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  2. Topics in Analysis: Example Sheet 3 Michaelmas 2011-12 N. ...

    https://www.dpmms.cam.ac.uk/study/II/TopicsinAnalysis/2011-2012/11-12sheet3.pdf
    15 Nov 2011: 3) Let Tj be the jth Chebychev polynomial. Suppose γj is a sequence of non-negative numberswith.
  3. 60 APPENDIX: ADEQUATE SUBGROUPS ROBERT GURALNICK, FLORIAN HERZIG,…

    https://www.dpmms.cam.ac.uk/~jat58/appendix.pdf
    29 Jul 2011: BJ,TJ)) is a Borel andmaximal torus in I (resp. J). (This follows from the fact that anysmooth connected soluble subgroup of (resp. ... isogeny of I onto its image I, which is a semisimple algebraic group.Note that T/Fl = TI TJ and that B/Fl = BI BJ
  4. Representation Theory Lectured by S. Martin Lent Term 2009, ...

    https://www.dpmms.cam.ac.uk/study/II/RepresentationTheory/2010-2011/Representation_Theory.pdf
    12 Jul 2011: Representation Theory. Lectured by S. Martin. Lent Term 2009, 2010, 2011. 1 Group Actions 1. 2 Linear Representations 3. 3 Complete Reducibility and Maschke’s Theorem 7. 4 Schur’s Lemma 10. 5 Character Theory 13. 6 Proofs and Orthogonality 17. 7
  5. EXPLICIT n-DESCENT ON ELLIPTIC CURVESIII. ALGORITHMS J.E. CREMONA,…

    https://www.dpmms.cam.ac.uk/~taf1000/papers/n-descent-III.pdf
    18 Jul 2011: where zi = z(Ti). Alternatively, since. r(Ti, Tj) =. {x ei if i = jy/(x ek) if {i, j, k} = {1, 2, 3},. ... ρ(Ti, Tj) =. {αi if i = jb/αk if {i, j, k} = {1, 2, 3},.
  6. 29 Jul 2011: 1n1 00 αj. )m. ]. One computes easily that V j = qjn1/2t(Tj). ... Note that since t is analgebra homomorphism, the fact that the Tj commute implies that theoperators V j must also commute.
  7. PART II REPRESENTATION THEORYSHEET 3 Unless otherwise stated, all ...

    https://www.dpmms.cam.ac.uk/study/II/RepresentationTheory/2010-2011/repex3.pdf
    24 Jan 2011: tr](this means that σ is a product of disjoint cycles of length t1 > > tr where n = t1 tr,and some of the tj may be equal to 1.

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