6
$\begingroup$

I'm currently studying the basics of Banach and $C^*$-algebras. Almost all the proofs i've seen so far are very simple but some of them are extremely tricky (in my opinion). This tricky interplay between the bits of analysis and algebra make some of the proofs seem unilluminating and disconnected from the flow of the theory.

Is there an approach to operator algebras that uses category theory to simplify and "trivilize" the elementary theory?

Ideally it would use input from analysis (category of banach spaces, hilbert spaces, topological vector spaces) and algebra (category of associative algebras over $\mathbb{C}$ for a start) only when absolutely (and obviously) necessary.

$\endgroup$
  • 3
    $\begingroup$ What is your question? $\endgroup$ – Prahlad Vaidyanathan Oct 6 '15 at 14:55
  • $\begingroup$ @PrahladVaidyanathan Better? $\endgroup$ – Saal Hardali Oct 6 '15 at 15:11
  • $\begingroup$ So are you asking if more advanced math will make current math easier to understand/solve? $\endgroup$ – Ali Caglayan Oct 6 '15 at 15:38
  • 1
    $\begingroup$ @Alizter In what sense is category theory more advanced than operator algebras? Are categories more advanced than homological algebra? or commutative algebra? Even though it's entirely self contained while the other are not? $\endgroup$ – Saal Hardali Oct 6 '15 at 15:43
  • 10
    $\begingroup$ It might help if you gave a specific example of a "trivial" fact in operator algebras which you want to see as "trivially trivial". People may also not realize that you're referencing a famous quote by Peter Freyd, along the lines of "Perhaps the aim of category theory is to see that the trivial is trivially trivial". $\endgroup$ – tcamps Oct 6 '15 at 16:22
5
$\begingroup$

While I cannot give an authoritative answer that such an approach does not exist, it would really surprise me if it did.

That "tricky interplay" between algebra and analysis is precisely what makes operator algebras interesting. In particular, the close relation between an algebraic notion (the spectral radius) and a topological one (the norm) is at the cornerstone of the theory. Every key theorem uses that interplay in a very smart (rather than "tricky") way.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.