First floor of the K-theory space

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  • Опубліковано 27 жов 2024

КОМЕНТАРІ • 7

  • @alexandershapiro28
    @alexandershapiro28 Рік тому

    I've been enjoying your series, and I saw the construction of an abelian group from semigroups in another talk by prof. Paul Baum after watching your series, and it all clicked! This group completion is like doing god's work, I would like to know if there are similar constructions for a magma to evolve through monoid and then semigroups. I have a few questions from the video, I know that it might not be answered since you have said that you may not be answering but here we go. How does a projective module would look geometrically? How does it look (trivially) for say the reals? Does this imply that the projective module is a semigroup? I know from wiki that it needs to have a basis in order for a module to become a proj. module. Is there a projective module counterpart for fields, what would this be ? I read that it's just called projective module since everything is nice and can be localized "everywhere", idk I guess by definition? Last question, the first homotopy group would be under what type of "classes"? I'm sorry if these questions don't make sense, I'm learning on my own after I fell in love with a ring theory course that I took last semester.

    • @saadslaoui8831
      @saadslaoui8831 Рік тому +1

      Some input: You can think of a projective module over a ring R as a vector bundle over a space associated to R, denoted by Spec(R) and called the Zariski spectrum of R-this is an affine scheme, a type of space in algebraic geometry.
      For a field F, Spec(F) consists of a single point, and a vector bundle over it consists of a single F-vector space V-i.e. all projective modules over F are free.
      For the ring C[t] of polynomials in one variable with complex coefficients, the space Spec(C[t]) looks like the complex plane, and it is a theorem that vector bundles on this space are always trivial, so that they look like a cartesian product V x Spec(C[t]), and all projective C[t]-modules are free. You can think of this as an analog of the fact that all vector bundles on a contractible topological space are trivial-though contractibility doesn’t quite make sense for affine scheme, a line certainly “feels” like it should be contractible.
      The existence of non-free projective modules over a ring R is an indication that the space Spec(R) is rich enough to support non-trivial vector bundles over it. As a simple example, if R is the direct sum of two smaller rings S and T, then Spec(R) will look like the disjoint union of Spec(S) and Spec(T), so you could have a vector bundle that has fibers of different dimensions over each component, corresponding to a projective R-module that is not free. Rings of an arithmetic nature also tend to admit non-free projective modules.
      Cheers!

  • @ProfessorJohnSmith
    @ProfessorJohnSmith Рік тому

    I dont see any previous video he refers to?

    • @math-life-balance
      @math-life-balance  Рік тому

      oh, sorry, let me add a link

    • @math-life-balance
      @math-life-balance  Рік тому

      he refers to the video "the magic of group completion", where K-theory space was introduced

  • @davethesid8960
    @davethesid8960 Рік тому

    How old is he?