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11:00 AM
@Gato Bonjour :P
 
@SohamChowdhury yes it is, it is in fact one of the given intervals
 
ad the union is the other of them
 
wait. what the heck was I thinking?
damn. I was thinking difference.
goodness!
 
@Hippalectryon ça j'aime
 
11:01 AM
Hi everybody,
I have an idea of wikipedia-like website, but it’s a huge project and I would like to know if it would really be useful before really launching it.
The idea came from this : I was looking for what is a Galois extension, and I got as a result (wikipedia) « In mathematics, a Galois extension is an algebraic field extension E/F that is normal and separable; », but then I needed to look at what an « alebraic field extension » was, then for « alebraic field extension » I had to look to another page, etc. At the end I had approximately 1247 tabs opened in Mozilla and did not know be
 
i was thinking "intersection".
 
I think I just wrote the biggest post on this chat ever :p
 
@Pholochtairze you mean something like this?
@Tobias, does this kind of thing happen to you any more? :P
 
@SohamChowdhury yeah, all the time
 
I'm actually writing out solutions to the ~45 exercises in Hatcher's notes.
 
11:04 AM
@SohamChowdhury good idea. It will be a good way to learn
 
@Pholochtairze I had a similar idea, which was to make a with linked elements, each "parent" linked to the necessary closest "childs" necessary
 
being diligent and not-handwavy is new for me.
 
Hello!! Is someone of you familiar with Fast Fourier Transform?
 
In fact my idea was : you search something on the website and the answer is self sufficient given your knowledge, so it means you don't have to find other sources of knowledge elsewer.
 
11:05 AM
@Pholochtairze: really funny story along those lines. (look at the end of the page)
(@Tobias, give it a look)
 
@Hippalectryon yeah I think we have the same idea ;)
 
@Pholochtairze I'm unfortunately not good enough at programming (and maths) for that :/
 
0
Q: FFT procedure for evaluationg a polynomial at $N$ Fourier points

Mary StarThe following is the recursive FFT procedure of Algorithm for evaluationg a polynomial of length $N$ at $N$ Fourier points. Algorithm (FFT - fast Fourier transform). Input arguments. $ \ \ $ integer $N=2^m$ $ \ \ $ polynomial $\alpha (x)=\sum_{i=0}^{N-1}\alpha_i x^i$ $ \ \ $...

 
@SohamChowdhury haha very good quote, it is a perfect example of what would solve such a website :)
 
except that it would probably not work well in practice.
it'd be like learning analysis and stuff from Bourbaki's books (I've never tried, so this is hearsay).
 
11:16 AM
Don't really know if it would work well. The problem would maybe be "too much definitions" in a row ... Maybe we could add a threshold. Or you could have several levels of definition for a given notion, like one more rigorous (would need more definitions to get to what you want) or more intuitive (you would get more quickly to what you wanted and with less details, easier to follow) ?
 
figured out the short exact sequence problem, @Soham?
 
nope, didn't think about it.
I'm $\LaTeX$ing up solutions to Hatcher's exercises.
 
have you actually solved 45 of them?
 
no. solving as I go.
 
ok.
 
11:23 AM
should finish in this week.
 
latexting takes too much time. i usually write them up.
@SohamChowdhury we'll see. knowing hatcher's style, that'd be hard.
 
@BalarkaSen well, that way people can make fun of my mistakes.
@BalarkaSen even then, a month, maybe, at most. :P
 
ok.
 
what then? can I go on to the One True Book?
 
which book?
 
11:25 AM
Hatcher's AT book.
 
definitely not.
do algebra, do analysis.
 
analysis?
 
yes, didn't SB tell you to do Rudin?
 
aww, c'mon.
 
listen to him, otherwise there's no point of having a prof.
 
11:26 AM
ah, that day I was almost paralyzed and couldn't talk at all.
 
if he tells you to do AT, do it. if not, don't.
personally, I wouldn't advise you to do algebraic topology yet. you need more top. than Hatcher has, and more algebra.
 
what, specifically, in topology?
 
Hatcher doesn't have tietze, uryshon, tychonoff, iirc.
@SohamChowdhury all of point set top. in Munkres will come handy at some point of time.
 
right.
 
$$\sum_{n=1}^{\infty} (-1)^n 2^{-4 n} \left(\zeta \left(2 n+1,\frac{1}{4}\right)-\zeta \left(2 n+1,\frac{3}{4}\right)\right)$$
 
11:28 AM
@BalarkaSen right. and algebra? modules seem important.
 
definitely all of what D-F has until galois theory. knowing homological algebra won't hurt either.
 
hhello @BalarkaSen
 
(I don't know homological algebra, ps)
hello @iwriteonbananas
@Soham oh, and you have to do linear algebra from Artin.
:P
 
if $p:Y\to X$ is a covering and $X$ a CW complex, then so is $Y$ and cells project homeomorphically onto cells.
 
11:30 AM
yeah, guys. start your engines. :P
 
so AT is far away, forget about it.
 
this is an exercise from hatcher
 
@iwriteonbananas yep.
 
im struggling w/ the details
basically
 
n-cells lift to n-cells
 
11:31 AM
if $f:D^n\to X$ is a characteristic map, then it has $d$ lifts if it's a d-fold covering
so for each cell in $X$ we get d cells in $Y$
 
that's it. and then you take pushout. the covering does same thing too.
 
w/ characteristic maps the lifts
what do you mean by "then you take pushout" ?
 
pushout of spaces is adjunction space.
 
that's precisely what you do when you glue a cell to a space with a char map
 
11:34 AM
@BalarkaSen right
so we found a cell structure for $Y$
why do cells project homeomorphically to cells?
 
well, interior of cells.
 
it's because the interior of the cell is an evenly covered nbhd for any point lying inside the int.
you should prove this.
point-set stuff.
 
i have to go, ping me when you have proved it.
 
11:38 AM
alright, i got a lecture in 20 min, hopefully i'll have it by then
i guess maybe we can do this:
nvm
meh, i gotta go
 
12:06 PM
Hello@TobiasKildetoft
 
@Rememberme Hi
 
Hi @TobiasKildetoft :)
 
@evinda Hi
 
@TobiasKildetoft How are you?
 
@evinda Good, thanks
 
12:08 PM
Hello@evinda it was a no in the referendum right?
 
@TobiasKildetoft Any news?
 
@evinda news about what?
 
@Rememberme Hi!!! Exactly...
@TobiasKildetoft Anything
 
@evinda Not really, no
 
sometimes, no news is good news :)
 
12:10 PM
@skillpatrol Yes, that is true...
 
12:26 PM
With a careful approach of my research today, we can show that $$\sum _{k=1}^{\infty } \text{sech}(\pi k)=\frac{1}{2} \left(\frac{\sqrt{\pi }}{\Gamma \left(\frac{3}{4}\right)^2}-1\right)$$
that is a pretty elementary approach.
 
12:40 PM
@Hippalectryon
 
How do you always manage to find polygamma O_o
 
@Hippalectryon :D
@Hippalectryon That integral looks pretty interesting. :-)
I have such fun playing with these integrals ... :-)))))
(laughing now)
Anyway, I wanna engineer it a bit.
 
the RHS is interesting
 
@Semiclassical Yeap. I can rewrite the integral as $$\int_{-\infty }^{\infty } \frac{e^y}{\left(e^{2 y}+1\right) \left(y^2+1\right)} \, dy$$
 
from that alone, my hope would be that one can do the indefinite integral as wwll
 
12:50 PM
I might add this form in my book.
Well, yeah, and I think I'm going to explain a nice trick less used in integration. So adding it to my book I have this possibility.
 
ah, sech over $1+y^2$. nice
 
hmm. is it just a matter of recognizing that $0<e^{-2y}<1$ for all $y>0$, and so (upon using symmetry to eliminate the negative half of the integral) expanding in a series of exponentials?
 
@Semiclassical Right.
 
which would lead to integrals like $\int_0^\infty \frac{e^{-n y}}{1+y^2}\,dy$
not sure i know how to do that off the top of my head, tbh
 
12:57 PM
Indeed.
 
though it does just amount to a laplace transform evaluated at integer arguments
 
Hello @Semiclassical
How are you doing?
 
oh, alright. not completely awake right now, despite having been up for a while already
how about you?
 
@Semiclassical Aha... I am ok... thanks :)
 
glad to hear it
 
1:03 PM
@Semiclassical Today it's windy here... How is the weather there?
 
it's pleasant enough outside in minnesota right now. but, unfortunately, with the big fires going on in canada there's a lot of smoke that's drifted down our way. so at that level it's not too great to be outside
for details
 
This song makes me feel boundless youtube.com/watch?v=y1vYtG_KTFQ :-)))))))
Only marvellous stuff today here!!!!
Incredibly very happy! :-)
 
@Chris'ssistheartist I got the full album on disk :P
 
@Hippalectryon Nice. :D
 
@Semiclassical :(
 
1:07 PM
it's a tad annoying, but it'll clear out
 
It looks like a goddess.
 
O_o sech
 
for my taste i like the form $\frac{1}{2\pi}+\frac{k}{4}$ for the arguments
 
In France we don't use sec etc so it feels weird lol
 
@Semiclassical You're right, it might look nicer.
 
1:14 PM
esp. since it then amounts to an appropriate definite integral on $\frac{d}{dx}\psi^{(0)}(x)=\psi^{(1)}(x)$ which is suggestive
so that you're down to showing the equivalence of two definite integrals etc.
 
Hi, I am reading "Mikio Nakahara - Geomoetry,Topology and Physics". I am stuck. I dont understand how g can be function and an element to GL(m,K). i.imgur.com/TtwDgKY.png
 
hello @Karl
 
Previous sections defines Dual Space.
 
@BalarkaSen hi
 
Hello @KarlKronenfeld
 
1:19 PM
how's it going?
 
i think it's just an abuse of notation. $g$ by itself is a matrix, but they also use it to denote the resulting inner product
 
i.e., what kind of mathematics are you thinking about?
 
@BalarkaSen Interested in learning about forcing
 
what's forcing? i didn't know that was even a thing.
 
Cohen "invented it" to prove the independence of the continuum hypothesis from ZFC
 
1:21 PM
one could instead imagine notation like $\langle x_1,x_2\rangle_g = \langle g x_1,x_2\rangle$. it's just notation
 
oh.
 
@KarlKronenfeld Are you a student?
 
It's now a heavily abused set-theoretic tool @BalarkaSen
@evinda yeah
 
i've seen that word 'forcing' when i've glanced at some of the set theory questions around here. can't say i understood any of it, though
 
@KarlKronenfeld interesting
 
1:24 PM
@Hippalectryon Might I scare my readers with questions like the last one? :D
Hehe they are safe in the journey with me. :D
 
Ah got it. Thank you @Semiclassical
 
@Hippalectryon I'm concerned about finding a way to calculate the squared sech version.
 
@KarlKronenfeld i am trying to develop an analogy for paths and homotopies in galois theory
 
@BalarkaSen hmmm
 
Huy
1:29 PM
@evinda: Are you satisfied with the results on Sunday?
 
@BalarkaSen does it have a precedent?
 
it has a motivation, sure.
if you're prepared to listen, i can tell you
 
@Huy We'll see what happen... In the afternoon we will know more...
 
Huy
Did you vote too, @evinda?
 
@BalarkaSen sure
 
1:31 PM
@Huy Yes, I did...
 
What about the line "g : V -> V*", which implies g(x) is a function. Also lhs has to be a function in <,> according to i.imgur.com/g5xVZEP.png
 
$\pi_1X$ can be defined in two ways : 1) as homotopy classes of loops in $X$ and 2) as deck transformation group of the universal covering $\tilde{X} \to X$. the latter is very similar to $\mathsf{Gal}(\bar k/k)$. if the two theories are similar, there certainly should be an analog for loops and homotopies in galois theory. that might give a better perspective to look at the absolute galois group, thus the hunt.
i have made an infinitesimal progress.
scrap the definition of path i said though. "correct" paths are commutative diagram consisting of maps $*_0 : k \hookrightarrow \bar k, *_0 : k \hookrightarrow \bar k$ and an aut $\bar k \to \bar k$. this is a path between $*_0$ and $*_1$.
all of these are $k$-morphisms.
 
bit more sensible
 
i found this by observing that isomorphisms $\bar k \to \bar k$ between (not nessesarily the same) algebraic closures gives an isomorphism $\mathsf{Gal}(\bar k/k) \stackrel{\cong}{\to} \mathsf{Gal}(\bar k/k)$ by conjugation.
now recall that isomorphism between $\pi_1$ of different basepts come from conjugation by a path between those points, too
 
it's a very general point (no pun intended), yes.
 
1:40 PM
i am lacking a notion of homotopies, though. the analogy, as of now, gives the undesirable conclusion that Gal(k^alg/k) is the loopspace.
in any case, this says (or makes me realize) that (1) is closer to Gal(k^alg/k) than (2). auts k^alg --> k^alg fixing k might be similar to deck transformations of the universal cover, but this is actually an analog for a loop.
that is, k^alg is somehow, mysteriously, "smaller" than k - thus is not the correct analog for universal covers
 
i guess all of these has been done already by grothendieck, but i don't want to look there. kind of want to find out by myself.
 
I am a fan of that kind of approach
 
anyway, i am not proving anything either. just making a page of analogies.
 
you learn more that way anyway
 
1:45 PM
yes, and it's more fun too.
 
so what, if anything, do you have so far for the actual homotopies?
 
i am not sure what you mean by that, @Karl
 
i mean what sort of ideas do you have behind a possible definition?
 
well, \pi_1(X, x_0) can be defined as homotopy (rel x_0) classes of loops based at x_0. that's the definition I am aiming to mimic for Gal(k^alg/k)
apparently, grothendieck's theory mimics the deck transformation definition, from what i've heard.
 
@BalarkaSen Yeah, I know. How do you currently think you plan on defining homotopy class?
Oh, so you still want to pursue the deck transformation route?
 
1:59 PM
@KarlKronenfeld I don't know. That's the problem.
no, I don't want to do the deck transformation defn.
I just said it to emphasize that my approach differs from that approach.
(if my approach is even sensible)
 
Hello All … Does anyone know whether this question has a solution ? math.stackexchange.com/questions/1336250/…
 
what I want right now is a notion of homotopy. this seems reasonable, as Gal(k^alg/k) can now be seen -- from my interpretation -- as group of loops based at the point k --> k^alg. however, the abs. galois groups seems to say absolute (no pun intended) nothing about an equivalence class of the loops, so i don't know what possible could the analogy for homotopies be.
 
It is a conditional expectation of the product of normal and log normal .. Seems simple but the conditional probability means I am not sure what density function to use?
 
that's the whole obstacle i am facing : i don't know where to look for homotopies anymore.
 
you could try to simplify it down and view homotopies as a restricted type of assignment of points along one path to points along the other.
then figure out the restriction
 
2:05 PM
@Owatch Sure !
 
f, f' : k^alg --> k^alg be two paths sharing the same basepts *_0 : k --> k^alg and *_1 : k --> k^alg. there might be a buckload of equivalence relations such that f ~ f'. how do I know what's the right analog for htpy?
hmm. the situation seems a bit familiar.
(i am thinking about chain homotopies)
i have to go, unfortunately. thanks for the suggestions and having a look, @Karl!
 
have fun!
 
2:20 PM
The closed form of a crazy integral I evaluated.
$$\frac{1}{2} \left(\gamma _1\left(\frac{3}{4}+\frac{1}{2 \pi }\right)-\gamma _1\left(\frac{2+\pi }{4 \pi }\right)+(\gamma +\log (2 \pi )) \left(\psi ^{(0)}\left(\frac{2+\pi }{4 \pi }\right)-\psi ^{(0)}\left(\frac{3}{4}+\frac{1}{2 \pi }\right)\right)\right)$$
 
@Chris'ssis, do you upload your solutions here just in case you lose the paper form at home?
 
Huy
@KhallilBenyattou: Have you done some more work with metrics?
 
@KhallilBenyattou Solutions? Which solutions? :-)
 
Nope! I've just been chilling out, @Huy. ^_^
You know, like closed forms, @Chris'ssis :-)
 
@KhallilBenyattou ahhh :-))))
 
2:32 PM
I've been looking at step functions and regulated functions as of late, @Huy. :-)
 
Huy
@KhallilBenyattou: What are your answers?
 
Just started looking at the question, but the step function will be constant on the open intervals, so there'll be $k$ constant values of $\psi$ on the open interval. All that remain are the possible discontinuities at the endpoints, so we can add at most $k+1$, so $2k+1$ is the max card, @Huy
 
Huy
@KhallilBenyattou: Note that a "squelch function" needs to be constant on ANY partition of $(a,b)$.
 
Hmm, I don't really understand what is meant by 'any', @Huy
 
Huy
@KhallilBenyattou: Think about it. :)
@KhallilBenyattou: You surely have seen statements of the form $\forall \epsilon > 0 \dots$
 
2:41 PM
Oh, I see
Then it'd be the cardinality of the segment of the real line $[a,b]$?
 
Huy
No, now you're going into the wrong direction. I think you misunderstood something. You were closer to the actual answer just before. :P
I need to go now. See you later!
 
Thanks for the help, @Huy!
See ya! :-)
(I'll keep thinking about squelch functions!)
 
3:02 PM
@Hippalectryon ^^^
 
Ugh O_o
 
@Hippalectryon and collecting the previous results,we have that
 
AAH
xD
Still no clues for the squared version ?
 
@Hippalectryon I'm almost done with it too.
 
3:14 PM
@Hippalectryon It's about using elementary identities.
I'm glad I'm just at the very beginning of my mathematical activities. I may guess the stuff will come in the next years will be really amazing. :-)
 
@Chris'ssistheartist $\gamma_1=?$
 
@Semiclassical Stieltjes gamma
 
mmkay
 
In mathematics, the Stieltjes constants are the numbers that occur in the Laurent series expansion of the Riemann zeta function: The zero'th constant is known as the Euler–Mascheroni constant. == Representations == The Stieltjes constants are given by the limit (In the case n = 0, the first summand requires evaluation of 00, which is taken to be 1.) Cauchy's differentiation formula leads to the integral representation Various representations in terms of integrals and infinite series are given in works of Jensen, Franel, Hermite, Hardy, Ramanujan, Ainsworth, Howell, Coppo, Connon, Coffey, Choi...
 
are you intending those as the generalized stieltjes constants given on that page?
 
3:20 PM
@Semiclassical yes
 
thought so, thanks
 
I met in the past such stuff involving the generalized Stieltjes constants, but I was a bit afraid of them, that is some years ago. :-)
It was about some tough integrals.
The answers (the closed forms) were supposed to be written in more lines.
 
is there anything nice about the first derivative of $\gamma_1(x)$? because that result has a similar structure to the one you gave before, which (per my earlier remarks) could be written as a definite integral of $\psi^{(1)}(x)$
 
Hi, can someone please answer this really simple question?
Does H^s = W^{s,2} continuous embed into L^2?
 
though i suppose that, if i've got a definite integral involving $\gamma'_1(x)$, doing integration by parts would work to eliminate the derivative
 
3:26 PM
it seems like it is the case from the example here:
https://en.wikipedia.org/wiki/Rigged_Hilbert_space
 
(i suppose what i'm doing right now is trying to reverse-engineer your integral from the result :P)
being more explicit, i can write your above result as $$\frac{1}{2}\int_{1/4+1/2\pi}^{3/4+1/2\pi}\left[\gamma_1'(x)+(\gamma+\log{2\pi}‌​)\psi^{(1)}(x)\right]\,dx$$
 
@DanielFischer what would be your argument to prove that $U$ and $[0,1]$ are not homeomorphic ?
 
@LeGrandDODOM Depends on what $U$ is.
 
@DanielFischer a unit 2D circle
 
3:41 PM
@LeGrandDODOM Unit disk [closed]? Probably I'd say that no point disconnects $U$, but there are points disconnecting $[0,1]$.
 
@DanielFischer I agree with this
 
@Owatch o/
 
Yes hello.
 
@DanielFischer I learned to show it by proving that any bijective map $S^1\to [0,1]$ must get discontinuous somewhere, iirc.
 
Would just putting "Hippalectryo" be fine, or your actual name? FYI, the screen looks like: i.imgur.com/gTC7uRL.png
(well, cut of the screen) @Hippalectryon ^
 
3:46 PM
@SohamChowdhury Works too. There are many ways.
 
@Owatch Just put "Hippalectryon" :-)
 
Cool!
 
4:06 PM
Hi @DanielFischer
 
@DanielFischer If you have a chance, see what you think of my answer to the post.
@DanielFischer How would you rate my proof out of 10.
 
@LucioD Saying "this can easily be shown" is always a very delicate thing in a proof. You should include the intersection property.
And when you assume that $\sup (C\cap B_1) \neq x \neq \sup (C\cap B_2)$, you next assume $y_k = \sup (C \cap B_k)$. But you are not guaranteed that that supremum exists if you only assume $X$ to be any partially ordered set.
 
@DanielFischer Damnit
 
Which makes me suspect, maybe we need some stronger hypotheses.
 
4:22 PM
@DanielFischer Yeah I suspect something like that as well. For a totally ordered set the proof should work.
 
We do. Let $X = \{a,b,c\}$, and let the partial order be $\{(a,c),(b,c)\}\cup \Delta$.
Then $\{a\}$ and $\{b\}$ are order-closed, but $\{a,b\}$ isn't, since $c = \sup \{a,b\}$.
 
@DanielFischer Could you elaborate on how that defines a partial order relation?
 
@LucioD We say $a \leqslant c,\, b \leqslant c$, and of course $x\leqslant x$ for all $x$, no other relations shall hold - so $a$ and $b$ are incomparable.
 
4:39 PM
@DanielFischer Yesterday when I asked if your argument could be modified to show that $$\int_{-\infty}^{\infty} \frac{x}{1+x^{2}} \sum_{n=0}^{\infty} a_{n} e^{inx} \, dx = \sum_{n=0}^{\infty} a_{n} \int_{-\infty}^{\infty} \frac{x e^{inx}}{1+x^{2}} \, dx$$ even in cases where the series only converges conditionally, you said that we could probably use the Fejér kernel instead of the Dirichlet kernel. What function would we then be integrating on the complex plane?
 
@RandomVariable In the end, the same, we just use another approximating sequence to justify the interchange of summation and integration.
 
@DanielFischer Oh yeah I see. So your example doesn't work since $\{a,b\}$ is topologically closed but not order-closed.
 
@LucioD Sort of. But we don't actually get a topology in that case.
 
@DanielFischer I was completely confused because I thought you meant we would need to modify the function being integrated.
 
@DanielFischer You don't get a topology if you define the closed sets as the sets which are order-closed.
 
4:49 PM
@LucioD Right.
 
@DanielFischer Thanks. I will think about this problem a bit.
 
@RandomVariable Sorry. It was totally clear to me what I meant, of course ;)
 
@DanielFischer At the risk of sounding stupid, how exactly was the Dirichlet kernel (or properties of it) being used in the case of the series converging absolutely?
 
@RandomVariable I just used the normal partial sums of the Fourier series (which is what you get by convolution with the Dirichlet kernel). If $\sum a_n$ is absolutely convergent, that converges uniformly on the unit circle/the interval $[0,2\pi]$/$\mathbb{R}$, and the uniform convergence was what we needed. If we take the Fejér kernel, we get the arithmetic means of the partial sums of the Fourier series, which converges uniformly to the limit function (when that is continuous).
 
5:15 PM
@DanielFischer It's slowly starting to sink in.
 
@RandomVariable Of course, since the coefficients of the trigonometric polynomials we use for the approximation depend on the "degree" of the trigonometric polynomial, we also need to verify that the residue sums converge to the right thing. But since the dependence on the degree is very benign - $(1-\frac{k}{n})a_k$ iirc - I expect that to be no problem. I haven't fiddled out the details, however.
 
5:33 PM
Are "harmonic forms" related to harmonic analysis (analysis on locally compact groups) in any meaningful way?
 
5:59 PM
hi @iwriteonbananas
@LeGrandDODOM Remove a point from the unit disk. It's connected. Remove a point from [0, 1]. Disconnected.
 
@Hippalectryon do you practice some sport?
 
@Chris'ssistheartist None other than running and occasionally tennis
 
@Hippalectryon I see. I was just doing some sport now and took a break.
 
This is my favorite approach because it uses $\pi_0$ of topological spaces. This generalizes to higher dimensions by using higher homotopy groups.
 
Pretty (or very) athletic.
BBL (jogging for some km)
 
6:19 PM
@BalarkaSen hi
differential forms are whacky
 
tell me about them
 
ok, let's go to the alg top chat room (even though it's not alg top)
 
sure.
 
@Balarka, cool fact I learned today: $\rm t:\sf{Ab\to Ab}$ taking every group $G$ to its torsion subgroup ${\rm t}G$ and every group hom $f:G\to H$ to $f|_{{\rm t}G}$ is a functor.
 
@iwriteonbananas Are you a Romanian, right? It's not that important, but well ...
Back.
 
6:35 PM
@Soham yep, it's the Tor functor.
if I recall correctly, that is.
oh, no, that's a different thing. derived functors of $R \otimes -$. ignore me.
 
I'm out for another session.
 
@Chris'ssistheartist thankfully not
 

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