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20:00
> [...] searches for a woman which would agree with main ideas of that book.
@MattN Oh wow.
I thought it was a joke at first. his list of requirements for a woman is funny
"Which"? You don't consider women as human?
i guess we're agreed on who will win a fields medal
20:02
@BillDubuque Might well be. Unfortunately, I never grokked quadratic reciprocity well enough to understand why higher analogues were such a holy grail. As a stepping stone to discrete logarithms somehow?
@PaulSlevin You mean you don't have a list like that? : O
@Eugene Yes, I just stopped laughing.
@MattN My list is $\emptyset$.
I'll take what I can get
@PaulSlevin : )
well in all seriousness i think bhargava is a heavy favorite for me
@Gigili glad i could repay the favor.
20:05
@Eugene Why are the conjectures he conjectured important? This new research area he created is important in what manner?
No matter how I hard I try I always forget this formula $\displaystyle\prod_{\alpha < \kappa} \sum_{i \in I_{\alpha} } u_{\alpha, i}= \sum_{f \in \prod_{\alpha < \kappa} I_\alpha } \prod_{\alpha < \kappa} u_{\alpha, f(\alpha) }$.
If I am triyng to integrate $sin^2 x cos^2 x$ and I turn it into $1 - cos^4 x$ is that progress or is that more complicated? I cant tell
@PeterTamaroff it's not. we're joking. you took it seriously which is pretty funny.
@Eugene DAMN YOUUUUUUUUUUUUUUUUUUUUUU! =)
i thought everyone understood
20:06
All that functoid and reloid seemed serious stuff!
didn't you see his wife page?
@Eugene Oh, no. Let me see it!
@HenningMakholm For an accessible introduction see Wyman's Monthly article What is a reciprocity law linked to on Wikipedia
@Eugene Well, that is serious, too.
@MattN my mistake. i shall begin looking for nominees.
20:09
@Eugene That might turn out to be difficult given the long list of requirements.
not requiring costly gifts is my favourite
@Eugene Ono, he got married? I'm out of luck.
@PaulSlevin: That reminds me of something I learned from Martin Hyland recently: the axiom of choice (in type theory) is a kind of distributive law...
@MattN that is true too. i'm between a rock and a hard place.
For example: how many women with a short neck, straight hair and a nice voice study maths?
20:10
@MattN i actually don't know
@ZhenLin cool. for me it just means that theorem in the end of chapter 7 of Jech
@Gigili i think the position for wife of porton is still open
@Gigili do you require costly gifts?
IS THAT EVEN SERIOUS?
@BillDubuque Seems to be paywalled.
20:11
@MattN Hey ... (I can't guarantee the straight hair part, though)
@PeterTamaroff Yes.
"smooth row of white teeths"
@PaulSlevin No, thanks!
@PeterTamaroff @MattN already verified that it is. are you questioning his sagacity??
@HenningMakholm I can email a copy if you like
20:12
@Gigili you should send him an email
@BillDubuque Thanks, but I'd probably never get around to read through it anyway.
@Gigili if you require no costly gifts you might be the right candidate.
@Eugene Damn you, I can't be using Google everytime you choose to use a word like sagicity, is it like "sagaz" maybe?
I think you meant "sagacity"
@PeterTamaroff his wisdom
@Eugene Yes, now I get it.
You meant sagacity, not sagicity.
20:13
fair enough
Sagacity, see Reykjavik.
@Eugene Okay, I'm on it. But ... parrot face?
@Gigili he meant vertically correct
@Gigili Sounds like he has his own brand of anthropometry too.
@Gigili if you are a woman of the mongoloid race it is acceptable.
20:14
I feel bad for him, he obviously has something wrong
F**K, I'm stunned by this guy.
I can't believe he actually put that page up.
I'm tempted to get a random pic from the internet and send him an email.
But if he knows Linux he might get pissed and hack me.
I haven't done so many steps to get my bachelor degree. It's so stressful, will I get accepted?
@PeterTamaroff don't poke a troll. it really isn't worth it
@PeterTamaroff Do you have too much time?
@Gigili What do you mean?
@Gigili it's $1 million he's going to receive. it should be worth it
20:16
@MattN How long can it take?
@Eugene He sounds too sincere (and non-provocative) to be a troll. I think he's just plain old deluded.
@Eugene LOL
@PeterTamaroff More than 0 seconds.
@HenningMakholm That's what makes me "sad"
The amount of work he's put into his papers suggest he is not a troll
20:17
@HenningMakholm fair enough
by work I mean writing time not actual mathematics
He's quite obviously not a troll.
what about the peer reviewed journal he was in
Maybe his moral ideas are just very different to ours.
So he thinks he can actually choose a woman.
I don't know.
I'm still baffled.
@MattN I mean, does he know he's extremely crazy? None of the people around him pointed that out?
20:19
@Gigili I don't think he's crazy at all. He just lives very far away from reality. I see nothing wrong about that.
Extraordinary facts about me

While being a teenager I for a limited amount of time was transformed into a superman with computer brain. Read this chapter of my book about my transformatiom.
Ultimate LOL
Yeah delusion doesn't really correspond to sanity
because he is obviously functional as a human being
@PaulSlevin Erm, yes? Iirc, being realistic is the same as being depressed.
So he's quite healthy, if one counts depression as illness.
I'll email him my avatar as the picture. I made a nice one just today:
Well, I wouldn't, really. Otherwise there would be too many sick people.
20:21
Yeah I agree with you , he's not crazy
@Gigili Hahahahahahaha
A foot divided against itself?
20:22
your middle toe is of the mongoloid race
against the middle toe really
and the pinky requires costly gifts
I'll be watching Regular Show now.
I'll be back!
This chat here is my regular show.
3
Never fails to amuse.
20:23
I should really go and learn some set theory
All my regular shows ended. :(
@ZhenLin me too
I should really go and learn category theory and commutative algebra.
I havent watched TV for so long
@ZhenLin Have you tried Misfits?
20:23
Commutative Algebra and Differential Geometry here.
apparently there is some important Jubilee weekend going on here and I have no idea
@MattN why bother when you can learn porton's work?
@Eugene ...
Aye, the Queen is celebrating 60 years of reign...
Monday and Tuesday are supposed to be bank holidays.
But exams stop for nothing, not even the Queen!
2
Sorry for being off-topic but can I ask a question about Abelian categories?
20:25
yeah people are shocked i have an exam
yeah ofc
well i should be off now to be productive and stuff.
@Gigili thanks for the laughs!
bye all
@Eugene Any time!
Bye!
Oh. An $Ab$-category is not the same as an Abelian category!? It's the same as a pre-additive category which is the same as a ringoid which is the same as an $Ab$-enriched category?
20:27
yeah
Tumeni words meaning the same thing. head asplode
enriched in abelian groups
No-one uses "ringoid"
Why do there have to be $n$ words to mean the same thing? : (
20:28
this is incredibly frustrating, I haven o idea what I am doing wrong on this problem I have done it 6 times now. If I am making u substitution for integrating can I just ignore the bounds until the end and the plug back in all my substitutions correctly and get a correct answer?
So which one should I prefer? I quite like pre-additive.
Ringoid not so much.
$\textbf{Ab}$-category is avoided precisely because it is confusing...
I just say enriched in abelian groups
because thats what it is
So enriched is always the same as enriched in Abelian groups?
@Eugene See you!
nope
20:29
Ab-enriched means enriched in abelian groups
Ah, ok.
The vast majority of categories are enriched in $\textbf{Set}$.
you can be enriched in pointed sets (which means you have 0 morphisms) for example
I'm very glad and grateful that you are around.
but not necessarily an additive structure
20:30
Ah, I guess the next thing I should look up is probably what enriched means exactly.
I hate how there are so many ways to do an integral right but only one or two that are the right answer even though I am doing the math right the answer is wrong somehow
And all this work just to understand Teddy bear's answer. sigh
(just being dramatic, of course I'd look up all these things anyway eventually)
it means $\mathcal{C} ( -, - ) : \mathcal{C}^{op} \times \mathcal{C} \to $ Set factors over something?
nope
apologies. that's how I understand factoring over abelian groups to meaan
20:33
The right notion of enriched category starts with a suitable monoidal category (or even bicategory).
I think I'll go with pre-additive to mean a category where the sets of morphisms are Abelian groups (I assume that means addition of morphisms) and such that concatenation is bilinear.
@Jordan What is the problem and what did you do which was wrong? In case you want help with it, otherwise you have my sympathy.
I posted the question already and got answers but trying to work through it on my own I can't get it
Yes, that works for $\textbf{Ab}$-enriched categories.
I have been doing this proble mfor over an hour, I have went though it half a dozen times and I can not find any mistakes, I have done it 3 different ways
20:34
@Jordan Okay, let me find the question then.
what is your question
0
Q: Integral of $\int \sin^2 x \cos^2 x dx$

JordanThis seems pretty simple to me but I can't get it. $$\int \sin^2 x \cos^2 x dx$$ $$\int (1-\cos^2 x) \cos^2 x dx$$ I know there is a rule in my book (with little explanation) that tells me when I had an odd and an even degree on two trig functions I should split the odd and convert it to an id...

I don't like the cheap trick answers that require an incredible amount of trig knowledge/memorization or weird manipulations I wouldn't be able to do
Hm... All categories I can think of are pre-additive. Surely, Set is pre-additive and so is Mod. Ah, Grp is not. Only if it's the category of Abelian groups. I see.
No, Set doesn't even have addition.
@Eugene Getting to "Residue classes and complete residue
systems"
(obviously following your teachs path)
So this definition only makes sense if we have some additive structure on the objects, right?
20:36
Well what does it mean to add objects?
you mean within the objects?
It means that $+$ is defined.
You add morphisms, not objects.
you don't really care about that, in category theory you care about the external structure
Although you can add objects in an abelian category...
so you add morphisms not the objects
20:37
I know but to add morphism (pointwise) I need addition in my objects, no?
no you just have to have a well defined morphism $f + g$
First rule of category theory: your objects are black boxes.
Like if $f,g : G \to H $ are group homomorphisms then $f + g (x)$ means something because we have addition on $H$.
@Jordan How did you get $$\int \cos^3 dx - \int \cos^ 5 x dx$$
You don't look into them, you don't talk about what's inside them, and for all you know, dragons lie therein.
20:39
I have no idea anymore, I haven't been doing the problem like that anymore
I have a much better incorrect solution to the problem now
@ZhenLin Then how do I give meaning to $f + g$?
I have gotten very efficient at getting the wrong answer, I minimzed the steps and time it takes now
It's given to you, like composition.
But composition already comes with a meaning.
Very nice.
20:40
So does the addition, it comes with the definition of pre-additive
Can you give me an example of an Abelian category without addition in its objects?
I have been doing $\frac{1}{4} \int 1 - cos^2 2x$
That seems correct to me
Sure. $\textbf{Ab}^\textrm{op}$.
What about Set?
It's not an abelian category.
20:41
From there I put aside the $\frac{x}{4} $and keep it out of the problem, I then work with just the $cos^2 2x$
@ZhenLin Do you have a more undergraduate example? Like Top?
That's not an abelian category either.
I'm running out of categories without addition : )
@Jordan $\frac{x}{4} $?
Matt N are you talking now about Abelian categories, or Ab-categories/enriched in abelian groups/preadditive
20:43
There exists an abelian category whose set of objects is $\mathbb{N}$, but it's basically a funny way of talking about $\textbf{Vect}^\textrm{f.d.}_k$.
@Gigili That is the integral of 1/4
According to wolfram my integrals are correct but I am not getting a correct answer
@PaulSlevin I'm looking for a pre-additive category that doesn't have addition in its objects.
@MattN: You are thinking about categories wrong.
Objects are just objects.
yeah you need to abstract a bit more
Think of a category as a kind of graph.
20:44
you are not interested in the objects, as Zhen lin said
But the categories I know are all not very abstract. They're just collections of mathematical objects.
@Jordan That's wrong. you should keep $\frac{1}{4} $ and multiply the final answer by it.
Category theory is all about the external structure, i.e. the morphisms between objects. thats why you are interested in universal properties
why?
no
because they tell you about the morphisms
20:45
@MattN: You understand an abstract group, yes? Then this is just a grown-up version.
What I am doing is correct, I am taking out the 1 from 1 - cos
I am splitting the problem up
Matt n. a finite set may be thought of as a category. the points in the set are jst little dots but they are the objects in the category and have no internal structure
@ZhenLin Yes, I understand as much. : ) But not enough, since I can't come up with my own examples.
@PaulSlevin And that category is pre-additive?
@ZhenLin And not pre-additive either, I assume?
A transitive undirected graph is a pre-additive category.
I have wasted so much time on this one problem and I only have 2 days left to study, I feel like I should just skip this problem because I can't ask it here anymore
20:46
In fact, even an abelian category.
@MattN well in this category there are only identity morphisms... so think about it
so each homset does have an abelian group structure, its just the 0 group right
Wait. But for each point there is only the constant map $f: \ast \to \ast$. What's $f + f$?
20:48
@Paul: No. A discrete category with more than one object can never be preadditive.
@PaulSlevin Oh. You define it to be that?
ah crap! haha ignore me
wel it was an example of a category without internal structure on the objects which was my original idea
@ZhenLin Does it not end up being a pre-additive category if you define $f + f = f$?
You can't make the empty set into an abelian group!
yes thats right of course, $Hom (A,B) $would be empty
thanks for pointing that out
20:49
The graph of any preadditive category, or for that matter, any pointed category, is strongly connected.
@ZhenLin But the empty set is not an object of that category. Paul said the objects are the points of the finite set.
but if you had one object, things would be ok
@MattN you need each set of morphisms between any given objects to be an abelian group
@ZhenLin So if you take a transitive undirected graph, what are the morphisms?
@MattN: We need to start from the beginning, I see.
DEFINITION.
for example if my set is $\{ x, y\}$ what is $hom(x,y) $ ...
ok i will leave Zhen lin to explain it to you
see you guys
20:51
@PaulSlevin See you!
Bye!
Sorry for my terrible explanations, I am just learning too so I thought I would have a try haha
Ill jst lurk in the background
That's no problem, I enjoyed that : )
Can I just ask the question again?
@Jordan Okay. You can either tell what you do or do what the first answer says.
I can tell you how to do it using that $(\sin x \cos x)^2$, but I didn't get what you are doing.
20:53
I don't see why we have $Hom(x, y)$ empty if the category is some points $\{x,y,z\}$.
No, let's just stop with the examples now, please.
@Jordan No!
If you mean on the main site.
A category $\mathbb{C}$ comprises the following data: a set $C_0$, a set $C_1$, a map $s : C_0 \to C_1$, a map $t : C_0 \to C_1$, a map $\rho : C_0 \to C_1$, and a map $\tau : C_2 \to C_1$, where $C_2 = \{ (g, f) \in C_1 \times C_1 : s(g) = t(f) \}$, and all these are required to satisfy various axioms.
20:54
Hey ho! Is there some name for basis of the form $e_{1} = b$, $e_{2} = Ae_{1}+b$, ..., $e_{n} = Ae_{n-1}+b$?
But first, some visualisation: we think of $C_0$ as the set of vertices of a graph, and $C_1$ as the set of edges.
$\int sin^2x cos^2 x$ = $\int (1-cos2x)/2 * (1+cos2x)/2$ = $\frac{1}{4} \int 1 - cos^2 2x$ = $x/4 + \frac{1}{4} -1 \int cos^2 2x$
I then work with just cos^2 2x
This is not an ordinary graph: the edges are directed, and there can be more than one edge between any two vertices, and loops are allowed.
$s$ gives the "source" of an arrow/edge, and $t$ gives the "target".
$C_2$ is then seen to be the set of pairs of edges which can be joined head-to-tail.
u = 2x $\frac{-1}{8} \int cos^2 u du$ = $\frac{-1}{8} \int \frac{1+cos2u}{2}$ = $ \frac{-1}{16} \int cos 2u$
We require $s(\rho(x)) = t(\rho(x)) = x$: $\rho : C_0 \to C_1$ is the witness of the reflexivity of this graph; in other words, $\rho(x)$ is a distinguished edge from $x$ to $x$.
20:57
then I just subsitute again and get sin but sin goes away with 0 and pi so that doesn't even matter
On the other hand, $\tau : C_2 \to C_1$ witnesses "transitivity", but not in the usual graph-theoretic sense.
@ZhenLin Sorry, shouldn't the set of edges then be $C_0$?
Please don't contradict me.
@PeterTamaroff Since you didn't backlink, I don't know to what you are referring.
@Nimza If you're free to choose $A$, then every basis has that form.
20:58
@HenningMakholm no, A is fixed
@Nimza I didn't even know it was broken :-)
Hi @robjohn!
@HenningMakholm Hey Henning! Long time, no type!
The main axiom for $\tau$ is associativity: $\tau(\tau(h, g), f) = \tau(h, \tau(g, f))$.

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