I don't know any special terminology for sections of ∧ⁱTX
#point-set-topology
sleek thicket
For the last question, your tangent bundle is trivial iff your cotangent bundle is
feral copper
Oh really ?
I wouldn't have thought !
But TX and T*X are not homeomorphic/diffeomorphic, right ?
sleek thicket
They are
feral copper
Oh x)
feral copper
Oh I see
sleek thicket
if you choose a riemannian metric, you get a bundle isomoprhism
and every smooth manifold has a riemmanian metric
but not a canonical one
but you can see this way more concretely
the tangent/cotangent thing is very special
but a vector bundle is trivial iff its dual bundle is trivial
what does a global trvialization of your bundle look like?
feral copper
Oh I should've thought that dual bundles are a thing...
Uhm it's TX≅X×Rⁿ
Oh so you use (Rⁿ)*≅Rⁿ
sleek thicket
well not quite
It's more subtle than that
what I was thinking is that a global trivialization is the same as a global frame
i.e. smoothly varying choice of basis on each fiber
sleek thicket
yes
so at each point you get a dual basis
and this defines a frame on the dual bundle, which might not be smooth
feral copper
Oooooh clever, you get the global trivialisation of T*X from the one of TX ?
feral copper
And conversely through V**≅V
sleek thicket
exactly
another good way to think about it, although more algebraic
a vector bundle is trivial iff its global sections are a free C^infty(M)-module
and the sections are always finitely generated
and the sections of the dual bundle are the dual of the sections of the original bundle
feral copper
a vector bundle is trivial iff its global sections are a free C^infty(M)-module
I'm actually working with C^k-manifold, not necessarily smooth ones tho
feral copper
Okay
sleek thicket
since the dual of a finitely free module is free, you get the result
okay, for your last question
the pullback
you have a smooth map f : X -> Y
feral copper
(Thanks already, you're helping me so much !)
sleek thicket
(np! I'm learning this stuff for the first time rn and so it helps to explain it to someone)
you want a map f* : T*Y -> T*X, right?
feral copper
Like the pushforward
sleek thicket
right
so we want a map f*_p : T*_p Y -> T*_p X
but by definition, T*_p Y = (T_p Y)*
and T*_p X = (T_p X)*
and we have a map f_* : T_p X -> T_p Y
feral copper
So it's just the transpose of the differential ? x)
feral copper
Omg it's so obvious when you see it x)
sleek thicket
more explicitly, if $\omega \in T_p^* Y$ and $v \in T_p X$ then
$$(f^\omega)(v) = \omega(f_ v)$$
feral copper
So it's just like f:V→W induces f*:W*→V* by f*(φ)=φ.f ?
sleek thicket
exactly
gentle ospreyBOT
shamrock:
sleek thicket
I mean that's literally what it is
with V = T_p X, W = T_p Y, and the linear map f_*
feral copper
Oh yeah sorry x)
sleek thicket
What book are you using to learn this stuff?
feral copper
Well, thank you so much for your answers, it was helpful !
sleek thicket
Introduction to Smooth Manifolds by Lee answers all of this
and once again, no problem!
feral copper
Erm x) W. D. Curtis_ F. R. Miller - Differential manifolds and mathematical physics, Wikipedia, StackExchange
nLab also, which I knew about thanks to categories
(now that I'm familiar with categories, I'm back to re-learning diff geometry, and I'm now further than my lectures went)
Ima download get a legal print of the book you adviced
wanton marsh
yes, something that is open in A isn't necessarily open in X
vocal wharf
is $\mathbb{C}^{n \times n}$ topologically closed in $\mathbb{R}^{2n \times 2n}$?
gentle ospreyBOT
Lochverstärker:
vocal wharf
i think i can show it by fucking around with some metric, but is there an easier way im missing?
considering $\mathbb{C}$ as a subset of $\mathbb{R}^{2 \times 2}$ ofc
gentle ospreyBOT
Lochverstärker:
uncut surge
Every linear subspace of a finite-dimensional subspace is closed, right?
e.g. find a complement of the subspace, define a continuous functional on the whole space by setting it to be nonzero on the complement and zero on the subspace (by choosing appropriate bases of the two things), then your subspace is the preimage of zero under this functional
(there's probably easier ways, it's just what pops into my head after working with infinite-dimensional banach spaces and the hahn-banach theorem for too long)
uncut surge
neat
yeah, I think most proofs of this will use either a scalar product, a functional or a complement of your vector space somehow
reflecting on exactly why infinite-dimensional spaces can be difficult to work with, because you might not have any of those three for a general closed subspace of an infinite-dimensional space!
vocal wharf
if i define $\mathbb{C}$ as the matrices of the form $\begin{pmatrix} a & -b \ b & a \ \end{pmatrix}$ in $R^{2 \times 2}$ i can define a continuous map $\mathbb{R}^4 \to \mathbb{R}^2$, $(a_{i,j}) \mapsto (a_{1,1} - a_{2,2}, a_{1,2} + a_{2,1})$ and $\mathbb{C}$ is the preimage of ${0}$
gentle ospreyBOT
Lochverstärker:
uncut surge
you can also use the sequential criterion for closedness: if a sequence of those matrices converges to some limit matrix, then this implies that all component sequences of the matrix sequence converge to some limit
from this it's then easy to derive that if your matrix sequence was in this special shape, then the limit sequence is in the same shape
hence, this space contains all its limit points, thus it's closed
vocal wharf
this was my motivation
to show that the limit of a converging sequence is in C^{n\times n}
uncut surge
oh, but like i said, that's pretty quick to deduce directly by just working with the component sequences
in any case there's a thousand ways to do it here 😄
vocal wharf
yeah, guess i was just overthinking too much
uncut surge
but yeah your functional is also good
unique wyvern
So when we take complements to get a topology, do we also take the complement of T?
vocal wharf
sure
the complement of T is the empty set
and vice versa
so those are always both open and closed
unique wyvern
so you take the complement of{T, empty set, .....}
you arrive at {empty set, T, ....}
vocal wharf
you take the complement of every element in the collection
with respect to T
then yes
unique wyvern
excellent thankyou
jaunty nebula
I'm supposed to prove that d_2(x,y) = |x^2-y^2| isn't a metric. I've been given 4 properties that a metric has to fulfil, but I can't get any of them to break.
vocal wharf
what's the space
jaunty nebula
R
vocal wharf
if you can't get the properties to break, try proving they hold
if you can't prove one, that is the place to look
jaunty nebula
I think I managed to prove that all of those 4 properties are fulfilled, that's just the thing.
jaunty nebula
oh wait yeah, I spent several pages on testing m4 and if d can be 0 only when x=y, then that breaks it
thanks
sleek thicket
Let M be a connected smooth manifold and X a vector field on M
Is it true that the image of any maximal integral curve for X is closed?
I think I can show that if there is a counterxample, X must be complete
err, wait no
I can show that for any p, the integral curve of X starting at p is either defined for all t > 0 or all t < 0
Proof: let $\theta$ be the flow of $X$ and $\gamma = \theta^{(p)}$ a maximal integral curve at $p \in M$. Suppose we have a point $x$ in the closure of $\mathop{im}\ \gamma$ but not in $\mathop{im}\ \gamma$. Suppose for contradiction that $\gamma$ is defined on a bounded interval $(a, b)$. We can choose ${t_n}{n=0}^\infty$ such that $x = \lim{n \to \infty} \gamma(t_n)$. If there are $A, B \in (a,b)$ such that $A < t_n < B$ for all $n$, we can pass to a subsequence and get $x \in \mathop{im}\ \gamma$. If not, we can pass to a subsequence such that $t_n \to a$ or $t_n \to b$. Since $\theta$ is defined on an open subset of $\R \times M$, we can choose a nbhd $U$ of $x$ and $\varepsilon > 0$ such that $\theta$ is defined on $(-\varepsilon, \varepsilon) \times U$. Then for some large $N$, we have $b - \varepsilon < t_N$ or $t_N < a + \varepsilon$ and $t_N \in U$. We can then translate the integral curve for $\gamma(t_N)$ to extend $\gamma$ to $(a - \varepsilon, b)$ or $(a, b + \varepsilon)$
gentle ospreyBOT
shamrock:
sleek thicket
I think we can get a little stronger and say that either $\theta^{(p)}$ is defined for all positive reals for all $p$ or $\theta^{(p)}$ is defined for all negative reals for all $p$
gentle ospreyBOT
shamrock:
sleek thicket
by some kind of like, connectedness+translation
sleek thicket
So here's my current conjecture:
Let $X$ be a vector field on $M$ and $\gamma : (a, \infty) \to M$ an integral curve of $X$ (for $a < 0$). Suppose we have $t_n \to \infty$ and $\gamma(t_n) \to x$ for some $x \in M \setminus {\mathop{im} \gamma}$. Does it follow that $X_x = 0$?
gentle ospreyBOT
shamrock:
sleek thicket
I guess my basic intuition is that if I like to backwards along the integral curve at x
I should reach some $\gamma(t_i)$
gentle ospreyBOT
shamrock:
sleek thicket
Unless it's constant
So by smoothness of things, $X_x = \lim_{n \to \infty} X_{\gamma(t_n)} = \lim_{n \to \infty} \dot{\gamma}(t_n)$
gentle ospreyBOT
shamrock:
sleek thicket
That's something
hmm so in coordinates around x this gives an actual sequence of vectors converging to another vector
If everything's nonzero then in coords I can think of this as a sequence of norms and a sequence of vectors on the sphere
The norms have to be bounded below if $X_x \neq 0$
gentle ospreyBOT
shamrock:
sleek thicket
So I guess what I want to say is that things go off to infinity
I feel like the idea I'm having only works if $\lim_{t \to \infty} \gamma(t) = x$
gentle ospreyBOT
shamrock:
stoic parcel
Suppose we have $n$ points $p_1,\ldots,p_n \in \mathbb{R}^d$ with the property that, for all $i,j$ we have that $|p_i - p_j| < r$. What is the maximum $\delta$ so that you can guarantee the existence of a ball $B$ of radius $\delta$, containing all the points $p_i$?
gentle ospreyBOT
brouwers_bane:
stoic parcel
I'm curious if anyone has a good answer/proof for this. The motivation is showing when the Vietoris-Rips complex is a subcomplex of the Cech complex for a certain parameter.
@gritty widget minimum* yeah thank you
stoic parcel
I think you can get a much tighter bound. Presumably the worst case scenario is when the points correspond to standard basis vectors, though I can't prove this
sleek thicket
If the points lie within a k plane then the minimal δ will be the same as for n points in R^k, right?
Not totally sure about that
but it would be nice to assume d = n and that they're linearly independent
why δ > sqrt(2)? Isn't (1,0,0) too far away from (0,1,1)?
oh lol sorry I was thinking of the unit cube for some reason
Ignore me
No idea where I got that from
that the inf of all δs is sqrt(2)?
unique wyvern
Hello I'm trying to find the significance of the remark
So its saying that not all open sets in the product topology are of the rectangular form
He does this by showing it on the usual metric on R2 then as a product
so then its saying that the basis for the topology doesnt cover all the open sets?
this covered what i was interested in
feral copper
Hello,
Earlier this week, I had a conversation with @sleek thicket that was very helpful about the cotangent bundle. Amongst all things, I asked myself : given a map f:X→Y, one get the pushforward f⁎:TX→TY. I wanted to know what's the cotangent analogous : how to relate f into a map f*:T*Y→T*X ?
So I was told I can do it punctually using :
Now I think like there might be something wrong about this, since, to define the pushforward, we did : f⁎(p,ξ)=(f(p),df(p)(ξ))
But this is not possible in the contrvariant way, since, if f sends every p∈X to q∈Y, then any q'∈Y that is not q won't give rise to a map from Tq'Y
stoic parcel
@feral copper what context are you talking about the cotangent bundle in? Like for schemes, varieties, smooth manifolds, complex manifolds?
feral copper
C^k-manifolds
stoic parcel
So what part are you confused about? The map on the cotangent spaces at a point (in the equation you linked) is just given by the vector space dual
feral copper
Yes, but I wanted to know how to turn this into a map on the cotangent bundles, and not just stick to the cotangent spaces
stoic parcel
Are you familiar with pullback and pushforward of bundles?
stoic parcel
Well so there's a couple things we could say - taking the differential of $f: X \to Y$ gives a map of bundles $TM \to f^\ast TN$ over $M$, and by adjunction of pullback and pushforward. Taking duals (or by adjunction, if you prefer) you get a bundle map $f^\ast T^\ast N \to T^\ast M$, so by abstract nonsense the bundle map exists
gentle ospreyBOT
brouwers_bane:
stoic parcel
Alternatively you could check that the map on fibers you described above patches together to give a genuine bundle map
Finally (and probably easiest) you can check what the induced map on cotangent bundles does on 1-forms
If you don't know differential forms, they are a great thing to learn and it will make understanding this bundle map a lot easier
feral copper
I have learnt about differential forms ^^ I know up to the De Rham cohomology ^^
This that you said and rendered in TeX... Well, I must say, I'm not too familiar with the categorical vocabulary, I only know the very basics about it, so I must admit it's quite obscure ^^"
stoic parcel
Oh awesome - so if you take a 1-form and precompose with $f$, you get a 1-form on the pullback of the cotangent bundle, and you can define the bundle map on these 1-forms: https://en.wikipedia.org/wiki/Pullback_(differential_geometry)#Pullback_of_cotangent_vectors_and_1-forms
gentle ospreyBOT
brouwers_bane:
Compile Error! Click the 
reaction for details. (You may edit your message)
brouwers_bane:
Compile Error! Click the reaction for details. (You may edit your message)
feral copper
So you're talking about the pullback giving rise to the map f*:Ω*(Y)→Ω*(X) ?
And particularly by restriction to Ω¹ ?
stoic parcel
Yeah basically
feral copper
But how does that transform into a map T*Y→T*X ?
The article talks about the pullback bundle, so I guess I've gotta learn this first ^^"
stoic parcel
The sections of the cotangent bundle are exactly one-forms
And giving a map of bundles is the same as giving a map of sections over each open set, subject to compatibility conditions
feral copper
Oh
stoic parcel
I would look at Bott and Tu's Differential Forms in Algebraic Topology, it's a really great reference for this kind of stuff 🙂
Especially if you already know de Rham cohomology
feral copper
Okay, Ima download legally get it
And giving a map of bundles is the same as giving a map of sections over each open set, subject to compatibility conditions
So, given a vector bundles π1:E→A and π2:F→B, you mean that a map φ:Γ(E)→Γ(F) can relate into a bundle morphism (E,A)→(F,B) ?
Cuz basically this is the part I'm missing ^^"
I can somehow conceive that it's possible, since sections contain all the information about the bundle, but I don't know the formalism ^^"
(and thanks already for the help ! :P)
stoic parcel
It's sort of the other way around --- a section of F will induce a section of the pullback bundle of F, which under certain conditions you can identify with E
So if I had a continuous map $f: A \to B$ and a bundle $F \to B$, then you get a map $\Gamma(F) \to \Gamma(f^\ast F)$
gentle ospreyBOT
brouwers_bane:
stoic parcel
Which is just given by precomposition
feral copper
Hmmm okay I think I get it
I suppose I'm looking for something that's not reallyfeasible after all, which makes sense tbh
stoic parcel
A good way to think about it is this (suppose we're in real vector bundles for a minute). Let $U \subseteq V$ be an open neighborhood, so then $f^{-1}(U) \subseteq A$ is open. The bundle $f^\ast F$ is defined over $f^{-1}(U)$ to be pairs $(a,x) \in f^{-1}(U) \times F$ such that $f(a) = \pi(x)$. In particular if you had a map of bundles $(E,A) \to (F,B)$ you always get a map of bundles $E \to f^\ast F$ over $A$.
gentle ospreyBOT
brouwers_bane:
stoic parcel
A section of $F$ we can think of as a map $F \to \mathbb{R}$, and we can precompose with $f$ to send $(a,x) \to x \to f(x) \in \mathbb{R}$, which defines a section of the pullback bundle $f^\ast F$
gentle ospreyBOT
brouwers_bane:
stoic parcel
So there are two things to think about --- (1) if you just have $A \to B$ and a bundle $F \to B$, you always get a map of sections $\Gamma(F) \to \Gamma(f^\ast F)$. (2) If you have a morphism of bundles $(E,A) \to (F,B)$ you get a map $\Gamma(F) \to \Gamma(E)$ by precomposition, but this in fact factors as $\Gamma(F) \to \Gamma(f^\ast F) \to \Gamma(E)$.
gentle ospreyBOT
brouwers_bane:
feral copper
Okay this makes sense to me now
Thanks for the explanations, I'm going to work on this now 🙂
stoic parcel
For sure, glad I could be of help! 🙂
gentle ospreyBOT
MaxJ:
velvet wagon
$\exists$
gentle ospreyBOT
gutgutgut:
velvet wagon
i think i forgot that
marsh forge
Ok I agree with you
but im not satisfied
how do you know such an epsilon exists
(i know it, but I also know how to prove it)
velvet wagon
if u give me any number in (0,1) i can find a small enough number so that their sum is in (0,1) as well cuz completeness (i think this is word?)
idk what that property is called but if you have like 0.00000000000000000000001 there's always a smaller number
uncut surge
archimedean?
marsh forge
Sorry back
theres an easy way to do this
let $x\in (0,1)$ and put $\delta = \frac{1}{2}\operatorname{min}(|0-x|,|1-x|)$
gentle ospreyBOT
MaxJ:
marsh forge
basically either x is closer to 1
or its closer to 0
take the midpoint between x and whichever its closer to and draw the ball with that radius
by construction it lives inside (0,1)
velvet wagon
oh yeah delta epsilon proofs
my bad
sorry i had a large gap between calc 1 and my current class
velvet wagon
yes because you get the difference of the number they chose
then you half it so it's guaranteed to be within the interval
i think u also choose the min so you get the smaller side length?
marsh forge
yes
thats all correct
and it proves that all elements of (0,1) are in the interior
velvet wagon
oh i see what you mean now
cuz irrationals are dense in R
u can't formulate a delta that will avoid the irrationals?
marsh forge
yeah thats why the stackoverflow example works
velvet wagon
thank you for all the help man
marsh forge
np
peak narwhal
What would be a good way to go about proving that if an isometry in R2 is odd, then it's a reflection?
I was thinking proof by contradiction, showing that if you have an even isometry that's a reflection you'd reach some kind of roadblock
We know that if it's even then it has to be the product of two reflections, so would I then just show that after two reflections you either get a rotation/identity/translation? I'm having trouble pinpointing a particular contradiction to use while doing this
gentle ospreyBOT
All functions are continuous:
floral gust
What did they mean by R2?
floral gust
Note that if T is an odd isometry in R^2* (typo in my solution)
peak narwhal
Oh I mean ultimately I'm trying to prove that if T is odd, then it's a reflection of some kind. But my initial assumption would be that there'd be a way to just contradict the idea of an even isometry being a reflection
floral gust
That doesnt follow. You have to show odd => reflection. Deriving contradiction for even being a reflection wouldnt do anything
peak narwhal
Ahhh I see
floral gust
You can argue not reflection => even but that seems much harder
peak narwhal
I'm still pretty new to when certain types of proofs are warranted lol
ivory dragon
theorem: all even numbers are prime
proof: suppose for contradiction that all odd numbers are prime. but 9 = 3*3, contradiction
peak narwhal
Now that I see it, it does make a lot more sense to just start with an odd iso and show that translation and rotation go away
floral gust
I'm still pretty new to when certain types of proofs are warranted lol
@peak narwhal That's alright. Do you see the mistake with why your approach was incorrect
peak narwhal
Yeah especially with oh no's example haha
floral gust
Perfect
peak narwhal
When you say that being an odd isometry fixes the origin, do you mean that it has to reflect through one of the points in the set such that the point remains stationary?
floral gust
I mean if T is odd then T(x) = -x. Set x= 0. Then T(0) = 0. This is termed as fixing the origin. (Or more generally, if T(a)=a for some a, then a is a fixed point)
peak narwhal
Ahhhh I see! That makes a lot more sense. Thanks
floral gust
Do you see why a translation wouldn't fix an origin?
peak narwhal
Yeah, because translations are distance-preserving and thus if you were to have a point at the origin, it would have to translate with respect to the rest of the points to preserve its position relative to the others
floral gust
Hmmm, a simpler argument is realizing that a translation is t(x) = x+v for some non-zero v. Thus we have t(0) = 0+ v = v ≠ 0 . Thus t doesnt fix 0.
But your idea is correct.
peak narwhal
Yeah, that too -- overcomplicated it a bit
warm mirage
I'm trying to show, using the fifth postulate, that "is parallel to" which will be denoted as "||" is an equivalence relation on the set of lines in the plane
Can you check if my proof is right and the author's is wrong or the opposite? Cause either their proof is wrong or I'm missing something and my proof is wrong
This is my proof
This is hers
Ping me when you answer please
Thank you
The problem I find is in the transitivity step
When she said A belongs to D'
Isn't that wrong?
Oops in my proof there is a little typo
I meant A not in D'
gritty widget
Is algebraic topology only used by like 100 people in the field of robotics?
tidal cedar
Wait, robotics?
gritty widget
@tidal cedar http://agt.cie.uma.es/~cata/robotics.pdf
gritty widget
@tidal cedar algebraic topological concepts such as cohomology are also used in crystallography https://www.jstor.org/stable/2322930?seq=1
zealous bison
hey I'm having a hard time with homeomorphic vs. homotopic... I understand homeomorphic is much stronger
Is it key that a homeomorphism needs to be bijective? Like a homeomorphism won't destroy any information but a tomotopy might?
like a homotopy is this kinda weak thing that explains how to squeesh one thing into another thing smoothly but it may map pieces of one space on top of the other making the map not bijective
wanton marsh
there aren't really the same thing ?
homeomorphism is a relation between two topological spaces
homotopy is a relation between two continuous functions from a topological space to another
marsh forge
I don't see why people are saying they aren't at all the same
A homotopy equivalence is called a 'weak equivalence'
Two homotopy equivalent spaces will share many of the same invariants
and often times we don't want to even bother distinguishing between homotopy equivalent spaces, we just consider them the same
When doing computations, it's often advantageous to swap out one space for a homotopy equivalent one
But homeomorphism is much stronger
i think they understand that
I don't know if the 'key difference' is bijectivity
but it certainly stands out as a big one
The best way to understand homotopy equivalence is as follows:
A map X->Y is a homeomorphism if it has a continuous inverse
a map X->Y is a homotopy equivalence if it has an inverse up to homotopy
wanton marsh
ah I forgot homotopy equivalence was a thing
wanton marsh
when I answered
marsh forge
no i see that i mean how did you forget that lol
nvm
anyway does that make sense @zealous bison
marsh forge
Continuous invertible deformation
zealous bison
the circle is homeomorphic to R through stereographic projection, for instance.
right?
marsh forge
no
what
Circle - pt is
but the whole problem w stereographic projectections is that theres one point that doesn't make sense
zealous bison
ah ok ok.
ahhh okay I guess the connectivity has to be preserved under a homeomorphism, right?
marsh forge
All topological properties
zealous bison
I apologize I'm not in any way a mathematician, I'm an experimental physicist
trying to cram AT in my brain
not enough it seems
ok!
marsh forge
Hatcher has an intro point-set set of notes
its basically like
the minimum required to do AT
zealous bison
is this Chapter 0 of AT or something else?
marsh forge
which is exactly the rgith amount to leanr
zealous bison
oh perfect. I'll check that out.
marsh forge
Also like just a heads up
dim meadow
chapter 0 of hatcher is the hardest part of hatcher tbh
marsh forge
AT is a complicated subject
and not something easily smashed into your brain
esp if by 'use AT for physics' you mean like Urs's stuff
zealous bison
yeah I'm having a rather hard time with it!
wanton marsh
yeah you need a bit of point set topology so that things like continuity and homeomorphisms make sense
wanton marsh
yes
marsh forge
I've never actually read them lmao but I'm told they work just fine
zealous bison
ok cool I'll check it out. Thanks!
yeah I mean I just need something with lots of concrete examples
yee thanks.
supple locust
If $\gamma$ is smooth curve segment in $M$ and $F:M\rightarrow N$ is diffeomorphism, then why is $(F \circ \gamma )'(t)=dF_{\gamma (t)}(\gamma ' (t)) $? Can someone explain this by unpacking the definitions?
gentle ospreyBOT
bertwit:
sleek thicket
what's your definition of prime, bertwit?
@supple locust
It definitely is just unwrapping definitions, but you I want to make sure we have the same definitions
my definition would be γ'(t) = dγ_t(d/ds|t), where d/ds is the standard basis vector on T_t R
And then this is immediate from the chain rule
supple locust
Actually, I think I got it! You please check if I am right. I started from RHS. $$dF_{\gamma (t)}(\gamma ' (t))= \gamma ' (t)(F)=\gamma _* {d/dt _t}(F)=d/dt _t(F \circ \gamma )=LHS$$
I am following Lee's definitions and notation
sleek thicket
You've lost me at the first step, sorry
gentle ospreyBOT
bertwit:
sleek thicket
I don't see how you can apply γ'(t) to F
γ'(t) eats smooth functions M -> R, right?
And spits out scalars
supple locust
γ'(t) eats smooth functions M -> R, right?
@sleek thicket yeah right
supple locust
yes i get that
sleek thicket
So what does γ'(t)(F) mean?
supple locust
sorry I am clearly confused in definitions
sleek thicket
And points p in M, q = F(p) in N
We want a map dF_p : T_p M -> T_q N
So let v be a tangent vector to M at p
Then dF_p(v) should be a tangent vector to N at q
What does that mean, by definition?
Also no problem, this book has so many fucking definitions and they're all abstruse
supple locust
we define it as $F_*(v_p) $
gentle ospreyBOT
bertwit:
supple locust
?
supple locust
v acts on smoth functions M->R. suppose g is smooth N ->R then F_*(v)(g) is v(g F)
gentle ospreyBOT
shamrock:
sleek thicket
That's the defintion of dF
supple locust
yes
sleek thicket
But really for your question you don't need to go down to definitions
You can just use the chain rule
$dF_{\gamma(t_0)}(\gamma'(t_0)) = dF_{\gamma(t_0)}\left(d\gamma_{t_0}\left(\frac{d}{dt}\Big|{t_0}\right)\right) = d(F \circ \gamma){t_0}\left(\frac{d}{dt}\Big|_{t_0}\right) = (F \circ \gamma)'(t_0)$
gentle ospreyBOT
shamrock:
supple locust
I get it what the confusion was I was used to writing $F_*$ for pushforward.
gentle ospreyBOT
bertwit:
sleek thicket
thanks lol
frozen token
ok ok guys dont bite me but
what does any of this have to do with geometry
i mean it's just some screwed up calc
sleek thicket
@gritty widget compose with the map M -> R×M sending x to (t, x)
For fixed t
The composition of g with that map is g^t
And a composition of smooth maps is smooth by the chain rule
@frozen token the answer strongly depends on your background
frozen token
wdym?
sleek thicket
do you know differential geometry or topology already?
frozen token
nopeee
sleek thicket
do you know calculus?
frozen token
yep
sleek thicket
Multivariable calculus?
frozen token
give me an example on what that is
sleek thicket
np slim
E.g. maximizing functions of two variables by looking at their gradient
Or taking integrals of functions of several variables over lines/surfaces/regions
sleek thicket
Well I guess the thing to start with is that calculus is very geometrical
the derivatives of a function f(x) describes the geometry of the curve y = f(x)
And also like, slope is a geometrical concept
As is area (= integration)
sleek thicket
spaces
frozen token
what exactly are the applications, what shapes, etc..?
sleek thicket
Although topology isn't about calculus
you need more structure
I'm not really sure how to answer "what shapes"
oh also I have an actual question, not directed at hellfire
frozen token
yes ofc slim
but from what im seeing you guys are just doing some equations and no shapes?
idek
:(
set of points in space?
sleek thicket
we actually work with things more general than that in topology
Sometimes the spaces we look at can't be embedded into Euclidean space (of any dimension)
For example, the set of continuous functions from [0,1] to R is "too big" to fit in Euclidean space, but we can give it the structure of a topological space
frozen token
ok this is really confusing
sleek thicket
Let $\eta$ be an $n$-form on $\mathbb{S}^n$ such that $\int_{\mathbb{S}^n} \eta = 0$. Why is $\eta$ exact?
If it's closed I see why it's exact
gentle ospreyBOT
shamrock:
sleek thicket
(this isn't relevant to the current discussion)
gritty widget
@sleek thicket i remember something like this gets proven in chapter 8 of munkres' analysis on manifolds
frozen token
ok ok soooo
gritty widget
although i dont know why off the top of my head
frozen token
assuming i know up to partial differentiations and vector spaces
what else should i study to be able to understand what toplogy is
ivory dragon
topology
gritty widget
topology will be unmotivated if you havent had analysis before, especially if you havent rigorously seen things like EVT, IVT, etc.
sleek thicket
if η is closed you can look at the de rham cohomology
frozen token
dont have access to uni :(
what books are suggested for analysis?
gritty widget
bruh
sleek thicket
@honest narwhal your flair says to ping gomez for help and you're a gomez alt
frozen token
whats wrong with rudin
gritty widget
nah rudin is good lol, its a bit terse and doesnt motivate a lot of topics though
i think its better as a reference
sleek thicket
lul
honest narwhal
Lol sure what's up
sleek thicket
Let $\eta$ be an $n$-form on $\mathbb{S}^n$ such that $\int_{\mathbb{S}^n} \eta = 0$. Why is $\eta$ exact?
gentle ospreyBOT
shamrock:
sleek thicket
If η is closed you can look at the de rham cohomology
Say it's cohomologous to a nonzero scalar multiple of the standard orientation form
frozen token
where actually is the differential part in that integral
what
sleek thicket
And so has nonzero integral
frozen token
ok im out i dont understand
gritty widget
@frozen token thats integrating differential forms, which you should completely ignore right now
sleek thicket
but I'm not sure how to argue it in the nonclosed case
honest narwhal
Hmm
sleek thicket
this is claimed as an exercise in Lee
sleek thicket
Sure. Are we still just assuming the integral of η over S^n is zero?
honest narwhal
Yeah
I've done the n=1 version of this case
Because if you know its integral over the unit circle is 0
sleek thicket
Yeah I see how to do that
honest narwhal
Then you know the integral over any loop in the punctured plane is 0
sleek thicket
Right
honest narwhal
And you can then explicitly build up a function for which you're df
sleek thicket
But I don't know a higher dimensional analogue of exact iff conservative
honest narwhal
Oh wait
Hmm
That's the direct proof of being exact
But proving you're closed is just Stokes'
sleek thicket
Oh really?
honest narwhal
The integral of $d\omega$ over any ball is $0$
gentle ospreyBOT
Daminark:
sleek thicket
I thought it was stokes but then I thought it wasn't stokes for some reason
ivory dragon
since that wasnt mentioned at all
gentle ospreyBOT
Daminark:
sleek thicket
I was thinking dω was zero
But we can lift up to R^n+1
Makes sense
Right so dami you're saying that if dω weren't zero, then it'd have nonzero integral over some punctured ball?
and by stokes we know every such integral is zero
honest narwhal
Yea
There might be a way to think of this "native to S^n" rather than playing games in R^{n+1}\{0} but
This makes sense to me
sleek thicket
Sorry why can't dω be like x^1 dx^1 ^ … ^ dx^(n+1)? Won't that have zero integral over every ball by symmetry?
but it is actually nonzero
honest narwhal
Is that guy exact?
sleek thicket
I don't know, I guess I'm just not sure why having zero integral over every punctured ball implies dω is zero
honest narwhal
Lemme pull up Lee actually
sleek thicket
Dami I'm a fucking idiot
n forms on S^n are closed
because n is the dimension of S^n
I still find your stokes thingy sus though
honest narwhal
So it's just cohomology lmfoa
honest narwhal
So the Stokes' thing is thinking of n-forms in R^{n+1}\{0}
sleek thicket
Right
honest narwhal
I guess the tricky thing is that we can't automatically write such a form as like
sleek thicket
and you get that the integral of dω over any punctured closed ball centered at the origin is 0
honest narwhal
f dx_I
sleek thicket
Right?
honest narwhal
So that makes life hell
honest narwhal
n-form, not n+1-form
honest narwhal
Oh oh wait no actually hmm
You're right I'm right
So assume d\omega isn't 0
Oh wait
Your earlier thing with symmetry doesn't apply I don't think
You're thinking over a symmetric ball around the origin
sleek thicket
Yes
honest narwhal
I'm saying over any ball
sleek thicket
But how do we get that by lifting from a form on S^n?
Whose integral over S^n is zero
I guess in my brain we could only assume the integral over spheres centered at the origin was 0
honest narwhal
Something something homotopy
Okay so wait now I'm confused about your example
So we have that n+1-form on R^{n+1} right?
sleek thicket
Yup
sleek thicket
Right
honest narwhal
Oh wait I was I think mixing up a few things rip
I was thinking that like
The integral of \omega over any copy of S^n in R^{n+1} that dodges the origin should be 0
Automatically
Because then you homotope it down to 0
But that's already assuming it's closed lol
So yeah nah this idea was doomed
sleek thicket
there is a lot of stuff going on with forms
and integration/cohomology/stokes etc
just like
lots of definitions and useful results
it's good shit but also I'm having trouble fitting it all in my brain
floral gust
Damn... I was planning to cover stokes from Lee in this upcoming summer but the above discussion scared me..... o.o
honest narwhal
Tbh I think I prefer math with a lot of juggling
Like compared to, idk math that's "bash this out hard"
Somehow the former feels more satisfying
dim meadow
I think math should be a lot of both
honest narwhal
I think it ends up being a lot of both
But the former part is definitely nicer lol
sleek thicket
I do like it when the definitions all just work together perfectly
Like, that's probably my favorite and least favorite thing about smooth manifold stuff
There's these great identities (which you have to remember 100 of) which are trivial from the definitions
But there's a ton of nested layers of definitions
It's easy to get twisted around ig
dim meadow
yeah
dusk heron
Does anybody here know about uniformization of surfaces with boundary?
clever lantern
hi guys i'm here for asking an help with this topology question: If a have a topological space X there is exist at least a subset closed and compact ?
my idea is to show that every open is a union of open and the complement of an open is a closed therefore placing an open set V as a union of open A of the topology the complement is an intersection of closed which form a family from which it is possible to extract a finite subfamily of closed with non-empty intersection and therefore this set is closed and compact as it is made up of a family of closed which possesses the property of finite intersection
sleek thicket
I don't think this is true but I'm having a hard time coming up with a counterexample
Your space needs to be really bad if this is false
I have no idea what your proof is trying to say
isn't R itself compact?
I did think about the indiscrete topology lol
also lower limit on R doesn't work
Or cocountable
I have a mental list of weird spaces
clever lantern
as it is not specified how it is made X I assumed the existence of an open set with which complement is closed and compact since the complement is equal to the intersection of closed which constitute a family of sets that enjoy the finite intersection property and therefore the complement is closed and compact and therefore it is always possible to find a closed and compact set in X.
sleek thicket
I'm thinking about whether you can get a space with no compact subsets @gritty widget
oh huh
lol duh
clever lantern
I think the question in the exam topic referred to a construction of this space, considering that it is a first year of mathematics
clever lantern
no no it's a generical topological space
you don't know how it's made
and you don't know what kind topology ithas
it has*
sleek thicket
oh it's true?
Oh lol
That was probably the intended answer
Yeah I'm curious about the interesting version
I'm looking at counterexamples in topology
Yeah
Found an example
#52 in counterexamples in topology
Yeah I'm posting
Take X = (0, 1) with open sets {U_n = (0, 1-1/n) : n >= 2}
As well as the empty set and 0
i don't like it
Open sets look like (-infty, a]?
clever lantern
I don't think the void is correct because it is both open and closed in X
clever lantern
ok but the question does not refer to a set that is only closed ?
clever lantern
hi guys i'm here for asking an help with this topology question: If a have a topological space X there is exist at least a subset closed and compact ?
closed and compact
sleek thicket
yes, the empty set is compact
clever lantern
but it's not only closed
clever lantern
ah ok
i understood the text as if it were asking me that the whole is only closed
the set ^
ok thank you for the help
i have another question :
consider R with the Euclidean topology and X the space identifying the interval (0,1) at one point. X have Hausdorff property?
sleek thicket
What do you mean by "identifying the interval (0, 1) at one point"?
clever lantern
i think it's mean there is a map that identify all (0,1) interval in only one equivalence class of X
an it is all X
i answered yes to considering X as a topological space with this relation x$\sim$ y if and only if x,y $\in$ (0,1)
gentle ospreyBOT
Black:
sleek thicket
yes, this space is hausdorff
clever lantern
is it correct the relation that i supposed ?
sleek thicket
If I understand you correctly, yes
clever lantern
i'm sorry for my bad englishbut it's not my first language
thanks for the help guys
little hemlock
can someone explain that? what if you take the point identified with 1 and the point identified with (0,1). What would be disjoint open sets containing those?
sleek thicket
Oh sorry, I read [0,1] for some reason
I think I need to stop doing math for the night
gritty widget
a necessary condition for a quotient to be hausdorff is for equivalence classes to be closed in the base space
that can be used here
although it may overcomplicate things especially for an example like this, its a good thing to keep in mind while studying quotients
clever lantern
thank you for the hint @gritty widget
clever lantern
@gritty widget the condition have you told me has to do with the pasting lemma and the saturation of subset of the base space ?
gritty widget
if the quotient is hausdorff then the preimage of equivalence classes under the projection map is closed - but those are precisely those equivalence classes
i guess that has to do with saturated sets (after some googling), although i must admit im not familiar with them so i avoided saying anything about it
floral gust
Can somebody explain why this holds true?
@marsh forge
This is the provided argument but I am not sure how this implies we should have a Ker f_*
rugged swan
it's bc you have the exact sequence Ker f -> H_*(X) -> H_*(point) and a section i : H_*(point) -> H_*(X)
floral gust
@rugged swan Ah, I have managed to see through one problem. That is if I am given Ker f → H*(X)→H*(point) is an exact sequence, since I know that there exists a map such that H*(point) → H*(X) →H*(point) is identity, thus we have that the exact sequence splits.
However, what still eludes me is how did you got Ker f at the starting position
floral gust
Isnt the induced exact sequence supposed to be
H*(point)→ H*(X)→H*(point)
Because otherwise, if I were to have any subset A of H*(X), then H*(X) = A ⊕ H*(point)?
rugged swan
no
I don't induce any exact sequence here. 0 -> Ker f* -> H*(X) is exact by definition of Ker f*. and H*(X) -> H*(point) -> 0 is exact because f is surjective bc you have a section. And because you have a section it splits
floral gust
Ah I see I see. So if I were to have any other subset say A, then 0 -> A -> H*(X) isn't necessarily exact. Can you remind me what is the definition of Ker f* here?
rugged swan
for a subset A, A -> H*(X) -> H*(point) isn't necessarly exact too
floral gust
Oh I see you are using this fact to get the required exact sequence.
rugged swan
It's very surprising that you don't know the definition of Ker f* while learning homology theory x). Ker fn is the subgroup of Hn(X) of all x that satisfy fn(x) = 0
marsh forge
This is an exercise you pretty much need to do: 0->X->Y being exact is equivalent to X->Y being injective and X->Y->0 being exact is equivalent to X->Y being surjective.
And I agree with Zak you should really know what a kernel is
floral gust
I was confirming to remember why exactly ker f* gives me an exact sequence 😅
marsh forge
an a cokernel
floral gust
But I located the proof. Thanks!
marsh forge
Do the exercise I sent, really
also show that 0->X->Y->Z->0 being exact is equivalent to Z=Y/X
floral gust
Yes, will do! Thanks! Actually I have attempted this exercise twice now, but been a while so I forgot the fact. Will revise my sequences.
honest narwhal
Yup
I mean really correct thing for your sanity is to not talk about diffgeo
But yeah
untold condor
I guess it's the life of doing differential geometry / pedagogy.
chrome dew
have you taken multivariable or vector calculus?
your question seems a bit too broad, like is there a general approach to drawing coordinate lines on a surface embedded in R^3?
do you have a more concrete question or example you have in mind
you also might have luck with some of the people who hang out in #multivariable-calculus
this channel is more topology oriented
but here is fine
so you're asking how to parametrize this?
chrome dew
hmm what happened to that guy, did he just implode?
sharp frost
he moved to a multivariable channel I think
chrome dew
not that I can see, he was in the middle of asking a question and I went to get more water and he deleted all his messages
whatever, his loss lol
unique wyvern
Hello does anyone have a good understanding of the heine borel theorem
I understand it up until one of the last statements if anyone could help would be greatly appreciated
bitter yoke
bitter yoke
What exactly that comes after that are you confused about?
unique wyvern
So is the jist of the contradiction, suppose c <b => we have an element greater than c in the set thus contradiction
bitter yoke
yes
tardy meadow
need a quick bit of help visualising the quotient space $$\frac {[0,1] \cup [2,3]} {1 \sim 2}$$. My aim is to show that it is homeomorphic to $[0,1]$. I suspect that the function $$f(x) = \begin{cases}\frac 1 2 x & x \in [0,1] \ \frac 1 2 x - \frac 1 2 & x \in [2,3]\end{cases}$$ is a homeomorphism $\frac {[0,1] \cup [2,3]} {1 \sim 2} \to [0,1]$?
gentle ospreyBOT
George!:
tardy meadow
(everything with the usual euclidean topologies)
is a good way to visualise it, just the interval [0,2]? since we glue together 1 and 2
or am I being stupid at some point here (to clarify, 1 ~ 2 meaning ~ is the relation where x ~ y iff x = y or {x,y} = {1,2})
tardy meadow
cool, thanks a bunch
it just felt a bit weird treating 1 and 2 as "the same" without it being modular arithmetic, but I think I'm getting used to this sort of thing
tardy meadow
isn't the inverse of that homeomorphism there sufficient for it to be path connected? I'm rationalising it is that because 1 and 2 are "stuck together" there's no longer a jump discontinuity as there would be without the 1 ~ 2
marsh forge
the homeo is sufficient but i think slim meant a visual proof
basically 1~2 is a 'bridge' connecting the intervals
Something worth keeping in mind: when you identify 1,2 via modular arithmetic, what you are actually doing is making that identification and then "changing everything that has to change so that its still a group"
Here we are only changing things that need to change to make it a topological space
so we don't care about modular arithmetic
and in particular, theres no reason to believe additive/multiplicative structures should have anything to do w the topology (and in this case they do not)
tardy meadow
basically 1~2 is a 'bridge' connecting the intervals
yeah that's what I had in mind
and in particular, theres no reason to believe additive/multiplicative structures should have anything to do w the topology (and in this case they do not)
what do you mean by this?
marsh forge
Which part, in our of the the parenthesis
tardy meadow
out
marsh forge
Ok so like
we can give R and its subsets a topology
we can also add elements in R
but theres no reason to believe that adding and topology play nicely together
(there actually is a way in which they do, but thats not my point)
So like, yeah we mess with the topology by gluing stuff, and no it doesn't make addition work
but we never expected it to make addition work
Basically once you are thinking of R as a 'topological space' you dont want to think too much about the arithmetic of R
Maybe another prespective is: 1,2 are not really "numbers" anymore as much as they are easily refrenced "points" in a space
tardy meadow
I get what you mean - surely it makes dealing explicitly with functions between the spaces awkward at times?
marsh forge
It depends
Not really
Although whenever you have a cont. function X->Y
and you take some quotient X/~
You have to prove whether you get a function X/~ -> Y
its normally not well defined
But for normal subspaces of R you just have to prove that the function always lands where you want it to land
pseudo crane
wicked trout
yeah i reckon so
paper saddle
Does anyone understand the Klein correspondence?
paper saddle
relating to projective geometry and exterior algebra
ivory dragon
do you have a specific question?
lyric quartz
So c) -> a) seems intuitively obvious since maps from circle to X can be viewed as a loop in X, but what would be the more formal justification of this?
floral gust
Can somebody help on part b?
floral gust
@gritty widget Like this?
marsh forge
You can take it further
floral gust
@marsh forge How so?
lyric quartz
@gritty widget , image of what exactly? Thanks!
floral gust
@floral gust well, what is the dimension of top homology for Y?
@gritty widget A finite dimension n
And yes I did what you wrote in the answer above
I dont have the answer 😅 I just have an attempt. The problem I have is in expressing int terms of character of X
And using that can you rewrite $\sum_{i=1}^n(-1)^i,\text{rank},H_{i-1}(Y)$?
@gritty widget I did this part though
gentle ospreyBOT
slimvesus:
floral gust
gentle ospreyBOT
slimvesus:
floral gust
Yeah I cant figure out beyond that
Oh I see
n-1?
Ah so they have the same characterisitic?
I am getting -ꭓ(Y)
gentle ospreyBOT
slimvesus:
floral gust
Ah okay okay, so I am guessing the induction would then be (-1)^l ꭓ(Y)?
floral gust
Although does having negative euler character makes sense?
Maybe there is a relation that relates negative ꭓ with some positive ꭓ?
Ah yes yes right
Indeed indeed
Yeah because we are not making any assumptions in our solution
Just straight, starting from defintiions
Yeah zero idea, maybe the Hi(suspension)= H_{i-1}(X) only holds in some specifc cases (although its true for sphere though)
floral gust
It resolves it no?
gentle ospreyBOT
slimvesus:
floral gust
Ikr xD
marsh forge
Oh the reason your thing is wrong slim is that its not reduced homology
I think theres some bad stuff in both formulas that happens at 0 here or smth
does it? maybe i made a mistake
oh rught
one sec
we did it reddit unless i cant do arithmetic
typo
first line should be suspension
oh on second look
looks like yours is correct but you simplified wrong?
oh no its a sign problem
gentle ospreyBOT
slimvesus:
marsh forge
yeah ahah
it happens lol
this is why problems like this should tell you what the formula should end up being
also, its shockingly hard to google that formula i couldnt find anything
i guess euler char is kinda a less used invariant tho
esp since it can be recovered from homology
unique wyvern
Having trouble showing a set in (X,d) is open in (X,dhat),
Becuase dhat maps to [1,inf) ? am I supposed to be taking scaled closed balls?
I tried saying let U be open in (X,d) this implies there exists an epsilon such that a ball centred at x radius epsilon is contained in U
All you can imply from this is that dhat(x',x)=1 for all x' in the ball?
I mean for that implication sorry restrict epsilon less than 1
dim meadow
@unique wyvern the max should be a min
unique wyvern
are you sure
unique wyvern
ok, time to rethink about this question lol
true
I was thinking about the mapping of it and it didnt seem right
dim meadow
The bounded metric stuff is very standard
unique wyvern
Thankyou anyways
meager python
@gritty widget @marsh forge shouldnt you use MV? On the sets SX minus the pointed points and then their intersection which is S x [0, 1].
Er nm I didn’t see the question
loud scarab
Hello, I don't understand the highlighted part, what does it mean that when a is not in the image, that it is a regular value of f? How can it be even a 'value' of f if f is only defined on U?
meager python
You can define the reduced characteristic. Express normal char in terms of reduced and the compute the suspensions reduced characteristic
The critical value and regular value live in R^m, not in R^n
small obsidian
@loud scarab
We define "regular value" as "not a critical value"
and we define "critical value" as "output of a non-surjective map"
meager python
A ”value” of f can be seen as f(x) for some point in R^n
small obsidian
So note that, by this definition, any point that isn't an output is a regular value
loud scarab
and i can only speak about the inverse f^-1(a) when a is in the image right?
even if a is a regular value but not in the image, i cant say f^-1(a) ?
loud scarab
im just trying to reach the conclusion of the level surface
the inverse of a regular value --> regular surface
small obsidian
What does your book define the theorem as?
loud scarab
this is it
its easy to prove and accept it as an abstarct theorem, but visually i cant tell how
$x^{2} + y^{2} + z^{2} = 1$ is an easy example
small obsidian
Swap that for "all points f(p) = a"
gentle ospreyBOT
khaled014z:
small obsidian
That's all it really means haha
loud scarab
swap which for which?
the problem i'm having is, what if f was surjective and injective
then the inverse cannot give a surface
small obsidian
What makes you say that?
loud scarab
If f is injective, then the inverse is just a one point
small obsidian
And that's not a regular surface?
I genuinely don't know myself lol, I'm actually studying similar material atm.
loud scarab
according to the definition of a regular surface, a one single point cannot possibly be a regular surface
small obsidian
But if a is a regular point, it fits the theorems
Oh yep, just looked at the def for that, and a single point definitely doesn't count
You've reached a contradiction. It would have to be the case that any point that is a unique output cannot be regular
loud scarab
Yes I dont know why is it not included in the statement of the theorem
small obsidian
No, like that's a result not a definition
loud scarab
do you mean that the statement: a is a regular value if its not critical
is not 100 % right?
small obsidian
Wat
No
It would have to be the case that any point that is a unique output must be critical
Any point that is unique is a max or min, and has a non-surjective differential. So it wasn't a regular level set
loud scarab
hard for me to come up with an example that would reinforce this
small obsidian
f = x² + y² + z² is the function
What's f-1(0)?
loud scarab
(0,0,0)
small obsidian
So that's just one point, clearly not a regular surface
Let's go on the other way. f is obviously a differential map, what's the differential at (0,0,0)?
.
It's quick to realize the differential there is, itself, (0,0,0) and thus not surjective. (0,0,0) is a critical point of f, and 0 is a critical value.
Ergo, this theorem has nothing to say about f-1(0)
But we know that's not a regular surface anyway
loud scarab
oh okay, this made some sense to me, thank you
hopefully on my way through more examples in the book i can understand it better
sweet wing
turbid halo
srry
wanton marsh
what shape can a singular conic be
warm mirage
I showed that the affine transformations from R to R are the transformations f(x)=ax+b, for a,b fixed in R and x varying in R.
I even showed that in the general case: the affine transformations from a vector space E to a vector space F, with their natural affine structure, are the transformations f(x)=a(x)+v, where a is linear from E to F, v fixed in F, and x varying in E.
But somehow, I am stuck on an easier problem: the linear transformations from R to R are the transformations f(x)=ax, where a is fixed in R and x varying in R.
Any hint?
sleek thicket
Suppose you already know f(x) = ax
How can you determine what value a is by using f?
warm mirage
f(x)/x ?
sleek thicket
That's true, but it's also not obvious that it's independent of x
Is there a simpler way to get it?
warm mirage
If x=0, nothing to do, f(x)=0=0.x
If x=/=0, and take x0 fixed =/=0, then there exists a=/=0 that x=a.x0
sleek thicket
Is that a the same a as above?
sleek thicket
Yeah exactly
warm mirage
Yes man
I tried that
But didn't put it in the correct way
That's what I get most angry about
Proving harder theorems, and not being able to prove this simple thing
sleek thicket
So can you see how this implies the original theorem?
warm mirage
About affine transformations from R to R?
sleek thicket
Linear ones
warm mirage
Yes sure
sleek thicket
Cool
sleek thicket
The proof I had in mind was f(x) = f(x * 1) = x * f(1)
gentle ospreyBOT
warm mirage
I showed these 2 theorems and lemma and couldn't show the one concerning the linear transformations from R to R
Couldn't notice that x=x.1
night parrot
hi all - i have a question regarding the 4-polytopes. i am writing a program to generate 4-D shapes, like the 120-cell, 600-cell, etc. using permutations and QHull (a library for n-dimensional convex hulls). the algorithm generally works, but what im most interested in is, the tetrahedral decomposition of these 4-polytopes. im wondering if anyone knows of any research / papers / starting-points for this? maybe a bit of a broad question, and perhaps more computational in nature. but just wanted to see if anyone had any thoughts. basically, the input is a bunch of permutations describing the vertices of the 4-polytopes, and im hoping to use this information to reconstruct a tetrahedral "mesh"
i should mention that it obviously doesnt need to be the "minimal tetrahedral decomposition," which i believe is np-complete
brave yew
bitter yoke
coarse kestrel
Can a topological space have no compact subset that is infinite?
marsh forge
finite topological space
coarse kestrel
hmmmm what about Z with the sets {2n,2n+1} as a basis?
dim meadow
lmao
dim meadow
yeah I guess modulo some reasonableness assumptions
I'm thinking stuff with limits
like if you have an infinite convergent sequence you should be fine
tbh my brain is fried from spending the last 2 hours doing finite injury stuff
yeah
I constructed a low simple set for extra credit on a test lmao
oh hmm
what's the harder version
ah ok
so obviously no closed points
hmm indiscrete topology is a dumb answer I guess
oh fuck
I'm dumb
okay I think I have a weird example
fuck it didn't work
nvmd
lol
yeah
dim meadow
oh lol
coarse kestrel
lmao
dim meadow
for a nonempty example but still trivial take the closure of a non closed point
wait maybe that doesn't work
was trying to get an indiscrete set
but yeah there are dense nonclosed points
so that was dumb
oh lol
I had a point set topology question recently where a counterexample was the sierpinski space but I don't know of any other counterexamples
so you take the set of continuous maps X \to X and assign to each map f its action on Top(X) by f^{-1}
if points are closed this mapping is injective
topology of X
idk what I'm writing anymore tbh
I was like "this seems like a good way to write it"
also I don't mean action really
I mean it assigns f to the map f^{-1}:Top(X)\to Top(X)
yeah
two points, 1 closed
the point is if all points are closed then you are injective
but I wanted an example where not all points are closed but the mapping was injective