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12:25 AM
@JackOhara I have shared the padlet with you.
It is incomplete but as I cover more of the course I will be adding more resources.
 
 
3 hours later…
3:53 AM
is terencd tao analysis good?
 
I don’t know it personally, but all reports are YES.
 
I was wondering which book to study analysis to improve my proof skill.
I really suck on it.
 
Consider an nxnx…n k-dimensional hypercube. Choose some integer C; let all cells whose coordinates sum to C mod n be "infected" cells. A cell becomes infected if it shares a hyperface (=codim 1) with k infected cells. Will the whole hypercube become infected?
 
@NotTfue Analysis is tough; maybe do proofs in linear algebra and modern algebra first?
DogAteMy, what do you mean by the coordinates of a cell? Those of its center?
 
Sure, if you align it so that those coordinates are integers
Or abstractly this is the set $[1\mathbin{..}n]^k$
 
4:08 AM
Yeah it seems much viable.
 
0
Q: $k$-dimensional variation on the infected squares puzzle

Akiva WeinbergerThe infected squares puzzle is a classic; you can read about it here. I'm interested in the $k$-dimensional version, specifically in establishing an upper bound. Consider an $n\times n\times\cdots n$ $k$-dimensional hypercube. Choose some integer $C$; let all cells whose coordinates sum to $C$ mo...

 
4:22 AM
If $\sigma=\sum_{i=1}^n (-1)^{i-1}x^idx^1\wedge\cdots\wedge\widehat{dx^i}\wedge\cdots\wedge dx^n$, $\omega = {1\over|x|^n}\sigma$, $\iota:S^{n-1}\hookrightarrow\Bbb R^n\setminus\{0\}$ and retraction $r:\Bbb R^{n}\setminus\{0\}\to S^{n-1}$, then showing $\omega = (r^*\circ \iota^*)\sigma$ directly by precomposing and computing differential is not a very good idea I think. It's quite complicated.
 
@NotTfue i liked classicalrealanalysis.info/com/Elementary-Real-Analysis.php as someone that started with only linear algebra
free download by the authors
i meant that about linear algebra, in the sense that it was all the proof experience i had, not to mean that that book needs linear algebra
but uh, now thinking retroactively, i would have started with spivak's calculus first
 
5:08 AM
@onepotatotwopotato It’s not bad if you think about differential forms efficiently.
@shin Quantifiers in analysis are just much more complicated than in algebra.
Potato: In particular, do not think about using coordinates in the sphere. Just restrict, which makes $|x|=1$.
 
5:24 AM
quantifier of epsilon delta proofs
bleh
i'm getting back into math in two weeks
last long school break i delved into first order logic in a vain attempt to make quantifiers used in limit proofs friendly
my working hypothesis is that everything in a proof must be friendly, otherwise there is mathematics out there that makes it friendlier
0
Q: Formally, how do I obtain $\lim\limits_{t\to 0}\frac{f(\vec a+t\vec v)-f(\vec a)-f'(\vec a)t\vec v}{|t|}=0$ from the definition of differentiability?

shintukuFormally, how would I obtain that differentiability at $\vec a$ implies $\lim\limits_{t\to 0}\frac{f(\vec a+t\vec v)-f(\vec a)-f'(\vec a)t\vec v}{|t|}=0$ if we define differentiability as $f:\Bbb R^n \to \Bbb R^m$ is differentiable at $\vec a$ if and only if there exists a linear map $f'(\vec a):...

^ tortured result of trying to make epsilon delta proofs more friendly
 
You don’t need epsilons for that. See my book or lectures.
 
well whatdyouknow, i was reading your book when I got into that question posted above
but i was just in general trying to make sense of the role of logic in proofs used in analysis
 
6:02 AM
Update: my variation on the infected squares question has been considered and answered before! imgur.com/a/tCJ5p5W
 
@TedShifrin I'm not sure of that restriction argument. $\omega$ is a form on $\Bbb R^n\setminus\{0\}$ not $S^{n-1}$.
 
6:33 AM
So what? Can’t you think of $dx$ as a $1$-form on any submanifold of $\Bbb R^2$ without pulling back by the inclusion map?
 
The equality should be hold on $\Bbb R^n\setminus\{0\}$.
 
I don’t know what you’re talking about.
 
So the problem is showing $\omega = (r^*\circ\iota^*)\sigma$ on $\Bbb R^n\setminus\{0\}$. I don't understand how the restriction helps.
 
 
2 hours later…
9:11 AM
@Jakobian @MartinSleziak this is bizarre, I checked today and in the recursive step Todorcevic's doesn't even check that (2) is satisfied, while he explains why all the other properties are
Todorcevic's book*
 
9:29 AM
@AlessandroCodenotti In case it helps, there is a chapter Grothendieck's theorem and its generalizations in Arkhangel'skii A.V. Topological function spaces (Kluwer 1992).
The proof seems to be similar to the one from Todorcevic's book. (It defines $x_n$, $x_\infty$ and some sequence of neighborhoods.)
 
9:40 AM
Thanks, having more references to compare is always useful, I'll talk a look
 
 
2 hours later…
11:28 AM
1
Q: Turning a tensor product from a $\mathbb Z$ module (abelian group) to an $R$ module, $R$ is not necessarily $\mathbb Z$.

KoroTheorem: Let $M$ be an $(R,S)$ bimodule, $N$ be a left $S$ module. Then $M\otimes_S N$ can be made a left $R$ module via the following action of $R$ on $M\otimes_S N$: $(r,\sum_{i=1}^tm_i\otimes_S n_i):= \sum_{i=1}^t(rm_i\otimes_S n_i)$. Here's a proof of this theorem that's along the lines of th...

 
11:50 AM
@Ajay Thank you AJAY ! i received the email now ! =)
 
@AlessandroCodenotti the proof is also in Banach space theory: the basis for linear and non-linear analysis, chapter 3, section 3.7 titled "weak compactness"
in fact they later prove that $C_p(K)$ is an angelic space
 
12:16 PM
@Jakobian yes I think this will be in Todorcevic's book as well. The goal is the Bourgain-Fremlin-Talagrand dychotomy
 
1:08 PM
@shintuku What do you mean by logic?
 
1:31 PM
@Yai0Phah the rules for manipulating statements, a difficulty i was having was making sure they correspond to the intuitions expressed by natural language
 
1:50 PM
@AlessandroCodenotti I'm don't know what Bourgain-Fremlin-Talagrand dychotomy is. My knowledge ends here.
didn't touch any functional analysis for a while, though I'm starting to learn it again, lots of cool theorems yet to be seen
In the proof of Phelps theorem that $c_0$ is not complemented in $l_\infty$ they used an existence of an uncountable ADF on $\mathbb{N}$, which I think is super cool
especially since I've learned what ADF's are for topology purposes
 
2:09 PM
Let $X$ be a compactly supported smooth vector field on $\Bbb R^{2n}$ and $\omega$ be a $2$-form $\omega = \sum_{i=1}^n dx^i\wedge dy^i$. Show that $\omega$ is invariant under the flow of $X$ if and only if for every $p_0\in\Bbb R^{2n}$, there exists an open neighborhood $U$ of $p_0$ and a smooth function $f:U\to\Bbb R$ such that on $U$ we have
$$\iota_X\omega = -df(X).$$
Here, $\iota_X$ is an interior multiplication
So $\omega$ is invariant under the flow of $X$ iff $\mathcal{L}_X\omega =0$ and by Cartan's formula, iff $d\iota_X\omega + \iota_Xd\omega = d\iota_X\omega = 0$ as $\omega$ is closed form. Since $\Bbb R^{2n}$ is contractible, iff $\iota_X\omega$ is exact.
Now if the direction is clear but I'm stuck in only if direction.
$\iota_X\omega = d\eta$ for some $1$-form $\eta$. Hmm anything I can do more?
 
Let $\epsilon>0$ and if $|x-a|<\delta$, then $|f(x)-f(a)|<\epsilon$. Suppose I have "only" $|x-a|\le \delta$. Does then $|f(x)-f(a)|<\epsilon$ hold?
 
2:41 PM
ignore my nonsensical question
 
3:13 PM
@NotTfue I like Jay Cummings book on proofs. You will probably learn very well from it.
 
I'm against learning from proof books
it's just not interesting
like, if I were to try and learn math, I'd like to actually learn it, and something exciting too
 
Why is a finite abelian group not projective module over Z?
How do I prove this?
 
do you know some facts about projective modules already?
also, you're learning lots of subjects in a short amount of time
 
what's your definition of projective
there's different ways of seeing this quickly depending on what your defn is
 
when I learned about projective modules first it was in homological algebra
 
3:27 PM
Let C be an abelian group (Z module). Given an exact sequence A-->B-->0 with morphism f from A to B, and a morphism g from C to B, I want to know if there exists a morphism h from C to A such that foh = g.
I am studying modules mostly from TS Blyth and trying to solve exercises from Dummit and Foote.
@Jakobian the diagram definition; and that every free module is projective.
 
ok, check the definition implies in particular that every surjection onto a projective abelian group has a section
 
So set $B = A = \mathbb{Z}/n\mathbb{Z}$ and $C = \mathbb{Z}$, $f = \text{Id}$ and $g$ being the canonical projection
 
then show e.g. that Z -> Z/2Z doesn't have a section
 
'The diagram definition' is basically a consequence of projective module definition- Given any short exact sequence 0-->A-->B-->C-->0 (let's call the morphisms (ab) and (bc)) , the induced Z morphism (denoted $(bc)_{*}$) between Hom_R (M, B) and Hom_R (H,C) is surjective.
 
I wrote something wrong
$A = \mathbb{Z}$, $B = C = \mathbb{Z}/n\mathbb{Z}$
so it boils down to noticing that $\mathbb{Z}/n\mathbb{Z}\to \mathbb{Z}$ is trivial
so for n > 1 we don't have a projective module
 
3:38 PM
what is section @Thorgott?
 
a section is the opposite of retraction
 
and what is a retraction?
 
a retraction is a map $r:X\to A$ for which there exists $s:A\to X$ with $r\circ s = \text{Id}$
you can think about retracts from topology if you know those
then section is this map $s$
 
@Jakobian: C need not be cyclic, only finite abelian.
yeah, I know these from topology. r composition r =r.
anyways, I'll think about this one later after going through more theory.
 
@Koro the "projection" definition is a good one, but it won't make sense in general
I think
actually idk. But I think there is some difference between defining retractions as endomorphisms which squares are themselves and what I wrote
@Koro Then map $\mathbb{Z}$ to $C$ by setting $1$ to some arbitrary non-zero element of $C$
 
3:47 PM
more generally, you can show projective -> implies torsion-free by splitting a surjection from a free abelian group to a projective one
 
I think we can just map $\mathbb{Z}$ to $C$ with $1\mapsto c$ where $c$ is non-zero and has finite order then
 
that's not a surjection if $C$ is not cyclic
 
4:06 PM
Does Google or other large tech companies hold coding competitions, math Olympiads or other contests among its employees?
 
4:22 PM
My book says "You can include edges of monotony intervals, so $(a, b) \implies [a, b]$" but then, say $f(x) = \frac{1}{x}, f'(x) = -\frac{1}{x^2}$, $f'$ is monotonic decreasing for $(-\infty, 0)$ and $(0, \infty) \implies $(-infty, 0]$ and $[0, infty) \implies (-infty, infty) \implies \mathbb{R}$ which is not true
How can we make this statement correct?
Would including "You can change (a, b) to [a, b) if f(a) is defined and (a, b) to (a, b] if f(b) is defined" make it always hold true?
 
4:47 PM
Yep, it seems so
 
@Ajay sure I will try it.
I didn't follow what jakobian mean by against learning proof from book.
I am comfortable with quantifies but not coming up with original clever proof when there is time pressure.
 
 
1 hour later…
6:11 PM
@ILikeMathematics Just work in the domain of the function, which is only where this all makes sense.
 
Converse of Hilbert basis theorem is true.
 
do you mean if $A[x]$ is Noetherian, then $A$ is Noetherian? Yes, that's obviously true
 
yes
I'm trying to prove this.
Alas, too tricky for me.
:(
 
do you know that a quotient of a Noetherian ring is Noetherian?
 
yes
 
6:16 PM
well, does that help?
 
I was thinking of A[X]/(X) ---> A and then using first isomorphism theorem.
 
yes
 
image of A[X]/(X) under this morphism shall also be Noetherian
So A is Noetherian.
 
yeah
 
let $d\ge1$. let $\mathbb{N}^{n+1}_d$ be the set of $n+1$-tuples of natural numbers whose entries add up to $d$. these naturally index a basis of the space of homogeneous degree d monomials over a fixed base field $k$. the Veronese embedding is the map $\mathbb{P}^n\rightarrow\mathbb{P}^N$, where $N=|\mathbb{N}^{n+1}_d|-1$, whose coordinates are given by the homogeneous degree $d$ monomials. clearly, the image of this map lies in the projective variety cut out by $y_Ay_B-y_Cy_D=0$ for $A,B,C,D\in\mathbb{N}^{n+1}_d$ s.t. $A+B=C+D$.
 
6:31 PM
proposition 4.2
 
robjohn's last message in here was 8 days ago; but, he did comment on the main site a few days back, i hope everything is ok?
5
 
Yikes, has it been that long? I know there have been health issues in his family.
 
yup
 
I'll check it out, thanks
 
6:54 PM
Am I correct in saying that a ring R is Noetherian if left R module R is Noetherian?
 
if R is noncommutative, then this would just be the definition of left-Noetherian
Noetherian means left-Noetherian + right-Noetherian
 
suppose that R is commutative with 1.
 
sure
 
sounds good to me, koro. if there are other kinds of objects, i don't want to know about them. i certainly wouldn't want people calling them R.
 
so you don't want to know about square matrices?
 
6:56 PM
this is because I know the definition of Noetheria modules; and I remember Noetherian rings being used in proof of PID implies UFD. But I thought how they were related.
 
lukas: just don't say that they're part of a ring
 
I think having coffee daily is NOT good/healthy.
 
if they are, that's their business and not mine
 
@leslie so you'd rather want to say that under addition, they form an abelian group, multiplication is associative has a unit element, and satisfies left and right distributivity? okay
 
I have noted that whenever I take coffee often I get cough or cold.
 
7:00 PM
koro: hmm, i'd worry more about the place where you're buying your coffee.
it's true that some people can develop a dependence on it, and it can be unpleasant if they don't get their coffee. but i'm not sure how much of that is physical vs. psychological. and some people never have issues with it. caffeine can not be good for some health conditions.
it's definitely closer to 'drug' than 'food' and for that reason i understand why in some traditions people just stay away from it.
 
These days I make it myself. But I was talking previous experience. Taking coffee with gaps doesn't make me sick.
I take when I want to stay up late.
 
@LukasHeger this is why people hate bourbaki, to reference a previous conversation
 
@Thorgott the only reason I mentioned this, is because koro talked explicitly about a left module
if he had just said "module" I wouldn't have mentioned noncommutative rings
 
koro: i tend to have it no more than once a week. if i drink it regularly it makes me feel woozy.
 
@Koro Research disagrees with you (provided it's a moderate amount).
 
7:04 PM
because when you talk about commutative rings, why worry about left/right modules?
 
perhaps there's a bimodule structure in the background :P
 
I think that taking coffee should be accompanied with lot of water.
 
I don't get the hate for noncommutative rings
there are situations where they are a natural thing to consider
 
this will help prevent dehydration due to coffee.
Thorgott: yes, there are bimodules too.
 
if you've made a convention that all rings are commutative in a book or lecture, then fine. But I refuse to automatically assume such a convention with no further context
 
7:10 PM
I studied about them during tensor products, wherein they are used to make them into say R- morphisms from Z-morphisms.
 
The word "bimodules" is slightly more ambiguous than "left/right modules".
There is a tacit base which might not be $\mathbb Z$.
 
yeah the proper way to talk about a bimodule is a $(S,T)/R$-Bimodule, which is just a left $S \otimes_R T^{op}$-module
where $R$ should be commutative (I think) and $S$ and $T$ $R$-algebras
 
Heyo
I had a close vote on my recent question, for lack of clarity, but no comment as to what needs clarifying
I made a guess and edited anyway, but I wondered if anybody had constructive criticism for me?
 
Often close votes occur because of total lack of effort.
 
@Jakobian ADFs (through Isbell-Mrowka spaces) are used a lot in set theoretic topology
 
7:23 PM
@AlessandroCodenotti Isbell-Mrówka spaces is exactly what I was reading about
more precisely, there was a construction which used them to show that from a separable metric spaces we can construct some counter-examples
in terms of dimension
but, I think they are used in functional analysis as well, since I remember one of the "legends" of functional analysis (his surname was Kania I think), used them for some of their answers
 
@TedShifrin research funded by who?
 
I'd be happy if I could know as much as them
 
I read medical stuff in Consumers Reports, for example. I don't care enough to spend hours looking into it.
 
Yeah, me neither.
But there's got to be a conflict of interest brewing in there somewhere 😃
Like everything these days.
 
I personally think questions like this are perfectly fine: math.stackexchange.com/questions/4589124/…
this probably wouldn't be closed on MO
 
7:35 PM
LOL, I actually don't understand the question at all. But I know nothing about the subject matter.
I wouldn't expect the average user (even professional ones) to understand some of the differential geometry or complex geometry questions I've answered.
 
@LukasHeger I don't understand anything
 
yeah, looks OK to me, although it has the issue common to advanced questions of not showing 'attempts.' as if the general reader would even be able to assess whether any given attempts are meaningful.
 
Yeah, if I were in the field, I might ask "what have you tried?"
 
my point is that beyond a certain level it's very unlikely that something is homework and it seems absurd to me that there are questions that could be closed on MSE, but probably would be fine on MO
 
I don't know enough analytic number theory to judge whether that question has any business on MO.
 
7:39 PM
let's say I've seen more elementary NT questions on MO not being closed
 
yeah, if i knew the subject, i'd be concerned about writing an answer that the asker couldn't understand because they didn't know X approach to the question.
 
Possibly because the researchers on MO aren't overwhelmed by all the "do my homework" questions.
 
MO seems in some ways more permissive about context or lack thereof, because - well ted's just written it out. lower volume.
 
@leslie Yes, that sort of thing happens to me frequently with graduate-level diff geo. My standard approach is using moving frames techniques, and relatively few people are comfortable/knowledgeable with that.
 
it happens enough on MSE that i've ceased thinking "gosh, a person who could pose that question must understand X tool for solving it." not a guarantee. maybe closer to a guarantee on MO.
 
7:41 PM
most of the "do my homework" questions are low-level questions and a lot of the time, dupes anyway
 
That's one reason I tell people they should show what they've attempted. That way we have some idea of their toolbag. It's often impossible to know what they know or don't know or what approaches their course is or is not taking.
@Lukas That doesn't matter. We're still overrun with them.
Early on in my "career" here I ran into an innocent-looking problem (whether it was homework in some obscure course was never established) that took me literally days to solve.
 
The tool kit around here is more like, "check Wikipedia"
 
ted: have you checked to see if someone published your solution? :)
 
I think people are redirecting others to google more often
 
@leslietownes Nah.
 
7:44 PM
@Jakobian LMGTFY
 
Huh?
 
Redirecting to google
 
@leslie This was that question.
 
@TedShifrin I remember posting one of the first problems here relating to euclidean geometry, and I didn't specify how I solved it, and that resulted in one answerer using Galois theory to solve the problem
 
Well, that's fair.
 
7:47 PM
Galois/field theory really is the only solution for some impossibility results in euclidean geometry that I know of
 
Probably should have specified my academic level and described my own method from the get go
 
I get annoyed when people post algebraic/differential topology questions and tell me I can't use techniques from either one.
@Goku Right.
 
@TedShifrin I've been downvoted for using Lefschetz fix point theorem to show something has a fixed point
 
@LukasHeger well usually the proof that you can't square a circle or can't trisect an angle or can't construct a heptagon using compass and straightedge follows fro! Galois theory. It reveals a lot about "non-constructible" numbers
 
Downvoted? Seriously? If the OP makes it clear they have no such tools to use, then a downvote is fair. Otherwise, not.
 
7:49 PM
no there was nothing made clear about tools in the OP at all
 
@Goku: No Galois theory at all needed for most of those. Just elementary linear algebra of degree of field extensions.
 
Galois theory also provides, in a similar manner, a proof of why there's no closed form solution, in general, to Quintic and higher degree polynomials
 
that's when it starts to get galoisian.
 
@LukasHeger :( racist against cool fixed point theorems
 
I remember seeing in graduate school a result about maximal tori in Lie groups that was proved with the LFPT. I was incredulous. Now I can't remember what it was.
 
7:51 PM
@TedShifrin I saw the heptagon proof in particular use Galois theory, it took me about a week worth of getting familiar with the basics for me to finally understand why I can't just construct a heptagon
I encountered a similar issue while solving a geometry problem where the answer was cube root of 2, I couldn't get the answer is the "normal" manner. Now I understand why
 
@Ted I think calculating the degree of $\Bbb Q(\zeta_n)$ can be facilitated by Galois theory. Either way, computing that degree involves some non-trivial algebra
 
Yeah, the heptagon may be complicated enough. But you had a long list of easier stuff there.
@Lukas Not for trisection or squaring a circle.
 
I was just talking about the degree of $\Bbb Q(\zeta_n)$ which relates to constructing a regular n-gon
I agree about those examples
 
Its annoying how there's stuff you just "can't do" in mathematics
 
How so?
 
7:56 PM
if you can prove that you can't do it, then the problem is considered solved for mathematicians
 
You can "approximate" trisecting an angle, I worked on such a technique myself with a decent error margin, but that's all.
 
Is there any book on module theory like Rudin's PMA?
 
I know good books that cover module theory, but I don't know if they are like Rudin...
if you mean by that really concise, then maybe Lang?
 
Most module theory leads to commutative algebra, and there are various books on that. Atiyah-Macdonald is the (tiny) classic. Eisenbud is a (huge) modern book.
 
(Lang is not an easy book, but Rudin isn't either, from what I heard)
 
7:59 PM
I tried Module theory from Dummit and Foote but found construction of tensor products very complicated.
for a beginner like me.
 
I think it's the standard construction, but I haven't looked recently.
 
the construction of tensor products is not important, just remember the universal property and that every element is a sum of elementary tensors
of courses, you need to justify at some point why the tensor product exists
 
Miles Reid's Undergraduate Commutative Algebra seems intuitive.
 
but it's not like the actual construction is super useful
 
You either do the standard manifold book construction with multilinear maps on vector spaces or you do the algebraic construction modding out the free product with the relations.
 
8:00 PM
Then I started studying TS Blyth and then I started understanding what is happening.
 
Lam - Lectures on Modules and Rings is a great book
 
This is meant for the US audience but, does anyone else feels like high school and even college tends to be "slow" at teaching mathematics?
 
Lam has a commutative algebra book, too.
 
But then I'm getting stuck at proving things like - $R/I\otimes M\simeq M/IM$
 
@TedShifrin I didn't know that
he's a great author
 
8:02 PM
@Goku Because the US does one-size-fits-all education. In Europe they track people before high school into science/engineering or not.
I never took a course from Lam in grad school, but he was great to chat with at tea :)
@Koro It's either Nakayama Lemma or universal property.
 
if you only worry about module theory, then commutativity hardly matters, by the way. Standard results about things like projective or injective modules don't become any easier if you assume that the base ring is commutative. Of course, the ring theory differs quite a bit between commutative/noncommutative
 
Yes, that is basically tensor products being right exact.
 
@LukasHeger definitely not for a beginner.
 
@TedShifrin I attended a year of high school in Japan and it instantly felt much "faster" compared to here and more of what I'd like despite struggling to understand Japanese, and yes they had slackers there too but still
 
as it would not have started straightaway with projective module.
 
8:05 PM
@PM2Ring thanks for replying
 
Oh, sorry, @Lukas. It was a non-commutative book he wrote. :P
 
yeah that book I know
 
Atiyah-MacDonald is nice
great exercises
 
argh
 
@Goku From what I know, Japan does a rigorous job at the elementary and middle-school levels, too.
 
8:05 PM
I agree about Atiyah-MacDonald
 
It's not for a beginner either.
 
I learned so much from the exercises
 
:(
 
I took a course from Atiyah-Macdonald as an undergraduate, having had the "Artin" algebra course.
 
for homological algebra, nothing beats just working everything out yourself, though
 
8:06 PM
An algebra book I like is Aluffi - Algebra: Chapter 0
@Thorgott have you done that Lang exercise :P
 
@LukasHeger I don't like that one
too much adhering to intuition
 
@TedShifrin are you familiar with the Kumon method of teaching math in japan, sir?
 
Not by name.
 
what? the proofs in Aluffi are rigorous
 
@TedShifrin and Japanese students tend to study far more rigorously than their american counterparts. I was part of one of these study groups. We went to the library and actually straight up studied for hours non-stop, and I'm talking actual studying, not studying+chatting+whatever other distraction
 
8:09 PM
@LukasHeger forgot how this reference goes
 
26
Q: "Pick up a homological algebra book and prove all of the theorems yourself" (exercise from Lang's Algebra)

Harry GindiThere's a famous story about an exercise from Lang's Algebra that says something along the lines of "pick up a homological algebra book and prove all of the theorems yourself". I cannot find it in the third revised edition, and I'm wondering if it's still in the third revised edition, if it's o...

 
@LukasHeger are the definitions rigorous?
I recall, Aluffi doesn't even introduce some of them
 
@Goku with the internet there's too much temptation for chatting + whatever in the library these days...
 
@LukasHeger right, amazing
 
@Jakobian I recall even the jokes being rigorous
 
8:12 PM
lets say I've done a fair share of that exercise
 
"Joke 1.1 Definition: A group is a groupoid with a single object."
 
heh. That's true. I close my laptop, turn on the lamp, suddenly I'm an efficient student
 
perfectly rigorous joke
 
@user2236 chatting+coffee+Tiktok videos and whatnot. Anything but studying
 
for definitions to be consistent, should a "category" be the same thing as a "monoidoid"
 
8:14 PM
@Jakobian I don't remember any missing definitions, but maybe that's because it wasn't my first algebra book
I just remember excellent exposition, an interesting mild cateogrical bent and great exercises
 
well, I remember I had some doubts about, I think kernels and cokernels in categories other than abelian groups
 
How does that book deal with size issues?
 
Aluffi didn't have a problem mentioning them and it was up to someone like me to figure out the meaning
 
Assuming a Grothendieck universe?
 
but yeah it definitely gets a bit shady in places, maybe you just can't see it at the first glance
 
8:17 PM
hopefully it doesn't deal with them
 
I don't think it deals with kernels/cokernels outside of modules, but I still had some other doubt about it
I remember some definitions weren't mentioned anywhere at all
the author wants you to "figure out" what he means
 
@Jakobian I just looked it up, Aluffi defines kernels for groups, for modules and then for additive categories
@Yai0Phah this is just an algebra book. IIRC all categories are locally small
 
Jakobian suggested that it might be non-rigorous, and the size-issue is the first thing that came into my mind in this direction.
 
8:42 PM
There is also a slight issue about viewing a group as a groupoid. The category of groups is not equivalent to the category of groupoids with single object, as a full subcategory of the category of small categories (the later is classified only up to categorical equivalence). To fix it, one should replace groupoids with single object by pointed groupoids with single object.
 
Tissot ellipses of a sphere via Desmos
 
@Yai0Phah I'm fairly sure that the 1-category of small categories makes sense and you don't have to consider categories up to equivalence and it's true that the category of groups is equivalent to the category of groupoids with a single object
 
@LukasHeger if that weren't true, the formalism is just set up wrong
 
yeah of course $\mathbf{Cat}$ is a 2-category and you can consider the homotopy category which is categories up to natural equivalence, but you don't have to do that
if you want to have something less "evil" then single-object groupoids, then sure you can talk about pointed connected groupoids
 
9:04 PM
Take three
This feels closer to uniform maybe
 
10:03 PM
0
Q: Infected cubes puzzle in 3D with threshold 4

Akiva Weinberger 3D infected cubes puzzle with threshold $4$: On an $n\times n\times n$ cube, some cells are infected; if a cell shares a face with $4$ infected squares, it becomes infected. What's the minimum number of initially infected cells required to infect the whole cube? The two-dimensional, threshold 2...

 
@Ted Shifrin Hello, I have been working on Exercise 2.3.14 (where you ask to identify $\mathcal{M}_{m\times n}$ with $\mathbb{R}^{mn}$) and I would like to have some feedback on my proof, since it is something new to me. If you don't mind checking it out, what follows is my work.
Identify $\mathcal{M}_{m\times n}$, the set of $m\times n$ matrices, with $\mathbb{R}^{mn}$ in the obvious way.\\
\textbf{(a)} Prove that when $n=2$ or $3$, the set of $n\times n$ matrices with nonzero determinant is an open subset of $\mathcal{M}_{n\times n}$.\\
 
I would not write out coordinates here. I would just observe that $f(A)=A^\top A$ is continuous. One can argue with polynomials in the entries or perhaps use continuity of matrix product by copying the product proof in the book.
P.S. You have remarkable LaTeX patience!
 
10:22 PM
@TedShifrin I explicitly wrote the function from $\mathbb{R}^{n^2}$ to $\mathbb{R}^{n^2} since it seemed to me that in the book the theorems are stated in the case $\mathbf{f}:\mathbb{R}^n\to\mathbb{R}^m$ and not for function which act between other spaces and I wanted to be sure about every step. So, even if perhaps pedantic, I hope the proof is correct.
Regarding my patience, years of self-studying several subjects (English included) have taught me that the only way to do things properly is to write everything up in full details and that is what I am doing with "Multivariable Mathematics" too.
 
Well, once you get to calculus in the next few sections and later in life, you'll find it advantageous to work with matrices as their own vector space. Indeed, there is a norm that you'll learn about that's different from the Euclidean norm (thinking of a vector in $\Bbb R^{mn}$).
 
and I guess that then the fact that the theorems stated for functions $\mathbf{f}:\mathbb{R}^n\to\mathbb{R}^m$ work also for functions between spaces of matrices is the identification $\mathcal{M}_{m\times n}\cong \mathbb{R}^{mn}$?
 
It's my own fault, really. I could have made the book more pedantic (and longer) and talked about all this stuff with an arbitrary normed vector space. Perhaps I should have, but since it's intended for a first- or second-year student audience, I kept things a bit more casual.
Sure, yes. You'll encounter the more natural norm for matrices (linear maps) in chapter 5.
 
I see; can you recommend some books for a more formal treatment of this topic? I think it is interesting how the theorems can be adapted to these spaces
 
You seem pretty sophisticated. You might enjoy Dieudonné's Treatise of Analysis (it has numerous volumes, but this stuff is in volume 1). I have it in French. You can get it in either English or French — although who knows what's in print or out of print.
 
10:33 PM
Math can be as intimidating (sometimes) as it is interesting. Sometimes I feel like I'm on a small rowboat navigating the pacific
 
Thanks
 
@lorenzo you'll learn that even for $\Bbb R^n$ there's more than one useful norm, but they are all equivalent in a precise sense (that only holds in finite dimensions)
 
@LukasHeger Thanks. It is interesting how there seem to be a few key notions (the norm of something being apparently one of them) that allow one to discover properties that span so many different sets of objects
 
yes, norms on a vector space are definitely a key concept
there are even infinite-dimensional examples that are not too difficult to understand. if you know that every continous function on $[0,1]$ takes a maximum, then you can define a norm on $C([0,1])$ via $\|f\|=\max_{x \in [0,1]} |f(x)|$
a detailed study of the infinite-dimensional case is something you do in functional analysis
so the concept of a norm has applications even beyond analysis on $\Bbb R^n$
 
10:57 PM
Thanks. I will read something about functional analysis then. Bye
 
@LukasHeger Can I email you a homework to take a quick look on ? latex wise and math wise, it is that question that we discuss and 1 more very short =)
you will ofcourse not do anything other than take a look
=)
 
@JackOhara I'm a bit busy atm sorry
 
okay thanks anyway !
 

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