Orthogonality of characters

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  • Опубликовано: 18 дек 2024

Комментарии • 6

  • @leilalovegood4131
    @leilalovegood4131 5 лет назад +2

    Thank you, your videos have saved my life!

  • @kanextension7076
    @kanextension7076 6 лет назад +1

    At 9:00, I don't understand how you went from average of the values of the character of Hom(V,W) to dim Hom_G (V,W). I see that the lemma gives that the expression in the second line is the dim of G-linear maps that are fixed by G. Why is this equal to dim Hom_G(V,W). Doesn't Hom_G(V,W) contain maps that commute with each rho_g? ie, rho_g (phi) = phi (rho_g) for all g in G. But maps fixed by G only satisfy rho_g(phi) = phi. Why are these the same?

    • @HarpreetBedimath
      @HarpreetBedimath  6 лет назад

      There is a small step missing here, one can make Hom(V,W) to Hom_G(V,W) by mapping \phi to 1/|G|\sum_g g\phi.

    • @davidt3039
      @davidt3039 6 лет назад

      @Kan : The expression "rho_g(phi)" has two different meaning in the end of your post. The first one refers to the action of G on W, the second one to the action of G on Hom(V,W). The point is that Hom_G(V,W) is equal to Hom(V,W)^G, in other words the equivariant linear maps from V to W are exactly the linear maps from V to W that are fixed by the action of G on Hom(V,W).
      Hope this helps.

  • @kanextension7076
    @kanextension7076 6 лет назад

    At 5:55, 1. in the definition of hermitian inner product, there is an inconsistency in notation. = \bar{a} + \bar{b} .

  • @davidt3039
    @davidt3039 7 лет назад

    Shouldn't you assume right from the beginning (or recall that hypothesis) that the group G is finite ?