Part 1: Sylvain Ribault: The Virasoro algebra and its representations

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  • Опубликовано: 14 янв 2025

Комментарии • 6

  • @symmetricshaman
    @symmetricshaman 5 лет назад +3

    Around the 22:00 mark, it should read L_1 |\chi> = 2 L_0 |v> = 2 \Delta |v>. Great lectures!

  • @antoinebrgt
    @antoinebrgt 5 лет назад +2

    Also at24:54 it should be 6 delta, not 3 delta.
    Indeed, great lectures !

  • @ANAND02120
    @ANAND02120 4 года назад

    Can you please tell us how would you come up with such change of variable? Is it by trial?

  • @ANAND02120
    @ANAND02120 4 года назад

    Why \mathcal{R}_{r,s}$ is irreducible? Is this due to fact that it's generated by a singular vector?

    • @alexeybychkov2742
      @alexeybychkov2742 3 года назад +2

      I’m not an expert by any means, so everything I'm about to say can be totally wrong.
      Consider the Virasoro module R_{r,s} = V_{Δ}/V_{Δ+rs}. That is, the initial module V_Δ has a singular vector v_{rs} on level N = rs. Suppose R_{rs} is reducible. Then it should have a submodule R', generated by a singular vector v' on some level N' with \Delta < N < \Delta + rs. But if it was so, in our procedure of finding singular vectors level-by-level we should have obtained v' first (and not v_{rs}) and stopped there.
      So, for me, the irreducibility of R_{r,s} follows more or less from the very construction of it: one finds a singular vector on lowest possible level, and takes a quotient with respect to the ideal it generates.
      Please, let me know if it makes sense.

    • @ANAND02120
      @ANAND02120 3 года назад

      @@alexeybychkov2742 I agree with you so by definition of \delta_{} the Verma module generated by the primary vector |v> with eigenvalue with respect to L_0 is \delta_{} have a Null vector at a level rs and in genric condition, it would not have a null vector in a level less than that. It depends on charge c (only) with a choice of generic c what you said is true. Thanks a lot for the reply.