Double hashing: Difference between revisions

From formulasearchengine
Jump to navigation Jump to search
en>WonderPhil
No edit summary
 
en>Mild Bill Hiccup
Line 1: Line 1:
== bladder Abercrombie Online Coupons ==
In [[mathematics]], a '''Young symmetrizer''' is an element of the  [[group ring|group algebra]] of the [[symmetric group]], constructed in such a way that the image of the element corresponds to an [[irreducible representation]] of the symmetric group over the [[complex number]]s. A similar construction works over any field, and the resulting representations are called '''[[Specht module]]s'''. The Young symmetrizer is named after British mathematician [[Alfred Young]].


Allow us to customize group transportation for your specific event and budget by calling 7815356344 today.Offpeak, local transfers, [http://www.qandaresearch.com.au/wp-content/plugins/akismet/soliloquy.php?page=106-Abercrombie-Online-Coupons Abercrombie Online Coupons] multiple buses and long trips qualify for discounts. Call 7815356344 for a detailed quote. See Rate Information tab above for additional important details.24Passenger Minicoach Exterior View24Passenger Minicoach Interior ViewAllow us to modify your reservation to fit your exact needs and budget. <br><br>The electronic service provided by FileLater generates a confirmation once you're approved and if your extension request is rejected, you will get a reason why. Thompson said more often than not the IRS rejects an extension just because a name doesn't match a Ssn, a mistyped SSN or a name [http://www.caloundrabridgeclub.com.au/results/unprocessed/frames.asp?headeventid=11-Nike-Shox-R5-Running-Shoes Nike Shox R5 Running Shoes] change due to marriage or other reasons. A FileLater client can correct and resubmit the extension form with no additional fee. <br><br>I answered their questions as well as I could. Pain would be much like strong menstrual cramps. Complications were rare. So the best way to get good ranking is thru good content. Content ca . (608 Views)Ease Your Vast Search With Free Articles Directory  How often have you felt yourself lost within the information jungle present on the internet? The answer will definitely come in two digits. <br><br>District Court in Central California a week ago against [http://www.drlindycrocker.com/AWStats/header.asp?n=91-New-Balance-Stockists-Sydney New Balance Stockists Sydney] Hunter Moore, former operator of. Does Brandi Schulz from storage wars have breast implants ? The KGB Agent answer: Brandi Passante, Jarrod Schulz's longtime girlfriend and partner within the reality. Pictures of Brandi Passante from Storage Wars in the Grand ReOpening of her and partner Jarrod Schulz's Now and Then Thrift Store in Orange, Californ.. <br><br>Within our [http://www.orgoneaustralia.com.au/cp/Scripts/Perl/Counter/setting.asp?i=79-Michael-Kors-Australia-Bags Michael Kors Australia Bags] tireless efforts to produce the best Corsos in the world, we have produced countless Champions, 2 Working Dogs of the season, set the record for quantity of Grand Championships achieved in 1 year, set the record for number of Reproduction Championships achieved in 12 months, produced the top Best in Show winning Female Corso of all time. It was a CastleGuard dog that won the very first weight pulling medal inside a National competition in the US. It was a CastleGuard dog who earned the very first BH title and that earned the very first Flyball titles in the US.. <br><br>KegelsKegel exercises are small internal contractions from the pelvic floor muscles that support your urethra, bladder, uterus, and rectum. Strengthening your pelvic floor muscles improves circulation for your rectal and vaginal area, assisting to keep hemorrhoids at bay and speeding healing after an episiotomy or tear, for those who have one during childbirth. Repeat ten times.<ul>
==Definition==
 
Given a finite symmetric group ''S''<sub>''n''</sub> and specific [[Young tableau]] λ corresponding to a numbered partition of ''n'', define two [[permutation group|permutation subgroups]] <math>P_\lambda</math> and <math>Q_\lambda</math> of ''S''<sub>''n''</sub> as follows:
  <li>[http://hbancai.com/forum/forum.php?mod=viewthread&tid=54613&fromuid=8432 http://hbancai.com/forum/forum.php?mod=viewthread&tid=54613&fromuid=8432]</li>
 
 
:<math>P_\lambda=\{ g\in S_n : g \text{ preserves each row of } \lambda \}</math>
  <li>[http://cert-centers.com/forum.php?mod=viewthread&tid=1265661 http://cert-centers.com/forum.php?mod=viewthread&tid=1265661]</li>
 
 
and
  <li>[http://www.christophegreffet.fr/blog/index.php?2013/07/09/81-une-semaine-cruciale-pour-le-finances-departementales/ http://www.christophegreffet.fr/blog/index.php?2013/07/09/81-une-semaine-cruciale-pour-le-finances-departementales/]</li>
 
 
:<math>Q_\lambda=\{ g\in S_n : g \text{ preserves each column of } \lambda \}.</math>
  <li>[http://www.middleeasttransparent.com/spip.php?article9467&lang=ar&id_forum=10581/ http://www.middleeasttransparent.com/spip.php?article9467&lang=ar&id_forum=10581/]</li>
 
 
Corresponding to these two subgroups, define two vectors in the [[group algebra]] <math>\mathbb{C}S_n</math> as
</ul>
 
:<math>a_\lambda=\sum_{g\in P_\lambda} e_g</math>
 
and
 
:<math>b_\lambda=\sum_{g\in Q_\lambda} \sgn(g) e_g</math>
 
where <math>e_g</math> is the unit vector corresponding to ''g'', and <math>\sgn(g)</math> is the signature of the permutation. The product
 
:<math>c_\lambda := a_\lambda b_\lambda = \sum_{g\in P_\lambda,h\in Q_\lambda} \sgn(h) e_{gh}</math>
 
is the '''Young symmetrizer''' corresponding to the [[Young tableau]] λ. Each Young symmetrizer corresponds to an irreducible representation of the symmetric group, and every irreducible representation can be obtained from a corresponding Young symmetrizer. (If we replace the [[complex number]]s by more general [[field (mathematics)|field]]s the corresponding representations will not be irreducible in general.)
 
==Construction==
Let ''V'' be any [[vector space]] over the [[complex number]]s.  Consider then the [[tensor product]] vector space <math>V^{\otimes n}=V \otimes V \otimes \cdots \otimes V</math> (''n'' times). Let ''S''<sub>n</sub> act on this tensor product space by permuting the indices. One then has a natural [[group ring|group algebra]] representation  <math>\mathbb{C}S_n \rightarrow \text{End} (V^{\otimes n})</math> on <math>V^{\otimes n}</math>.
 
Given a partition λ of ''n'', so that <math>n=\lambda_1+\lambda_2+ \cdots +\lambda_j</math>, then the [[image (mathematics)|image]] of <math>a_\lambda</math> is
:<math>\text{Im}(a_\lambda) := a_\lambda V^{\otimes n} \cong
\text{Sym}^{\lambda_1}\; V \otimes
\text{Sym}^{\lambda_2}\; V \otimes \cdots \otimes
\text{Sym}^{\lambda_j}\; V.
</math>
 
For instance, if <math>n =4</math>, and <math>\lambda = (2,2)</math>, with the canonical Young tableau <math>\{\{1,2\},\{3,4\}\}</math>. Then the corresponding <math>a_\lambda</math> is given by <math> a_\lambda = e_{\text{id}} + e_{(1,2)} + e_{(3,4)} + e_{(1,2)(3,4)}</math>. Let an element in <math>V^{\otimes 4}</math> be given by <math>v_{1,2,3,4}:=v_1 \otimes v_2 \otimes v_3 \otimes v_4</math>. Then
:<math> a_\lambda v_{1,2,3,4} = v_{1,2,3,4} + v_{2,1,3,4} + v_{1,2,4,3} + v_{2,1,4,3} = (v_1 \otimes v_2 + v_2 \otimes v_1) \otimes (v_3 \otimes v_4 + v_4 \otimes v_3). </math>
The latter clearly span <math> \text{Sym}^2\; V\otimes \text{Sym}^2\; V</math>.
 
The image of <math>b_\lambda</math> is
:<math>\text{Im}(b_\lambda) \cong
\bigwedge^{\mu_1} V \otimes
\bigwedge^{\mu_2} V \otimes \cdots \otimes
\bigwedge^{\mu_k} V
</math>
where μ is the conjugate partition to λ. Here, <math>\text{Sym}^i V </math> and <math>\bigwedge^j V</math> are the [[symmetric algebra|symmetric]] and [[exterior algebra|alternating tensor product spaces]].  
 
The image <math>\mathbb{C}S_nc_\lambda</math> of <math>c_\lambda = a_\lambda \cdot b_\lambda</math> in <math>\mathbb{C}S_n</math> is an irreducible representation<ref>See {{harv|Fulton|Harris|1991|loc=Theorem 4.3, p. 46}}</ref> of ''S''<sub>n</sub>, called a [[Specht module]]. We write
:<math>\text{Im}(c_\lambda) = V_\lambda</math>  
for the irreducible representation.
 
Some scalar multiple of <math>c_\lambda</math> is idempotent, that is  <math>c^2_\lambda = \alpha_\lambda c_\lambda</math> for some rational number <math>\alpha_\lambda\in\mathbb{Q}</math>. Specifically, one finds <math>\alpha_\lambda=n! / \text{dim } V_\lambda</math>. In particular, this implies that representations of the symmetric group can be defined over the rational numbers; that is, over the rational group algebra <math>\mathbb{Q}S_n</math>.
 
Consider, for example, ''S''<sub>3</sub> and the partition (2,1). Then one has <math>c_{(2,1)} = e_{123}+e_{213}-e_{321}-e_{312}</math>
 
If ''V'' is a complex vector space, then
the images of <math>c_\lambda</math> on spaces <math>V^{\otimes d}</math> provides essentially all the finite-dimensional irreducible representations of GL(V).
 
==See also==
* [[Representation theory of the symmetric group]]
 
==Notes==
<references/>
 
==References==
* William Fulton. ''Young Tableaux, with Applications to Representation Theory and Geometry''. Cambridge University Press, 1997.
* Lecture 4 of {{Fulton-Harris}}
* [[Bruce Sagan|Bruce E. Sagan]]. ''The Symmetric Group''. Springer, 2001.
 
[[Category:Representation theory of finite groups]]
[[Category:Symmetric functions]]
[[Category:Permutations]]

Revision as of 22:17, 22 December 2013

In mathematics, a Young symmetrizer is an element of the group algebra of the symmetric group, constructed in such a way that the image of the element corresponds to an irreducible representation of the symmetric group over the complex numbers. A similar construction works over any field, and the resulting representations are called Specht modules. The Young symmetrizer is named after British mathematician Alfred Young.

Definition

Given a finite symmetric group Sn and specific Young tableau λ corresponding to a numbered partition of n, define two permutation subgroups and of Sn as follows:

and

Corresponding to these two subgroups, define two vectors in the group algebra as

and

where is the unit vector corresponding to g, and is the signature of the permutation. The product

is the Young symmetrizer corresponding to the Young tableau λ. Each Young symmetrizer corresponds to an irreducible representation of the symmetric group, and every irreducible representation can be obtained from a corresponding Young symmetrizer. (If we replace the complex numbers by more general fields the corresponding representations will not be irreducible in general.)

Construction

Let V be any vector space over the complex numbers. Consider then the tensor product vector space (n times). Let Sn act on this tensor product space by permuting the indices. One then has a natural group algebra representation on .

Given a partition λ of n, so that , then the image of is

For instance, if , and , with the canonical Young tableau . Then the corresponding is given by . Let an element in be given by . Then

The latter clearly span .

The image of is

where μ is the conjugate partition to λ. Here, and are the symmetric and alternating tensor product spaces.

The image of in is an irreducible representation[1] of Sn, called a Specht module. We write

for the irreducible representation.

Some scalar multiple of is idempotent, that is for some rational number . Specifically, one finds . In particular, this implies that representations of the symmetric group can be defined over the rational numbers; that is, over the rational group algebra .

Consider, for example, S3 and the partition (2,1). Then one has

If V is a complex vector space, then the images of on spaces provides essentially all the finite-dimensional irreducible representations of GL(V).

See also

Notes

References

  • William Fulton. Young Tableaux, with Applications to Representation Theory and Geometry. Cambridge University Press, 1997.
  • Lecture 4 of Template:Fulton-Harris
  • Bruce E. Sagan. The Symmetric Group. Springer, 2001.