Ok, I'm not sure I'll be able to answer it as I haven't actually taken a course on this, but I'm sure someone will have something to say
#point-set-topology
limpid mural
Ok thank you :)
cursive flume
@limpid mural if you define connection as this maybe you can parallely transport covector fields too?
🤔
note:i'm not studying mathematics,this is a mathphys lecture and lecturer said this connection can be called covariant derivative too
limpid mural
Sadly I've defined the connection only on vector fields and I am using the parallel transport to define the covariant derivative for tensor fields
cursive flume
is this definition sloppy from a mathematical PoV?
limpid mural
I think your definition is easier to work with
Because you already know how the connection behave on tensors
Without using parallel transport to define it(that is awful tbh)
cursive flume
he defined parallel transport with the help of this: $\nabla_{v_{\gamma}} v_{\gamma}=0$
where v is the tangent vector(field) to the curve $\gamma$
ivory dragon
is this definition sloppy from a mathematical PoV?
well
the spelling of "Leibniz" is certainly sloppy
cursive flume
ivory dragon
i suppose thats more of a linguistic POV though
limpid mural
so my problem is this
I know, by the equation of parallel transport that
$$\dot{\bold{w_i}}(t) + \gamma'(t)^j w_i^k \Gamma_{ij}^l \bold{e_l}(\gamma(t)) = \bold{0}$$
fml
limpid mural
that at $t = 0$ gives $$\dot{\bold{w_i}}(0) + \gamma'(0)^j \Gamma_{ij}^l(0) \bold{e_l}(\gamma(0)) = \bold{0}$$
gentle ospreyBOT
emme:
limpid mural
now I would like to deduce that if $\bold{w}^i(t)$ is the dual basis of $\bold{w_i}(t)$ then $$\dot{\bold{w^i}}(0) - \gamma'(0)^j\Gamma_{jl}^i(0) \bold{e}^l = \bold{0}$$
gentle ospreyBOT
emme:
limpid mural
cursive flume
how would one prove that isometries preserve geodesics?
I read this here,but i don't properly understand it https://math.stackexchange.com/questions/109423/isometries-preserve-geodesics
unkempt bolt
@cursive flume I would imagine that the easiest way to show it, is to first show that an isometry preserves distances and then deduce it from that. Do you want to do show it that way or to go through the explicit calculation? I'm not familiar with most of the symbols so no guarantees I'll be able to help to much
@limpid mural I'm not familiar with the equation for parallel transport using the Christoffel symbols, but your statement about the dual basis should follow from the previous definition of parallel transport of covectors you gave. If you define $w^i(t)$ to be ${\Gamma(\gamma)_{t}^{t_0}}^{\star}(e^i)$ then $w^i(t)$ is the dual basis to $w_i(t)$ by just applying the definitions.
gentle ospreyBOT
The_Vman:
unkempt bolt
If you know how that parallel transport equation is derived I might be able to help, but otherwise I haven't found anything so far
limpid mural
It's derived by applying the equation of parallelism to a vector field, which is $D_t X = 0$, probably I need to derive a similar formula for covectors by using the definition of connection on tensor fields
gentle ospreyBOT
emme:
limpid mural
my only information is that I have defined $\nabla$ on tensors $T$ as $\nabla_v T = \frac{d}{dt}\bigg(\Gamma(\gamma)t^{0}(T(\gamma(t))\bigg)\bigg|{t = 0}$, where $\gamma: I \to M$ is smooth such that $\gamma(0) = p, \gamma'(0) = v \in T_pM$.
gentle ospreyBOT
emme:
limpid mural
where $\Gamma(\gamma)$ is (abusing notation) the parallel transport map induced naturally by the parallel transport of vectors on tensors
gentle ospreyBOT
emme:
unkempt bolt
So then the idea is given basis $e_i$ at the point $\gamma(0)$, and some vector fields $X_i$ which satisfy $\nabla_{\gamma'(t)} X_i = 0$ for all $t \in I$ and $X_i(\gamma(0)) = e_i$, if $\omega^i(\gamma(t))$ are the dual basis to $X_i(\gamma(t))$ then $\nabla_{\gamma'(t)} w^i = 0$ for all $t \in I$ using the above definition of $\nabla$ on tensors?
gentle ospreyBOT
The_Vman:
limpid mural
yes
unkempt bolt
hmm
Ok so I'm just going to throw out an idea, and so be warned that it may be complete nonsense
limpid mural
don't worry
unkempt bolt
So to show $0 = \nabla_{\gamma'(t)} \omega^i = \frac{d}{ds}\bigg(\Gamma(\gamma){t + s}^t(\omega(\gamma(t + s))\bigg)\bigg|{s = 0}$ it is enough to show that the term
$$\Gamma(\gamma)_{t + s}^t(\omega(\gamma(t + s))$$
is actually independent of $s$.
gentle ospreyBOT
The_Vman:
unkempt bolt
Which should follow from
$$\Gamma(\gamma)_{t + s}^t(\omega^i(\gamma(t + s)) \big(X_j(\gamma(t))\big) = \omega^i(\gamma(t + s))\bigg(\Gamma(\gamma)_t^{t + s}(X_j(\gamma(t)))\bigg)$$
gritty widget
notation 
gentle ospreyBOT
The_Vman:
unkempt bolt
$= \omega^i(\gamma(t + s))\big(X_j(\gamma(t + s))\big) = \delta_{ij}$
gentle ospreyBOT
The_Vman:
unkempt bolt
As the $X_j(\gamma(t))$ form a basis at the point $\gamma(t)$ and so determine the covectors
$$\Gamma(\gamma)_{t + s}^t(\omega(\gamma(t + s))$$
gentle ospreyBOT
The_Vman:
unkempt bolt
Yeah my LaTeX abilities took a big oof today
limpid mural
ok that seems true
ok from that I should be able to deduce the equation, I just need to investigate the christoffels' symbols for the dual basis
thank you very much @unkempt bolt
unkempt bolt
No problem, I learned some stuff while doing it
cursive flume
my professor stated this and I am confused. can one not reconstruct the global geometry from the local one?
@cursive flume I would imagine that the easiest way to show it, is to first show that an isometry preserves distances and then deduce it from that. Do you want to do show it that way or to go through the explicit calculation? I'm not familiar with most of the symbols so no guarantees I'll be able to help to much
@unkempt bolt just saw your answer,yes, it seems intuitive, i'll try to do that. thanks ^_^
onyx crow
Did he mean to say motion doesn’t make physical sense in the limiting case? 🤔
fickle linden
Hello can someone help me with surfaces and 3d curves
gritty widget
just ask
sleek thicket
fickle linden
yeah sorry so i just need the logical idea how the 3d curve is designed and how to calculate its parametric equation
our book sucks and its explenation is pretty vague
where gamma is the 3d curve
gritty widget
what do you mean by "designed"?
gritty widget
ah
fickle linden
like
for a straigh line
you need 1 point and a vector
for 1 curve you need the intersection of 2 planes or surfaces
i get the idea
but idk how to actually use that in any exercise
idek what that picture means
what does F1 (x,y,z) = 0
yes its a function
with 3 var
but why it goes to 0
i dont get it
gritty widget
the picture means that Gamma is the set of points (x,y,z) where both F_1(x,y,z) = 0 and F_2(x,y,z) = 0
for example
fickle linden
yes that but what is the meaning of F_1(x,y,z) = 0 in particular
i have a book example and ill translate it
here you have 2 surfaces
a sphere and a cylinder
ik they have that shaped cause ik how they are logically defined
fickle linden
but idk how the book came up with the parametric form
yes take ur time just giving more info
what should i actually do
to come up with that sin cos curve equation
if u wanna voice chat that would be great like have a conversation about it let me know
gritty widget
the sphere given by x^2 + y^2 + z^2 = 4a^2 can be thought of as the zero set of the function F_1(x, y, z) = x^2 + y^2 + z^2 - 4a^2. similarly, the cylinder given by x^2 + (y - a)^2 = a^2 can be thought of as the zero set of the function F_2(x,y,z) = x^2 + (y - a)^2 - a^2.
to get the set given by the intersection of the sphere and the cylinder, you therefore look at the set of points (x,y,z) satisfying F_1(x,y,z) = 0 and F_2(x,y,z) = 0. this would be the set Gamma, in this example.
basically, we write Gamma as the intersection of the zero sets F_1 and F_2 because many curves arise in this manner. geometrically we can think of Gamma as obtained by the intersection of the surfaces given by the zero sets of F_1 and F_2 (which is almost always the case, ignoring regularity issues - if you don't know what that means yet, don't worry about it)
that's just explaining what Gamma means, for your very first picture
i cannot use vc right now, no mic
but i think some of the people in there can explain this
rokabe def knows this stuff
now, as for actually getting a parametric equation out of this
fickle linden
no it's fine
just didnt want to bother u typing
so basically
both surfaces intersact
and they leave a trace
on the coordiantes 0 ?
like the root
of the equation
i dont think that question made sense xD
gritty widget
i feel like you're approaching the right idea
fickle linden
its my bad my mathematical terms suck in english
gritty widget
here's a super simple example
fickle linden
kk hit me
gritty widget
for real constants a through h,
F_1(x,y,z) = ax + by + cz + d
F_2(x,y,z) = ex + fy + gz + h
the zero sets of the functions F_1 and F_2 are planes (assuming the coefficients a through h are nice enough). if we assume that these planes are not parallel, then their intersection forms a line. their intersection is precisely the set Gamma defined by F_1(x,y,z) = F_2(x,y,z) = 0
you might've seen this example if you took a linear algebra class
gritty widget
what's happening is: both surfaces (the planes) are defined by the functions F_1 and F_2 being zero. (by defined i mean they're exactly the set of points where that thing happens). their intersection is defined by two functions being zero
dw when i first saw this it took a moment to wrap my head around
but it is a very convenient way of specifying curves and surfaces
fickle linden
okay let me try to explain what i learned
fickle linden
so we have this sphere
and its eq is x^2 + y^2 + z^2 = 1
basic sphere
its easy to say that x^2 + y^2 + z^2 = 1 and x^2 + y^2 + z^2 - 1= 0 will give us the same object?
now if we get a cylinder
x^2 + y^2 = 1
its the same as x^2 + y^2 - 1 = 0
gritty widget
yup
fickle linden
the intersaction of these 2 its pretty easy to even imagine
its a circle
with the coordiantes of where you cut the sphere in half horizontally
gritty widget
(well it's two circles unless you restrict to positive z, if that's what you mean)
gritty widget
oh oops i was imagining a thinner cylinder
fickle linden
they intersact on the half of the sphere
oh ye that aswell if it was thinner
ye so okay i can easily create a function for it
gritty widget
well if the cylinder has a smaller radius, you'd get two circles. my bad
gritty widget
yes
fickle linden
but
if it's more complicated
and u cant imagine that
how would i come up with all those
sin cos
stuff
gritty widget
that's a good question
fickle linden
the book easily said
yee this is my problem i dont get
our teacher was pretty vague and weird when explained it i had no idea what he even said
gritty widget
i don't think there's a general method for doing so
fickle linden
well how do u analyze it then
gritty widget
i also haven't needed a parametric equation for a curve in a while, so i might just be rusty
fickle linden
to get to the answer
gritty widget
give me a sec to think
fickle linden
ye sure
u know whats weird
our teacher said
you will have to learn cylindric surfaces conic surfaces and surfaces of rotation
didnt even mention how the exercises would look like...
gritty widget
oh while i'm at it, sections 3.2 and 3.3 of this book offer a nice (english language) overview of the basics of curves and surfaces. i found it very helpful when i was taking mvc, so i'll share. (it was an assigned textbook but not the main one)
i think there's an answer to parametrizing curves of the form F_1 = F_2 = 0 in here. i'm checking it cause i'm too smoothbrained to remember off the top of my head
fickle linden
alright will check it rn
slim surge
Are you looking for a method to parametrize the intersection?
gritty widget
they are
fickle linden
@slim surge idek man
slim surge
Set the two equations equal to each other
fickle linden
actually let me find the chapter in my book to see what type of exercises there are in there
gritty widget
i just question if there's a truly general method to doing so, so i'm hesitant to say anything
gritty widget
that's what i'm thinking
slim surge
You could easily create 2d parametric equations whose intersection is any function
So choose that intersection function be something unsolvable
slim surge
Like the integral of (e^-x^2)
fickle linden
show that (x-1)^2 + y^2 + z^2 = 36 and y + z = 0 define the curve with parametric equation: x = 1+6cost y= sqrt18 * sin t z -sqrt18 sint
slim surge
First, solve for y in terms of z ( or vice versa)
slim surge
Oh lol sorry
That first steps the easiest anyway
Actually, my first step was wrong anyway if you want to show there equal
fickle linden
wait
slim surge
I was thinking you had to create the parametric curve
But merely showing the parametric curve works is easier
fickle linden
i should turn the parametric into cartezian ?
nah that makes no sense xd
is it allowed to take same value for x y z
and if they give the same number we good >
?
like take a number t
for the parametric
and we will get 3 x y z
if these full fill the equations its the same ?
slim surge
Your on the right track
slim surge
You don't want to show that the parametric is equal to the equation: you want to show its equal to the intersection
fickle linden
ok so first the intersaction is the system between the two
y=-z
we put them on the first one
and we geyt
get*
(x-10)^2 + -(z)^2 + z^2 = 36
slim surge
I think the easier way would be like this:
y+z = 0
(x-1)^2 + y^2 + z^2-36=0
then
(x-1)^2 + y^2 + z^2-36=y+z is when they intersect
fickle linden
oh
ye xD
ok so after that
we get t = something x
and t= something y
and t = something z
and put these on our thing
equation *
(x-1)^2 + y^2 + z^2-36=y+z
slim surge
You don't have to solve for t, if thats what your doing. You just replace, x for example, with x=1+6cos(t) [I'm not exactly sure what your saying, you might already be doing this]
slim surge
Yep, good luck
fickle linden
wait so
the book clearly expected that and it got factored totally
its 36(cos^2 t + sin^2 t) = 0
so cos ^2 + sin^2 isnt = 1 ?
slim surge
You forgot the -36 part I believe
You should have started with this:
(x-1)^2 + y^2 + z^2-36=y+z
And simplied down to
36cos^2(t)+36sin^2(t)-36=0
fickle linden
oh
so it goes to
1 = 1
which proves the point ?
that's why i messed up
since for every x y z
which is expressed with t
we need to have an equal solution ?
and we got it
so it proves our question
damn my mathematical english sucks
slim surge
Its sort of difficult to explain, I feel you
fickle linden
so basically if we got
1=0
or something else
that makes no sense
so our x
of the equation
wouldnt have the same soltuion
asx y z expressed with t
slim surge
When we set our curves equal, we got the intersection of those curves. The function for the curve represents the values that are on that curve
Think about x^2+y^2=1
The only values that work are those that are on the circle, if we choose, for example x=1, y=2, we don't get that x^2+y^2=1
fickle linden
ye
u can get in voice ?
well thats what i kinda mean
basically
when u equal those
u need to get equal simplification
so their intersection and the parametric
is the same
slim surge
It shows that your points are on the curve, not somewhere else
fickle linden
and when we get 1=1 that means we are talking about the same curve
slim surge
Exactly
slim surge
Yep
fickle linden
if we got 1=0
then they dont express the same thing
EASY
xD
thanks @slim surge
u the goat man
fickle linden
hmm u wouldnt get annoyed if i asked again xd?
slim surge
Any time my friend
fickle linden
i mean i think i got it but i have the whole book ahead
thanks alot
life saver doing god work
and @gritty widget thanks aswell
that explanation fullfilled the reasoning i need
needed
gritty widget
glad i could help
if this is a proper differential geometry class, then i probably should have elaborated on when F_1 and F_2 define surfaces and why i used the words "almost always the case"
but
for getting the basics down i hope that helped
fickle linden
this is a general first year electronic engineering book
called
linear algebra and analytic geomtry
it touched every topic slightly
so it's new for me thats why i dont know much here
really helped a lot
i need the logical reasoning for every new subject to memorize it
so yea
feel way more confident now
gritty widget
if it's electronic engineering then i don't think i should clarify too much
fickle linden
yeah i mean i would like to know it but i dont have time tbh
i really love math
but we have 7 other exams in 1 month
i need 2 learn em all
gritty widget
if you're interested though, the keywords are:
"regular level set theorem" for when a function defines a surface
"transversality" for when the intersection of surfaces gives a curve
"sard's theorem" for the meaning of "almost always the case"
gritty widget
that's a lot of exams, good luck
gritty widget
all you need for a basic understanding of the things i just posted is like, a decent exposure to (theoretical) multivariable calculus and linear algebra
even less maybe, i'm not sure
btw for future reference, questions like these might belong more in #multivariable-calculus, but imo this channel works too
fickle linden
hehe i was a bit confused where to ask i knew i chose the wrong one since im totally not advanced XDD
gritty widget
well it is geometry, but definitely not at the #geometry-and-trigonometry level
so the next reasonable place to ask is here
fickle linden
alright
fickle linden
@slim surge hey man
i got a question xd
i need to show that this equation is a cylindric equation and define it's orienting curve and the vector
slim surge
This looks like a problem of completing the square
fickle linden
ye i did that
and i got 2(2x-y)^2 + (3x-z)^2 -2 = 0
so how do i prove that this equation is infact a cylindric surface
it needs to have 1 curve
in it
and 1 vector
that projects this curve in that orentation
orientation*
so this is the part im totally lost
how do i come up with these
slim surge
Well first, I checked and you got the completing the square correct, so thats good
slim surge
(I have done this type of problem before, so I'll have to work with you on this one)
My understand is that we want to find an vector in which changing that value doesn't change the value of the cylindeer
slim surge
Like, when we have a normal cylinder, such as x^2+y^2=1
fickle linden
mhm
slim surge
Z doesn't matter
fickle linden
yes
slim surge
So thats why it looks like a bunch of circles with different z
So we want to find that direction where this cylinder doesn't change
slim surge
For the normal cylinder, but not for our cylinder
fickle linden
ye
for our cylinder
we have x
that changes with z
srry
x changez with y
and x changes with z
thats where i lost it a bit
if you can get on voice chat let me know that would be faster
typing kinda sucks
anyway
slim surge
I can get into a voice chat, but let me first figure out how to solve the problem
slim surge
Sure
fickle linden
so basically that on top is 1 of the straigh lines
equation
which you get the vector we needed
1 2 3
and to find a curve
the book says we need to find a trace of the cylinder
which is possible by making x = 0
and OK i understand the book
i logically understand why the curve part stands
but i lost it how he came up with the line equation
he just took what was inside the squares
and equaled it to 0
o shit
i think i got it
...
lines are parallel to the cylinders
so -2 is not necessary
and the multiplayer on 2(2x-y)^2 is not necessary
since it just squashes it
doesnt change any direction
orientation*
slim surge
I think the reason that they set x=0, is to see what it looks like when it is intersected by a plane
fickle linden
so we know that the direction
we can get it from the lines
that are parallel with the cylinder
slim surge
Yep
fickle linden
since the 2 multiplayer just squashes the cylinder
and -2(constant)
just pushes it in a way
they dont change the direction
its the same as (2x-y)^2 + (3x-z)^2 =0
slim surge
Unfortunately, thats not quite true. The (2x-y) and (3x-z) reorients its direction
slim surge
But, you had a great idea with doing (2x-y)^2 + (3x-z)^2 =0
And, actually, thats the way to get the answer, now that I think of it
fickle linden
was about to continue that
since they are squared
they need to both be = 0
in order for that to be true
slim surge
If we ignore the -2 part, its like we talking about a cylinder with radius 0
fickle linden
oh
slim surge
Which is like the center of the cylinder
fickle linden
wait
2 multiplayer
is no problem
lol
let me continue xddd
2(2x-y)^2 + (3x-z)^2 =0
so we got this
since they are squared
they need to be both 0
so their sum is 0
slim surge
Yeah!
fickle linden
and we can devide after that
the 2
and we will get those 2 things on the book
damn
slim surge
We don't need to worry about the 2 part
What we need, is to have 2(2x-y)^2 + (3x-z)^2 =0
Now, notice that both parts of squared
fickle linden
yee
slim surge
So the only thing that works
fickle linden
is when they both are 0
slim surge
is (2x-y)^2 = 0 and (3x-z)^2 = 0
fickle linden
slim surge
Yeah, exactly, your right
slim surge
Oh yeah, you've already got this
fickle linden
i still dont get why we had 2 square them tho
slim surge
I makes it easier to solve
fickle linden
like yeah it's practical but
alright i guess
so
this was just the example of the book
slim surge
Otherwise, we couldn't seperate them into the two parts
fickle linden
theres this exercise
slim surge
Do you see where we go to solve it from here?
fickle linden
well we have the direction 1 2 3
slim surge
Yeah
fickle linden
and when we do x = 0
slim surge
Oh wait, you had already solved this lol
fickle linden
yee xd
slim surge
lol mb
fickle linden
just had to understand the logixc
so
come to voice chat
on mathematics
exercise 8
@slim surge you said u cant hear me?
?
slim surge
(Let me disconnect and reconnect, I can't hear you)
fickle linden
i can hear you tho
lets try 256kbps
(x+y)^2 -4x -3y +z + 2 = 0
2(2x-y)^2 + (3x-z)^2 -2 = 0
x^2 + 2xy -2x + y^2 + y - z = 0
(x+y)^2 -2x + y - z =0
x+y = 0
-2x + y - z = 0
x = -y
-3x = z
x=z/-3
y=x/-1
b
b
b
b
b
b
x+y = 0
-2x + y - z = 0
z = (x+y)^2 -2x + y
0i + 0j + 1k
x^2 + y^2 = 1 and z = 0
0i + 1j + 1k
x^2 + y...
z= 0
slim surge
i+-j-3k
fickle linden
x^2 + 2xy -2x + y^2 + y - z = 0
(x+y)^2 -2x + y - z =0
x+y = 0
-2x + y - z = 0
x = -y
-3x = z
x=z/-3
y=x/-1
slim surge
fickle linden
(x+y)^2 -4x + y - z =0
x^2 + y^2 =0
(x+y)^2 = 0
-4x + y - z =0
(x+y)^2 = 4
-4x + y - z = -4
(x+y)^2 = 0
-4x + y - z =0
s
s
s
s
s
(x+y)^2 = 0
-4x + y - z =0
x
y
z
x = -y
-3x = z
x=z/-3
y=x/-1
x-x0 = t a
y-y0 = t b
z-z0 = t c
unkempt bolt
@dim meadow If you are interested, I think I have a proof to show that $I^\bR$ is not sequentially compact
gentle ospreyBOT
The_Vman:
dim meadow
Please share
unkempt bolt
Ok so since $\bR$ has the same cardinality as the power set $P(\bN)$, we can consider $X = I^{P(\bN)}$. Then define the sequence $f: \bN \to X$ by
$$\pi_A(f(n)) = 1 \text{ if } n \in A$$
and $0$ otherwise, so like an indicator function. Above $\pi_A$ is the projection onto the component corresponding to a subset $A$ of $\bN$.
gentle ospreyBOT
The_Vman:
unkempt bolt
Then a subsequence is choosing some injection $g: \bN \to \bN$, and then considering $f \circ g$. Then setting $B = {g(0), g(2), g(4), \dots}$ the sequence $f(g(n))$ will oscillate in its $B$ component and thus not converge to any point in $I$. Therefore the sequence $f(g(n))$ as a whole cannot converge.
gentle ospreyBOT
The_Vman:
unkempt bolt
I guess g is technically an increasing injection not any random injection but that doesn't change anything
nimble cipher
Is the X here the space that contains A or just an arbitrary set? Thanks!
<@&286206848099549185>
gritty widget
(2) leads me to believe that X is a space containing both A and B @nimble cipher
nimble cipher
Oh because in the statement of the problem, $A\subset X$ and $B\subset Y$. so now im confused with this proof
gentle ospreyBOT
emphatic_wax:
gritty widget
probably a typo then
try writing in the correct expression for (A x B)^c at the bottom and see if the proof still works
should be (A^c x Y) u (X x B^c)
nimble cipher
Okay I will try this
gritty widget
yeah the proof should go through just fine if you put Y in the right places
nimble cipher
Yeah it works perfectly. Thanks for the help!
gritty widget
i like how it looks like the product rule for derivatives
the end result, i mean
nimble cipher
Yeah! That's the version of the proof I am yet to read
I read this easy one first because it's on our topology homework
tough imp
So Hartshorne II.4.6 says that a proper morphism of affine varieties over a field is finite. I'm pretty sure any proper morphism of affine schemes is finite so that this is trivially implied by it??
Any morphism of affine schemes is affine, and integral = affine + universally closed so f is integral, but it's also finite type since it's proper so f is finite
dull wave
@tough imp Affine morphisms are only introduced in the exercises in the following section, so you're 'cheating'
west spindle
is SL^+(n) := {A in R^nxn | det(A) = 1} connected if given the subspace topology inherited from R^(nxn)?
vocal wharf
yes
it's path-connected
you can check generators for SL
it's generated by transvections
which are "obviously" path connected
the alternative is induction on n
and using some weird argument: you can show that every element in SL\{-1} is the product of 2 matrices with eigenvalue 1
@west spindle
west spindle
transvections?
vocal wharf
shear mappings
although apparently it's not that simple, you need to specify some relations
west spindle
i mean i am not looking for a presentation now am i
vocal wharf
in this case, yes
the other proof is "easier", just write A = BC with B and C having eigenvalue 1, then find a path from 1 to A
but both B and C are similar to a matrix with a 1 in first position, so you can use induction
the hard part is seeing that you can write any element of SL as the product of two elements in SL with eigenvalue 1
which i have not yet found a nice proof for
floral gust
Can somebody explain how does the basis {X} generates the {X, ∅} topology? Is it because topologies are arbitrary unions of base elements and so if we consider the union of no elements we get ∅?
vocal wharf
yes
floral gust
Fnx
vocal wharf
also the empty intersection is the whole space
so the empty set generates the same topology
floral gust
But {∅} cannot be a base right? Because it doesn't satisfy the x in ∅ for all x in X.
fading vale
brain bad
I'm pretty confused by the second part of this problem with the Phi([f]) = Phi([g]) implies [f] is conjugate to [g]
[f] and [g] are conjugate implies Phi([f]) = Phi([g]) i think i understand
can we say like if [f] and [g] are equal under Phi then they may be based at basepoints x_0 and x_1
and x_0 in [g] and x_1 in [f] so we can construct some kind of loop
vocal wharf
But {∅} cannot be a base right? Because it doesn't satisfy the x in ∅ for all x in X.
@floral gust oh yeah, you are right
i was thinking of subbases
vocal wharf
i think it's fine
fading vale
maybe im dumb and misinterpreting this
actually i go more general and ask a question for my intuition
whats an example of two conjugate elements of pi_1(X, x_0) that arent homotopic
@marsh forge please give me intuition i am so hungry and my brain is so bad
dull wave
a and bab^{-1} on the torus @fading vale
fading vale
hmm
maybe i just dont have good intuition for why forgetting the basepoints makes these homotopic
fading vale
oh
but yeah i feel like i again just dont see why ignoring the base points makes these homotopic
maybe im just an idiot lol
and this is simple and im overcomplicating it
dull wave
I think you're confusing things
The conjugation is just to move the base point, right?
dull wave
Well draw a little picture
fading vale
i mean i understanding it when you're working with change of basepoint homomorphisms
and x_0 and x_1 are distinct points
but in this case wouldnt this be changing the basepoint to like... itself lol
dull wave
Yeah but you're on the wrong side of the picture lol
There's a correspondence here
Maps [S^1,S^1 v S^1] and conjugacy classes in Z * Z
dull wave
@fading vale Did you consider that you can change of basepoint to the same basepoint along non contractible paths?
Exercise 3. For a path connected space X, show pi_1(X) is abelian if and only if all basepoint-change homomorphisms depend only on the endpoints of the path
Well S^1\wedge S^1 has very non-abelian fundamental group
Now take the path a, and basepoint change it along b (this changes the basepoint to the same point)
marsh forge
@fading vale
Just one heads up
There is the change or basepoint ‘conjugate’
And there are actual conjugates
The former looks like a conjugation
But isnt one
The path between two points isnt in pi1
marsh forge
Tbh the terminology messed w me at first
fading vale
ok i think i understand it now
also i yearn for the end of all things
@marsh forge misery.
sleek thicket
Whitney embedding and noether normalization are so weird
They're like
Okay something insanely nice is true
And the proof is that this is just generally true for projection onto linear subspaces
you just need to do some work to show that this happens generally
Proofs which are "pick one and you have to be insanely unlucky to get it wrong" are hilarious
marsh forge
Do you find whitney embedding surprising?
sleek thicket
I'm not sure
I find the method of proof a little weird
I.e. "pick a codimension 1 linear subspace at random and project onto it"
I think I find it unsurprising that we can embed manifolds at all but a little surprising that there's such a nice bound on the dimension of the space we embed into
dull wave
There is the change or basepoint ‘conjugate’, And there are actual conjugates. The former looks like a conjugation but isnt one. The path between two points isnt in pi1
Except when it is @marsh forge, like when you change the basepoint to itself along a loop.
dull wave
It is helpful @marsh forge if you read the previous discussion.
You get a (non-trivial) automorphism of pi_1(X,x_0)
gentle ospreyBOT
Lartomato:
honest narwhal
Damn that's ballsy
uncut surge
the statement or me saying that the statement is garbage
Because it does feel like something that might be true, but it's just missing e.g. the assumption that gamma is a closed loop
And the integrality of omega does nothing...? I'm really confused. Physicists frustrate me.
honest narwhal
Oh physics okay lol
I guess symplectic should've been a hint
But yeah I also kinda for some reason thought it was the main thing
And I was like damn finding a thesis and being like "Yeah the main statement here is bullshit" just made me be like damn
uncut surge
Eeeh, it's not a main statement I think, it's something that shows up in the more introductory part, and they're probably gonna be fine in the end
like, from what i've read in the next couple sections, they always have "canonical" 1-forms theta that they can choose
But yeah, this is a very mathematical thesis so I had some hope that they're not bullshitting me, but I do think they are
shadow charm
Ive started coxeter's book on non euclidian geometry but im having a good amount of trouble understanding the foundations of real projective geometry, especially the notions of duality and double orientation
Does someone think they could explain?
shadow charm
<@&286206848099549185>
marsh forge
It might help if you have a specific question
shadow charm
Yeah I get that thing is I dont have a specific question. The text is quite abstract and I usually get the worked "examples" if you can call them that way, but I guess I have trouble generalizing? Definitely the main problems are understanding the notion of duality and double orientation of a line. Most of the other stuff Ive read so far I've slowly struggled through but I generally get it.
marsh forge
Do you have a specific definition or statement youre struggling with
uncut geyser
can I create a diffeomorphism from the "self intersecting" 2-torus to the 2-torus with a bigger(major) radius to show it is diffeomorphic to S1 X S1 ?
I just scaled the coordinates in the plane the major radius is defined on. is my reasoning ok?
or is that diffeomorphism not even possible
nvm. thats not possible
patent ridge
What is "self intersecting" 2 torus?
uncut geyser
its like
(sqrt(x^2+y^2)-a)^2+z^2=b^2, but b \ge a
basically the outer ring is smaller than the inner ones
patent ridge
Well, self intersecting torus has property that I can remove two points and result is disconnected. So no way can it be homeomorphic to normal torus. Thus no way can it be diffeomorphic. It is image of immersion of normal torus into S^2 though, right? I never studied differential geometry, so I am mostly guessing though. Are you trying to do anything in particular?
tidal forge
This might be a really trivial question but I just want to make sure.
Is R^2 \times R = R^3 ?
floral gust
Shouldn't this be "topological subspace of R^n+1"
gritty widget
yes
shadow charm
@marsh forge here is what I struggle with. The unfinished sentence at the top is "In projective geometry of three dimensions, we might define a sense-class by means of the vertices of a complete pentagon, but it is easier to use Veblen's notion of a doubly oriented line." I just get so lost here....
As for the concept of duality it's not a specific sentence more than the repeated usage of the idea as if we understand it, while the definition was very loosely just toed to examples.
uncut geyser
@patent ridge was trying to show the self intersecting torus is not a manifold basically
I guess that idea should be enough
carmine delta
fading vale
ok if someone could check my work real quick that would be cool
R^3 - X should deformation retract onto S^2 minus 2n points
if we let one of these points be our north pole we can then retract onto R^2 - {2n - 1} points
which gives us the wedge of 2n - 1 circles
so pi_1(R^3 - X) is the free group on 2n - 1 generators
haughty vale
Your argument sounds good, though I wouldn't use the word "retract" for the stereographic projection
fading vale
yeah actually i think its just homeomorphic
so project onto R^2 - {2n - 1} points
so project onto R^2 - {2n - 1} points
south geode
I like the argument. I think i did this exact problem last year!
cursive flume
how can I porve this?
I can't see how 8 is similar to 3 😦
i agree with the first term of RHS on eqn 8,but i don't see how i get the second term
cursive flume
or knowing this algebra,can one reproduce the metric?
magic stump
surreal flintBOT
pastel linden
is this just a shitty quirk of notation? this seems trivial if U_\alpha is allowed to be equal to U
actually wait I'm being dumb this is super simple
every neighborhood contains a smaller neighborhood right?
yeah you can take a homeomorphism \Phi from U to the open ball at 0, then by continuity any smaller ball at 0 is taken to a subset of U, and since that homeomorphism restricted to the smaller ball is a coordinate chart for the subset, then it is by definition a part of the maximal atlas
pastel linden
an open subset of a topological manifold is always homeomorphic to an open subset of some n dimensional euclidean space, right?
patent ridge
@pastel linden No. For example, take your topological manifold to be S1.
gritty widget
lol its no big deal
pastel linden
quite odd
my book lied to me, or I'm missing something here
I think this rules out the manifold itself, although I guess it is technically an open subset of itself
gritty widget
you can't necessarily cover that open set V by a single chart tho
pastel linden
I see, so the open set is just locally euclidean
dim meadow
@pastel linden it's always a submanifold of Euclidean space for some n, not an open submanifold
gritty widget
because any (topological or smooth) manifold embeds into some euclidean space right?
or is it something else
pastel linden
can someone confirm this is correct now?
because I didn't notice that the problem said "any open set U" so I kinda just tacked the intersection on
I'll fix that
the book tends to use balls because they're pretty straightforward but euclidean just means locally homeomorphic to an open set of R^n
I see the overcomplication though
I just drew out a nice diagram on scrap paper
homeomorphisms are just so nice to work with
elfin patrol
Hi, I’m trying to solve this topology problem, we’re given the set X={(0,y): 0<= y<1}U {(x,0): 0<=x<1} and told to show it isn’t a smooth manifold. I’m doing this via contradiction, but I’m stuck at showing that the Jacobian would have to be 0 at the “corner” at (0,0) but I’m stuck trying to figure out how to show it’s meant to be zero, I thought about using intermediate value theorem but that doesn’t make any sense since the derivative ranges between infinite and 0.
elfin patrol
I don’t know, the hint just says to assume it’s a smooth manifold and from that show the Jacobian ends up being zero
pastel linden
I think it's asking you to use the identity mapping on the respective x and y sets, and then at their intersection {0}, the functions aren't C^\infty compatible?
it basically gives you an "atlas"
marsh forge
[incorrect, redacted]
wanton marsh
but isn't it homeomorphic to (0;1), which can be given a smooth structure
marsh forge
Oh youre totally right
That question is poorly phrased
Smooth submanifold would be better i think
gentle ospreyBOT
notwhale:
dire rapids
elder yew
Rick Miranda has a pretty comprehensive text on Riemann Surfaces
dire rapids
thank you!
elder yew
Also if you look up Terry Tao 246C lecture notes, the first set of notes is on Riemann Surfaces and Riemann Roch
dire rapids
nice, thank you!
elder yew
No worries
warm hedge
a thesis for bachelor degree on general topology would be something ok to do or better to look for something else ? 😛
warm hedge
nice
this year i will have to do my thesis but i dont know yet in which field i should look 😂
marsh forge
I wish i got to write a thesis
elder yew
Are there research opportunities at your school max?
e.g. working with someone to get a paper out?
marsh forge
Not formally
elder yew
I just kept jumping schools/cross enrolling till I found someone to work with
Since everyone I wanted to work with was busy
deft creek
is there anyone willing to help me with some topology problem
deft creek
sure one sec
I want to make sure I have the correct options hopefully i'm not missing one. i believe the statements 1 3 5 6 7 8 are true. I could be wrong.
let's say X = {a,b,c,d} and T = {empty set, {a,b}, {b,c}, {a,b,c} then the options I got I think worked out.
patent ridge
i believe the statements 1 3 5 6 7 8 are true. I could be wrong.
@deft creek
You are right.
deft creek
i'm not sure if the last two options are true or not
marsh forge
Look at the defn of a topology
And try to think about the real numbers
For the last one
Thats my hint
pastel linden
consider singleton sets in real space under the usual topology and take the complement to get open sets
and then fuck around with them
pastel linden
that's my go to for weird examples, since if you're talking about closed sets, taking the union of specific points makes it easy to construct sets
gritty widget
I'm planning to take an algebraic topology course at the level of Hatcher soon and I figure I should review a bit of point-set topology. Are there any good & relatively short resources for that someone would suggest (other than just skimming through Munkres or Lee)?
honest narwhal
Hatcher's got notes on point-set
south geode
Hatcher doesn’t like Munkres’ book
I like both of them although I didn’t like Hatcher the first couple of times I read him
Dugundji also has a good point set book that’s worth skimming
honest narwhal
I don't like Munkres in general lmao it's way too long
chrome thistle
anyone know likea good vid series/youtube channel that i can learn diff geo from
prof and book arent really working out for me
prof is using shifrin notes
he told me to read de carmo but its kinda hard to understand at times
south geode
XylyXylyX has some good lecture vids on manifolds and geometry. He fears it towards being able to teach relativity but his manifold and tensor videos are purely mathematical
Here’s his channel
chrome thistle
thanks ill look into it
uncut geyser
You are right.
@patent ridge what?
patent ridge
I could be wrong.
@deft creek
This statement is right.
uncut geyser
oh :v
feral copper
Hi ! I have a question regarding relative homotopy !
Our teacher defined it using the Jr, as in https://www.math.ru.nl/~mgroth/teaching/htpy13/Section04.pdf.
However, I find it more visual to define (absolute) homotopy as homotopy classes of pointed maps (Sⁿ,∗)→(X,p). How is it possible to define relative homotopy in a similar fashion ? What is the spherical equivalent for the Jr construction ?
feral copper
The link works for me :/
Which word ? 😦
gentle ospreyBOT
MaxJ:
feral copper
marsh forge
Yeah so there are a few different constructions of the fundamental group
the best one is homotopy classes of pointed maps $S^1\to X$
gentle ospreyBOT
MaxJ:
feral copper
Of course ! But I'd like to see the details of the construction in the case of spheres, which my teacher did not detail !
(that's the point 😉 )
marsh forge
Oh
So
the multiplication is given as follows
take S^n
there is a copy of S^n-1 called the 'equator'
there is a map
S^n \to S^n v S^n
given by squishing the equator to a point
then to compose f and g
you just do f on the first copy of S^n
and g on the other
feral copper
I have already worked out the proofs for the absolute homotopy groups in this setting ! I proved that πn is abelian for n⩾2, and I proved functoriality, but now I am willing to do a similar thing for relative homotopy groups 🙂 I was wondering how to define them by using maps from pointed n-spheres and a subspace to pointed spaces and a subspace 🙂
marsh forge
oh sorry I was misunderstanding
feral copper
Don't worry, perhaps I wans't clear ! And it's already nice to give me some of your time 😉
feral copper
Oh ! I mean, yeah, I feel dumb now, for not having even cared looking at the Wikipedia page... ^^"
Thanks ! 😉
marsh forge
np
shut moat
weird question, but how important is this theorem?
turns out I was reading an old edition of this book and missed out on some theorems like this 
shut moat
yeah it's fairly intuitive. I'll probably need to do some problems so I don't forget it tho lel
low key salty that this book has more cool shit lol
ty!
gritty widget
@shut moat what book is this?
shut moat
Hubbard & Hubbard- Vector Calculus, Linear Algebra, and Differential Forms
gritty widget
thanks
wind portal
Multvariable Calc prof taught out of this for a lil bit.
It's like a vector calc book but also tries to prep you for multivariable analysis.
uncut geyser
thats characterizing being Cp on manifolds
pretty important imo
so you can basically locally inherit the topology used to determine the continuity of the extension
nimble cipher
how do I prove that the infinite product of bijective functions is a bijection
patent ridge
Sounds like an axiom of choice thing.
patent ridge
No. I don't think you need axiom of choice. I think you can prove it without.
Infinite products get weird without axiom of choice. But in this case, I think it is okay.
wanton marsh
can't you just show that the infinite product of the inverse functions is the inverse of the infinite product of the functions
patent ridge
I agree.
nimble cipher
Thank you @wanton marsh and @patent ridge
fallen canopy
I have a topological space G and a subspace H which also happens to be a subgroup of G. I want to show that H is a topological group - to do this I imagine I need to show that the map (x,y) to xy^{-1} is continuous. But I’m not sure why i should expect this function to be continuous- any hints?
gritty widget
the group operation on H is the restriction of the group operation on G to the subspace H
maybe thinking along those lines will help
patent ridge
Well, are you assuming G is a topological group?
patent ridge
Maybe Michael is only assuming G is a group and a topological space. (In which case his statement is false.)
gritty widget
i interpreted it as "G is a topological group" (yay for skimming); i guess it's up to them to clarify
patent ridge
I agree. That is how I interpret it. But why is it then hard to show that H is a topological group?
Kinda weird to consider non closed subgroups of topological groups. But doesn't matter, I guess.
fallen canopy
Sorry I mean topological group for G!
gritty widget
i have a good feeling you can come up with a good argument by thinking of restricting the group operation
fallen canopy
The restriction of a continuous map is continuous? I would imagine yes now that I’m thinking about it
gritty widget
to a subset given the subspace topology, yes
fallen canopy
wow that’s kind of obvious now that i think about it lol tyty
gritty widget
there is one small detail to note
when you restrict the domain of the group operation, you'll have something of the form H x H -> G but you want something H : H -> H; fortunately for you, H is a subgroup, so the image of this map will be a subset of H, so restricting the codomain presents no issues
sturdy basalt
Show one can define a distance function on an arbitrary set X by d(x, y) = 1 if x != y and d(x, x) = 0. What topology does d give to X?
Its easy to prove that its metric. but how do you proceed after that?
patent ridge
Next step is to answer question "what topology does d give to X?"
sturdy basalt
which is what I'm not able to answer
little hemlock
by definition, every open set of a metric space is a union of open balls, so it might help to know what the open balls look like.
sturdy basalt
every individual point in X seems to be an open ball
little hemlock
yea || so what are the open sets? ||
sturdy basalt
Every set
which Ig would make it discreet topology?
little hemlock
yep. indeed, the metric d(x, y) = 1 if x != y and d(x, x) = 0 is called the discrete metric
sturdy basalt
Which send me to another question. I thought I'd understood discrete topology, but I seem to have issues in it now
So, in Armstrong Topology, A topology and discrete topology seem to have definition that are the same
A topology on X has been defined as Non empty collection of open sets
and discrete topology also as a topology where all sets are open
both seem to be the same. Idk what detail I'm missing in there
gritty widget
all sets are open vs. some sets are open
sturdy basalt
ohk. Thanks
gritty widget
a topology on a set is just a collection of its subsets containing the empty set, the whole set, and which is closed under arbitrary union and finite intersection
so long as you have any subset of the power set satisfying those properties, you have a topology
the way the discrete topology is defined is that you take literally every possible subset of that set
it should be obvious that this gives a topology
now, the definition of a topology allows for a lot more than just that
on the opposite extremal end, you can just take the collection {empty set, whole set}
the defining property of the discrete topology is that all the sets are open
it would probably suck if the only topology on a set were the discrete one
sturdy basalt
thanks for clearing it up
ivory dragon
i wonder if the use of the term "open" when referring to elements of a topology presents an intuition barrier for students
i never really struggled with it but i could see it being an issue
since it marries students to the idea of openness in R^n
which is an important case, sure
and indeed the motivating case
but still
wind portal
It did in my topology class.
People can't thinking of it as balls for one of the assignments and had a hard time deviating from that to the topological definition.
ivory dragon
i wonder if we had a special term instead of "open/closed"
like idk
"topic and cotopic" or something
yeah im misusing the "co" prefix but
fuck it
gritty widget
i took topology before taking mvc/"analysis in R^n"
so i actually saw "an open set is an element of a topology" before "an open set (in R^n / metric space) is a union of balls"
there wasn't really an intuition issue, but i can see how someone taking the other way around (what i presume is the more traditional route through these things) would be confused
ivory dragon
i mean the idea in my eyes would be to prove that the open sets in R^n (defined as sets-that-dont-contain-boundary or whatever, theres a billion equivalent ones) form a topology rather than the other way around
ie constructing the standard topology on R^n via applying open set axioms to open intervals
¯_(ツ)_/¯
floral gust
yeah im misusing the "co" prefix but
co-vfefe
sturdy basalt
i wonder if the use of the term "open" when referring to elements of a topology presents an intuition barrier for students
@ivory dragon yes
I have had a few issues with it over time
also there are some intutive issues at the start about spaces that you'd think as closed but also works as open. Like when you intersect a Open set from a space with a closed subset to get an Induced topology
floral gust
I guess open works very nicely when you consider metric topologies but it becomes a problem when you try to consider abstract topological spaces.
marsh forge
I feel like its fine
deft creek
Anyone want to help me with this question?
rocky wave
$xy = K$ for $K \in (1,2)$ can be sketched by Realizing this is $y = K/x$
gentle ospreyBOT
phao:
rocky wave
It's a family of hyperbolas, it's the union of the graphs of tehse functions for $K \in (1,2)$. It's open because m is continuous and (1,2) is open in R
gentle ospreyBOT
phao:
rocky wave
Hmm.. Not sure what you mean.
Consider y = k/x for all k in (1,2). YOu know how to plot each of these. It'll be a "band like" region.
deft creek
what does R^{2} -> R mean
rocky wave
m: R^2 -> R means a function from R² to R
it takes pair of real numbers to a single real number, and m does that by multiplying the two given numbers.
deft creek
oh i think im getting the idea
rocky wave
You can be more formal about this sketching.
deft creek
I would now have to sketch the preimage of that interval
rocky wave
You can show that the preimage of (1,2) through m is the union of these two sets {(x,y); x > 0 and 1/x < y < 2/x} and {(x,y); x < 0 and 2/x < y < 1/x} and
So, essentially, you should draw the graphs of x mapsto 1/x and of x mapsto 2/x, and fill what is in between them without the border.
deft creek
got it, Well this really helped. I will now try it and ill show results right now
rocky wave
@deft creek I'm sorry. I don't mean to be rude, but if you had trouble understanding the notation R² -> R, maybe you shouldn't be studying topology right now.
Have you gone through standard introductory analysis course?
timber estuary
uum
does my answer for (d) look okay ? heres the definition of subbasis S:
and heres my proof
fading vale
I'm not totally sure if im interpreting this correctly
gentle ospreyBOT
Moth:
marsh forge
So subgroups of pi1A correspond to covers of A
so the kernel of that map gives us some cover
you want to show that this is exactly the cover Atilde
Yes I think what you wrote is sufficient
fading vale
okay
so ig we have to show that if f is in ker(i*) then f lifts to A tilde
i guess
actually wait hmmmm 
fading vale
omg im so fucking dumb
its not even hard i just misread and didnt realize that X tilde was simply connected
floral gust
Can someone assist in how can I give a smooth structure on R2 using polar coordinates?
Some solutions that I saw, in attempting to give a diffeomorphism from S1 x R to R^2-{0}, just say that "using polar coordinates construct a map (r, theta) to e^{i theta}, ln r) " but I dont see what do they explicitly mean by "using polar coordinates"
Are we endowing R^2-{0} with some new smooth structure different from the standard one?
signal venture
let $E : y^2 = x^3 + 1$ be an elliptic curve over some field $F$ and let $R \in E$ with $|R| = 2^n$ for some $n > 1$. let $G = \langle R \rangle$ be the group generated by $R$. every nonzero point $Q$ in $G$ can be written in affine coordinates as $Q = (x_Q, y_Q)$. can the sum $\sum \limits_{Q \in G \setminus \lbrace 0 \rbrace} x_Q \in F$ be evaluated into any nice expression? is it ever 0?
gentle ospreyBOT
Auvera:
signal venture
i know that Q and -Q have the same x value, so you can simplify the sum slightly that way. but beyond that i'm stuck
given some x_Q you could find the point on E with x coordinate being -x_Q, but i don't know if that point is actually in G
also i'm looking at positive characteristic, forgot to mention