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12:57 AM
youtube stop recommending me art of dying challenge
my neck can only take so much this early in the day
 
No wonder you're a pain in the neck, Soham :P
 
 
2 hours later…
2:40 AM
off on a picnic with everyone from the gym
how's everyone else's Sunday going
 
Still Saturday, for me
 
 
2 hours later…
4:42 AM
@Loch

If $\{U_i\to X\}$ is an etale cover, and $\mathcal{F}$ is a sheaf for the etale site $X_{et}$, how would you interpret $$\mathcal{F}\otimes_{\mathcal{O}_X}\mathcal{O}_{U_i}?$$
Would you take it to mean $\mathcal{F}\otimes_{\mathcal{O}_X}f_*\mathcal{O}_{U_i}$ where $f:U_i\to X$ and the $\mathcal{O}_X$-module structure given to $\mathcal{O}_{U_i}$ given by $f^\#:\mathcal{O}_X\to f_*\mathcal{O}_{U_i}$?
 
@TedShifrin hey Ted!
finally getting back to the work this week
finals are over
 
I'm a different Ted I'm afraid
 
@TedE hahhahaha yes I'm aware
 
Oh good, I wouldn't want to disappoint you :-)
 
always welcome more people to SE :)
 
4:44 AM
hmm i'm not 100% sure
what's the context?
 
The context was a seminar talk (which I have bad notes for), but I had found similar uses in an etale cohomology text, let me see if I can find a specific one
That being said, I can't see any other way to interpret it atm
Two alternative questions for you. Are you familiar with the formalism of ringed sites? Does the structure sheaf $\mathcal{O}_X$ or the structure sheaves $\mathcal{O}_{U_i}$ give etale sheaves on the site $X_{et}$ in a natural way? Rather than these just being zariski sheaves on their appropriate zariski sites
I'm looking at page 190 of Lei Fu's etale cohomology, looks like we turn a quasi-coherent sheaf $\mathcal{M}$ of $\mathcal{O}_X$-modules into an etale sheaf on the site $X_{et}$ by taking $U\mapsto \Gamma(U,\pi^*\mathcal{M})$ for any $U\to X$ in $X_{et}$, where in particular we take $\mathcal{O}_X=\mathcal{M}$ to get $\mathcal{O}_{X_{et}}$
 
5:03 AM
wouldnt say im familiar but yep that sounds right
 
I have you down as my etale cohomology guy :P
 
that's a bad idea - i never studied it carefully :p
 
Uh oh :D
Denis Nardin in the math overflow chat seems very knowledgeable, but I'm trying not to ask to many things there (lest I wear out my welcome).
 
im sure they're all nice people
but homotopy theorists do speak a different language
 
He answered my question I had the other day
in Homotopy Theory, Dec 8 at 15:18, by Ted E
Just to confirm:

Given an $S$-scheme, $f:Y\to S$, such that the structure morphism $f$ is affine, and a sheaf $\mathcal{G}$ on $S$, by making use of the isomorphism $H^i(Y,f^*\mathcal{G})\cong H^i(S,f_*f^*\mathcal{G})$, we can get a well defined homomorphism $H^i(S,\mathcal{G})\to H^i(Y,f^*\mathcal{G})$ by making use of the unit of adjunction applied to $\mathcal{G}$, i.e. we have a morphism of sheaves
$$\mathcal{G}\to f_*f^*\mathcal{G},$$
which induces a morphism:
$$H^i(S,\mathcal{G})\to H^i(S,f_*f^*\mathcal{G})\cong H^i(Y,f^*\mathcal{G})$$
That you also answered, but your remark seemed slightly cryptic at the time
I guess I'm just having one of those days where I feel retarded, and nothing makes sense
 
 
2 hours later…
7:48 AM
Does $\int (a(x)(\int b(x)c(x)dx)dx)=\int(\int a(x)b(x)dx)c(x)dx)$?
Actually, nevermind, I'm pretty sure it doesn't
Well, it does hold when $a(x)=c(x)$, but that's just a consequence of commutativity
 
8:03 AM
Hello!!
How can we check the below sequence for convergence and determine the limit if it exists?

$\displaystyle{g_n:=(-1)^n+\frac{\sin n}{n}}$
The limit of $(-1)^n$ doesn't exist. Does this mean that the limit of $g_n$ also not exist?
 
 
2 hours later…
9:39 AM
 
10:06 AM
How can we find the equation to an ellipse with arbitrary center?
 
What actually you want?
 
@yuvrajsingh Did you read the question?
 
Sorry for my ignorance.
Does the de Rham-singular cohomology comparison map commutes with cup products on the nose?
 
Hello!
I have a function $\frac{2x}{\sin{x}-x}$
How can I get the inverse of this?
I mean how can I transform this into $\frac{\sin{x}-x}{2x}$?
What is the result of $\lim{x\to{0}}\frac{2x}{\sin{x}-x}$.
 
10:28 AM
I assume that you know $\lim_{x\to0}\sin x/x=1$.
 
I can't find any solutions for this, I tried multiplying it with $\frac{1/2x}{1/2x}, \frac{\frac{1}{x\sin{x}}{\frac{1}{x\sin{x}}\dots$.
@Yai0Phah Yeah, I know it. I tried to transform this limit into it but everything gives me 0 on the denominator.
 
But you know $\lim_{x\to0}1/x$
 
@Yai0Phah It gives $\lim{x\to{0+}}\frac{1}{x}=+\infty$ and $\lim{x\to{0-}}\frac{1}{x}=-\infty$. So there's no limit.
Am I correct?
Wait a minute.
 
Yeah, you need to determine in fact the one-sided limits.
 
Oh, let me find the one sided limits of this expression
 
10:36 AM
That's not necessary. I mean, one-sided limits of $1/x$ when $x\to0$.
But you know that $\sin x/x\le1$
 
Ax = b has a solution for every b iff A has a full row rank?
 
I didn't know $\frac{\sin{x}}{x}\le{1}$
But I'm confused right now.
 
You said that you get 0 on the denominator, but the nominator is nonzero
which is essentially $\lim_{y\to0}1/y$
where $y$ is that "zero" you get
 
$\lim{x\to{0+}}\frac{2x}{\sin{x}-x}=\lim{x\to{0+}}\frac{\frac{2x}{x}}{\frac{\sin{x}}{x}-\frac{x}{x}}=\lim{x\to{0+}}\frac{2}{0}$
I get this.
 
You need to put the $x\to0$ in the subscripts.
 
10:40 AM
I don't have any $x$ on the denominator :/
 
You need to show a proof of the second equality
 
$\lim_{x\to{0+}}\frac{2x}{\sin{x}-x}=\lim_{x\to{0+}}\frac{\frac{2x}{x}}{\frac{\sin{x}}{x}-\frac{x}{x}}=\lim_{x\to{0+}}\frac{2}{1-1}$
I mean I get this but it doesn't have $x$ on the denominator, so what does this mean?
 
You cannot simply "replace" the $\sin x/x$ by $1$
 
Hmm, why?
 
Why can you replace it?
You need to show your proof.
 
10:49 AM
I don't know :D
There is one thing I know is $\lim{x\to{0}}\frac{\sin{x}}{x}=1$
 
 
1 hour later…
12:02 PM
Huh, proof of a big conjecture in diophantine approximation announced
 
dsm
Is it valid to define the set of dual vectors $f_i$$^j:(1,1)\rightarrow\mathcal{C}$ such that $f_i$$^j(T) = T_i$$^j$ for any $T\in\mathcal{T}_1$$^1(V)$?
Seems like I need that set of duals in order to define the "trace" of a $(1,1)$: $\text{Tr} \equiv \sum_if_i$$^i$ $\Rightarrow$ $\text{Tr}(T) = \sum_iT_i$$^i$
 
12:43 PM
Does anyone know what sum rule is used in the following

$$ \sum_{i=1}^{n^2} (i) = \frac{n^2 (n^2+1)}{2} $$
 
$\sum_{i=1}^n i = {n(n+1)\over 2}$
 
I mean, it doesn't make intuitive sense that by increasing the upper sum limit by multiplying by n, you also multiply the n terms in the formula it is equal to.
 
$n^2$ is just another integer
 
you're just plugging $n^2$ in, no multiplication
 
If you set $m = n^2$, you use $$\sum_{i=1}^m i = {m(m+1)\over 2}$$
Then substituting $m\to n^2$
$1 +2 +\dots + (m-1)+ m = 1+2+\dots + (n^2-1) + n^2$
 
12:54 PM
Ahh, thank you! That makes a lot more sense now
 
 
3 hours later…
3:42 PM
i did absolutely nothing today
ill read some representation theory to make up for it
 
@BalarkaSen take a break
 
From nothing?
You should do what I do -
- actually, scratch that, you shouldn't do what I do, I have no healthy response to days like this
You should do what I do and watch one of the many hour-long videos on your Watch Later playlist (jk)
Thought -
"The process of doing your second draft is the process of making it look like you knew what you were doing all along" - Neil Gaiman
I think this can be harmful when writing proofs
When something's written down differently than how it was thought up, it might end up written down differently than how it should be read
 
4:15 PM
Hi to all.
Anyone could help me understand what this means:

$$ \dfrac{d \log a}{d \log b}$$

? Thanks
 
In what context ?
 
looks like a fraction to me
 
surely it's a derivative
 
4:32 PM
If it's a derivative it equals $\dfrac{d\log a}{da}\dfrac{da}{db}\left(\dfrac{d\log b}{db}\right)^{-1}=\dfrac ba\dfrac{da}{db}$
equivalently $\dfrac{da/a}{db/b}$
 
@ÍgjøgnumMeg what conjecture
 
4:48 PM
@BalarkaSen Enjoy the act of doing absolutely nothing. Sometimes people deserve breaks
 
Rehi @Ted, sorry I disappeared yesterday
 
rehi, demonic @Alessandro
 
How did you tell at a glance that the form from yesterday was closed? I computed its exterior derivative and got zero, but I doubt that's how you decided it's closed
 
Yes, that's how you decide. I knew it because I know it. :) If you think it through in spherical coordinates, it's clear. In 3D, you have $dA$ for the $2$-sphere is $\rho^2\sin\theta\,d\theta\wedge d\phi$ (using European coordinates), and you're dividing through by $\rho^2$, so what's left has no $\rho$ dependence (we already discussed that), and so taking $d$ you have to get $0$.
 
21 hours ago, by Alessandro Codenotti
So I have $S^{n-1}$ as a Riemannian manifold with the metric and orientation it inherits from $\Bbb R^n$ with the standard ones. I want to verify that the volume form on $S^{n-1}$ is $\omega=|x|^{-n}\sum_{i=1}^n x^i dx^1\wedge\cdots\wedge\widehat{dx^i}\wedge\cdots\wedge dx^n$
this form?
 
4:58 PM
Yes.
 
I see, the spherical coordinates thing makes a lot of sense
 
You can generalize easily to all dimensions.
You can also make a local argument with Stokes's Theorem, based on my observation that the flux across (i.e., integral over) spheres centered at the origin is independent of radius.
@ConstantineBlack If you've ever graphed data on log-log paper, you're computing the slope in that plot.
 
5:22 PM
@Alessandro: Hast thou disappeared yet again?
 
Nope
I was computing stuff (trying to decide whether that form is exact or not)
 
It all boils down to my comments about spheres.
 
@TedShifrin Hmmm I'm not seeing what you mean
 
I'm saying you can make a Stokes's Theorem argument on a little region. If $d\omega\ne 0$ at some point, then it won't change sign nearby. But you can choose the region to be a "spherical" frustum (flat sides on which $\omega=0$ with little pieces of spheres for ends).
 
Hmm but that helps me seeing that the form is closed, not to decide whether it's exact, or am I missing something?
 
5:30 PM
Oh, yes, that's correct. If it were exact, what would the integral over a whole sphere have to be?
 
@Soham The Duffin-Schaeffer conjecture
 
But we know this is the volume form for the sphere, so the integral is very nonzero, @Alessandro.
 
Indeed, it is, well, the volume of the sphere as the name suggests :P
 
Done.
 
5:36 PM
Well that was easier than I thought
 
Stokes's Theorem is powerful, man.
 
@TedShifrin Thank you. I figured out, there was a typo in the notes I was checking and I couldn't make sense of the result I was getting using the formula and what the notes where having. Thanks again.
 
That's fair
 
Oh, no problem, @ConstantineBlack.
 
@Soham youtube.com/watch?v=1LoSV1sjZFI a video about it :)
 
5:38 PM
Greetings, @ÍgjøgnumMeg.
 
Funny thing: as I was searching Numberphile on youtube I automatically typed Number field by muscle memory
 
Now I'm asked to prove that the divergence of a vector field on an orientable Riemannian manifold is independent of the orientation, but it's not clear to me what exactly that means. I can write down a local coordinates expression for $\mathrm{div} X$ depending only on $X$ and $g$ but not on the volume form, isn't that enough?
 
Hi @Ted and @Alessandro
 
Hi @ÍgjøgnumMeg
 
@Alessandro: I don't think so. Suppose you switched two of the coordinates. Could you tell that the formula stays the same?
 
5:40 PM
@Alessandro I haven't heard anything back from that prof but my friend is doing his bachelorarbeit under him and said that he's really bad at replying rofl
 
@TedShifrin Ah, I guess not because the components of $X$ might change sign in local coordinates if my new charts have the standard basis for $T_p M$ oriented differently
 
Yes, lots of things will change. I think it's clearest from the definition with the Hodge star.
 
Hello guys!!
If we have the conjunction $x\wedge y$, how are $x$ and $y$ called?
 
@TedShifrin We defined $\mathrm{div} X$ as $\star^{-1}(\iota_X dV)$
 
Like $1+2$, then $1$ and $2$ are called the addends
 
5:45 PM
@Alessandro, OK, so orientation change puts a sign on $dV$ and what to $\star$?
 
Hi chat
 
Salut, @Astyx.
 
It also puts a sign to $\star$, by definition $\star f=fdV$, so if $dV$ changes sign so does $\star$
So they cancel out
 
Yeah.
 
I always try to make things harder than they need to be in diffgeo
 
5:48 PM
It's because you haven't done enough computations, I think.
 
You developed your intuition in algebra, for example, by doing lots of concrete examples as an undergraduate.
 
I have one more week of classes before winter break, I want to get a lot of explicit exercises done during the holidays
 
@TedShifrin hello! Can you help me about tagging the $x$ and $y$, please?
 
I have no idea, @manoooh.
 
5:49 PM
Ok, thank you for answering!
 
conjuncts?
 
You didn't give context. If it's mathematical logic, maybe Alessandro knows.
 
@AlessandroCodenotti oh, so if we had $x\wedge y\wedge z$ we have 2 conjunctions but 3 conjuncts right?
 
yes
I'm not sure whether they're commonly called conjuncts but it seems like a reasonable guess to me
 
@AlessandroCodenotti yes for me is also right!! Now my problem is to translate "Conjuncts" to Spanish (my native language)... :D but don't worry, you've done your work
Now if we have $x\vee y$, $x$ and $y$ are disyuntors?
 
5:53 PM
No idea, I don't think they have commonly used names
 
Conjunction = conjunción, disjunction = disyunción
I've never heard of a name for what you want though
 
Hi, DogAteMy.
 
@AkivaWeinberger yes, but what is the translation for "Conjuncts"? Conjuncts = Conjunciones, so then a Conjucnt will be a Conjunción, which is equal to Conjunction
 
I've never heard the word "conjunct" even in English
Maybe "component (en)/componente (es)"
 
@AkivaWeinberger that's what @AlessandroCodenotti told me (which I accepted)
 
6:11 PM
I want to say conjunto but that means set doesn't it
 
Yeah :/
 
tengo una soluccion
no hagas matematica en espanol
 
Leaky has never met a language he doesn't try to maul. :D
 
"Soluccion"?
Soluction lol
 
solucion?
 
6:16 PM
So, DogAteMy, how did your first semester go?
 
Good to hear!
 
Did well on Japanese tests… less well on attendance
Something I need to work on
 
Yeah, at the university level, professors don't relish students who skip.
 
Or show up late
 
6:18 PM
I had some colleagues who would lock the door one minute after class started.
 
When students explained to me that they had a professor across campus who traditionally kept them late, I told them not to worry, just to do their best. But oversleeping is NOT a valid excuse.
 
@LeakyNun @AkivaWeinberger Solution is translated as Solución!
 
Although I never required attendance, I usually was aware when students were missing (since I had 30 or fewer students, typically) — and since I like a lot of class participation, having students missing might have a deleterious effect.
 
hrmp, didn't skip classes, but i stopped taking notes towards the end of undergrad, although some profs insisted on referring back to their notes
i prefer a good book
 
6:21 PM
Sometimes books just don't do an adequate or tasteful job. Of course, when I wrote my own I could no longer complain, but I still did a lot of examples and heuristics that weren't in the book.
 
your book is a masterpiece
 
I try to put myself in the professor's shoes. If they didn't show up, there might be students who would want to learn something but can't. Class topics would fall behind, it would just be bad all around.
And if the professor has to come to class, why should I skip out?
 
Flattery accepted, @JoeShmo, but unnecessary.
@Rithaniel: Sadly, it was the students who needed the most help — with guidance on homework, clues about exams, etc. — who tended to miss.
 
@manooooh Yeah I know but Leaky put in an extra 'c' by accident
 
well, considering you're not grading me, no harm done :)
 
6:24 PM
Hahha ok
 
@JoeShmo What a relief!
 
Yeah, makes me wonder which causes which, or if there is only a correlation.
I have a bad habit of doing other things during lecture, though
 
I guess if you're going to be rude and not pay attention/participate, I'd just as soon you were not there. Having said that, if you've got a high grade in my class and I know you're bored, I will tend to let it go. ... Of course, this is all moot now, since I've retired.
 
I try to participate if the professor attempts it. It's just when the lecture is purely lecture that I'll start doing things with my laptop
 
Yeah, the European teaching style dies hard. All the classroom flipping in the US is probably an encouraging thing.
 
6:29 PM
Plot twist: Rithaniel is in a lecture right now
 
Sunday sermon?
 
Oh I forgot it's a Sunday today
 
Lol, nah. Last exam was last Friday. I have finished my Undergrad
 
Félicitations, Rithaniel.
 
Nice
What are your plans next?
 
6:30 PM
Congrats @Rithaniel
 
The college I'm going to has this "Bachelors to Masters" program where you automatically get accepted to the graduate school after you graduate
 
Automatically? This is assuming sufficient GPA or recommendations?
 
So, I'm doing that, but the plan is to use this time to go to as many conferences as I can manage
Oh yeah, but if you're in the program you probably have the GPA and faculty involvement already
 
I mean, you have to earn acceptance at some point.
 
After the masters, I plan to know what college I want to go to for my Ph.D. and that will be the next step
 
6:33 PM
"College" = undergraduate, typically.
 
hrmp, is the torus -- $\{(x_1, x_2, x_3, x_4) \in S_3 | x_1^2 + x_2^2 = \frac{1}{2} = x_4^2 + x_4^2\}$ open in $S_3$?
 
Is the term "University" then?
 
@Rithaniel, yes.
@JoeShmo: How could it be?
 
Alright, I'll remember
 
i didn't think so..
but now i have a problem
 
6:34 PM
I'm looking at places to apply to for a phd at the moment, the situation is "meh"
 
Have to learn another language, @Alessandro?
 
math.ucla.edu/~radko/191.1.05w/john.pdf page 8 computes the fundamental group of a torus knot using VK
 
You thicken it up to make it open, @JoeShmo.
 
but VK assumes $A, B, A \cap B$ are open..
ok, cool
 
You always do these games with MV and VK.
 
6:36 PM
Nah, English is enough pretty much anywhere in Europe except maybe France
 
yes, that's what i thought
 
Interesting, @Alessandro.
 
Hatcher did that to compute the fundamental group of $S^2$
IIRC
 
@TedShifrin But I'm looking mostly at places in Italy or Germany (and English is often enough in Germany)
 
OK, @Alessandro. So why "meh"?
 
6:39 PM
(well I'm also looking at places in Finland and Austria, but English is fine)
@TedShifrin There's few places with set theorists and not a lot of phd positions in general
 
And even fewer jobs in that field.
 
I have a representation of a Lie algebra that looks like $(\Bbb Z/2\Bbb Z)^N$ in the sense that all states are of the for $r(q_{i_1}q_{i_2}\dots q_{i_k})x_0$ for a ground state $x_0$ with $q_{i}^2 = 0$ and $[q_i, q_j] = 0$ and there are $N$ such $q_i$. How do I word the similarity with $(\Bbb Z/2\Bbb Z)^N$ ?
 
what is MV?
 
Also a lot of places wants you to find a supervisor before applying and I'm not sure how do they expect people to do so
 
You read papers and contact people ahead, @Alessandro ... or go to conferences.
Mayer Vietoris @JoeShmo.
 
6:41 PM
@JoeShmo Mayer-Vietoris
 
Specifically I'm thinking of a superalgebra of a single massless particle
 
@TedShifrin I mean should I just email professors out of nowhere "hey are you looking for a phd student?"
 
I know nothing about this, @Astyx, but it looks like quantum- or super-algebra.
 
It is a superalgebra
 
@Alessandro: Preferably having read a paper and having some specific comments/interest.
 
6:42 PM
Makes sense
 
But I was wondering how to word the fact that they look the same
 
"look the same" ordinarily is "are isomorphic"
 
That might be it. Thanks
 
7:05 PM
Well, I guess chat has gone dormant.
 
well, speaking of $\pi_1 S^2$
or dormancy
$\pi_1 S^2 = 0 * 0$ amalgamated along $\mathbb{Z}$
 
Yup.
 
so in paritcular i suppose $0_a = 0 = 0_b$
but everything else in $\mathbb{Z}$ we throw away?
 
Not that I have any idea what your subscripts mean.
 
the left copy of $0$ and the right copy of $0$.
in $0 * 0$
and so my question stands
 
7:11 PM
Write down carefully the definition of free product with amalgamation.
 
OK, I think I get it
any homomorphism $\mathbb{Z} \rightarrow \{0\}$ annihilates $\mathbb{Z}$ anyway
 
Yup.
 
7:28 PM
I met a guy a couple weeks ago that knows you. a friend of a friend
Matthew something. Sends his regards
former student of yours
 
Hmm ... I have a guess ...
 
he was a little confused when I said I knew one Ted Shifrin from UGA..
 
confused?
 
/surprised
I didn't go anywhere near UGA
 
Jewish guy, I assume. Then I know who it is.
 
7:33 PM
yup
works as a software engineer. Did grad school at UGA for a while, then again in Europe?
rings a bell?
 
Yeah, I knew who it was without all that.
 
Hi
 
Hi, @tigre.
Friendly and smart fellow.
 
Why thank you
 
very bright. i liked him
 
7:35 PM
(I know you're talking about someone else lol)
 
I don't know you well enough @tigre :P
 
Well, that wouldn't help in that regard
Since I'm about to ask dumb questions probably
 
Well, we can direct them to the people who answer only dumb questions.
 
$R-S$-bimodules correspond to left $R\otimes_{\Bbb Z}S^{op}$-modules right?
 
$R$ on the left and $S$ on the right for the bimodule structure?
 
7:38 PM
In one direction it seems reasonable. I can use the $R\otimes_{\Bbb Z}S^{op}$-module structure to define a left $R$-module structure, and a right $S$-module structure, such that they are compatible
Yep
 
Yeah, looks reasonable.
But you'll have to call for the algebra experts soon.
 
i like your commentary along the way, ted
its the second time i chuckled
 
To be more precise, given a left $R\otimes_{\Bbb Z}S^{op}$-module $M$, I can define left $R$-module structure on $M$ by taking $r\cdot m = (r\otimes 1)\cdot m$, and left $S$-module structure by $m\cdot s = (1\otimes s^{op})\cdot m$
 
you mean right $S$-module structure.
 
Yes, typo
Then using the bilinearity of the tensor, I can check all the module axioms
 
7:42 PM
Sure.
 
Of course I'd have to write it all up to be rigorous. For the other direction I want to take an $R-S$-bimodule $M$, and give a left $R\otimes_{\Bbb Z}S$-module. So a potential choice is to take $(r\otimes s^{op})\cdot m := r\cdot m \cdot s$
 
Yes, agreed.
 
And then extend by linearity I guess
 
Yes, you have to check some things are well-defined ... in particular, why do you need the 'op'?
 
An alternative viewpoint is that an $R$-module $M$ is a unital ring hom $R \to \operatorname{End}_\Bbb Z(M)$
so the correct way to glue two such things together is with the tensor product
since tensor product is the coproduct in the category of rings or something like that
 
7:50 PM
Then I can go through and check carefully:

1) $$(r\otimes s^{op})\cdot (x+y)= r\cdot (x+y) \cdot s = r\cdot x\cdot s + r\cdot y \cdot s = (r\otimes s^{op})\cdot x + (r\otimes s^{op})\cdot y$$

2) $$(r\otimes s^{op}+t\otimes u^{op})\cdot x = r\cdot x\cdot s + t\cdot x\cdot u = (r\otimes s^{op})\cdot x + (t\otimes u^{op})\cdot x$$

3) $$(r\otimes s^{op})(t\otimes u^{op})\cdot x = (r\otimes s^{op})(t\cdot x\cdot u)=rt\cdot x \cdot us= (rt\otimes (us)^{op})\cdot x = (rt\otimes s^{op}u^{op})\cdot x$$
 
It's #3 that is more interesting. Most of the things are boring.
 
so you don't need to check any axioms :P
 
Well, surely you need something @Leaky ... otherwise leaving out the 'op' would violate things.
 
@TedShifrin a right $S$-module is a left $S^{op}$-module
 
Tensor product is coproduct in the category of commutative rings right?
 
7:53 PM
<--- deserts the conversation now that Leaky is here.
 
I don't think the argument fails without commutativity
 
But you wanted to make sense of gluing with the tensor, but the tensor isn't the coproduct in this setting
Thanks for your help deserter Ted :D
 
they are still Z-modules and the universal property of tensor product for modules apply
the fact that the glued map is a ring hom should be naturality
 
Could you be more precise sorry, I don't see it yet
 
sure
the point is that if you have ring homs $f: A \to C$ and $g: B \to C$, then $A \times B \to C$ given by $(a,b) \mapsto f(a) g(b)$ is bilinear
so by the universal property of tensor product for modules, you obtain a linear map $p: A \otimes_\Bbb Z B \to C$
now we need a ring structure on $A \otimes_\Bbb Z B$
 
7:59 PM
Okay, your first assertion is that a left $R$-module $M$ is given by $R\to \text{End}_{\Bbb Z}(M)$. Where I suppose you send $r$ to the endomorphism it defines $r\cdot -$, which is $\Bbb Z$-linear since all abelian groups are $\Bbb Z$-modules via $n\cdot x = x+\cdot +x$, and $$r\cdot (x+\cdots+ x)=r\cdot x +\dots r\cdot x= n\cdot (r\cdot x)$$
and similarly for a right $R$-module
 
so we need a tetralinear(?) map $A \times B \times A \times B \to A \otimes_\Bbb Z B$ given by $(a_1, b_1, a_2, b_2) \mapsto (a_1 a_2 \otimes b_1 b_2)$
this should be tetralinear
 
@LeakyNun Did you just use the universal property for coproducts for a product?
 
no I just gave you the map explicitly
 
Ok
So that's claimed to be a ring hom
 
if you want to use universal properties, I'm taking the product of the two maps to get $A \times B \to C \times C$ and then composing it with the ring multiplication $C \times C \to C$
 
8:02 PM
But that implies you have two maps $A\to A\times B$ and $B\to A\times B$
 
what implies?
 
Sorry, I mean that you are making use of those two maps
Are these just $a\mapsto (a,1)$ and $b\mapsto (1,b)$
This looks very much like a coproduct argument, which is confusing to me
 
product is a functor $\mathcal C \times \mathcal C \to \mathcal C$
functors take objects to objects and maps to maps
a map in $\mathcal C \times \mathcal C$ is a pair of maps in $\mathcal C$
$(f, g)$ is a map in $\mathcal C \times \mathcal C$
 
Okay, I now understand
 
the functor "product" takes it to the map $A \times B \to C \times C$ given by $(a,b) \mapsto (f(a),g(b))$
 
8:11 PM
@LeakyNun Thats the wrong multiplication though I guess
 
why?
 
@LeakyNun As a $\Bbb Z$-module?
@LeakyNun Because we expect $B^{op}$ to appear right?
Just to be sure, when you took the product functor, did you mean $Ring\times Ring\to Ring$?
 
yeah
we don't need Bop to appear
it will appear when you transition from right module to left module
 
You were taking $M$ to be a left module with respect to both $A$ and $B$ (i.e. $C=\text{End}(M)$ and $A,B\to C$ are both giving left module structure)?
 
a right $S$-module is a left $S^{op}$-module
 
8:20 PM
I suppose I proved that above implicitly
@jcora after clicking into your profile to see the picture enlarged, it's clear you aren't, but in the little picture, it looks like you're holding an ak-47
 
hi all how do I find the differential for the scalar product? specifically I have f : R^n x R^n -> R, f(x, y) = (x|y), I need to find Df(x, y)(h, k)
@tigre lol
 
You just have to compute the Jacobian matrix, treating your map as $\Bbb R^{2n}\to \Bbb R$ I think, which is a $1$ by $2n$ matrix
$\begin{bmatrix}y_1&y_2&\cdots&y_n&x_1&x_2&\cdots &x_n\end{bmatrix}$
 
we usually have examples in R^n -> R^m, wasn't sure how to approach this one. I'll try it out to see if I can understand treating R^n x R^n as R^2n in this case
 
Well if you do that, you should just get what I wrote, so you can check with that
 
Better than a matrix is to write down the limit definition of $Df(x,y)(h,k)$ and just do it.
 
8:29 PM
yay I like that
 
@TedShifrin don't you need a candidate solution to place it in the limit, and see if =0? and then conclude that this is the unique linear operator? I didn't see an explicit definition of Df that would let me compute it directly
 
No, compute the directional derivative by the definition.
 
ok I'll try that then
 
$Df(a)(v) = \lim\limits_{t\to 0} \dfrac{f(a+tv)-f(a)}t$.
(Then, if you have to prove differentiability by the definition, you can use this in the limit definition of the derivative.)
 
8:51 PM
okay goofed up first time, but now I got D(x, y)(h, k) = (x|k) + (y|h) is that correct?
 
Yup.
 
neat! we usually defined directional derivative as $\partial_v f(x, y) = \nabla f(x, y) \cdot v$, and this definition makes sense to me too, but I've never seen it before
 
the more you know
 
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