition. How can we find that number (n+1)C2? Thanks
From: GeometricGroup on 20 May 2010 05:54 The detailed Coxeter graph of my question is found in http://www.combinatorics.org/Volume_5/PDF/v5i1r18.pdf (In page 5, the first figure in table 1) I just started learning this stuff, so it confuses me a lot. Thanks for any help.
From: Derek Holt on 20 May 2010 16:54
On 20 May, 14:54, GeometricGroup <ggx...(a)gmail.com> wrote: > The detailed Coxeter graph of my question is found in > > http://www.combinatorics.org/Volume_5/PDF/v5i1r18.pdf > > (In page 5, the first figure in table 1) > > I just started learning this stuff, so it confuses me a lot. > > Thanks for any help. You have removed the context! Properties 1 and 2 that you mentioned in your earlier post (corrected): > Let \tilde{S_n} be affine symmetric group. Elements x \in \tilde{S_n} are bijections from Z to itself satisfying: > > 1. x(i+n)=x(i)+n for all i \in Z. > 2. \sum_{i=1}^n x(i)=(n+1)C2, where nCr denotes "n choose r". are Lemma 4 and Proposition 9 of the paper you cited above. Derek Holt. |