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00:22
@Jakobian: Thanks! Another question: if $a_n>b_n$ for each $n \in \mathbb{N}$, can we say that $\sup_{n\in\mathbb{N}}a_n > \sup_{n\in\mathbb{N}} b_n$? I think it is true because from $\sup_{n\in\mathbb{N}} a_n \ge a_n>b_n$ we deduce that $\sup_{n\in\mathbb{N}}a_n$ is an upper bound for $b_n$, hence $\sup_{n\in\mathbb{N}}a_n \ge \sup_{n\in\mathbb{N}} b_n$ because by definition the supremum is the least upper bound.

If by the sake of contradiction we assume $\sup_{n\in\mathbb{N}}a_n = \sup_{n\in\mathbb{N}}b_n$, we have $\sup_{n\in\mathbb{N}}b_n=\sup_{n\in\mathbb{N}}a_n \ge a_n>b_n$, so $\sup
00:32
I'm playing with the coproduct in the category of idempotent monoids. It's quite puzzling.
Basically, the coproduct employs some equivalence relation on the free product.
For consider the boolean set $B = \{\bot, \top\}$ endowed with logical or ($\lor$), and consider the set $3 = \{0, 1, 2\}$ endowed with $\max$.
Then in the coproduct $B + 3$, $\top 1 \top 2 = \top 1 \top (1 2) = (\top 1) (\top 1) 2 = \top 1 2 = \top 2$. It's not directly obvious how to generalize this procedure.
01:27
@ZaWarudo no
and this is wrong
 
4 hours later…
05:46
@SayanDutta please come to chat
It's rare that I find someone interested in the same kind of approaches
 
1 hour later…
06:51
consider the quotient space $\Bbb{R}^2/\sim$ where $v\sim w$ iff $\text{span}v=\text{span}w$
$[0]$ is the only point that is closed in this space, no other point is closed.
So this space is not $T_1$, hence not metrizable.
Is it correct to say that if a space is not metrizable then there is no geometric picture of that space? because having a geometric picture means the space is embeddable in some $\Bbb{R}^n$, but then the space inherits the Euclidean metric.
07:13
@SoumikMukherjee if thats what you want it to mean then its tautological
For me geometric picture means can be imagined geometrically, so if I can identify open sets as some geometric sets then its good enough
@Jakobian For the example above, can you imagine the space?
I have some vague picture of this, this is circle where $[0]$ is in every nbd of every point.
07:56
@EE18 this means that R is not very large.
08:09
@EE18 What was the context?
 
2 hours later…
10:32
@SineoftheTime I don't know if you saw point 2, if so, is the procedure correct?
 
1 hour later…
11:56
@Pizza I can't see the full solution of point $(2)$
@SineoftheTime i did this
it asked that: $(2)$ determine the line through $A$ incident to $r$ and $s$.
12:21
@Pizza the line you found is parallel to $s$
@SineoftheTime to solve point 2, can i start by finding a point $B$ of intersection between the line $r$ and $s$?
they don't intersect
actually, it's the first time I see this kind of exercise
@SineoftheTime how do you understand it?
@Jakobian Thanks, can you point out where is the mistake, please? And with "this is wrong" you mean that the result is false in general?
@Pizza from $s$ you deduce $x=z$, but $r$ is defined by $x=2$ and $z=0$ so they can't intersect
if there's a point of intersection $P$, then $P(x,y,x)$ i.e. $x=z$ but that's impossible
12:32
so the line doesnt exist
or you can solve the sistem
@Pizza the point of intersection does not exist
the line might exist
try to find all the lines through $A$ that intersect $r$ and then choose the ono which intersect $s$
It should work
Can it be useful to me to determine the point $B$ of intersection between the line and the plane $\pi$?
or, if you have the plane through $A$ and $s$ (or $A$ and $r$), find the point of intersection with the other line
@Pizza what is $\pi$?
12:35
honestly, I can't find this point in the slides, I have no idea
@SineoftheTime the plane passing through point $A$ and line $s$
$\pi:x+2y-3z=0$
that's the best way since you've already found $\pi$
find the intersection between $r$ and $\pi$
okok
I can replace the variables $(x,y,z)$ of the line $r$, in the plane $\pi$
wait, It might be wrong
try with the lines throgh A
12:45
from the equation of the plane $\pi$ I obtain $t=-1$, substituting it in the equation of the line $r$, I obtain the intersection points $B=(2,-1,0)$
I don't know if there's a solution, I'll let you know
yes, then you get a line parallel to $s$ if you compute it
what I said does not work
@SineoftheTime I'm trying to find a solution
this is what I tried to do at the beginning, but I honestly didn't know if it was what I should do, so I asked you
Your method is correct, now I understood what you did
however the line you found is parallel to $s$
Yes, I skipped the calculation steps
@SineoftheTime do you mean the one in the link?
yes
and you used the same method
12:54
yes i checked the guy in the comment
because they are things that are in the slides, so I knew what they were
yes
you can draw on geogebra 3d to verify your work from a geometric point of view
ok
ca i ask you a question
13:00
have you ever participated in, like, math competitions?
@Jakobian Ok, I was reckless indeed: a simple counterexample (for a finite supremum) is $a_n=1-1/(n+1)$ and $b_n=1-1/(n+2)$. We have $a_n<b_n$ for each $n\in\mathbb{N}$ but the supremum in $\mathbb{N}$ of both the sequences is $1$. Similarly, for a infinite supremum we have $n<n+1$ but the supremum of both the sequences is $+\infty$. So I assume that the mistake in my "proof" is when I write $\sup b_n > (\sup b_n+b_n)/2>b_n$, but isn't this a property of the real numbers?
I mean, when we have a strictly inequality $x<y$ of real numbers isn't it true that the midpoint of the segment $(x+y)/2$ is such that $x<(x+y)/2<y$?
@SineoftheTime I've seen some videos, MIT integration bee, where you have to solve integrals in 4 minutes, and like I see some who use trigonometry I guess?
it depends on the integral
what are you currently studying?
13:10
differential geometry
Are you studying it on your own, or do you have to take some exams?
I'm preparing a couple of exams
13:26
@ZaWarudo The inequality is true for each $b_n$ (assuming everything is positive and $sup b_n>b_n$). But when you take the limit, you are getting $ \text{sup } b_n=(\text{sup } b_n+\text{sup }b_n)/2 $ which is trivially true. So you are not getting any contradiction by proceeding this way
14:10
@SoumikMukherjee not really. I have hard time visualizing the projective space
14:27
But that example was S^1 with a weird point. I just can't properly imagine a point being in nbd with every point in the circle
0
Q: Find all possible Jordan forms of a real matrix $A$ of order 7 whose minimal and characteristic polynomial is $(x-2)^3(x+3)^2$ and $(x-2)^4(x+3)^3$

Thomas FinleyOur professor recently taught us minimal polynomial in a Linear Algebra course. At the end of his lecture, he wrote on board just how does a Jordan matrix looks like and told us to consider that as the apparent "definition" for the same. He didn't teach us anything in 'detail' about Jordan matric...

I need some help with this...
15:03
A question from a passage in my book if possible
TBH, probably related to Jakobian and leslie's comments that topology abstracts many of the notions I was asking about a few days ago
Two questions in increasing order of pedantry: (1) Should the two equalities of sets (of the individual unit balls and of the set of neighborhoods) be understood as meaning that the image of one under the bijection is the other (i.e. using "=" is really a shorthand)
@EE18 The passage you quote is about topological spaces, not metric spaces. I don't see any references to "balls" in that text.
(2) given that $\Bbb C$ actually is $\Bbb R^2$ when it comes to sets and ignoring the operations thereon, is there even anything to be said here? Or (and this is my suspicion) is the semi-non-trivial point that even though the two inner product spaces $\Bbb C^m$ and $\Bbb R^{2m}$ strictly speaking have different inner products defined thereon, there is something to be said in terms of showing that the two norms induced lead to the same balls and sets of neighborhoods in the sense of (1) above
@XanderHenderson My bad Xander, $\Bbb B_E$ is notation for the unit ball in the normed space $E$
Okay, in that case, yes. With respect to the canonical isomorphism between $\mathbb{C}$ and $\mathbb{R}^2$, balls in one are identical to balls in the other.
They are not, strictly speaking, "equal to" each other, and yes, the "$=$" is a shorthand.
@EE18 I don't remember discussing topology with you
Perhaps I'm misremembering Jakobian, I had some vague recollection of some theorem I asked about a week or so ago and a discussion which ensued about much of it being independent of metric spaces or something like that
15:14
@EE18 but this is a canonical bijection
@EE18 If $f:\mathbb{C}^m\to\mathbb{R}^{2m}$ is such canonical bijection then $\|f(x)\| = \|x\|$ so this is indeed an isometry. Not just topological properties are preserved but also all metric properties
they are different as vector spaces (so obviously as inner product spaces)
Got it, so I guess the answer is affirmative to my (2): that is, the nontrivial part is that they are different vector spaces (even though the underlying sets are precisely the same)
"nontrivial" is maybe stretching it
well yes, everything is fairly trivial here
nonetheless important
Well taken
Thank you very much for all the help Xander and Jakobian. I didn't appreciate this passage at first brush
Off to work now. Have a great day all!
Could someone who knows about NFA help me with this question? math.stackexchange.com/q/4924750/390226. It's about the implication of transition functions' value being an empty set. The thread has become a bit long, and I (unfortunately) suspect that the existing (only) answer hasn't resolved my confusion.
@SoumikMukherjee That a resistance was not infinite (and it was meant to be implied that it was "relatively small"):)
i.e. @Koro was spot on in the interpretation
15:23
Also another way to look at it, is that if $\mathbb{C}^m$ is reduced to a $\mathbb{R}$-vector space, then as normed spaces, $\mathbb{C}^m$ and $\mathbb{R}^{2m}$ are "precisely the same" in the sense that the canonical bijection above is an isometry (but this time instead of isometry of metric spaces I mean isometry of normed spaces) @EE18
15:39
Consider all homeomorphisms $f_i : S^2 \to S^2$ which fix distinguished antipodal points $p=(-1,0)$ and $q=(1,0).$ What questions can we ask about the space of these surfaces?
@zetaspace sounds like nonsense
$p$ and $q$ belong to $\Bbb S^2$ (?)
Understood Jakobian
yes @SineoftheTime
15:49
there should be three coordinates
@leslietownes man I'm not gonna lie, that conversation yesterday really bummed me out realizing that unless I'm planning on going the pure route/academia, learning even applied math won't net me a job.
One question which occurred to me on my bike ride: obviously this canonical bijection is not an isomorphism of vector spaces. Is there anything more pithy to be said than it’s a map between inner product spaces over different fields such that the two “equalities shown” hold
zzz I have to learn programming
When we consider their respective induced norms
15:50
$p=(-1,0,0)$ and $q=(1,0,0)$
@EE18 "obviously this canonical bijection is not an isomorphism of vector spaces"
I just said it is
Also (and this is obvious but just want to confirm) the point to be noted is that we can use this canonical bijection ad hoc to show that anything true about convergence in one space is true in a corresponding sense in the other
@EE18 Actual bike ride or is there a different meaning of this phrase?
Sure, if we restrict to R. But without that then no right?
isn't canonical bijection the automorphism of categories
15:52
Actual bike ride lol
without that the question makes no sense
you can't have an isomorphism of objects from different categories
there is no such thing as vector space without an underlying field
Agreed, that’s what I wanted to confirm. So then my question was about what we say when they have different fields in this way, ie ther s only some metric space preservation in the end?
oh, maybe you should not think about math while driving/walking on road
Hello everyone...is any moderator here??
It’s mostly a trail, but you are still probably right :)
15:54
@MathStackexchangeIsNotSoBad check out 'Math Mods' Office' room
@Jakobian I made a minor correction. Now it should make sense to you
@EE18 the point in the book is probably about convergence but topology is really about continuity albeit in case of metric spaces those are pretty much intertwined. But the bijection, as I said, is not just a homeomorphism, nor isometry of metric spaces. Its something stronger, an isometry of normed $\mathbb{R}$-vector spaces
saying this is ad hoc is misunderstanding something, however
the bijection is extremely natural to make, and this is in no way ad hoc
@SineoftheTime thnx let me see it rq
Do you know the sherlock scene where he fires a gun to quickly call the police. We can bring the mods by similar methods.
I agree with you, but I’m just saying notationally/as a matter of convention ive been told in this book to interpret it as a C vector space
15:57
The usage of equality $=$ itself tells you that the author wants you to think of this as the same thing.
flags messages
@SineoftheTime I can't find it..pls help
second page

 Math Mods' Office

For informal chat with the site moderators about moderation, s...
@EE18 you are misunderstanding the book
$\mathbb{C}^m$ is primarily a $\mathbb{C}$-vector space, but so what?
that's not to say that you can't think of it in any other way
16:00
@SoumikMukherjee thnx
this is just extreme case of pedanticism
I agree that you can think of it in other ways, but having flipped forward I know that’s not what they intend to do. They intend to interpret it as a C vector space in the future and to use this “theorem” in a few proofs about C^m as a C vector space
That is true lol
@EE18 you don't know that
they're not introducing this just so that they can talk about something
sure that might be part of it
"Consider all homeomorphisms $f_i: S^2\to S^2$ which fix points $p=(-1,0,0)$ and $q=(1,0,0)$."

Is the above statement nonsense?
but you don't know their actual intentions, and from what's natural they probably just put it there because it was convenient
we can treat those two technically different objects as one and the same if we squint our eyes a little bit, and this is our way of introducing it before we are going to need it to do something else
its not in any way ad hoc because this is not a method you're introducing just for the purpose of this one thing and skipping ahead proves literally nothing
its just a case of your pedanticism
16:08
I think the statement is quite clear...but then again I could be missing something silly
@zetaspace that's correct but isn't it dishonest to change your sentence
@zetaspace I'm not sure what you want us to consider, but the set / space of automorphisms is fine. I have no idea what $i$ is supposed to denote.
No it's not dishonest to make a necessary correction
you didn't just change points $p, q$ but left out an entire part, and that I thought looked like nonsense
you're making it seem like I'm talking about above, so its dishonest
or am I simply asking whether the statement makes sense? just asking a question
if you are perceiving that to be dishonest, sorry for the misunderstanding
my only intention was to get feedback
16:20
@SoumikMukherjee it's just projective $1$-space together with a closed point whose only nbhd is the entire space
yes i get that
@EE18 that's what identification means
@XanderHenderson I want to allow for maps that preserve $S^2$ as a topological surface and fix p,q so wouldn't the automorphisms you speak of be a subset of those maps?
also as far as I'm concerned, $\mathbb{C}=\mathbb{R}^2$ as sets, so this would be literal equalities, but I guess your definitions may or may not differ
I am not understanding the language here, what will be the basic elements? for any sequence (x1,x2,...) we can get an open set U_n in R^n containing (x1,x2,...,xn)
16:23
@zetaspace Sure.
But you haven't asked a question.
The space of homeomorphisms which preserves those points is a thing. Yes. So what?
And I still don't know what $i$ is.
@SoumikMukherjee is that an issue?
I put $i$ there to denote a number of mappings where $i$ was supposed to index those mappings. I should have also included the index set
@Thorgott what will be the open sets in this space?
the passage very explicitly states that, what part isn't clear to you?
@XanderHenderson If I may, my question is does the space of homeomorphisms which preserves those points form a group under composition?
16:33
Okay so does the passage mean that we choose an open set U_n from each R^n and consider those elements (x1,...) such that each (x1,...,xn) belongs to each U_n?
@zetaspace Does it satisfy the axioms of a group?
no, you choose one $n$ and then one open $U\subseteq\mathbb{R}^n$
I've been skimming through the book Calculus on Manifolds by Spivak and other books on the same topic. The change of variables theorem seems to be an important theorem in that book and in that subject, however, without having read anything very carefully, I'd like to know how is it related to the change of variables theorem stated in measure theory books?
My impression is that calculus on manifolds and measure theory deal with different types of integrals. The reason I'm asking is because I'm reading about transformations of vector-valued random variables and how to find their transformed distribution. I'm not sure where I should look for this theorem.
@XanderHenderson I think so but I don't know what the identity element should be
@zetaspace Is, or is not. Do not think.
16:44
@psie just to make sure, this isn't what you're after right?
@XanderHenderson Yes it's a group!
Okay, then.
@Obliv yeah, I believe that is what I'm after (in my book it is stated slightly different with other notation). The key ingredient is the change of variable theorem, but there seem to be variations.
@Thorgott Thanks, now I understand. Also can you suggest a learning roadmap for rational homotopy theory?
@SoumikMukherjee Depends on where you live, which institution you plan to study at, and whether you are waling, taking public transit, biking, driving yourself, or using some other form of transportation.
I would suggest that Google Maps or MapQuest might be more useful for this.
:P
16:53
@SoumikMukherjee I don't really know, I guess the answer is read Sullivan
@psie not sure if this helps, but what book are u talking about
@XanderHenderson hehe
@Thorgott Okay, thanks!
@psie I think this is right but @Thorgott ought to know
@psie this should be in a good probability book by the way
yeah, unfortunately, it's not in the book I'm reading currently, which is Gut's An Intermediate Course in Probability (@Obliv)
The Wikipedia page refers to Folland, and since I'm reading about vector-valued random variables, i.e. measurable functions to $\mathbb R^n$, this is probably where I should look
17:11
@psie In Gut's, page 41 cites pages 407-408 in Apostol (1969) for the change of variables in multiple integrals
Idk if you already checked that out
@psie sorry for img spam not at home so I can't upload and embed easily
@Obliv nice, but are you sure it's page 41 in the book I mentioned? I can't find it. Maybe it is his more advanced book? Or how did you find that it is on page 41?
I have the second edition by the way
maybe different edition
ok
omg XD this was not Gut's
some random lecture notes that had gut's as link name for some reason
well no it had the same "an intermediate course in probability and statistics" title
but it was some chinese lecture notes :D here if you want it
ok :)
17:18
google must have just picked up the reference
thats what I get for blasting through covers,prefaces,etc looking for table of contents.
@Thorgott what is enough to publish a paper?
I think I've proven something interesting but I've never published anything
@Jakobian result is correct and isn't already published :D
I don't know why I'm pinging you specifically Thorgott, I guess its because its about topology so I automatically thought of you
@Jakobian this should show that you can publish anything lol
the conway one is pretty funny
I don't know how to check if its not published already
17:31
but I guess the results are still of substance despite the length.
or if its enough for me to really write a paper about
I don't know what it is you've shown or the implications (maybe it's worth exploring any interesting things about the result for a while, to make the paper more "complete")
apparently this paper was written by one person but they credited the other (dead) person because they visited them in their dreams detailing the steps of the proof.
average mathematician I guess
17:54
I know nothing about publishing
18:14
@Jakobian do you have an advisor? Talk to them about it
@s.harp I left university
Are there people you are still in touch with that are connected to the field?
You would want to figure out if your result is something that is unknown and potentially interesting
yeah I'm thinking that maybe I should email my topology professor
though I've asked him before about a similar problem and he didn't know so I doubt he knows of any results
Or someone here, like thorgott, could be a good collaboration partner
since you guys seem to talk about topology a lot
also balarka but they don't visit as often
Balarka is not a topologist
from what I know
18:30
his top tags on MSE seem to be somewhat related
yeah his profile also says he's interested in geometry and topology
I will try to find something on this since I did find a slightly related theorem, albeit about something different
there seems to be some (implicit) misunderstanding in the air
I'm not a publishing mathematician
18:47
By the way @Thorgott you'd agree with me that Balarka is not a topologist but a differential topologist if anything? I believe there is a difference
I believe he mainly considers differential manifolds
I would most definitely consider Balarka a topologist
to me, "topology" is the term for the overarching subject
Indeed. Topology is very broad.
hm. I guess so
still, our interests are very different
so the hypothetical "collaboration partner" is not really realistic even if it was possible and we did that
🐷 - 3,14 = 9.8
which is basically what I was thinking of
18:56
can anyone guess it?
@Jakobian hi
@BinkyMcSquigglebottom Yes, but your equality should be only an approximation.
Since $\pi \ne 3.14$.
You've also left out the units.
my interests are related to general topology which deals with more abstract things, where Balarka's interests are well, mainly manifolds. I believe he considers any space gnarly enough as not interesting to him, even if subspace of $\mathbb{R}^n$
I know he was joking but I think he's being mostly serious about it
@XanderHenderson 🐷=?
🐷/3.14 ≈ 9.8
I think it should be
🍫so 🐷=? 🍫
19:05
that is true
I also mostly care about nice spaces
but speaking homotopy-theoretically, niceness is a bit broader
@Jakobian would you agree with this answer
I don't study either yet so I'm curious what the differences are. I've always felt like topology and geometry were very related
but I guess not always
@Obliv what subjects do you like?
Does ted know topology? Maybe you can email him and see what he says about it
I like them all tbh, but I haven't studied many formal maths yet beyond an intro abstract algebra course :P
are you studying on your own?
topology is a sort of bridge between analysis and algebra
yes and no, I don't need any more math classes for my physics degree but I will eventually become like jakobian and just study math for no good reason lol
19:11
Anyone know the Karnaugh maps?
@Obliv a lot of people have some grasp on topology
Please don't spam the chat like that.
but yeah I think math without physics is blind, physics without math is pointless
@BinkyMcSquigglebottom 💀
19:13
@Obliv definitely not
it doesn't take too much time to study the basic concepts of general topology
geometry is about properties that are stronger than topological ones
@SineoftheTime ?
so differential topology isn't that close to general topology then huh
I'd say differential-type properties are also not strong enough to consider them geometry
the place where geometry begins is probably when we consider distances
something more rigid
that's what I think when I think of geometry
differential topology is still too soft, bendy and stretchy
19:20
lol
weird. I had this feeling as if music was playing but when I got back to my seat it felt out of place
wat
what felt out of place?
because there was no music playing. And there never was. And it felt flat
all of a sudden
what is "it" in those sentences lol
like you mean something felt out of place? "it" implies something you referred to earlier felt out of place, like the music which doesn't make sense to me
not something
19:26
I don't know, "geometry" is somewhat of a loose term
oh the seat?
geometric topology is a thing and most definitely not about metrics, distances, curvatures, etc.
there's no it
hi
19:28
@Thorgott I feel like the word geometric is there to adhere to something closer to geometry than general or algebraic topology
does anyone know the p-adic numbers?
yea @XanderHenderson loves p-adics
rather than describing that the field is about geometry
I tend to call anything geometric when I can draw a sort of picture
I thought geometry necessarily dealt with distances
funny game
you know, I talk about geometric intuition of objects, especially when they live in $\mathbb{R}^n$
this seems hardly related to topology
the first section on "deformation theory"
it was linked from the wiki page on topology
But why, if i write @ Ted Shifrin, is there no way to tag him?
19:38
I got curious because geometric topology studies manifolds and maps between them which sounded like topology but I think topology deals with maps on a broader range of topological spaces besides just manifolds
I think that when we say the word geometric we mean it in a more loose sense
Has he left the channel?
he hasn't been here in a long time so you'd have to invite him to the room via MSE, but I don't think he'll come
@Jakobian is geometric topology not just a more specific branch of general topology?
definitely not a branch of general topology
general topology doesn't encompass every subfield of topology
that's not really what "general" means
looks like the origin is from this distinction in 1935
LOL the reference directs to meta stack exchange
oh no, we've come full circle. What can we even know..
16
Q: What is geometric topology?

Willie WongI see that we've recently just created the tag geometric-topology. Considering that the subject has its own arXiv subject code, I don't object to the tag's existence. But for me, geometric topology sort of lies in the fuzzy area between differential topology, differential geometry, and low dimens...

19:43
@SineoftheTime I was browsing Ted's site, I found this pdf, maybe it could be useful to you (if you don't already have it) alpha.math.uga.edu/~shifrin/ShifrinDiffGeo.pdf
Imagine if the answer was actually thorgott or jakobians. I'd actually think I'm in the truman show
@Pizza if you go to ted's profile he has a more recent version idk what the difference is but the date is 2023
@Obliv ah ok I didn't know, I was browsing, thanks a lot anyway!
what did you mean by via MSE?
I think wikipedia is not entirely accurate on what geometric topology is, perhaps
@jakobian another option could be to check if your result is on math overflow
@Pizza math.stackexchange
I've tried googling it and the answer I've been given is that the methods are geometric
so it seems like geometric topology is not about geometry but about studying things using geometry @Thorgott
which makes sense in comparison to such fields as geometric group theory
difference between methods and objects of interest
all I have is bunch of citations so not entirely credible, plus geometry is somewhat of a loose definition
but I think that geometry is when distances enter the picture is a somewhat accurate description of what we think when we talk about geometry
19:54
time to consult nlab
idk how credible nlab is but ncatlab.org/nlab/show/geometry they also don't mention geometric group theory hmm I wonder why
@Pizza thanks, this is Ted's book on differential geometry
it's a gem
it's bsasically what I'm studying right now
Do you sometimes have difficulty with English? Or do you know it well?
I don't know if I speak well
I guess because it's in group theory duh, not geometry.
but I try my best
I understand almost everything though
what about you?
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