@hierarchon uhh hold on let me re-read my algebra texts and i'll get back to you

@KitRedgrave @hierarchon I suspect you need some post-graduate mathematics to even understand the question being asked here. I have a minor in math and I'm sitting here going "I recognize most of these words and symbols but have no idea why they're in that order"

@hierarchon @KitRedgrave Firstably how dare you have a hamburger and not use it for the Hom set

@CoronaCoreanici @KitRedgrave i didn't make this, i just found it on some image search while i was looking for something else :v

@hierarchon @KitRedgrave OK, got nerd-sniped to fuck, and here's my conclusion

whoever wrote this is an asshole, because the index variable i in the exact sequence isn't actually specified anywhere, and I don't think this actually holds for all possible i (try i = n+1, for instance - the middle term vanishes, but the second and fourth don't, so this can't be exact)

@hierarchon @KitRedgrave Admittedly I'm not 100% sure that $\banana$ is actually the Tor functor. Also, $\mathscr{C}$ isn't actually defined anywhere, either, so unless that's meant to be a comment that Hom-ing into Set from any other category is contravariant, this problem has some issues

@hierarchon @KitRedgrave ...hush, abstract predators are allowed to show their power level little a, as a treat.

@CoronaCoreanici @KitRedgrave i don't know nearly enough about this kind of thing as i want to, but i never get around to learning

@CoronaCoreanici @hierarchon @KitRedgrave yeah that's what I'm confused about, a lot of stuff isn't defined at all. I'm not convinced it's even asking a well-formed question(although I'm also garbage at anything to do with category theory)

@zardoz @hierarchon @KitRedgrave The question it's asking is actually decently well-formed apart from that. It's asking about the nature of a graded ring over Z/2 in terms of the nth cohomology groups of (I think?) the real projective n-space. (the dimension of which is ALSO unspecified)

@CoronaCoreanici @hierarchon @KitRedgrave yeah but there's also all that shit about H that looks completely pointless

@CoronaCoreanici @zardoz @hierarchon @KitRedgrave hmm, yes, from this i can deduce that these are definitely words referring to concepts, indeed

@CoronaCoreanici @hierarchon @KitRedgrave I think most of it is superfluous information though. The object they're asking about isn't very complicated.

@zardoz @hierarchon @KitRedgrave Yeah, now that I think about it, the exact sequence isn't used anywhere in the actual question, and neither is Tor.

@CoronaCoreanici @hierarchon @KitRedgrave I'm gonna take a total shot in the dark and say that the answer is (Z/2Z)[x]/<x^n+1>

@zardoz @hierarchon @KitRedgrave That definitely can't be right - they're looking for a graded cohomology ring, and that'd have to have indexed components over k \in \nat.

@zardoz @hierarchon @KitRedgrave Oh, wait, are you saying that the ring is \oplus_{n \in \nat} (Z/2)[x]/<x^n + 1>?

@CoronaCoreanici @hierarchon @KitRedgrave no, just (Z/2)[x]/<x^n + 1>

plus a bunch of zero rings because H^N (S,R) = 0 for N > dim(S)

@zardoz @hierarchon @KitRedgrave Ah, I see. But won't some of the homologies for RP^n in Z/2 coefficients be nontrivial for k < n?

@CoronaCoreanici @hierarchon @KitRedgrave yeah. The individual cohomology groups would be {0,x^i} for some fixed i

@zardoz @hierarchon @KitRedgrave Then the direct sum of all of those would be your graded ring of cohomology rings... I think? There's something to check about the actual grading on the ring but that looks to follow immediately.

@CoronaCoreanici @zardoz @hierarchon @KitRedgrave
annoyingly I think the sequence given is stated incorrecly. It's the universal coefficient sequence for cohomology but the role of the burger and the sandwich should be swapped

@hierarchon Hah, such an elementary problem is nothing to my advanced intellect.

However, it would be unfair of me to spoil the answer. I will leave it to the rest of you to sort it out.

No, no, my friend. You need to use the correct terminology.

"The solution is left as an exercise to the reader." :blobcoffee:

@hierarchon i know for a fact that burger is 1/2 but that's as far as I got

@hierarchon contravariant? derived functor? When did this leave math and enter programming?

@moonbolt the way mathematicians use 'functor' is very different from the way the c++ nerds use it (but closer to how the haskell nerds use it)

@moonbolt a 'functor' in c++ means an object you can call (i.e., that implements `operator()`)


H^*(🌭;🍔) = 🍔[🍟]/(🍟^(n+1))

where 🍟 is of degree 🍇

@hierarchon Okay listen I'm dumb as fuck so while I can appreciate the surface-level joke, is there another, secret layer where this is actually a solvable thing? Or is it just nonsense?

@hierarchon it has no solution because cookie is never defined and could be zero, which would cause hamburger to also be undefined.

Sign in to participate in the conversation
inherently digital

The social network of the future: No ads, no corporate surveillance, ethical design, and decentralization! Own your data with Mastodon!