midnight jewel
a one-to-one function (also called an injective function or injection) is one which keeps distinct elements distinct (no two separate elements get mapped to the same place).
the other keyword here is onto, which means that every element in S is actually mapped to. in my opinion, they should have highlighted that much more clearly, because it’s really important
injective = one-to-one
surjective = onto
bijective = both of the above
zinc tinsel
Isn't bijective one to one correspondance?
midnight jewel
so, to state the definition more clearly:
A set $S$ is finite if one of the following equivalent conditions is met:
\begin{enumerate}
\item There exists a bijective function between ${1, 2, \dots, n}$ and $S$ for some $n \in \mathbb{N}$
\item There exists an surjective function from ${1, 2, \dots, m}$ to $S$ for some $m \in \mathbb{N}$
\item There exists a injective function from $S$ to ${1, 2, \dots, k}$ for some $k \in \mathbb{N}$
\end{enumerate}
Convince yourself that each of these is reasonable
yes, tywin, as I said the onto there must have been intended as a keyword, I really dislike how they worded it
I dislike the use of “one-to-one” and “onto” anyway
zinc tinsel
For the third, could you say k is smaller than or equal to the cardinality of the set S
gentle ospreyBOT
Sascha Baer:
midnight jewel
I swapped surjective and injective
midnight jewel
yea
the other two aren’t really necessary, but I wanted to include them fore completeness’ sake
zinc tinsel
Surjective is where n is smaller than or equal to the cardinality of S, and injective is where the n is larger than or equal to the cardinality of S
whoops
Sorry
Just realised
I meant to use m and k respectively
So that's all?
Wait
What about infinite sets
Can't you say they have one to one correspondance
Since they are of the same size
midnight jewel
in both 2 and 3, m,k ≥ |S|
and what do you mean by that?
not all infinite sets are bijective to one another
which is why it makes sense to state that not all infinities are the same
zinc tinsel
Isn't the cardinality of the set X less than that of set y in the injective non-surgective function
And the bijective function is where the cardinality is equal
midnight jewel
notice how I reversed the directions
zinc tinsel
What do you mean?
midnight jewel
I’m not gonna explain it. you need to slow down and read more carefully what others have written
midnight jewel
yes
midnight jewel
they are
zinc tinsel
Might I ask what do you mean by direction in this context?
midnight jewel
f: X→Y and g: Y→X go in “opposite directions” so to speak
midnight jewel
let me guess: you didn’t read carefully enough
zinc tinsel
Yes, as you said before
midnight jewel
I don’t know how many times we can emphasize it
but math books assume the ability to read carefully (especially in a topic like topology whihc is not usually the first “rigorous math” topic students tackle)
“mathematical maturity” really just boils down to being able to read correctly what the author wrote, not miss details, and be able to follow logical argumentation (and spot flaws if there are any)
so that’s what you have to practice
the topics you‘re working at are good at practicing that
because there’s often important details
midnight jewel
S can map injectively to {1,2,3,4}, but not onto it (surjectively)
zinc tinsel
But
midnight jewel
note of course can
zinc tinsel
{1,2,3,4} can map surjectively to S
midnight jewel
yes
zinc tinsel
cool
I need to sleep now
Thanks!!!!
I know it's difficult having to deal with someone like me, so thank's for persevering :D
Very nice
sleek forum
midnight jewel
idk there’s plenty of points in a circle
ember maple
Wow this reminds me when I first learning topology. I think I needed to reread definitions hundreds of time each.
Learnt how to formalize statements with quantifiers too as a bonus.
midnight jewel
Can someone give me an example problem of a space whose fundamental group can be computed with Van Kampen where the answer’s actually interesting? We’ve done Torus and Klein Bottle in class, something along that complexity would be nice
I suppose projective plane is also along those lines, but the approach to doing it seems to be exactly the same as with the torus (I’m sketching it rn)
I kinda wanna see an example where neither of the two subspaces have trivial fundamental group
honest narwhal
You mean where the intersection has non-trivial fundamental group?
midnight jewel
basically all three involved parts. like, in the torus, one of the two open sets is just a disk
double torus seems to fit the bill tho
workin gon that now
I don’t know the answer either, so we’ll see how this goes
so the double torus is the union of two tori with a disk removed each, overlapping in a cylinder. torus minus a disk is homotopic to a rose with two petals so it has the free group with two generators as its fundamental group. cylinder is homotopic to a circle, so it’s ℤ.
now I need to look what happens to a loop upon embedding. if the generators of T1 are a,b and those of T2 are c,d then the inclusion of a simple loop (the “1”) into T1 would be aba⁻¹b⁻¹. and the embedding of the inverse of “1” into T2 would be dcd⁻¹c⁻¹, and so I have to quotient out the normal subgroup generated by aba⁻¹b⁻¹dcd⁻¹c⁻¹, right?
so the answer is ⟨a,b,c,d | aba⁻¹b⁻¹dcd⁻¹c⁻¹⟩, whatever the heck that is
well, the answer appears correct
I was kinda hoping it’d be prettier :P
honest narwhal
Yeah that's right, and I mean it's not bad, noting that each guy is a commutator
Inductively this holds for the genus g surface, the fundamental group is <a_1,b_1,...,a_g,b_g | prod_i [a_i,b_i]>
dim meadow
@midnight jewel every group is a fundamental group, and groups are rarely pretty
midnight jewel
yea, but one might hope that a "pretty" space would have a "pretty" group. double torus is reasonably "pretty", but its group is much uglier than that of the single torus
like, sure, every group is a fundamental group since you can always construct some complex but those are generally, shall we say, weird
actuallty this got me thinking tho. if you take a disk D^2 and you identify the edges like for the projective plane, but in n segments, then you get a space with fundamental group Z/nZ, right? are those spaces ever "interesting" (worthy of closer inspection/study)?
dim meadow
I think the infinite holed torus is pretty
Actually yes
Well, what I'm thinking of is when you take an octagon and identify the edges to get the 2 torus or something
Look up the classification of surfaces and you'll get your answer
Cause that's what they do
midnight jewel
yea but there you do pairwise identification. I mean take an n-gon, and identify all edges with each other
(in such a way that the "arrows" all point tge same way)
doing that with two sides gets you RP2
midnight jewel
yea I know
dim meadow
You would probably get a connected sum of projective planes or something
midnight jewel
hm I gotta think about what the fundamental group of the sum of two proj planes is
(no spoilers)
dim meadow
Lmao, the shape is very familiar
It's a bit of a meme
Yeah I'm wrong
That's not what you want
midnight jewel
it's a sphere isn't it
cause unless I'm mistaken one way to do it would be to just remove the boundary circle and glue there, which gives you two hemispheres glued together, ie a sphere
and a vague, nonrigorous trying to think it through with van kampen also tells me that (if I didn't miss sth) the group is trivial
no, wait
I talked to my topo prof about this
sum of two RP2 is a klein bottle isn't it
I don't recall the justification but it rings a bell
dim meadow
No
Klein bottle
I mean, you wouldn't expect the connected sum of 2 non orientatable things to be orientable
In fact it can never be orientable
@midnight jewel
dim meadow
The argument for the Klein bottle is very simple using the surface symbols (which I forgot what they're called because I'm very tired)
midnight jewel
the polygons?
midnight jewel
mhm, I'm gonna read munkres over summer as part of exam prep
we used hatcher's notes and they don't go much into detail
dim meadow
I like rotman now
There are 10 million different algebraic Topology books lmao
On the intro level
And none on an intermediate level
It's a bit of a meme
midnight jewel
I've had meh experiences with rotman
honest narwhal
Rotman AT?
midnight jewel
his modern abstract algebra is not to my liking
dim meadow
Ah ok
midnight jewel
that was our text for algebra I
dim meadow
I haven't looked at his group theory book
honest narwhal
Ah I haven't read that though I think jacobian said it was good
midnight jewel
and the prof was like "this was a mistake"
dim meadow
Also some books are better if you've had more experience
honest narwhal
Or at least what I read of it
dim meadow
Eg lang
honest narwhal
His group theory book I remember being real nice
Or at least what I've read of it
Also lmao is there actually a good intro to algebra book out there?
I feel like D&F doesn't do anything until the heat death of the universe, Lang seems a bit too tough to jump into and apparently doesn't do enough group theory
Hungerford is apparently a watered down rewrite
Artin and Herstein seem fine for somewhat easier ones I guess
midnight jewel
my prof has the following opinions:
We can also recommend the following two books for the course:
N. Jacobson, "Basic Algebra I" (excellent)
I. N. Herstein, "Topics in Algebra" (good)
I have never opened either
dim meadow
Have you looked at Jacobsen?
Yeah, but it might be easy in retrospect
I've suggested it as a first book often, and have been told that it's too hard
So that's what I assume
honest narwhal
I mean not even, just that when I was reading it for the first time (that was how I learned group theory) it was pretty smooth going, like it strikes me as strictly easier than, say, D&F
Though while I very much believe we should switch to x(f) notation, it's not common anymore and a bit of a disservice to teach people. I had to unlearn that way of multiplying cycles
I mean people say it's hard when they're asking for something like Fraleigh for a first book
But if you're not a toddler Herstein's fine lmao
dim meadow
That's good to know, I guess I would feel okay recommending it then
It's hard for me to tell what's hard now lmao
honest narwhal
But yeah I feel it's a bit old school, covers less than the others, and again you don't want to get in the habit of multiplying cycles backwards so I'm hesitant. Probably Artin is your generic "Intro to Algebra" book
dim meadow
But if it was your first book and you found it okay it's probably fine
honest narwhal
Anyway I'll check out Jacobson then
honest narwhal
Yeah kinda, summer after first year I did my REU on Sylow stuff
And Herstein + Keith Conrad's blurbs were basically my sources
honest narwhal
"first courses don't usually treat group actions"
If an algebra class doesn't talk about group actions it's not an algebra class lmao
honest narwhal
Rings before groups? That feels odd
dim meadow
I mean, I think rings are very natural by themselves
honest narwhal
I mean it's not necessarily a bad order (though you wanna talk at least a bit about groups before modules and it's strange to do rings->groups->modules), really it just feels strange.
But yeah have you guys used Jacobson? How does it compare to D&F?
(I'm gonna review some algebra for grad school to do the qual)
midnight jewel
I've never really learned with books, gonna try to change it up this summer and go through munkres and a measure theory book
sooner or later I'll have to learn how to learn from books
honest narwhal
Munkres topology? Feels a bit overkill
midnight jewel
I mean it is recommended reading and I like what I've seen. I'm probably only gonna do exercises on the parts that I
a) don't find trivial yet and
b) are relevant to my exam
and just read the rest out of interest
there's some stuff we didn't cover, like metrization
dim meadow
I learn exclusively from books, arguing with friends, and seminars/talks
midnight jewel
so far I've just atteded lectures, done homework and worked through some old exams
but this semester I can't do the last part cause they're oral exams
for the first time
honest narwhal
I mean yeah it's probably a good book, I just mean that point-set topology is sorta expendable
honest narwhal
Do these notes cover most of what your exam does? https://pi.math.cornell.edu/~hatcher/Top/TopNotes.pdf
honest narwhal
Rip
midnight jewel
we did some extra things, there's also just stuff I wanna see more detail about or a second perspective
I'm not gonna waste too much time on topo anyway
the other subjects are all higher prioroty
topology's my best subject this semester
honest narwhal
Nifty
midnight jewel
measure theory i've fallen behind, algebra I need to review tons, P(probability is fine)=0.6ish and numerical analysis can suck some non-PG words
honest narwhal
Soon you'll hit differential and more advanced algebraic topology and hopefully then topology will be a high priority if only out of interest
midnight jewel
I mean that's the case already
I really enjoy the subject and the intro to algtopo we did at the end now has me excited for more
what's the difference between difftopo and diffgeo btw?
honest narwhal
I'm somewhat surprised lmao, idk point-set feels straight up unlikable to me. But maybe in that case the fact that according to jan, operator algebras involve the stuff won't be as depressing
Differential topology is basically trying to understand the topology of smooth manifolds, often using said smoothness
One thing you end up showing is that a continuous map between manifolds is homotopic to a smooth one, and if there's a continuous homotopy between two smooth maps, then there's a smooth homotopy
midnight jewel
hey we sketched that in complex analysis
dim meadow
I love that theorem
honest narwhal
Yeah it's nice, tubular neighborhood + convolution + I think for the homotopy one you need to play a bit with bump functions
dim meadow
You can also preserve a closed subset if the function is smooth on that subset
honest narwhal
But yeah if you define some invariant of smooth maps under smooth homotopy then you get it for continuous maps under continuous homotopy
In fact the main homotopy groups I know how to compute are in the smooth case
dim meadow
Tubular neighborhoods are great
honest narwhal
Sard to show pi_i(S^n) = 0 if i < n, and Hopf degree for pi_n(S^n) = Z
That + stuff like LES of a fibration and I guess Van Kampen though I've got very little practice with that
midnight jewel
does van kampen generalize to higher homotopy groups?
honest narwhal
Nope
midnight jewel
aw :(
honest narwhal
Yeah it's kinda heart breaking
Oh I forgot, so yeah differential topology is smooth manifolds
Differential geometry, usually you're putting some extra structure
dim meadow
Is the abelianization of a higher homotopy group the corresponding homology group
honest narwhal
"Connection" seems to be the main word here, a fairly common example being something called a Riemannian metric
LMAOOOOO my AT prof wouldn't have a job if that were the case
But yeah the idea of a Riemannian metric is that on the tangent spaces of a manifold you put an inner product
dim meadow
midnight jewel
I've tried reading about what homology is but it seems to require twenty pages of definitions and motivation before you can even describe it so i've not dug deep enough yet. but eh, I'll learn about it next semester
dim meadow
It really doesn't though
honest narwhal
So you now have angles there, and the notion of lengths of curves as $\int_a^b |\gamma'(t)| dt$ now makes sense. So that's more amenable to doing wonderful geometric things like Christ Awful Symbols
gentle ospreyBOT
Dami:
midnight jewel
I assume those awful symbols are mostly variations of the letter d?
honest narwhal
I think he's asking about Christoffel symbols
midnight jewel
so squiggly d
honest narwhal
No?
midnight jewel
I'm asking about what you wrote
dim meadow
Oh lol
honest narwhal
Yeah those are usually denoted $\Gamma^{ij}_k$ I think or something
gentle ospreyBOT
Dami:
dim meadow
I thought he was talking about homology
honest narwhal
I dunno I've avoided diffgeo for the past couple years
west spindle
christoffel symbols 
ember maple
Christofel symbol is a variation of d.
Well at least kind of.
Lot of things in diff geo feels like variation of d anyway. Since that's kinda the whole point of differential in differential geometry
heady epoch
working with the christoffel symbols is a pain...
frigid patrol
is this the right place to ask about de Rham's theorem?
If so could someone explain it to me
zinc tinsel
Just to clarify
If you have a topological space (X,T)
And for every x that is an element of X
There is a singleton for that in the T
That makes it a discrete topology
midnight jewel
yes @zinc tinsel because then you can write any subset of X as a union of those singletons, so it has to be in T
thus all subsets of X are in T
thus T is discrete
zinc tinsel
Great!
What if a question with that space was asked
But it was asking you if the topology is a discrete, but only gave you the information that T contains X, the null set, and the singletons? Would that imply that the second condition for a topological structure is applied? Or would it have to be phrased "Prove the topology T on X is discrete if"?
Sorry if I worded that terribly.
west spindle
you kinda did
zinc tinsel
Sorry
west spindle
if you are only given that T is a collection of subsets of X and aren't given that it is a topology then no you cannot conclude that it is the discrete topology just from it containing ∅, X and all the singleton subsets of X
zinc tinsel
Ok cool
midnight jewel
but if you’re given the further condition that T is a topology, then yes
see my proof just above
midnight jewel
and only those?
zinc tinsel
No.
midnight jewel
then word it better
state precisely what T is
what you just said is that T does in fact consist of exactly those five sets
zinc tinsel
midnight jewel
yea, that’s not what you said
zinc tinsel
Sorry!
midnight jewel
what it says here that you didn’t say is:
•T is a topology (very important bit of info!)
•there may be other sets in T than what you mentioned
zinc tinsel
What if it does not mention T is a topology
midnight jewel
you said
T = {∅, X, {a}, {b}, {c}}
they say
T is a topology on X and {a}∈T, {b}∈T, {c}∈T
if it doesn’t say T is a topology, then T could be anything
you would know nothing about it except what it says explicitly
zinc tinsel
What if it said prove T to be a topology
midnight jewel
then there would be enough other information for you to do so
zinc tinsel
Instead of discrete topology
midnight jewel
if it said that then T would be fully described in some other way
like stating exactly which sets are in it
zinc tinsel
cool
ember maple
@frigid patrol I think there is someone in physics server in #old-network that knows a lot about de rham cohomology, ask there since there are things I want to know too
west spindle
this is entirely false. tau_1 and tau_2 are topologies, while tau_3 is not
@zinc tinsel
sleek canyon
@frigid patrol what do you want to know about de rhams thm
given a closed p-form, you can integrate it on a p-simplex
so any such p-form gives a function from Hp(M) to R
that is, an element of H^p(M)
(since integrating over boundaries is 0 by stoke's thm)
de rham thm says this is an isomorphism
zinc tinsel
@west spindle I'm confused :p
That's what I though
Because if you union {e, f} with {a,f} you get {a,e,f} a subset which is not there right?
west spindle
yeah. τ_3 isn't closed under union
minor stag
T2 isnt under intersections 
minor stag
T2 {a b f } intersect {a b d}
west spindle
oh. yeah. rip
tough hamlet
so
west spindle
i'm dumdum
tough hamlet
none of them are 
minor stag
None of them seem like topologies
zinc tinsel
ok coolio
west spindle
@myself
gritty widget
lol
tough hamlet
so logically this isn't topology 
tough hamlet
missing {b,d}={b,d,e} intersect {a,b,d}
tough hamlet
looks like it
west spindle
yup
bitter yoke
I wasn't arguing that c wasn't not, I just wanted you to explain yourself lmao
west spindle
triple negative 
bitter yoke
Feel like it's justified here
frigid patrol
@sleek canyon thanks
Wikipedia says "Stokes' theorem is an expression of duality between de Rham cohomology and the homology of chains."
How is this so?
ember maple
@frigid patrol
honest narwhal
Don't quote me on this but I think integrals of closed forms are the same on homologous chains
ember maple
Sorry I guess that was not too relevant. Though I'm looking about it too
honest narwhal
Also integrals of exact forms on closed chains is 0
The latter holds by Stokes' theorem, if you integrate d\omega on a closed chain, that's like integrating omega on the boundary, but if it's closed the boundary is 0
So you have the pairing (\gamma,\omega)-> \int_{\gamma} \omega
And I think by the above facts you can replace gamma with a homology class and omega with a DR cohomology class
ember maple
Is the duality between $\omega$ and $\gamma$ where $\int_{\gamma} d\omega = \int_{\partial \gamma} \omega = 0 $?
So to say we are looking at pairing between form and chain for which the above expression = 0. Kind of like a vector space duality? Is it isomorphism?
honest narwhal
Uh so the way it works is this
gentle ospreyBOT
Raysena:
honest narwhal
(At least I think, I'm really just piecing this together)
So by universal coefficients theorem, $H^k(M,\mathbb{R}) \cong \text{Hom}(H_k(M),\mathbb{R})$. So now let $[\omega] \in H^k_{\text{DR}}(M)$. Then the map $H_k(M) \to \mathbb{R}$ given by $[\gamma] \mapsto \int_{\gamma} \omega$ should be well-defined by Stokes' theorem
gentle ospreyBOT
Dami:
honest narwhal
That's what I was saying above, since gamma is closed the choice of form in a cohomology class doesn't matter, since they'll differ by an exact form which integrate to 0, and if we swap with a homologous chain then closed forms shouldn't change the integral
Now, we have a map $H^k_{\text{DR}}(M) \to H^k(M,\mathbb{R})$ given by $[\omega] \mapsto \int_{\cdot} \omega$. De Rham I think says that this map is what's an isomorphism
gentle ospreyBOT
Dami:
midnight jewel
"Today we're gonna talk about why there is a 1 in π₁"
dim meadow
They should first say why there is a 0 in pi_0
midnight jewel
why is this a course in diagrams now instead of in drawing toroids
sleek canyon
how are diagrams not rigorous?
midnight jewel
diagrams are perfectly rigorous once you’ve proven that you can use them, essentially
which in this case we had done
(though still note: “sketch”)
ember maple
Is it algebraic topology course?
midnight jewel
last lecture of an intro topology course
the last third or so was an intro to alg topo tho
(on the fundamental group and some stuff on covering spaces in order to prove the fundamental group of S¹)
ember maple
Huh I thought it's an early semester for you since you posted your course schedule a little while ago
midnight jewel
I’m planning my next semesters out
the times for next semester got released a few weeks ago
my semesters are september–december (exams in january/february) and february–may (exams in august)
I like to think ahead a lot
get an idea for what to do, then think about it for a while, change stuff up…
I’m already thinking about my master’s now even though there’s a whole year separating me from it
midnight jewel
the ideal student would be doing homework instead of procrastinating that way :P
ember maple
Hearing that makes me more tempted to ditch a job offer and just attend class on this professor I'm planning my master with
midnight jewel
hearing what?
ember maple
The ideal student and procrastinating student
midnight jewel
here’s my plan for now, btw:
- BSc Math
+ 5th Semester
Algebraic Topology I (10 KP, pure core course)
Differential Geometry I (10 KP, pure core course)
Theoretical Computer Science (7 KP, applied core course)
Geometry (3 KP, additional subject)
Human Learning (2 KP, GESS)
Communication in Mathematics (1 KP, elective)
+ 6th Semester
Algebraic Topology II (10 KP, pure core course)
Differential Geometry II (10 KP, pure core course)
Seminar (4 KP)
Bachelor Thesis (8 KP)
- MSc Applied Math
+ 1st Semester
Introduction to Lie Groups (8 KP, core course)
Functional Analysis I (10 KP, core course)
General Relativity (10 KP, application area)
+ 2nd Semester
Representation Theory of Lie Groups (8 KP, core course)
Theoretical Astrophysics and Cosmology (10 KP, application area)
Semester Project (8 KP)
+ 3rd Semester
Quantum Mechanics I (10 KP, applied core course)
Numerical Methods for PDE (10 KP, applied core course)
Seminar (4 KP)
+ 4th Semester
Master Thesis (30 KP)
Do you normally get AT and DG in undergrad? 
midnight jewel
(seminars, projects and theses are left vague)
ye
the offered pure math core courses for bsc math next semester are
diff geo I
alg topo I
func ana I
dynamical systems I
commutative algebra
intro to algebraic number theory
these are meant for 5th semester students
though they can also be taken during the master’s
you have to take one of them
at least
ember maple
Wow nice those geo, top and algebra course are ones I would like to take
And func analysis too
I wonder if you need math BSc to specialize in mathematical physics
midnight jewel
I suppose you could also do it with a phys BSc but you’ll have to take some extra math courses
ember maple
Well I'm self studying right for now
6 years in physics BSc program and the last 3- years were spent reading math book
ember maple
midnight jewel
you need to pass first year within two years, and graduate within 5 (regular study time: 3 years)
ember maple
Well there are lot of unrelated things and field work in my place 3 year is probably impossible
Like religion indoctrination course
On the other hand that's also partly reason I can read math books from time to time
Max 7 years (regular 4) or drop out here btw
Though I admit my math probably all shaky and have lots of gaps in it
midnight jewel
quick! what is the dual of $L^p$?
gentle ospreyBOT
Sascha Baer:
ember maple
What was lp again? |f|^p Summable?
midnight jewel
so, $\mathcal{L}^p$ is the set of all functions (over a given domain) where $|f|^p$ is integrable
gentle ospreyBOT
Sascha Baer:
midnight jewel
and then $L^p$ is $\mathcal{L}^p$ but you declare all functions equivalent if they are almost everywhere the same
gentle ospreyBOT
Sascha Baer:
midnight jewel
(according to some also given measure)
ember maple
Okay so how is the dual construction goes?
midnight jewel
so e.g. over ℝ with the lebesgue measure, the 0 function would be equivalent to a function which is 0 everywhere except at the integers
the dual is $L^q$, where $q$ is such that $\frac{1}{p} + \frac{1}{q} = 1$
gentle ospreyBOT
Sascha Baer:
midnight jewel
in particular the dual of $L^2$ is $L^2$, which allows you to define an inner product on it
gentle ospreyBOT
Sascha Baer:
midnight jewel
by integrating the product of |f| and |g|
hence quantum mechanics works
(I’m being deliberately vague here, in part to mask my own ignorance, in part because a detailed discussion would take too long anyway)
ember maple
actually some people use rigged Hilbert space which is probably just say you don't need V* = V. I think
midnight jewel
but the idea is basically that for every g in L^q and every f in L^p, the integral over f*g is well-defined
and defines a linear, continuous functional
ember maple
I think I have read it somewhere before. Probably wiki and a strange QM book that used rigged Hilbert (use L^p and L^q). But would never remember it again if you didn't point it out
And there a big hole
One of the peril of self education is no one is kicking your butt to work off. So the foundation is really shaky and lots of missing or forgotten definitions.
ember maple
Though it can still be reinforced with time
Like you need years to be able speak topology
More reason to staying around here I guess 
Seeing people much better than you give you that little bit of social butt kicking
Are fundamental group and H_1 actually just circles mod homotopy but the other one is free group while the other is free abelian ?
Sorry I mean if you self study. Without ever done analysis course. Much less formal logic. Without mentor
Yeah I guess you can do faster than years
Singular?
Oh since you actually use it to calculate integral I guess it's not homotopy right
I mean the formal sums of map of circles. Not the H1 itself
Like start with a mapping from a simplicial complex K to X. Define measure on K. Then we can define integral on function f on X by taking its pullback to K. Though the sums maybe should be defined ala Riemann sum as limit of barycentric subdivision.
That's how spivak did it on Calc on manifold I think
If the value of integral on a circle is mod homotopy of the circle it would be really weird
I guess they didn't really mean it quite literally when saying H_1 is simply abelianization of fundamental group
ember maple
I was just confused on what abelianization means. I guess there is this definition (no 2). I thought you need to have surjection.
ember maple
Is it a surjection? Since pi1 is loop mod homotopy. While H1 is circle mod boundary. I thought the codomain f is strictly smaller than H1
Okay what's f so that it's abelianization?
If pi1 is loop mod homotopy while H1 circle/loop without homotopy. I think you need to pick for each loop mod homotopy class in pi1, one of its loop. H1 itself seem strictly bigger than group f maps into.
cedar pebble
what
I think you are mistaken about the definition of H_1. Certainly \pi_1 surjects onto H_1 by taking the homology class of a loop
^
cedar pebble
if you like you can do this with simplices rather than spheres
ember maple
^
Okay so using simplices you map a complex K (not mappings of all complexes of K) surjectively to X and generate free abelian group over it?
cedar pebble
what
ember maple
Okay what's surjection?
Okay I will just take the definition of H1 1-cycles modulo boundary that also an abelianization of p1
Much less trouble that way
cedar pebble
what
ember maple
I think I will just read this for now https://math.stackexchange.com/questions/1949774/the-first-homology-group-is-the-abelianization-of-the-fundamental-group to clarify definitions
I feel like I'm confused on the construction of H1
ember maple
Nah. I'm really sure my core problem here is confusion of construction of H1. Though I do need to review group basics, that's probably for other reason.
I can see now for simplicial case why it's surjective homomorphism
clear jackal
Sometimes I feel like Geometric Group Theory is a little too good for explaining things
honest narwhal
Yo I so wanna learn GGT
clear jackal
I've invented a knot invariant with it and would like to know if anyone has seen a similar structure
Basically a natural permutation group for a projection of a knot. Since you can get very arbitrary with how a knot is projected, the most common permutation group is the smallest one, the trivial group
But among knot (and link) projections with a nontrivial permutation group, there can be a powerful invariant: the biggest group. Because, I've conjectured that if a biggest group is known to exist, then all other projections must have a permutation group that is a subgroup (usually satisfied trivially) of the biggest group
and I don't even know what to call any of this to research more into it because I wasn't inspired by anything similar to this, I just kinda noticed this pattern and have been going a little crazy computing symmetries of various knots that are predictably nontrivial.
real notch
so now just prove it
clear jackal
yeah that's the problem 😂
real notch
So, does a biggest group in this context exist for a knot?
I mean like is it properly invariant i suppose
clear jackal
it's more powerful of a conjecture when the group is finite, which is not always the case, but I believe that, most knots have simply a trivial group because they are too asymmetrical so to speak, so no matter what projection you have there is no nontrivial symmetry, as I've defined it
I believe that all of the knots listed as N_1 for odd N have a largest symmetry in their canonical projection.
Which is Z_N
*isomorphic to
real notch
Hm, so can you prove that largest symmetry group?
clear jackal
I don't know how
that's why I'm reaching out
I've been thinking about this for around 8 weeks so far
I can TeX my definitions, if that would help
real notch
You could probably prove it for some certain knots and work from there as a guide?
clear jackal
I don't know to prove maximality for even the trivial knot
bitter yoke
So how do you actually compute the group for a given projection of a knot
clear jackal
I believe that it is isomorphic to the circle group
which contains all finite cyclic groups as subgroups
real notch
and yeah would be nice to have definitions formally done
also, maybe don't do the trivial knot, but rather a knot for which the maximal group would have to be finite
clear jackal
Best candidate is trefoil
real notch
Maybe trivial would be a good starting point, but you said yourself it's finite that it's good for
clear jackal
actually
trefoil is prime which makes the subgroup conjecture irrelevant
as in, the group is prime. The knot is also prime by coincidence but that's not what I meant
so the figure eight is better, because I can easily find subgroups of a 4 group, they all have to be 2
one sec let me open up by original TeX document to copy
real notch
So, is there one which is simple enough to work with but has a strong enough group to give you some actual fuel to work with?
clear jackal
yes the figure eight, but I haven't proven maximality, I've only guessed, but I have a good reason to make such a guess
real notch
so prove that
real notch
So, does the maximal group thing remain invariant for the figure 8?
real notch
So, reversing the orientation doesn't change it, does it?
clear jackal
no and I've proven that
or taking the mirror image, for that matter
same proof essentially for both
real notch
So, does the crossings thing count all crossings of that particular one or of the prime-ier maximally detangled version?
clear jackal
it's on a particular projection, but, if my conjectures are correct, and you manage to find the maximal group for a link, you can distinguish it from any projection of a knot or link that does not have a symmetry group as a subgroup of the known maximal group
so it's very powerful as long as you have the maximal group of one
this is a side project of mine so even though it's been in the back of my mind I haven't been able to give it full thought because I have to take my actual courses as a higher priority
real notch
aw
clear jackal
but I'll have almost 2 weeks of break soon between semesters where I can hopefully develop this further, but I'd like to know if someone has seen something similar to this before first
ironic how a math degree can get in the way of your math projects
real notch
But uh, it's not something i've seen before
The thing I'm most interested in is the invariant part for a maximal permutation thing
clear jackal
Conjecture 1:
For a link $L$, if there is a finite group of symmetries for a projection of $L$ that is maximal, then the group is unique up to isomorphism, denoted by $\textsc{Sym}_{\textsc{max}}(L)$.
Conjecture 2:
If a link $L$ has a finite maximal symmetry group, then $\textsc{Sym}(P) \le \textsc{Sym}_{\textsc{max}}(L)$ for any projection $P$ of $L$.
gentle ospreyBOT
datorangeguy:
real notch
Now you just need a Conj 3 or so that says there is a maximal symmetry group (allowing infinite and quite possibly uncountable cardinalities)
clear jackal
Well the thing is, a knot must have a finite number of crossings
in any projection
so the only reason to assert the existence of an infinite maximal symmetry group is in the case where no projection could possibly faithfully represent the group via group action
clear jackal
Well trivial links as well
so you could get a baker's dozen of donuts and still find infinite groups that nicely describe what the other projections have to be a subgroup of
😃
And I can also give a good argument for why those trivial links in their canonical, no-crossings projection, should be assigned those infinite groups
But like, I'm also kinda the one defining everything in my paper so I have no real rigorous approach for why it should be the case. I think it's just a matter of establishing convention, whether or not that kinda thing should be included in this approach towards developing an invariant is kinda hard.
So it feels wrong to jump right into that first.
The one thing I've tried to prove is that you can not alter the trivial knot so that its symmetry group is not cyclic
I feel like I'm close to a proof, and it will certainly get me somewhere, but it is still merely evidence towards my main conjectures.
Something about, for a single crossing, being able to tell the output of the permutation for each crossing only based on the image of your one chosen crossing
because beginning to travel on any of the crossing components, you pretty much know that for any valid symmetry, the order of the crossings by traveling along the string has to be preserved,
desert vortex
I have an emergency related to hartshorne
On page 70 where he defines structure sheaf for a scheme he says that s(q)=a/f where f is not in q
But a/f is defined to be in A_q and thus f is in q
by the definition of localization of a ring
honest narwhal
Uh, usually you localize a ring at the complement of a prime ideal, no? @desert vortex
For me that's what A_q would denote, (A\q)^{-1}A
desert vortex
I suppose I got the wrong idea from this wikipedia article
I just substituted S for q
honest narwhal
I mean the article says that a multiplicative set contains 1
Prime ideals don't contain 1
desert vortex
But I see in the example section exactly what you said, thank you for clearing it up
honest narwhal
No problem! But yeah that's why you restrict to prime ideals, dropping the 1\in S condition would make any ideal work but only when the ideal is prime is the complement multiplicative
desert vortex
Yes, I made some tea and thought about what you said and it makes a lot of sense. Thanks again.
gentle ospreyBOT
mathbb:
Got a question for you all. If $\alpha:[0,\infty) \rightarrow \infty $ is a maximal integral curve of a smooth vector field V on a manifold M and $\alpha $ is extendible (I.e. there is a point $q\in M $ such that lim_{t\rightarrow\infty} \alpha = q), I’m trying to show that V must be zero at q.
```Compile error! Output:
! Missing $ inserted.
<inserted text>
$
l.11 ... there is a point $q\in M $ such that lim_
{t\rightarrow\infty} \alph...
I've inserted a begin-math/end-math symbol since I think
you left one out. Proceed, with fingers crossed.
(/usr/local/texlive/2018/texmf-dist/tex/latex/ucs/data/uni-32.def
File: uni-32.def 2013/05/13 UCS: Unicode data U+2000..U+20FF
)
coarse kestrel
I'm currently reading Topology without Tears. It says any two open interval (a,b) and (c,d) are homeomorphic. But the definition of homeomorphism is between two topological spaces, and open intervals are just elements of the usual topology on the real numbers, so how can they be homeomorphic?
gentle ospreyBOT
Whoever:
gritty widget
the intervals are being interpreted as topological spaces in their own right
ember maple
Exercise : find a topology on the interval
tough hamlet
the set containing only the interval and the empty set?
gritty widget
lol
ember maple
My gut feeling tell me that some novice might not even notice such topology.
coarse kestrel
Oh I see
The definition of the euclidean topology can be defined on an open interval too
xD
I do feel like a novice @ember maple
dim meadow
@coarse kestrel did you read about the subspace topology?
Usually that's what this sort of thing is referring to
coarse kestrel
Not yet
I literally started reading 2 days ago
I wanted to start learning something on my own in this month
dim meadow
Read up on it
That should answer your question
Basically you take your topology on (a, b) to be every open set in R intersect (a, b)
Btw it doesn't take much work to show that every (a, b) is homeomorphic to R itself
@coarse kestrel
dim meadow
so I was thinking about the wedge of n circles and defining a multiplicative structure using something similar to multiplication in the fundamental group
So we say a function from the infinite earring to some space Y is "compactly supported" if it is constant on all but a finite number of earrings
This means that it is basically a function from the wedge of n circles to Y for some n
You can compose these functions by saying f*g=f on the first n earrings and g on the next m earrings, where f is constant after n earrings and g is constant after m earrings
west spindle
can i ask for a clarification
what exactly do you mean by the infinite earring
countably many circles of the same size wedged together?
coarse kestrel
What do you mean by wedge of a n circle?
west spindle
wedge as in wedge sum
west spindle
basically gluing them all together at one point
dim meadow
Nope
coarse kestrel
I have no idea I haven't started learning the actual "fun" part of topology
dim meadow
Oh yeah it is
west spindle
wait
what
hang on
is it compact
i feel like i can think of an unbounded cts function from it to R
coarse kestrel
What I mean the not fun part is topology with sets
dim meadow
I think there's two different topologies
coarse kestrel
The fun part I'm referring to is like 3d surfaces
dim meadow
That you put on it
dim meadow
I don't mean the Hawaiian earring actually
Oops
I mean the rose with infinitely many petald
I messed them up cause they look the same
Sorry
keen cliff
west spindle
ok so just a wedge of countably many circles then?\
aight
so reading again what you wrote
this seems to give you some sort of monoid
assuming Y is a pointed space ofc
and we're talking about pointed maps
dim meadow
Yeah
That's the assumption
So the question is if modding out by homotopy will make it into a group
We can assume that Y is path connected
Since we're mapping into a path component anyway
west spindle
dim meadow
And also that Y is not a single point
Cause that wouldn't be interesting
Why the thonk Ann?
west spindle
this is definitely interesting
dim meadow
The construction is modeled after the tensor algebra
I think commutivity of the monoid follows from the fact that the wedge product is commutative
@gritty widget you up?
Can you read what I wrote above and tell me what you think?
So basically we want to multiply maps from.the wedge of n circles to some space Y
We say a map from the wedge of infinitely many circles is compactly supported if it is constant on all but a finite number of circles
Such a map has a number n associated to it which is the number of circles it is not constant on
Yes
No, but it looks like it
Lol
Hawaiian earring is compact, this is not
Whatever, not the point
The point is we can define a multiplication of f and g in an obvious way by doing f on the first n circles then g on the next m circles
Then we mod out by homotopy
What do you mean?
Oh
We're doing pointed spaces
Sorry
Forgot to mention
Yes
And some point in Y
So the point of wedging is mapped to that point
Same as fundamental group
So the question is if this gives you a group
f is nonconstant on the first n circles
Lol
We're associating some number n with the map
The minimal number so that f is constant on every circle after n
That's where homotopy (may) come in
Or rather where I want it to come in
Yeah sure
That's actually good
Hmm
Oh ok
What do.you mean?
Maybe
We might lol
The obvious inverses would be g is just f going in the opposite direction on the first n circles
But I'm not sure that works
Lol
Aw, I thought this would be an interesting invariant
I guess you could take the grothendieck group
Lmao
Why?
Hmm, maybe
I guess if Y, Z have the same fundamental group they would have the same associated monoid?
But this thing is commutative
So that's weird
That's true
dim meadow
I guess there's no reason you can't do the same thing with wedges of other topological spaces
honest narwhal
Have you heard about H spaces? @dim meadow
I wonder if that's what you have in mind
Or maybe H cospaces, yeah
Basically they're spaces which have a map X->X\times X which behaves almost like a group up to homotopy
And if X is an H-space, then [X,Y] is a group for any Y
Or sorry not X->X\times X, rather X->X\vee X
So yeah turns out the suspension of a space is always an H cospace since you can kinda pinch down along the center
So for example, S^n is an H-cospace
H spaces are in reverse, there's a group-like map X\times X-> X
Loop spaces are always H-spaces, idea is the same as multiplication in pi_1
Turns out this gives neat stuff
For example, if X is an H-cospace and Y is an H-space, then you have two products on [X,Y], one coming because X is in the first slot, and one coming because Y is in the second
And they distribute over each other
So this business called Eckmann-Hilton duality tells you the products agree and give you something abelian
Now, also throw in the fact that [\Sigma X, Y] is the same is [X,\Omega Y]
This gives a proof that higher homotopy groups are abelian, since [S^n,Y] = [\Sigma S^{n-2},\Omega Y]. Also fundamental groups of topological groups are abelian (also fundamental groups of H spaces in general)
So yeah idk the stuff is slick
dim meadow
@honest narwhal thanks, I'll read up on them
This is actually really amazing, thanks!!
dire warren
Guys do you all think point set topology is really great?
oak vapor
Anybody here taken non-euclidean geometry?
visual grove
non euclidean geometry is what math teachers take lul
dim meadow
Wtf
I mean, the term non euclidean geometry is more of a historic one than a practical one I'm think
marsh forge
Well I assume it would just be the study of various non-euclidean geometries which is a well defined thing
marsh forge
They're kinda neat I guess?
ember maple
Point set (general) topology is useful, and maybe can be quite interesting the first time seeing it, but in the end it's just another basic thing you are supposed to know and use imo. It's no different to calculus for engineers.
sleek canyon
highly true
lucid turret
Can someone help me pls?
I know for each n>=1 : H^n(X v Y) = H^n(X) + H^n(Y) and for n=0 H^0(X v Y) = Z
Does it follow directly from this fact?
lucid turret
@ivory dragon @steel needle @gritty widget hey sry to bother, do you have any idea how to solve this?
From MV I'm getting the isomorphisms between the H^n 's for all n
But idk how to turn this into cohomology ring isomorphism
lucid turret
I can apparently use the fact that cts map f:A -> B induces cohomology ring homo
Ok that's enough for today. I'd still appretiate if some kind soul helped me in the future tho 😃
sleek canyon
yeah looks fine
what he said
he has explicit descriptions of things
it's just computing
seems he didn't realize that the cohomology ring was functorial
at first
real notch
Probably a poor channel for this, but anyone know any good books for introductory (and later for once that's done) topology?
real notch
Might I bother you for a bit more on the title, i'm still rather new to the whole finding texts thing
sleek canyon
Topology/fundamental of topology/topology and geometry
real notch
this i'm assuming
sleek canyon
Yes
small obsidian
@real notch
"Topology without tears" for a very gentle intro to point-set
Which is a pdf online
real notch
added to my collection
midnight jewel
and allen hatcher’s notes on pointset topo are linked in #books-old, I can vouch for them but they’re fairly short and assume knowledge of set theory and proofs
fairly short may be good or bad, of course
midnight jewel
I didn’t do the exercises (we had our own assignments), are they good?
I feel like topology is helped by a hands-on approach even more than many other mathses
just cause you need to get an intuition for the counterexamples and all
honest narwhal
I'm not sure about its exercises, but I'm hesitant about the importance of the counterexamples
Like, we don't care about stuff like axioms of separation or countability because we're interested in stuff like the order topology given by the dictionary order on [0,1]\times [0,1] or the long line or something
There are actual spaces that give reason to wonder about such notions
For example, in AG you have Zariski topology (on varieties and on Spec(R))
honest narwhal
That isn't Hausdorff, and in the Spec(R) case it isn't even T_1
(Spaces which aren't T_0 are bullshit as far as I know, tbh we should just append T_0 as one of the axioms for a topology)
midnight jewel
what was T1? for every pair there are open sets such that one of them but not the other is contained?
(in both ways)
honest narwhal
I remember T1 as "points are closed"
midnight jewel
and T0 was that there are no two points with exactly the same set of neighborhoods?
honest narwhal
Yup
You can just identify all points with the same neighborhood set
Boom you have a T0 space
But yeah functional analysis is probably the main source that I know of for worrying about countability stuff. Banach spaces under the norm topology aren't locally compact, many aren't separable, and if you go to weak/weak* topology, they're not metrizable, in fact not even necessarily first countable
The one time I ever had to think about countability was in this one problem from my functional analysis class which said to show that if a Banach space is reflexive or has separable dual, then there's a sequence x_n such that ||x_n|| = 1 but x_n converges weakly to 0
I was talking to friends and we were kinda confused, like wait a sec we already proved in class that the weak closure of the unit sphere is the unit ball, how does that not kill this problem/what are with the assumptions?
The point is that if you're not in a metric space you no longer have the guarantee that for any point in the closure there's gonna be a sequence in your set converging to it
Maybe a net or something
And the point of this problem is that if the dual space of a Banach space is separable, then the weak topology on the unit ball is metrizable, so now you can use sequences. If it's a reflexive Banach space, you can take a separable subspace. That'll be reflexive (subspace of reflexive space) and separable, so its dual will be as well
And then you apply the theorem to that subspace
midnight jewel
fun anal sounds painfully hard
honest narwhal
Lol, it takes some time but it's fun, as the name suggests
Point is these are the examples that matter
But they take forever to define and work with
Hence why they're part of AG and functional
But point-set topology book writers were like "hMMM we need cOUnteRexAMPLES" and made up some shit
naive flint
if Tᵏ: X→X is a contraction, does that imply T is a contraction for the same space?
and if this is true
what assumptions are required on X for it to be true
because my friend thinks maybe a counterexample exists in infinite dimensional spaces
dim meadow
By contraction, you mean d(T^k(x), T^k(y)) < d(x, y)?
@naive flint
I don't think this is true in a general metric space
Consider a continuous function T so that T^k is constant
naive flint
let us work in the Rⁿ space
dim meadow
Sure
dim meadow
It may be true in R^n
Because if T^k is constant than it's automatically a contraction
naive flint
that's no fun
dim meadow
Lol
You asked a question about metric spaces
The answer is no
I couldn't give a shit if my answer is fun or not
Consider the 3 point space {x, y, z} with the discrete metric
T maps x, z to z, y to x
Then T^2 is constant
But d(T(y), T(z))=d(x, z)=1
@naive flint that's my shitty counterexample
gritty widget
I thought contractions were strictly less than?
dim meadow
I forget
dim meadow
But yeah
gritty widget
i meant to the definintion earlier
real notch
oh that
dim meadow
Yeah I edited it
real notch
yeah the identity map is def a contraction
dim meadow
To be wrong
Reedited it
To be correct again
Maybe you can use banach?
To do a nicer thing
Doesn't really matter tbh
gritty widget
wait, you said T(x) = T(z) = T(y) = z, did you mean somethingn else?
gritty widget
@naive flint I think that if T is Lipschitz continuous, your result holds
T^k being a contraction implies T is a contraction
dim meadow
um
gritty widget
since d(T(x),T(y)) < K * d(x,y), this means that d(T^2(x),T^2(y)) < K^2 * d(x,y), and if T^k is a contraction, K^2 < 1 and so K < 1
dim meadow
T is lipschitz in my counterexample
you need restrictions on the space
your argument doesn't work @gritty widget
because K^2 is just a upper bound
on the lipschitz constant
marsh forge
I'm trying to compute the homology of T^2 quotient S^1 where the quotient is done along the handle part of the torus
I keep getting an answer that seems to check out, but implies H1 of this space is F^2 for a field F
But since we collapsed one of the loop generators to a point I'm confused as to why its not just F
I'm pretty sure the fundamental group would only have 1 generator for this space
dim meadow
would the picture for that be that you take a tube whose ends are points and join the two ends?
marsh forge
yes
sleek canyon
why are you doing it for a field?
do it for Z
you know that H1 is pi1ab
so that's a way to check your work
marsh forge
Well original reason is that the final tomorrow won't ask for anything but over a field
But it already seems like my answer is wrong for pi1ab
As the torsion-free part would just be Z^2 right
marsh forge
Ok so I was trying to approximate my problem but it turns out it was wrong
I'm actually computing the homology of T^2 mod S^1 twice
two different places along the handle
I get the right answer for just taking mod S1
but not for modding twice
Figured it out, I'm dumb
marsh forge
I was practicing les of a pair but yeah it works out nicely both ways
marsh forge
Does anyone know of any important theorems with (surprising) topological proofs
Algebraic topological*
Other than fundamental theorem of algebra, Brouwer for Dn, and hairy ball
wanton bone
So, I believe this is just a quick question, but I just started going over a book on differential geometry with a prof of mine, but he's been sick this week and I haven't had a chance to ask him: a 1-form is just a linear operator from a vector space to the real numbers, and a differential form is just a 1-form with differentials dx_i of the inputs x_i, right?
then with a tangent vector v_p at point p, we get a differential in the direction of v at p (a directional derivative) by applying the differential form to the tangent vector with the x_i's given by p? or am I off base there?
sleek canyon
A differential form is a form that varies smoothly
wanton bone
That last part is where I am confused, I think. Is it like df = f_i(t,p) dv_i + ... ?
sleek canyon
Locally its given by a smooth 1 form
wanton bone
right, but when actually manipulating it, applying it in a tangent space
(the concept of a tangent space is a bit iffy for me too, i think)
gritty widget
a 1 form is $\omega: U \to \cup_{p\in U}T^*_p(U)$
gentle ospreyBOT
grade a failure:
wanton bone
I thought it was the other way around?
gritty widget
that T* is the dual space to the tangent space in p
gritty widget
it builds a point in your open set to a contangential vector in the dual space
wanton bone
I remember being taught a bit about dual spaces in operations research, but the formal idea of them wasn't really gone over (bad prof didnt like to teach)
gritty widget
so to a map from the tangent space to R
if you have a vector space $V$ over a $\mathbb R$, the dual space is the space of linear functionals $V \to \mathbb{R}$
gentle ospreyBOT
grade a failure:
wanton bone
Ohhhhh
gritty widget
yea
u can sum functionals and multiply them by scalars so it itself is a vector space over the field R
or whatever ur field is
midnight jewel
(linear and continuous, which only matters in the infinite dim. case)
wanton bone
can you give me a bit of an example, grade a?
midnight jewel
on ℝⁿ, you could see linear functionals as row vectors after choosing a basis
wanton bone
ok
gritty widget
ya if your vector space is $\mathbb{R}^3$ over $\mathbb{R}$ then a functional is for example $(x,y,z)\mapsto x+y+z$
wanton bone
yeah thats how i think of them
gentle ospreyBOT
grade a failure:
midnight jewel
you could represent that function there as $\begin{bmatrix} 1 & 1 & 1 \end{bmatrix}$
gentle ospreyBOT
Sascha Baer:
gritty widget
its basically a linear operation on the vectors component that gives u a real number
wanton bone
by $R^3$ over $R$ you mean that associated with each point in $R$ there is a vector in $R^3$, yes?
gentle ospreyBOT
chrstfer:
midnight jewel
no, it means that the field of scalars is $\mathbb{R}$
gentle ospreyBOT
Sascha Baer:
midnight jewel
as opposed to, say, $\mathbb{Q}$
gentle ospreyBOT
Sascha Baer:
gritty widget
or C
wanton bone
Ok
midnight jewel
(over which ℝ^3 is an infinite-dimensinoal vector space)
well, ℝ³ isn’t a ℂ-vector space
wanton bone
I'm not sure I follow
midnight jewel
in a vector space you always have two things
you have vectors
but you also have scalars
wanton bone
right
wanton bone
so "over" just means the scalars are in the given set
midnight jewel
yea
gritty widget
ya
midnight jewel
a structure where addition, multiplication, subtraction and division all work
gritty widget
the scalars have to come from a field otherwise it doesnt work so well 😃
midnight jewel
and you can take the same vector space to be over different fields
wanton bone
alright. So there are vector spaces where the vectors are in $R^n$ but the scalars are integers, or rationals? what makes that infinite dimensional?
gentle ospreyBOT
chrstfer:
gritty widget
they cant be integers
wanton bone
ok
midnight jewel
e.g. you can see ℂ as a 2-dimensional vector space over ℝ, or a one-dimensional one over ℂ
gritty widget
the integers arent a field
midnight jewel
or an infinite-dimensional one over ℚ
wanton bone
I get how it would work with $\mathbb{C}$
gentle ospreyBOT
chrstfer:
midnight jewel
of course a one-dimensional vector space is always just the field itself
gentle ospreyBOT
grade a failure:
gritty widget
yeah liek padiacs or field extensions
midnight jewel
at least I’ve not really come across them
gritty widget
but not relevant idthink 😃
wanton bone
At least, I think i do
midnight jewel
you could also use skew-fields like the quaternions, but let’s ignore that
wanton bone
yeah now were a bit into the weeds i think
midnight jewel
either way a field is a thing where all the operations, in particular also division, work
gritty widget
you can use rings but then theyre not called vector spaces anymore
we clal them modules
midnight jewel
(a ring is where division can fail)
(the integers are the “classical” ring)
but we’re getting too deep in probably the wrong direction
I mean it’s interesting stuff
but I don’t think it’s relevant rn
wanton bone
yeah i get what a field is
gritty widget
in your case id stick to R as a field and R^n to vector spaces
wanton bone
at least, the idea of it. A set of scalars closed under the algebraic operations?
gritty widget
yeah its the best thing u can get
gritty widget
2 operations with inverses a neutral element and distributivity
wanton bone
but in euclidean space
wanton bone
yes
midnight jewel
at least that’s how I know the word
wanton bone
that is what i mean, and thats the context of the chapter im talking about
midnight jewel
a euclidean vector space is a finite-dimensional space over ℝ with an inner product
and finite-dimensional vector spaces over ℝ are always isomorphic to ℝⁿ (where n is the dimension), so you can think about just those
wanton bone
so, a 1-form on $R^3$ is a linear function from the tangent space (which is a vector field, right?) to $R$
gentle ospreyBOT
chrstfer:
wanton bone
and a differential form is a smooth 1-form?
gritty widget
no
wanton bone
right
gritty widget
1 form is a from an open set in R^n to the dual tangent space
gritty widget
it builds points to functional
wanton bone
in my latex setup I normally have \R as \mathbb{R}, sorry
honest narwhal
Did someone say differential forms?
wanton bone
yes im trying to wrap my head around them
meager pagoda