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12:25 AM
@geocalc33 given $f$, you're going to have an affine hyperplane of solutions, so a very-infinite-dimensional space.
 
 
2 hours later…
2:08 AM
Can anyone take a look at my question about proof of Riesz theorem for C([0,1])? math.stackexchange.com/q/3869735/792125
1
Q: An alternative proof for Riesz Representation Theorem for $C([0,1])$

MikeDuring my class in real analysis, my teacher mentioned an alternative method to prove Riesz's theorem for $C([0,1])$. He just mentioned the method, but did not prove it. (A usual way of doing this is provided here: Riesz representation theorem for $C([0,1])$) Step 1: Let $\Lambda$ be a nonnegati...

 
2:20 AM
Can anybody take a look at my question?
0
Q: Integral transform reduced?

geocalc33I discovered that the following integrals are equal: $$ \int_0^1sx^{s-1}\exp\bigg(\frac{t}{\log(x)}\bigg)~dx=\int_0^1\exp\bigg(\frac{st}{\log(x)}\bigg)~dx $$ Let $f^s(x)=x^s,$ then the LHS can be written as $$ \int_0^1\frac{d}{dx}\bigg(x^s\bigg)\exp\bigg(\frac{t}{\log(x)}\bigg)~dx$$ It reminds me...

 
 
1 hour later…
3:49 AM
Is there a difference between a combinatorial family and the sequence generated by a generating function, or are these the same thing?
Basically trying to understand this video (youtube.com/watch?v=R6szFb4tIow) but the conceptual explanation in the first 90 seconds isn't very clear and I can't find other resources on the subject.
 
 
4 hours later…
8:06 AM
0
Q: Which mathematician traveled to and moved in with each collaborator?

uhohI remember a video, perhaps a Numberphile episode where a mathematician was described who would simply move in to the home of a collaborator with which they were engaged to so that they could work more closely together, essentially sleeping on their sofa and going to work with them and keeping a ...

 
9:28 AM
@JoséCarlosSantos , is there a congruence property like $x \equiv y (\mod{(x-y)})$ ?
 
@Spectre I have no idea whatsoever about the meaning of that question.
 
@JoséCarlosSantos Well, this thing arose when I was trying a hand at Brocard's problem...
I am not very much sure whether I did it right or not, but still here's how it came to my mind :
As we all know, n = 4,5,7 and m = 5,7,11 are the only possible solutions currently available.
 
@Spectre This room was created to discuss this answer. What has Brocard's problem to do with it?
 
@JoséCarlosSantos No other chatroom for math was live.. and I required immediate help to see whether I am on the right track..
Sorry if I intruded into your discussion...
Well, here's what I did :
I took each pair of $(m,n)$ (for eg., $(m,n) = (5,4)$), then calculated the number of partitions for $n!$ and $m^2$, and took their difference.
On checking the remainder when the number of partitions were divided by the difference, the remainders in both cases ($n!$ and $m^2$) were the same...
@JoséCarlosSantos , why did you leave ?
I am sorry .. I can't but intrude....
Hello ?
 
9:58 AM
Does the colimit coincide in all the usual "algebraic" categories ?
 
@Astyx no, but the limits do
 
I am a beginner, that's why I stumbled upon this thing..
Anyway, thanks
I got it cleared..
 
So does the definition of Stalk of a sheaf varies with the "codomain category" ?
A sheaf $X\to Ring$ would not have the same stalks as its image through the forgetful functor $Ring \to Set$
 
@Astyx aha but filtered colimits coincide
 
@skullpatrol Thanks and thanks!
@skullpatrol Found it! after 03:00
 
10:45 AM
@LeakyNun Oh ok. Is that hard to prove ?
 
@Astyx bah si tu sais comment construir le colimite filtré alors ça serait trivial
you can basically take "union" (or more rigorously, quotient of disjoint union)
and define a ring structure thereon
 
But when you restrict categories you also restrict the morphisms between objects, which I believe means the universal property that defines the filtered colimit is not the same
I've only seen the contructive approach to colimits done on Sets
And it's not obvious to me that it would work with other categories
 
@Astyx but the functor Ring => Set is faithful
anyway you can just check it by hand
 
 
2 hours later…
1:13 PM
@SayanChattopadhyay Let me know when we're doing 1.5 next
 
@BalarkaSen chess?
 
Nah
 
 
1 hour later…
2:31 PM
Hey there, chat
So I'm taking Linear Algebra 2 and we have to present project on applications of primary/cyclic/Jordan/Frobenius decompositions of linear operators
the professor said we could take any topic, possibly the one we may have if in scientific research
 
I can't really see an obvious application of these topics in Galois theory or Frobenius algebras, but I don't wanna talk about the first application we've ever seen (in matrix exponentials when solving ODEs). I want to study algebraic applications. Does anyone know anything of this in representation theory, group theory, ring theory or something like that? Thanks!
 
What sort of applications? How long will the presentation be?
 
2:52 PM
@BalarkaSen literally any application, like "you can apply complex numbers to solve integrals" or "you can use group theory to solve number theory problems"
I'm not sure about how much time the presentation will take, but I think the article is more important than the presentation (possibly we'll just send our individual projects and, if we want to, read another one's)
 
Learn how the Smith normal form works over PIDs (this is closely related to primary decomposition, just a bit more organized). Use it to prove the invariant factor decomposition of finitely generated modules over PID's (a sweeping generalization of fundamental theorem of finitely generated abelian groups).
Alternatively you can gun for smaller, cuter applications. Jordan normal form proves that finite-order matrices over $\Bbb C$ (algebraically closed fields of characteristic 0 in general) are diagonalizable. Can you prove a generalization: Every representation of a finite group over $\Bbb C$ is unitary?
Smith normal form can be used to efficiently compute abelianization of a given group using its presentation. Do it for $\text{SL}_2(\Bbb Z)$. Can you show this otherwise/interpret this?
You can use the rational canonical form to come up with necessary sufficient conditions for two operators to commute. I forget what this is called but you can find this in standard texts, or try to prove it yourself.
 
@BalarkaSen that's much more elementary
 
I don't know what that sentence means :P
 
wagwan
 
I can give more ambitious things you could do with these decomposition theorems that I don't know the answer to but maybe I shouldn't.
In particular there's a deep theorem in integer programming called Parry's theorem that for a long time I thought could be done using invariant factor decomposition. I couldn't write a proof though.
 
3:10 PM
A nice consequence of the Smith normal form is that a homomrphism $f\colon\mathbb{Z}^n\rightarrow\mathbb{Z}^n$ has finite cokernel iff $\det(f)\neq0$ and, in that case, $|\coker(f)|=|\det(f)|$. This is useful to calculate the order of free abelian groups in terms of a presentation.
 
@Thorgott What was your "much more elementary" proof, by the way, whatever it means?
 
minimal polynomial splits in distinct linear factors => diagonalizable
 
Oh, sure. Note that this plus Jordan normal form proves spectral theorem over $\Bbb C$
Hm, that's maybe not what I want to say. It's clear that gen. eigenvectors of normal operators are actual eigenvectors.
Then Jordan normal form proves spectral theorem, period.
@Thorgott You don't need this fact to prove JNF, and you can deduce this fact from JNF.
That's what Artin does
 
it's trivial to deduce this from JNF, I'm just saying you can also prove it with a lot less machinery
 
Why? Factorizing minimal polynomial factorizes the vector space as a $k[x]$-module is something like the primary decomposition already.
"less machinery" is an illusive concept
Everything follows from everything in linear algebra
 
3:17 PM
everything follows from everything in mathematics
JNF is a pretty technical result
 
No, it's very easy.
There's a reason Artin does away with minimal polynomials completely.
 
Hey... why was I invited here ?
Did anyone tell you that I was after a proof ?
@uhoh I am after this....
 
Did you speak to a shady looking man in a bar around 4.30 am?
 
@EdwardEvans IDK....
 
easy doesn't mean not technical
 
3:20 PM
Who is that shady guy ?
@EdwardEvans Was that question for me ?
 
I do mean that. It's not technical at all.
 
@Spectre yeah it was just a silly joke
 
@EdwardEvans Haha...
Well, I think it's all math experts here, right ?
I am not an expert, but may require guidance...
 
no just degenerate math enthusiasts
but some people know a lot of stuff
 
@EdwardEvans I don't think so...
@EdwardEvans But yes that's gonna help me
I am trying to formulate a proof either to prove or disprove (I haven't taken my stand) the existence of more Brown numbers....
But that's kinda hard if I don't assume anything correct, but that's how I wanna do it.
 
3:25 PM
eh that sounds like recreational number theory to me
 
@EdwardEvans May sound so, but we need something to help those hard working people operating computers to work out that problem to numbers to the order $10^8$ or ahead....
@EdwardEvans And I am yearning to possess a Fields Medal in my name (lol)
 
If $f,g$ are linear maps, $\dim\ker(g\circ f)=\dim\ker f+\dim\ker(g\vert_{\im f})\le\dim\ker f+\dim\ker g$. Assume $f$ is an endomorphism s.t. $m_f=\prod_{i=1}^r(X-\lambda_i)$ with $\lambda_i$ distinct eigenvalues of $f$. Then $\dim V=\dim\ker\bigcirc_{i=1}^r(f-\lambda_i\id)\le\sum_{i=1}^r\dim\ker(f-\lambda_i\id)\le\dim V$, so $V$ is the direct sum of the eigenspaces of $f$ and $f$ is diagonalizable.
Now convince me JNF doesn't take significantly more effort
 
just read the proof in artin
 
skimming through it and it looks like the standard proof?
 
@LeakyNun i have a little question. i'm asking to you since you speak french. in linear algebra course i take, we talked about "normes qui provient d'un produit scalaire". for example we defined $N_\infty((a,b)) = \operatorname{max}\{|a|,|b|\}$ and said that "$N_\infty$ ne provient pas d'un produit scalaire". and we gave the requirement as follows: "norme provient d'un produit scalaire <=> il satisfait l'égalité de polarisation"
 
3:38 PM
thats the only proof i have read
 
he deduces it from the structure theorem for modules, yes?
 
what no
lol
 
and i've looked at wikipedia articles etc. but couldn't find any plausible terminology in english. do you know how those are studied in english? @LeakyNun
 
what are you reading man
 
@sevdaicmis norm induced by an inner product, polarization identity
 
3:40 PM
give me a page number
 
induced! ah
 
Am french and confirm
 
i dont have it with me rn, you can find it from the index
 
"provenir" means "to come from"
 
I'm looking at the chapter where it is first mentioned and he interprets the decomposition of the vector space as module over the polynomial ring in terms of the linear endomorphism to get the rational canonical form
 
3:44 PM
Artin doesn't talk about the rational canonical form
What are you talking about
This M. A. Artin's Algebra book by the way
 
yes, that's what I'm looking at
 
not Emil Artin
Totally confused.
 
yes, actually i would say i've understood it when it is phrased in french @Astyx though the verb induce didn't come to my mind :)
 
so am I
 
are you reading the module theory part
this is in linear algebra not module theory
2nd edition
 
3:47 PM
yes, this is in the module theory segment, but it's the only mention of Jordan form in the index
 
oh then youre reading some messed up edition
in my edition he does a very simple proof in the linear algebra chapter
just by induction and choosing explicit jordan basis
ill tell you the idea later if you cant find it; trying to get some work done rn
i took one of the classes at the end of my linear algebra course because the instructor was away on a seminar and did this proof and a bunch of applications
 
I think I might know that proof
trying to find a proper edition rn
 
4:26 PM
I certainly remember rational canonical from Artin. But walking out the door now, so I'm not going to look.
Hi, bye @Thor, a @Balarka.
 
4:40 PM
Hi, everyone. Does anyone here have an idea of the Laplace method for the expansion of an integral?
$F(\lambda) = \int_{1}^{2} e^{- \lambda (t-3e^t)} dt$
Need to find a leading order term in the expansion as $\lambda \to infty$
 
5:30 PM
Suppose $f(z)$ is analytic at $z = 0$ and satisfies $f(z) = z + f(z)^2$. What is the radius of convergence of the power series expansion of $f(z)$ about $z = 0$?
From the conditions, it's clear that $f(z) = 0$ or $f(z) = 1$. I tried to use $f(z) = z + f(z)^2$ to deduce a (recursive) formula for the $n$-th derivative of $f(z)$, so I could calculate the $n$-th coefficient in the power series expansion, but this didn't seem helpful.
I was reading a solution and it says "Near $0$ the function coincides with one of the branches of $(1 \pm \sqrt{1 - 4z})/2$. The radius of convergence of the power series of either branch is $1/4$, which is the distance to the singularity at $1/4$" But this explanation doesn't make any sense to me at all.
 
5:48 PM
0
Q: Determine whether or not the given $*$ is a binary operation on the given set S.

user1993$S = \mathbb{Z}, a * b = a+b^2$ Commutative: $a*b = b*a$ $a*b = a + b^2$ and $b*a = b+a^2$ and they aren't the same at all. Associative: $(a*b)*c = a*(b*c)$ $(a*b)*c = (a+b^2)* c = a+b^2+c^2$ and $a*(b*c) = a + (b+c^2)^2$ and they aren't the same at all. therefore it's not a binary operation on t...

need help on this
 
6:15 PM
In the definition of a metric space we require semipositive-definiteness of the codomain of the metric, does this mean that spaces like Minkowski space used in physics $(\Bbb R^4,\eta)$, $\eta=\text{diag}(-,+,+,+)$ is not a metric space? Or do metric spaces without semipositive-definiteness have a different name?
 
@Charlie The Minkowski space is not a metric space, specifically for the reason you gave
 
ah
does it have a different name? or is it just a lorentizn manifold?
 
The latter yeah
You can make sense of some kind of Lorentzian distance but only for causally linked points
 
6:30 PM
-+++
more like
+---
 
we do not use the godless signature in physics
 
lol
 
in The h Bar, Mar 27 at 17:51, by John Rennie
In God's own signature we write $ds^2 = -c^2d\tau^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2$
 
6:44 PM
Would someone have a good reference on Hilbert Polynomials and Projective schemes ?
Is Hartshorne good ?
 
uhoh
hIlBeRt PolYnomIalS
 
:(
 
i feel like i have seen that in
Weibel
but skipped it all
i think a homological perspective is the right perspective for that shit
 
Found a task in my Professors exercises which I am not sure I understand:
Find $lim_{x \rightarrow a+0} \frac{1}{(x-a)^{2k}} \, \, \, k \in \mathbb{N}$
How would one proceed with such a task?
 
I'll check it out thanks
 
6:51 PM
I was thinking L'Hopital but that wouldn't get me very far
 
What do you think the limit is ?
 
Not defined?
Really lost on this one, sorry
 
try graphing the plot of the function
 
Why do you think it's not defined ?
A first step would be to figure out the limit of $(x-a)^{2k}$
 
Wouldn't it be zero?
if x approaches a
 
6:58 PM
Indeed
 
Which means that the limit I have got will give a division by zero, which is not allowed
 
It depends if you allow your limit to be $+ \infty$ or $-\infty$
 
That I do
 
Ok, so what do you mean by "not allowed" ?
 
If I insert $a$ for $x$ I will get $\frac{1}{0}$ which is not possible ( I think )
 
7:04 PM
limit =/= plugging in
very fundamental difference
 
That is a good point
 
You are correct on (I think) @BobPen
do you know why?
 
It means that the function is undefined at that point?
 
We have skillpatrol and skullpatrol now ?
 
Specifically, why can you never divide by 0?
 
7:07 PM
p sure he just changed his name?
 
yup, same person guys :-)
 
It just doesn't make sense?
 
new account ?
 
Yes. Why?
 
Because there is nothing to divide on
 
7:09 PM
Sort of.
What about dividing on negative numbers then?
That^ is allowed
 
That is possible, but I don't really know the correct answer to why that is allowed and dividing by zero is not allowed
Let us say $\frac{1}{0} = x$ but $1 \ne 0*x$
 
Dividing by 0 would mean multiplying by the reciprocal of 0; but 0 has no reciprocal because 0 times any number is 0, not 1.
a/b = a * (1/b)
 
Indeed
Thanks for the explanation
 
But I am still a little lost on how I can show that my limit goes to infinity
 
7:15 PM
I gotta run, cya
 
@BobPen How do you usually show something goes to infinity ?
 
just remember a * (1/a) = 1, for all nonzero a.
::runs::
 
@skillpatrol With epsilon-delta?
 
Kind of
That's for when the limit is a real
 
Any hints hehe?
 
7:20 PM
Instead of $\delta$ you use $M$
conventionally
 
Indeed!
Thank you!
 
7:54 PM
@Thorgott @BalarkaSen, I asked a close professor about it and he said I could study the connectedness of $\mathrm{GL}(n,\,\mathbb{C})$. I haven't had a lot of topology, so it looks fun.
 
That's a fun one for sure
 
@LucasHenrique go ahead
 
I don't feel like that requires the full power of JNF, but I'd have to think about it
It's a nice result in any case
 
you definitely don't need JNF to do it
 
It helps to have a decomposition though
JNF might be overkill, but the mindset of the proof is roughly the same in all the ones I've seen
 
8:02 PM
you only need basic linear algebra
 
maybe you don't need it, but I could use some help
 
como posso eu ajudar voce?
 
Ask away
 
@LeakyNun o que eu quis dizer é que esse artigo tem bastante informação sobre como demonstrar e essencialmente usa matrizes "quase diagonais" (teorema de Jordan)
 
@Astyx do you feel like Brazilian Portuguese and French sound very similar?
I'm only talking about the sounds
 
8:09 PM
No
 
like they both have the same r sound and they both have nasal vowels
 
@LeakyNun I don't, lol. Never seen a single Brazilian pronouncing menu correctly
 
Let me listen to brazilian and portuguese TV first actually
 
ah right, French has more vowels
 
oh, I feel stupid
 
8:12 PM
Portuguese "rolls" its r's way more
 
I saw that the triangular matrices are path-connected and every matrix is triangularizable, but I was trying to think of a good path from a matrix to its conjugate rather than just trying to conjugate the path in the triangular matrices lol
 
You can just path-connect any matrix to $I_n$
@LeakyNun I get where you're coming from though
 
@LucasHenrique yeah?
 
did you get any of it? :p
 
yeah all of it but that's not a question
so I didn't know what to respond
 
8:17 PM
You don't even need linear algebra to see GL(n, C) is path-connected.
It follows from pure topology. GL(n, C) is C^(n^2) minus det = 0, which is a real codimension 2 subvariety. Join any two things in GL(n, C) by a path in C^(n^2) and use transversality.
 
Do you even need GL(n, C) ?
 
Nope :)
 
do we even need math :P
 
8:20 PM
@BalarkaSen nice
 
great
just saw this one, $t\mapsto e^{t\log A+(1-t)\log B}$ is a path joining $A$ and $B$
 
trivial
 
yeah but you have to show exp is surjective
thats a JNF exercise
(its not true for GLnR)
 
(otherwise GLnR would be connected)
 
I think you can do that with some complex analysis magic
 
8:29 PM
@Astyx I mean it's not surjective to GL+(n, R)
@Thorgott Maybe
 
8:50 PM
Oh by the way @LucasHenrique
 
I saw the same on with spectral sequences
 
Nice
 
It's a propos given the arrival of Halloween as well
 
9:12 PM
Hi again, a @Balarka, @Astyx.
 
Hi @Ted
 
Hi @Ted!
 
@Balarka: The machinery for your proof (transversality) is probably more sophisticated than the easy linear algebra, but I like it.
 
Yeah
You can con someone into believing transversality as obvious though
 
Well, you're better at bullying than I am.
4
 
9:16 PM
:D
 
Depends on which sort of bullying. I've been losing patience with a lot of the crap on main these days.
 
I got played a few days ago when I told a probability person that obviously two random walks on $\Bbb Z^3$ will intersect each other with 0 probability
Because of transversality ("how can random paths intersect in 3 space with positive probability?")
 
Well, Brownian motion doesn't sound too smooth.
 
Right haha
Brownian motions are Hausdorff dimension 2, as I rightly guessed immediately afterwards I was pointed out to be wrong.
They are really more like planes
 
I have no intuition for these things.
Oh oh .. someone starred that.
 
9:22 PM
How do I show this apparently elementary result? Let $A$ be a compact set with positive Lebesgue measure and let $h$ be constant. Define $A + h = \{a + h: a \in A\}$, and suppose $G$ is an open set with $A \subset G$. How do I know that $A + h \subset G$ for $h$ "sufficiently small"? (If I'm missing anything additional, let me know and I'll add in more facts about these sets.)
 
What's the distance from $A$ to the boundary of $G$?
 
Apparently it's positive. Thanks.
I was asking my classmates if they had justified this result, but they told me they didn't. That I find concerning.
 
Yeah, you're right, but it's easy if you remember basic metric space stuff.
 
you don't need positive Lebesgue measure, do you
 
As far as I can tell, you don't, but I only knew that because of the question Ted asked me
 
9:32 PM
laybayesgooay
 
If the result were false, there'd be a sequence $(a_n)_n$ in $A$ such that $a_n+1/n\not\in G$. Since $A$ is compact, let it WLOG converge to $a\in A$, then $(a_n+1/n)_n$ converges to $a$ as well. $G$ is open, so $G^c$ is closed, hence $a\not\in A$, contradiction.
 
I really hope that some day, I can take a proper course in topology
Another apparently elementary question I'll throw out there: math.stackexchange.com/questions/3871367/…
0
Q: With $(X, \mathcal{B}, \mu)$ a measure space, $\mu$ nonatomic, if $0 < k < \mu(X)$, exists increasing sequence of sets with limit of measure $=k$?

ClarinetistIs the following claim true? Let $(X, \mathcal{B}, \mu)$ be a measure space with $\mu$ nonatomic. If $0 < k < \mu(X)$, there exists an increasing sequence $B_1, B_2, \dots \in \mathcal{B}$ (i.e., $B_1 \subset B_2 \subset \cdots $) with $\lim_{n \to \infty}\mu(B_n) = k$. I'm honestly not sure if...

 
cute question
like an intermediate value theorem for measures, eh?
 
In a way, yeah
 
What does atomic mean ?
 
the section right below says its true and it was first proved by Sierpinski
lol
 
surprise Zorn's lemma lol
 
seems like a result worth knowing
 
@BalarkaSen Not quite. I'm looking for an increasing sequence of sets that have such a measure. Also, I edited the post to clarify that I'm not allowed to use that result.
 
(i didnt know this was true)
 
9:45 PM
Yeah I find it actually surprising
 
@Clarinetist in the proof sketch those $S(t)$'s give you an increasing sequence, no?
 
@BalarkaSen Oh, lol
I see
I've been doing this stuff for too long today. Going to eat dinner.
 
Bon appetit
 
Bon appétit
 
Enjoy!
Very nice result. Love it.
 
9:47 PM
BTW for those of you who have made it through PhD coursework... any tips for studying for a measure theory exam? The scope just seems ridiculously vast, and my prof is definitely not relying on a textbook
 
Ok back to drawing pictures
 
Look at past exams and work lots of problems.
 
Yeah, I've bought an AMS problems book, so as soon as that comes, I'll be working through that
 
cannot imagine taking a phd measure theory course
ill be screwed
 
You're talking about your course or a departmental qualifying exam?
 
9:49 PM
The course in particular
 
Then it depends far more on the personality of the individual professor teaching/writing the exam.
Look at his past exams.
 
AFAIK this is the first time he's taught this class
 
I mean, if you have a set with that measure, you can also take the constant sequence of that set
 
LOL, oh well.
Any hour exams?
 
captain tautology is here
 
9:50 PM
None that I can find
 
@BalarkaSen LOL
 
I feel bad for the undergraduates in my class. I get the impression this is their first grad-level course after the two-semester Baby Rudin sequence
 
That's not so bad, although I would still recommend a point-set topology course for more sophistication.
Berkeley (used to have?) (has) an undergraduate measure theory course that lots of undergraduates took. Usually used Royden.
 
I'm more or less used to this after going through my MS degree. I think my undergrad classmates were expecting the homework assignments to be more... straightforward. Had to break it to them that this has been my grad-level class experience.
 
My long experience teaching/advising is that graduate complex analysis is a better place to start.
 
9:53 PM
I wish I had taken complex analysis period. Haven't done any.
 
Algebra-inclined students often did the graduate algebra.
 
I have Gamelin sitting in my shelf and have been intending on reading it.
 
Gremlin
my complex analysis course sucks lmao
Conway's functions on one variable is a terribad book
its complex real analysis, not real complex analysis
 
It's a more readable rewrite of Ahlfors.
 
I've never liked complex analysis
 
9:55 PM
mr sheaves is here
 
I love complex analysis. Loved teaching it.
 
coherent sheaves over TOPOS
 
You're amassing nicknames with pace!
 
french people are perverted. even their mathematics is
@TedShifrin what nickname?
 
mr sheaves, captain tautology
 
9:57 PM
oh
lol
 
I shudder to think what mine will be.
 
algebraic approach
 
Bourbaki is not a phase, it's a way of life
 
V. I. Arnold: Life- 10, Weapon- Physics, Special- ADE
Bourbaki: Life- 10, Weapon- Homological algebra, Special- garbological sheaves over coherent schemes
FIGHT!!
 
00:00 - 22:0022:00 - 00:00

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