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12:20 AM
I think I'm missing something that should make this question easy
I'm trying to compute the $\|T\|$ where $Tf(x)=k(x)f(x)$ under the $L^2$ norm on the space $C[a,b]$
@Thorgott There are various notions of the dimension of a topological space which don't require too much in the way of setup which distinguish R^n from R^{infinite set}. You can also compute using local homology that they are not locally homeomorphic.
Note that if V is a vector space, (V + R) - 0 has the homotopy type of the suspension of V - 0. Since an infinite dimensional vector space has V + R ~ V, it follows that if V is an infinite dimensional vector space the local homology groups are zero.
12:38 AM
which notion of dimension do you have in mind? covering?
I remembered liking a treatment of inductive dimension too
How does infinite inductive dimension work?
For every $n$, there is a point $x$ and an open $U$ containing $x$ such that for all open $V$ containing $x$ and whose closure is contained in $U$, the boundary has inductive dimension $n$?
1:11 AM
For each of the following $R$-modules $M$ determine whether they are free. If free find a basis and rank: $\Bbb{Z}_{10}$-module $(2,1,2) \Bbb{Z}_{10} + (4,1,0) \Bbb{Z}_{10} + (0,0,1) \Bbb{Z}_{10}$...Question: Should I interpret those tuples as living in $\Bbb{Z}_{10}^3$?
@user193319 yep
1 hour later…
2:32 AM
Learning generating functions but occasionally come up with two expressions for the same sequence which aren't equivalent. Would appreciate help detangling it!
Eg. the sequence $1,1,1,2,3,4,5\dots$ seems like it could be either $\frac{x^3}{1-x}+\frac{1}{1-x}$ or $1+x+\frac{x^2}{1-x}$.
(Learning on my own from Concrete Math)
2:57 AM
a sequence isn't determined by finitely many of its terms, so I'm not sure what you're looking for
3:25 AM
Maybe I've been doing this stuff for too long tonight, but let's consider the following... let $X_1, \dots, X_n$ be numbers such that $0 < X_i \leq 1$, and assume $X
= \sum_{i=1}^{n}X_i$ satisfies $0 < X \leq 1$ as well. How do I know that $X_j \leq \dfrac{1}{n}$ for some $j$?
try contradiction
@Thorgott Thanks, that was simpler than I thought. Is this a standard name for this result?
not that I know of
@MikeMiller is there a name for what you get by identifying the hemispheres of a sphere, but with a twist? Like say you had an "L" on the north pole, and you drag it to the south pole, then take its mirror image. There is a glide reflection that preserves this pattern, so I want to know if its orbifold (quotient under the action of this single symemtry) has a name. It looks like a cross-cap sort of thing, but "flattened" out?
Sounds good
3:37 AM
we all awake at 5:30 again
3:53 AM
@Clarinetist it's a variant of the pigeonhole principle, I suppose.
Why is $C_{p}^{\infty}(V)\neq C_{p}^{\infty}(M)$
4:17 AM
@EdwardEvans ye we are
the dream
But only if the "the dream" includes being flung out of your dreams by an early alarm
can't have dreams if you don't sleep big brain
I seriously doubt that I'll be able to stuey o
Study on the flight
Bigthink time
where you off to?
4:21 AM
Germany, back from a trip to malta. Funny thing is that the infection rate in my area is higher than in malta lol,but at least I'll get a free test
In Berlin?
Cologne atm. I wouldve moved to berlin but that didnt happen due to a pandemic
Ah sad
Mannheim is a Risikogebiet now too
Nahh izza okey
More time with le gf
ajo nicht schlecht
4:22 AM
Haha yeah basically half of germany is
Why you up? Classes havent started yet, boie
I'm reading about Langlands and can't stop
or rather, I think I'm going to be completely überfordert in this seminar and I'm trying to get as far ahead as I can before the semester starts so I have some buffer space in case I don't understand shit
Ahhh yes the "i'll try to get some buffer space" tactic where you get hardly any buffer space but in exchange you have a semester that starts way earlier
yeah man
works every time
i've been working on the material for the first talk for like 2 weeks
4:27 AM
I do that almost every semester ans never feel like I've bought much time but as I do it every time, I dont know how hard the semester would be if I didnt
that's fair rofl
One of the courses I'm doing is just a special case of the seminar I'm doing
overlap I guess
Getting into the mode of thinking is useful
or I just get a more general perspective on the shit I'll learn in the lecture course and not actually help myself
But you save some time researching resources
yeah true, I've just been watching a bunch of lectures of local Langlands for GL_n even though we're only doing GL_2
4:32 AM
@EdwardEvans i'm pretty happy that the cpurses in my coming semester overlap nore than in the past
yeah that's probably a good thing
I underestimated how long it takes to switch between courses in the last semester
With regards to studying
As in, to switch to a different set of tools/area of mathematics?
yeah I get that, I'm fairly confident in algebraic number theory but venturing out of number theory scares me
might drink a bunch of red bull and write out a huge map of dependencies for the theorems in the seminar instead of actually doing any mathematics
4:41 AM
Hahahaha knowing the relations helps tho
Gotta shower and get to the airport now, bye bye again
5:05 AM
@Thorgott True, that's why there was an ellipsis :p It's the counting numbers offset by two, with the first two elements set to 1.
@EdwardEvans what does
47 mins ago, by user2103480
Oh shit here we go again_german.jpg
just said some words in German
I see. Thnx
5:29 AM
"If $(\pi, V)$ is a smooth irreducible rep of $G$ then $\dim_{\Bbb C} V$ is countable" that sounds weird to me
that's not a set
i assume they mean $V$ has a countable basis
just weirdly worded
5:58 AM
what even is linear independence in an infinite dimensional space
6:36 AM
@EdwardEvans pros of malta airport toilets: they play paul hardcastle on loop
Cons of malta airport loos: they keep flushing on loop while you sit on it
The fundamental group of malta airport loos is generated by paul hardcastle and flushing
7:00 AM
@user2103480 what's the relation between paul hardcastle and flushing tho
7:17 AM
@EdwardEvans well it's a cardinality right
the only countable cardinality is $\aleph_0$ but presumably they wanted to avoid that
@EdwardEvans it says every finite subset is linearly independent
oh, apparently finite is also countable
anyway I'm more interested in the proof
the proof that smooth irreducible rep of G has countable dimension?
for it to work you need to assume that $G/K$ is countable for any compact open subgroup $K$, which holds for all the groups that are relevant in the book
and then $V$ is spanned by $\lbrace \pi(g)v : g \in G/K\rbrace$ lol
1 hour later…
8:47 AM
what's the easiest example of a $C^k$ function?
or what
nah obv not
what a mong
9:08 AM
9:35 AM
let's say I want to make a $C^k$ function that is not $C^{k+1}$, the idea is to integrate a function that's not $C^k$ but is $C^{k-1},...,C^1$, does anyone have any idea on how to do this?
you can start with k=0 and then just integrate it
and I suppose you know of a continuous function that is not differentiable
yeah i know how to do it for $C^1$ but not $C^2$ just $\int t^{1/3} dt$ for example, but how to do extend it for higher $k$?
9:56 AM
hello people
theorem 2.34
the problem is first he write very sloppy $V_q$ and $W_q$ are neighborhood of p and q but I think why book means is $V_q$ is neighborhood of $p$ and vice versa
now problem with $1/2d(p,q)$ so does it means this neighborhood V_q and W_q be less than this $0.5d(p,q)$?
I find it notation abuse if my first statement is true they use q in both V_q and W_q
Now first Why V is defined that way? What is purpose of intersection ?
and why V and W intersection is empty?
how does V being subset of K^c which means, not K is is open set?
I have tried drawing these thing but I can't grasp the theorem
how do visualise it?
10:22 AM
hi chat
10:49 AM
For each of the following $R$-modules $M$ determine whether they are free. If free find a basis and rank: $\Bbb{Z}_{10}$-module $(2,1,2) \Bbb{Z}_{10} + (4,1,0) \Bbb{Z}_{10} + (0,0,1) \Bbb{Z}_{10}$...Question: Since $\Bbb{Z}_{10}$ is a PID, and the module generated by those three vectors is a submodule of the free module $\Bbb{Z}_{10}^3$, that submodule should also be free, right?
@anakhro I don't understand the description. Can you give me a formula?
It sounds like this is supposed to be an involution, but every involution on the 2-sphere is equivalent to (x,y,z) mapsto (pm x, pm y, pm z) for some choices of plus or minus.
When they're all + the quotient is the identity. When 1 is - the quotient is a disc (the hemisphere). When two are - it's the suspension of RP^1, so it looks sort of like a football in my mind. And when all three are - you get RP^2.
Does the statement that all finite dimensional vector spaces are isomorphic hold for vector spaces defined over arbitrary fields?
I'm changing my name from "Balarka Sen" to "Gromov Hausdorff"
11:07 AM
@BalarkaSen tagging you will feel weird :p
@Charlie yeah
thanks :)
pick $V$ and $U$ as $\mathbb{F}$-vector spaces. If $\{v_i\}_{i\in I}$ and $\{u_i\}_{i \in I}$ are bases, then choose the linear map that gets $v_i \mapsto u_i$ and extend linearly. this is a bijective homomorphism, i.e., isomorphism
Ah I see, ty @LucasHenrique
11:20 AM
@BalarkaSen on MSE or legally?
The correct answer
2 days ago, by skillpatrol
I think it started with his Gromov
2 days ago, by Balarka Sen
imagine this Arnold complaining that Gromov was too sloppy a mathematician as a postdoc student of Rokhlin
Do $(4,1,0)$ and $(0,0,1)$ span $(2,1,2)$ in $\Bbb{Z}_{10}^3$, where the coefficients are coming from $\Bbb{Z}_{10}$?
I don't think they do, and here is my reasoning (but correct me if I am wrong): If there were $n_1,n_2 \in \Bbb{Z}_{10}$ such that $(2,1,2) = n_1 (4,1,0) + n_2 (0,0,1)$, then we'd have $n_1 = 1$, so $4n_1 = 4$ which is not equal to 2 in $\Bbb{Z}_{10}$.
Does that sound right?
11:40 AM
well ignored
I am trying to find a basis for the $\Bbb{Z}_{10}$-module $(2,1,2) \Bbb{Z}_{10} + (4,1,0) \Bbb{Z}_{10} + (0,0,1) \Bbb{Z}_{10}$. I know that $(4,1,0)$ and $(0,0,1)$ are linearly independent over $\Bbb{Z}_{10}$, but I don't think they span the module for the reason I gave above. I could use help finding a basis.
11:57 AM
@LeakyNun oh ok i got it, we just take $\int t^{1/3}dt$ and integrate it again, isn't it?
1 hour later…
1:09 PM
If $e\in\Gamma(X, O_X)$ is an idempotent section of a locally ringed space $(X, O_X)$, why is the set $\{x\in X, e_x=1\}$ open ?
or closed
@Astyx Does $e_x$ mean the residue?
$e_x$ is the germ at $x$
I know $e_x$ is an idempotent in a local ring, so it's either 0 or 1
So if I prove that it's open, then it follows that it's closed, and vice-versa
aha it follows because if you're 0 in the colimit then you're 0 in one of the components
A component would be an open set of $X$ right ?
1:23 PM
containing x
Yes ok
Why is that then ?
that's an algebraic fact
how do you construct colimit
or rather, do you know that two germs are equal iff the sections are equal on a small enough neighbourhood
Ok yeah
Is that a fact in any cocomplete category ?
don't think so
Ok that makes sense
1:26 PM
that's not a categorical statement
yeah, objects don't have elements
I meant to ask if that was true for a sheaf on any category, and not just on an algebraic one
godless people man
how can I get the backward slash to render?
1:31 PM
now the under-score doesn't render :-/
I'm trying to show that if $F$ is a free module, then $Hom_{R}(F,F)$ is also a free module. If $F$ is free, then it has a basis $\{m_i \mid i \in \}$. I was thinking that the functions $f_i : F \to F$ defined by $f_i(m_j) = m_j$ if $i = i$ and $f_i(m_j) = 0$ if $i \neq j$ would form a basis for $Hom_{R}(F,F)$, but I think this forms a basis only in the case that $F$ is a free module of finite rank.
How do I show that $Hom_{R}(F,F)$ is free?
yeah F needs to have finite rank
Wait...what? So $Hom_{R}(F,F)$ is generally not a free module?
yeah, shame
But Thorgott and some other person said it would be...
Hmm...I guess I need to rethink my proof of the fact that if $P$ is finitely generated projective module, then $Hom_{R}(P,P)$ is projective too.
I wanted to say that if $P$ is finitely generated projective, there exists a module $Q$ such that $P \oplus Q$ is free, then $Hom_{R}(P \oplus Q, P \oplus Q)$ is free, and then use this to argue that $Hom_{R}(P,P)$ is projective.
1:39 PM
you can make P (+) Q free of finite rank
Oh, really?
But what if $Q$ is "huge"?
Are can $Q$ always be chosen "small" in this case so $P \oplus Q$ is free of finite rank?
you can make it not huge
P is finitely generated m'dude
Yes, but why couldn't $Q$ be infinitely generated, so that $P \oplus Q$ is a free module of infinite rank? What precludes this?
nothing precludes it
1:42 PM
you have an abstract definition of projective modules
think concretely
there are many $Q$s such that $P\oplus Q$ is free
we're saying there is some $Q$ such that $P\oplus Q$ is free of finite rank
So there are many choices of $Q$ for which $P \oplus Q$ is free, but you are saying that we can choose $Q$ such that $P \oplus Q$ has finite rank?
1:42 PM
Oh, okay...hmm...
for example if $Q$ is such that $P\oplus Q$ is free of finite rank, then $P\oplus(Q\oplus R^{(\mathbb{N})})$ is free of infinite rank
very boring example, but the point is that making it larger is easy
If $\{p_1,...,p_n\}$ generate $P$, I guess the most naive thing would be to take $Q$ to be the free module generated by the $p_i$ are something like that...But would $P \oplus Q$ be free...hmmm....
you dont want to take P + this module. you have a map from your free module to P
No, maybe I need to look at a map from it to $P$
1:47 PM
Okay. Let me think about this a bit. Thanks guys!
2:24 PM
Okay. I think I have a proof (sorry it took so long...had to attend to other things). Let $P$ be generated by $\{p_1,...,p_n\}$, consider $R^n$ with standard basis $\{e_1,...,e_n\}$, and define $g : R^n \to P$ as the $R$-linear extension of $e_i \mapsto p_i$. Then the sequence $0 \rightarrow \ker g \rightarrow R^n \rightarrow P \rightarrow 0$, where the first map is inclusion and the second map is $g$, is exact and therefore $R^n \cong P \oplus \ker g$.
2:35 PM
the last conclusion holds because the sequence is split, not only exact
but yes, that's correct
The vector space of $2\times 2$ traceless Hermitian matrices is isomorphic to $\Bbb R^3$ right? This relates to proving the double-cover relationship between $SU(2)$ and $SO(3)$
you mean $\mathbb{C}^3$?
The book I'm reading states $V\cong\Bbb R^3$ where we earlier defined $V$ as the traceless Hermitian 2x2 matrices
It is $\Bbb R^3$
And we define the inner product on $V$ as $\frac{1}{2}\text{Tr}(X_1X_2)\forall X_1,X_2\in V$ which, if the above is true makes them isomorphic as inner product spaces
ty @Astyx
2:48 PM
It is generated as a $\Bbb R$-vector space by $\begin{pmatrix}1&0\\0&-1\end{pmatrix}$, $\begin{pmatrix}0&i\\-i&0\end{pmatrix}$ and $\begin{pmatrix}0&1\\1&0\end{pmatrix}$
Does a regular topological homomorphism preserve connectedness?
ah, I missed the Hermitian part
the continuous image of a connected space is connected @Charlie
Nice, thank you :)
that can be seen as a very broad generalization of the IVT
3:09 PM
A map that takes every element of its domain to a single element in the codomain is trivially a continuous map right?
I keep on doubting my use of Cauchy-Schwarz
Is this an appropriate use of Cauchy-Schwarz or am I missing something? $$\sum\limits_{n=1}^\infty\big\vert\frac{x_n}{2^n}\big\vert^2\leq \sum\limits_{n=1}^\infty\big\vert\frac{1}{2^n}\big\vert^2\sum\limits_{n=1}^\infty\big\vert x_n\big\vert^2$$
every map takes every element in the domain to a single element in the codomain
do you mean it takes all points to the same element, i.e. that the map is constant?
Ah yeah that's what I meant lol
Just a trivial all-to-1 map
what does any preimage of this map (so in particular, the preimage of an open set) look like
it's the entire domain
which must by definition be open?
so it must be continuous
3:16 PM
it's not necessarily the entire domain
but it's either that or..
um I'm not sure
the empty set?
ah nice, ok thanks :P
which subsets of the codomain have the entire domain as preimage and which the empt set?
the non-zero elements of the codomain and the empty set respectively?
3:26 PM
consider the map $f\colon\mathbb{R}\rightarrow\mathbb{R}$ that sends everything to $0$. What is $f^{-1}((2,3))$?
the empty set
So if $f:A\rightarrow B$ is a constant map then the subsets of $B$ that have the entire domain as a preimage are $f(A)$ and those that have the empty set are $B-f(A)$
almost, but those are just two possible sets
3:48 PM
Okay. I am trying to understand the hom functor. For example, I want to understand one of the form $Hom_{R}(_,R)$ and how it "acts" on sequences. If I have a morphism $f : A \to B$, how does this induce a map from $Hom_{R}(A,R) \to Hom_{R}(B,R)$?
it doesn't, the hom-functor is contravariant in the first variable, meaning the map will go in the other direction
compose with $f$ on the appropriate side
ahh, woops
so a morphism $f\colon A\rightarrow B$ induces a map $\operatorname{Hom}(B,R)\rightarrow\operatorname{Hom}(A,R)$
you can guess what that map looks like
Ah, you're right...In the statement of the problem I'm working on, I didn't even realize that the sequence "flipped" after applying $Hom_{R}(,R)$.
on the other hand, $\operatorname{Hom}$ is covariant in the second variable, a morphism $f\colon A\rightarrow B$ induces a map $\operatorname{Hom}(R,A)\rightarrow\operatorname{Hom}(R,B)$, you can also guess what that one looks like
3:52 PM
Ah, I see. Is there a standard notation used to denote the induced maps on the Hom level?
for the first example
it's sometimes also denoted $\circ f$ or $-\circ f$ for obvious reasons, but I believe that notation can get misleading
Ah, I see. Thanks. Okay, so I am trying to show that $Hom(,\Bbb{Z})$ is generally not exact. Is the following a counterexample? Consider the sequence $0 \rightarrow \Bbb{Z} \rightarrow \Bbb{Q} \rightarrow \Bbb{Q}/\Bbb{Z} \rightarrow 0$, where the first map is the inclusion map, and the second is the canonical projection. Applying the Hom functor we get the sequence $0 \rightarrow Hom(\Bbb{Q}/\Bbb{Z}) \rightarrow Hom( \Bbb{Q},\Bbb{Z}) \rightarrow Hom(\Bbb{Z},\Bbb{Z})$.
But this cannot be an exact sequence because the middle term is $0$ and the last term is $\Bbb{Z}$, and there cannot be any surjection from $\{0\} \to \Bbb{Z}$. Does this sound right?
Cool. Thanks!
4:11 PM
@user193319 Simpler, try $0 \to \Bbb Z \to \Bbb Z \to \Bbb Z/n\Bbb Z \to 0$
does that work?
yes, you get $0\rightarrow0\rightarrow\mathbb{Z}\rightarrow\mathbb{Z}\rightarrow0$ but the non-trivial map is multiplication by $n$
ah nvm I forgot to flip it
and was very confused
Leaky Nun is finally becoming a geometer
only geometers get confused with contravariance
4:28 PM
the short gander I took at Kolar-Michor-Slovak suggested otherwise
$M\mapsto C(M,\mathbb{R})$ is a fully faithful contravariant embedding of the category of differentiable manifolds into the category of commutative $\mathbb{R}$-algebras
Michor is a mutant geometer
@Thorgott reference?
the reference is Kolar-Michor-Slovak
somewhere in chapter 35
I think 35.10.
this was an exercise in Milnor-Stasheff or something iirc
If g is continuous function on a compact support (g: R^n\to R) then how can I show $\int |g(\delta x)-g(x)|<\epsilon$ for given $\epsilon>0$?
OH $\delta>0$
as $\delta\to 0$
nono $\delta\to 1$
4:44 PM
so you want to find a neighborhood of $1$ for $\delta$ in which this holds, given $\varepsilon$?
Yes basically.
why not use dominated convergence if you want to show $\lim_{\delta\rightarrow1}\int|g(\delta x)-g(x)|=0$
sounds like a massive point to do by hand
So you mean if we consider $g(\delta x)$ as a sequence of function that converges to $g(x)$, then as $g$ is continuous and have compact support, it's bounded so by DCT, $\int |g(\delta x)-g(x)|\to 0$ as $\delta\to 1$?
I'll leave the details to you, but yes
Ok thank you for your input
5:45 PM
@Rithaniel, looks appropriate
What is an example of a complex manifold $M$ for which $H^1(M, O^*)$ is nontrivial
$O^*$ = sheaf of nonzero holomorphic functions
6:00 PM
When we deform a loop in a topological space to define homotopy classes can the loops pass through themselves?
@Charlie sure
ok ty
That's part of why it's important that paths are functions $f:[0,1]\to X$ instead of their images
@Lelouch $\Bbb{CP}^1$
The loop can even be a single point
6:13 PM
@Astyx That would have been my first complaint if the answer was no :P
A loop homotopic to a point is called null-homotopic, and the homotopy class is the zero element of the homology group
Ah I haven't reached homologies yet
someone remind me why we require manifolds to be second-countable rather than just paracompact
Q: the equivalence between paracompactness and second countablity in a locally Euclidean and $T_2$ space

Alexsuppose $M$ is a locally Euclidean Hausdorff space, show that $M$ is second countable if and only if it is paracompact and has countably many components. This is Problem 2-15 p.59 (or 1-5 p.30 in the new version) in: Introduction to smooth manifolds by John M. Lee. In the hint he said if $M$ is...

@Astyx Fundamental group, not homology group.
6:21 PM
Isn't the fundamental group the first homology group ?
yeah, I know this
My knowledge of terminology for this is lacking
requiring paracompact is the same as requiring each connected component is second-countable, but there may not necessarily be countably many
@Astyx the first homology group is the abelianization of the fundamental group
why do we not want to consider "manifolds" with uncountably many connected components
6:22 PM
@Astyx I believe the first homology group is the abelianization of the fundamental group.
you got sniped
No. Fundamental group is generally non-abelian: first homology is the abelianization.
Whoops didn't see that.
Too slow typing on iPad.
@Thorgott because we don't want to equip R^2 with the horizontal lines topology and call it a manifold
6:23 PM
$H_1$ is the abelization of $\pi_1$
JK :-)
everyone is lining up to get sniped
sorry, nvm
@Lelouch: Huh?
yeah, but why not
@Lelouch what's wrong
6:24 PM
it's not like allowing uncountably many connected components makes classification any harder
@TedShifrin Sorry, I thought the $H^2(M, Z) = 0$ which is clearly not the case
then you'll start calling the projection onto the y-axis a fibration
What's an example of a manifold with non-abelian fundamental group ?
that's the example Vakil gave anyway
Indeed, the space of holo line bundles on $\Bbb P^1$ is classified by Chern class.
6:25 PM
@Astyx R^2 remove two points
@Astyx Take a torus with two hahndles
I mean, yeah, that doesn't sound too wrong
Isn't it true that the fundamental group of a topological group is always abelian?
6:26 PM
Can every abelian group be realized as the fundamental group of a topological group?
Oh ok I see
@Lelouch Oh, cool! Thanks!
I think you can start with $\pi_1(S^1) = \Bbb Z$ and do some Galois theory
but I'm not sure
Is everything settled
6:31 PM
What is an example of a complex manifold $M$ with $H^1(M, O)$ nontrivial ?
I see a bunch of deleted messages
@BalarkaSen Yeah, sorry about that
I thought $H^2(M,Z) = 0$ for some reason twice
I think $H^*(M, O) = 0$ always for Stein manifolds
@TedShifrin will tell me I am wrong
But most Riemann surfaces will not satisfy $H^1(M, O) = 0$
@BalarkaSen ok, if I assume that then the noncompact riemann surfaces are stein manifolds from wikipedia page, so $H^1(M,O)$ for that is zero. But is it zero for compact riemann surfaces as well ?
Just compute some
6:36 PM
How do people compute them ? I only know the exact sequence $0 \mapsto H^0(M,Z) \mapsto H^0(M, O) \mapsto H^0(M,O^*) \mapsto H^1(M,Z) \cdots$ and the fact that $H^n(M,Z)$ is same as $n$-th simplicial homology group
But I don't know any easy way to compute either of $H^q(M,O)$ or $H^q(M,O^*)$
OK, so I see $H^0(M,Z) = H^2(M,Z) = Z$, $H^0(M,O) = \mathbb{C}$ and $H^0(M,O^*) = \mathbb{C}^*$, and $H^1(M, Z) = Z^d$ for some $d$
harmonic archepelgeogsog
but how to compute the middle terms ?
OK from dolbeault's theorem $H^1(M,O) = H^{0,1}_{\overline{\partial}}(M)$ but idk how to compute that as well
OK nevermind, the above is same as genus from main theorem
meant to say "harmonic archipelago" if anyone likes topology
By Dolbeault isomorphism, this is $H^{(0,1)}$. For compact Kähler, this is the "same" as $H^{(1,0)}$, whose dimension is geometric genus.
Sounds like you figured this out. I'm just late on everything today.
6:51 PM
is the space euclidean of dim 2 a projection of it's dim 3 rep?
@Lelouch Ah yeah correct, I forgot this.
@BalarkaSen For $*>0$
I guess if $\Bbb R^3$ was projected down to $\Bbb R^2$ you'd lose some valuable information
anyway, so that implies if $M$ is a torus then $H^1(M, O)$ is nontrivial. What is an explicit example of a generator ?
6:55 PM
What do you mean by explicit? Are you taking an open covering and asking for a Cech-cocycle?
@TedShifrin Yes, I am asking for an cech cocycle which is not a coboundary
Should be able to get something explicit out of the pairing with $H^0(M,\Omega^1)$.
But you have to specify an open cover, then. I'm not going to have time to go through this now. Have a zoom call in literally a few moments.
I think you can do a sneak computation of $H^1(X, O)$ for a torus $X$. Run the exponential sequence $0 \to O^*(X) \to H^1(X, \Bbb Z) \to H^1(X, O) \to \text{Pic}(X) \to H^2(X, \Bbb Z) = \Bbb Z \to 0$
Also, where I'm going wrong here: Since $0 \mapsto O \mapsto M \mapsto PP \mapsto 0$ is exact, a data $\tau$ for Mittag Leffler is solvable iff $\delta \tau = 0$ in $H^1(M, O)$. For $M = \mathbb{CP}^{1}$, $H^1(M,O) = 0$ so this then implies any mittag leffler problem is solvable, which is very wrong
The penultimate map is the degree map from the Picard group to $\Bbb Z$, so the kernel is $\text{Pic}_0$
6:59 PM
I would recommend looking at $H^1(M,\mathscr O)\otimes H^0(M,\Omega^1)\to H^1(M,\Omega^1) \cong H^{1,1}(M)$. Both the other two spaces have explicit generators.
$O^*(X) = 0$
So $0 \to H^1(X, \Bbb Z) \cong \Bbb Z^2 \to H^1(X, O) \to \text{Pic}_0(X) \to 0$
But $\text{Pic}_0(X) \cong X \cong \Bbb C/\Bbb Z^2$ by the Abel-Jacobi map
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