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Stream: deprecated: algebraic geometry

Topic: stokes' theorem as a coend


view this post on Zulip Tim Hosgood (Apr 01 2020 at 19:36):

There's a nice little story about how you can understand Stokes' theorem as a coend (see e.g. https://twitter.com/ququ7/status/1169048827703287814), but I can't figure out how it's really a co_end_ and not just a co_wedge_, i.e. what does the universality condition of the coend tell you in all this?

Fun fact: Stokes' theorem is a result about coends. (credits to Tim Campion for having told me that a few years back!) https://twitter.com/ququ7/status/1169048827703287814/photo/1

- Fo Uche (@ququ7)

view this post on Zulip Amar Hadzihasanovic (Apr 02 2020 at 16:05):

The “coend” part is the “Bonus point” part of what Fosco wrote, isn't it?
The existence of the Stokes cowedge induces a universal map from the coend of CΩC \otimes \Omega to the constant R\mathbb{R} functor. I have not tried to solve the exercise to see what it amounts to, though.

view this post on Zulip Tim Hosgood (Apr 02 2020 at 17:58):

ah, that would make sense! maybe i'll try to have a think about this bonus exercise then