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2:00 PM
sorry didnt have that written correctly
 
@LeakyNun so the points are (-3,-3,-2)
and it looks like of a plane.
 
@KasmirKhaan so what is your question?
@Dodsy but a point cannot define a plane
 
No question leaky =P just wanted to see if I got that right
f :G-->G' had a kernel the subgroup H
which is normal
but then it H collapsed to 1 element
in G/H
basicly what we did was , taking the image of all fibers
and made a group out of those
 
That doesnt sound quite right to me but i only learned it yesterday
 
lets work thu this one piece by piece
 
2:03 PM
@LeakyNun "Provide a general-form equation of any plane which includes both the origin and the point (-3,-3,-2)."
 
oh ok
 
we have a surjective group homomorphism f : G--->G'
the kernel is the subgroup H
 
@KasmirKhaan its the collaspe to one elment im no following the way you have said it
 
is general form something like ax+by+cz+d=0? @Dodsy
 
Right.
 
2:04 PM
so d=0 by substituting the origin
 
if you look at the G-->G'
the kernel is a subgroup of G
with many elements in it
H= {e,a,b,......}
now consider G/H --->G'
the kernel of that is not one element
that is H
 
@Dodsy and by substituting (-3,-3,-2) you get -3a-3b-2c=0 so c=(-3a-3b)/-2, which gives you ax+by+((-3a-3b)/2)z=0
 
Hi @BalarkaSen what happened to your herculean (or some other guy ?) profile picture ? :P
 
because our elements now are cosets not elements in G
in other words, imagine we split a group G
 
@LeakyNun Can we continue the discussion regarding fields ?
 
2:05 PM
into subgroup
 
@AlexKChen sure
 
or to see it better , let G be a set
split it into disjoint sets
A,B,C,D say
 
The order of a complex number looks 36 (not 49 lol) modulo 7.
 
@AlexKChen not really
 
in G, the the elements of the set A was the kernel
 
2:07 PM
@AlexKChen which complex number?
 
@LeakyNun I'm 100% sure if my python code is not botched up.
 
@AlexKChen show me
 
now when we consider the set of cosets G/H
 
Okay, lemme run the code again.
 
@LeakyNun ah, thank you
 
2:07 PM
what it does , put the elements into sets that split G
 
@Dodsy do you know how to describe a plane using linear algebra?
 
shrinking it in other words
 
this is for lin alg.
 
seems you have the right idea?
each equivlance class of G is mapped to an element of G'
 
yes exactly !
instead of working with each element in G seperat
we create G/H the set of equivalence classes
and surprisinly we can have a group structure on those sets, ie cosets
now the subgroup H is just 1 element in G/H
where in the original map from G--> H , it was consisting of many elements
 
2:12 PM
brb everyone
 
@Faust the idea is , when we have an epimorphism we can collapse it to an isomorphism by goin to G/H
 
u mean homomorphism?
 
i mean surjective homomorphism
if f:G-->G' with full image
ie surjective homomorphism
 
the theorem doesnt actually require that
 
i did not mean it that way
full image was wrong of me to sa
 
2:15 PM
no i mean the first isomorphism theorem
it doesnt require
surjective
it trivially implies it by construction though
 
how else you gonna keep the structure?
if not all elements are hit ?
 
you creat a new G" where all elements are hit
 
the map from G-->G' , is kinda the same as G/H -->G'
 
we went over both of them
 
exept the latter is an isomorphsm
all right maybe you split that into two cases
anyway
how is ur homework ?
 
2:17 PM
uh
 
btw do you get graded for those?
 
i have done some of my abstract algerbra hw
i do
worth 40% of final grade
 
wow
nice
we get at max 3 pt
 
yeah and there pretty easy
 
on the exam if we do them all
and the exam is on 30 points
like you could do homework and solve 2 qustions and u pass
in ur case =P
 
2:19 PM
yeah midterm is 20%
final is 40%
 
@LeakyNun Hey the order is divisors of 48 :)
 
so can pass the class before the final
 
very nice ._.
 
(1, 1) 24
(1, 2) 48
(1, 3) 48
(1, 4) 48
(1, 5) 48
(1, 6) 24
(2, 1) 48
(2, 2) 8
(2, 3) 16
(2, 4) 16
(2, 5) 8
(2, 6) 48
(3, 1) 48
(3, 2) 16
(3, 3) 24
(3, 4) 24
(3, 5) 16
(3, 6) 48
(4, 1) 48
(4, 2) 16
(4, 3) 24
(4, 4) 24
(4, 5) 16
(4, 6) 48
(5, 1) 48
(5, 2) 8
(5, 3) 16
(5, 4) 16
(5, 5) 8
(5, 6) 48
(6, 1) 24
(6, 2) 48
(6, 3) 48
(6, 4) 48
(6, 5) 48
(6, 6) 24
 
we have only 1 exam
3 points from hw, and 30 on exam
 
2:20 PM
Very nice.
 
but those 3 points dont count to 100 %
 
@LeakyNun Does there exiss similar stuff where I get the answer to be $7^3 - 1$ ?
 
pretty test heavy
 
if you have 27 in the exam you get only 1 point out of the 3
but if you get 12 /30
then the 3 points count as full and u pass
anyway faust ill keep studying and ill come back with more awesome ideas shortly :D
good luck @Faust
 
i have one 4th year class that is 55% hw 15% per take home midterm and theres 3 of those
ok gl
 
2:26 PM
@AlexKChen right
@AlexKChen sure
 
@LeakyNun can you find the normal vector from this?
 
@Dodsy there are many normal vectors
 
@LeakyNun So well this looks amazing ! In particular, can I make that $p^k -1$ for any prime $p$ and integer $k$ ?
 
how to find
 
but a characterisation is that they are perpendicular to ((-3,-3,-2) - (0,0,0))
@AlexKChen yes
there exists a field for any p^k
and all fields of order p^k are isomorphic
and there is no field otherwise
oops
corrected
 
2:30 PM
okay I'll think on it, thanks leaky.
 
@LeakyNun What does "isomorphic" means in elementary terms ? (Eg give example when $k = 1$, the normal integers modulo 7)
 
@AlexKChen it means "have the same structure" informally
and more concretely, yet still informally, the fields are the same except for the names of the elements
e.g. you can call it "1+3j" instead of "1+3i"
you can even call it "asdfghjkl"
 
Okay, so that mean if number that $p^k$ elements A,B,C, ... then A*A = B (mod 7) for all such structures ?
 
@AlexKChen exactly
 
Is that what finite fields actually mean (generalizing to arbitrary $k$ and prime $p$ ?)
 
2:33 PM
@AlexKChen is what?
 
Well I mean is finding such structure where order divides $p^k-1$ is all about finite fields ?
 
you can say so
where addition, subtraction, multiplication, and division are unique and well-defined
(except that you can't divide by zero)
that's an informal description of field: something in which you can do +-x/
 
division ???
 
1/2 = 4, because 4x2=1
 
Oh you mean inverses
 
2:38 PM
right
 
Ya I'm stupid.
2
 
2:51 PM
The numbers are completely werid if you replace $p$ by a composite.
$40$ gives order of $[0,2,4]$
$35$ gives $[0, 2, 4, 6, 8, 12, 16, 24, 48]$
 
@AlexKChen I wonder how you are doing it
output?
 
Okay I am giving the code (But works perfectly fine for primes)
def mul(a,b):
	return (a[0]*b[0]-a[1]*b[1], a[0]*b[1]+b[0]*a[1])

def mod(z, p):
	return (z[0] % p, z[1] % p)

def order (z, p, k):
	a = z
	if a[0] == 1:
		if a[1] == 0:
			return 1
	for i in range(2,k):
		a = mul(a, z)
		a = mod(a, p)
		if a[0] == 1:
			if a[1] == 0:
				return i
	else:
		return 0

def pr2(a):
	d = []
	for i in range(1,a):
		for j in range(1,a):
			z = mod((i,j), a)
			y = order(z, a, a**2)
			if y not in d:
				d.append(y)
	d.sort()
	return d
Now typing pr2(a) will give you the possible orders modulo `a``
 
should have totally used class :P
@AlexKChen I have to point out, you can't get the field with order 25 this way
compare [8, 16, 24, 48] for 7 but [0, 4] for 5
 
The notation for relational algebra is so confusing
 
@JennaSloan how so?
 
3:06 PM
It uses Greek letters as operators
 
@LeakyNun In general, I learnt about class long ago, but never use them (because I think they're unnecessary), and as result my code are not understandable one hour later. How will using class make the code elegant ?
Wait wait what the hell. WHY 5 IS SHOWING [0,4] ?!?!
Hey you told me eventually they'll reach 1+0i modulo prime for nonzero complex number, but modulo 5 some numbers cycles (so are invertibe, like zero) without reaching 1+0i (for example 4+2i)
 
 
2 hours later…
4:45 PM
@EricSilva: The geometric analysis lectures are now on YouTube. Here's the link: youtube.com/playlist?list=PL4ji5DBtd6LyAR4OrsJJoAxp5soTgkuqI (I haven't looked yet)
 
5:19 PM
Sweet! Thanks @Ted
 
$hi$
 
5:36 PM
(removed)
 
Hi @Eric
 
sniped
 
@AlexKChen the point is that "i" represents the solution of "x^2+1=0", which is already there in 5^2, i.e. 2^2+1=0, but is not there in 7^2. You can only establish "i" as a new dimension only when what it represents (i.e. x^2+1) doesn't have a solution yet
 
hi @Balarka @Jasper @Leaky
 
hi
 
5:42 PM
Hi @Ted
 
Whats a good notation for the directed or un- line segment from complex number $x$ to complex number $y$?
my textbook uses $[x,y]$ which I'm unhappy with because it looks like an interval and $x,y$ might be wholly real
though i guess an interval is a line segment of sorts
but could $[1,0]$ be the line segment from $1$ to $0$?
 
@GFauxPas sure
@GFauxPas $x + [0,1](y-x)$ :P
 
but $y$ shouldn't move
 
I occasionally do write $[\vec a,\vec b]$ for the line segment going from $\vec a$ to $\vec b$. The vector symbols make the context clear.
 
okay, this should be fun: let $f: z \mapsto z^3$ on $\mathbb C$, prove there does not exist a $c \in [1,i]$ s.t. $\dfrac{f(i)-f(1)}{i-1}= f'(c)$
 
5:53 PM
Hi @Ted
 
so $LHS = i$
$RHS = 3c^2$
 
Yo @Alessandro
 
then, uh
 
@TedShifrin I'm having trouble reading through the algebraic functions section.
 
hint please
 
5:57 PM
Not on principle, but it's hard to read for some reason
 
Solve $3c^2=i$, @GFauxPas. Is either solution on the given line segment?
I never taught that section, @Balarka. I was limited in time and wanted to do a bit more with concrete stuff that's not in the book, so I started in the second main section.
 
Ahh
 
If you have a specific question, probably @Danu (who took this class from Forster) or I can address it.
 
Ok, thanks, I'll push through and see if I get stuck
 
not sure how to tell what the solutions are to that without using polar form, and that wasnt taught in class yet so i dont think i can use it
oh, but I can square both sides
 
5:59 PM
Well, then parametrize the line segment by $z(t)$ and look at $z(t)^2 = i/3$.
 
what if i take the modulus of both sides and get $1/\sqrt{3} = |c|$
 
It seems like the main upshot of the section is if $f : Y \to X$ is a meromorphic map, then the monomorphism $f^* : \mathscr{M}(X) \to \mathscr{M}(Y)$ is a field extension and how $\text{Deck}(Y/X)$ is isomorphic to $\text{Gal}(\mathscr{M}(Y)/\mathscr{M}(X))$ under certain regularity assumptions perhaps
 
is that sufficient Ted?
 
Not without saying more ?
 
the line segment is equivalent to $x + y = 1$, and the modulus tells us $3x^2 + 3y^2 =1$
and, uh
 
6:05 PM
Funny how I think geometrically and distance and or length :) But go on.
 
so $3x^2 + 3y^2 -1 = 3x^2 + 3(1-x)^2 - 1 = 3x^2 + 3 - 6x + 3x^2 -1 $
which has a negative discriminant
 
OK, now what's the immediate way to see what's going on?
 
yeah, that did seem like more work than necessary
 
Draw a picture of the line. What point is closest to the origin?
 
$x = y = 1$
err no
 
6:13 PM
Um, not quite.
 
1/2
 
OK, and what is the distance to that point?
 
$\sqrt{2}$
 
Um, no.
 
ugh im not thinking correctly
$1/\sqrt{2}$
 
6:15 PM
Right. Done.
 
much easier
thanks Ted
 
Yuppers. :)
 
I knew the closest $x + y = 1$ gets to $0$ is at its halfway point, and I know i can show it with partials, but is there a way to show its closest to $0$ without Calculus?
 
Of course. Remember basic vector geometry.
Use dot products.
Or, use Pythagoras to know you need a point where the vector from the origin is orthogonal to the line.
 
thats more or less equivalent :)
Ted my friend was having trouble with diff. geom. and I sent him a link to your book and videos
he was psyched
 
6:22 PM
Well, yes, the normal vector to the line is $(1,1)$. You need that either way. :)
Well, the videos aren't particularly diff geo. Dunno what your friend is studying precisely.
 
me either :P
okay, problem $5.9;(i)$
construct a function holomorphic except at $\pm 1$
well, holomorphic implies continuous, right?
 
Did they not want a continuous function everywhere?
 
not yet, we just learned about derivatives so i assume theyre starting easy
take $\dfrac 1 {1-z^2}$, easy
$(ii):$ an $f$ holomorphic everywhere such that $1/f$ fails to be holomorphic at precisely six points
${1-z^6}$
okay, $iii$ is the interesting one
$f = u + iv$, neither $u$ nor $v$ is constant, holomorphic nowhere
so something that fails the CRE's everywhere
 
@TedShifrin o/ hi
 
Howdy @Danu
What is the verdict on the magnum opus?
 
6:30 PM
$\sin x + y - \sin y$
 
Defending on Thursday @Ted!
 
oh wait, not enough
 
Preparing the preseentation now, though I'm a bit bored/uninspired
 
Oh, I met a grad student at UCSD who's working on combinatorics who is from the Netherlands and went to uni there (and got his masters there), but I missed which uni.
 
I guess I'm mostly just going to reproduce my PhD applications talk
adding something here 'n' there
 
6:31 PM
@Danu Take inspiration from xkcd. The best defense is a good offense.
 
What's the audience for the defense?
 
$e^x + i(e^y + y)$
 
@TedShifrin Supervisor, second reader and anyone from the uni who wants to show up. Probably just my friends.
 
Hey everyone!
 
that works, right Ted?
because $u_x \ne v_y$
 
6:32 PM
Hi @Daminark
 
How goes it?
 
So it's supposed to be intelligible basically to your adviser and the second reader (the friends don't count), so that means you might want a more advanced talk than you did earlier.
Wayyyyy too complicated, @GFauxPas. Why did you put the $+iy$ in?
(Even so, wayyyyyy too complicated.)
 
@Danu What did you write about? And what sort of thing is this?
 
$x + iy$ is enough:?
 
Um, close but no cigar.
 
6:35 PM
$x - iy$ rather
 
@TobiasKildetoft Complex/differential geometry of certain homogeneous spaces
 
There you go.
 
This is a master's thesis
 
@Danu Neat
 
im just not used to thinking of functions like that as not diffable, they "feel nice"
 
6:35 PM
BTW, @Danu, please send me a .pdf when you're all done. :)
 
I am all done, I'll send it now
 
@GFauxPas: They are nice, but they involve $\bar z$, so that makes them naughty.
Well, they might ask for a change or something, @Danu.
 
No, that's not how it works here
 
LOL, oh.
 
You hand in the final version before the defense
Only then does the supervisor really read it
 
6:36 PM
So he already asked you to make changes?
 
Yeah, tiny tiny ones
he seemed content about everything mathematical
 
The system makes no sense to me, but great.
 
Oh well... I think the point is that this minimizes supervisor effort. 'Cause he doesn't have to read it at all now.
(and he may very well not :D)
 
I'm actually not sure Chern ever read my Ph.D. I know a few people who did, but I don't think he did.
 
Things never really change ^^
 
6:38 PM
But I had presented the main stuff to him.
And he trusted my math/writing skills.
 
Yeah. I've talked enough to my supervisor about all this for him to know that/what I know and I think he's fine with that.
 
Anyhow, @Danu, I'd make it a bit more sophisticated for them than for your visiting talk.
But your choice, obviously.
 
Why? It's still a 30-minute talk...
 
let $f$ be holomorphic on $D(0,1)$. Prove $g(z) = f(z^*)^*$ is holomorphic on $D(0,1)$
 
Because you can assume they know the basic set-up and no need to bore them with that.
You have only one way to do this so far, @GFauxPas. Just do your work and stop posting your homework here for us.
 
6:40 PM
@TedShifrin I'm not sure... I think that many presentations are more about the basic stuff... You think I should just skip over it?
I'm not so sure to what extent the presentation is really for the supervisor... Idk really :p
 
oh, I was doing it on my own, but the last one and this one just seemed interesting enough to post here oto. Sorry
 
@GFauxPas: Presumably you don't post every "interesting" homework problem ...
 
sorry
 
@Danu: Obviously I do not know your system. In the US it's meant to be a bit more for the general audience, but in the end one makes it technical and the committee can ask questions. (Plus our talks are usually at least 1 hour.) I always encourage people to do something explicit (like examples/counterexamples), too.
 
Also, I know for a fact that my second reader will not actually have looked at the thesis so if I skip past the basic definitions I'll have lost him in the first five minutes too... @Ted
 
6:43 PM
Oh, so if the second-reader isn't an expert, yeah, you have to do that.
 
@TedShifrin Right... It's a bit weird here... Usually Kotschick lets his students do a 90 minute talk where they can give full proofs of the main results.
However, I'm in a special study program where it has to be a 30-minute talk.
 
Ohhhhh ...
Dumb.
 
He prefers it longer, so I think I'll sorta push it to like 40
But I definitely don't have much time for full proofs and stuff
 
They can always ask for more details in questions after.
 
Right
 
6:45 PM
I think at this point you know stuff pretty well, so no need to fret too much.
 
Let $f:\Bbb R \to \Bbb R$ be a real function such that $f \circ f = -\operatorname{id}$. Show that $f$ is not continuous.
 
In that vein, there is one claim that I'm not 100% sure about: When I have two almost complex structures on $M$ and they are homotopic, the total spaces of the (complex) tangent bundle are supposed to be diffeomorphic. How do I prove that? @TedShifrin
 
I am confused, how does a different choice of the almost complex structure change the topology of the complex tangent bundle?
 
I hear choice
 
This was exactly what I got confused about hahaha
 

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