How long does it take for an answer to be deleted?

@BalarkaSen @Thorgott On second reading, I've realised that I don't know why one knows that $M_g$ is generated by the $n$ elements $\{m_i\}_{i=1}^n$ as an $A_g$-module. I can see that the generators of $M$ map to elements in $M_\mathfrak{p}$ that can be expressed in the form $\sum a_i\cdot m_i/ g$ for $a_i\in A$, but even still, we end up allowing $A_\mathfrak{p}$-linear combinations of these generators (in $M_\mathfrak{p}$)

Every element is generated by a linear combinations of the $b_i$'s

(I.e. it's not clear to me that one can use the first half of their argument to show that $M_\mathfrak{p}$ treated as an $A_g$-module by restriction is isomorphic to $M_g$, which I think they are using?)

@Astyx Sorry can you be more precise by saying $A_\mathfrak{p}$-linearly generated, or whichever ring you mean (and/or say "every element of <module>")

@Astyx Note that there are k of these, and we want to reduce this to n in the potentially-exact-sequence in question, and we only re express these in terms of the basis of $M_\mathfrak{p}$ as an $A_\mathfrak{p}$-module, which is part of my contention

I mean the $b_j$ generate M as an A module, so the $b_j$ generate $M_g$ as an $A_g$ module

@Astyx Yeah, but there are $k$ of those, not $n$ of those

@Thorgott lmfaoooo

I like that it was accepted as correct

If you look at the post history of the answerer it's the second time they answer this particular guy

@tigre We don't want to show injectivity, just surjectivity

@Astyx ?

@Astyx I wanna know the story behind it

@Astyx Why is $A_g^n \to M_g$ defined as $e_i\mapsto m_i$ a surjection (as $A_g$-modules), that's my question

because every element in $M_g$ is of the form $\sum {a_i\over g^n}m_i$, because the $b_j$ generate M as an A module, hence generate $M_g$ as an $A_g$ module

Does anyone know a good way to learn theory behind integer linear programming?

@Astyx I don't understand

which part ?

Currently that just reads like asserting what they're claiming...

I would like to understand how to solve problem in ask.sagemath.org/question/55877/…

Any $b_i$ can be written in the form $A_g\{m_i\}_{i=1}^n$ over $A_\mathfrak{p}$, but you're now claiming it can be done over $A_g$

Yes, that's what the computation above in the answer gives

You put everything over a common denominator g

I thought that at first, but I don't think so

An arbitrary element takes the form:

And get $b_i = \sum {a_{ij}\over g}m_j$

$$\sum_{i=1}^k \frac{r_i}{t_i}\left(\sum_{j=1}^n \frac{(\prod_{j\ne i} \prod_{j=1}^n t_{kj}s_j)t_{ij}\cdot m_j}{g}\right)$$

Where that $r_i/t_i$ part has $t_i\in A\backslash \mathfrak{p}$

(Arbitrary element of $M_\mathfrak{p}$)

We're not looking for $M_p$ but for $M_g$

I just don't understand why you know the m_i generate M_g as an A_g-module

@Astyx You just assert it here for example

Because the $m_i$ generate the $b_j$ and the $b_j$ generate $M_g$

The m_i generate the b_j over $A_\mathfrak{p}$?

Oh, an element of $M_g$ is by definition of the form $m/g^n$ for an element $m\in M$ and $n\in \Bbb N$, is that what's confusing you ?

@tigre Over $A_g$

Everything I'm doing is in $M_g$ over $A_g$

@Astyx Nah I get that

$b_i = \sum {a_{ij}\over g}m_j$ <- this shows that the $m_j$ generate the $b_i$ in $M_g$ as an $A_g$ module

right?

Where you've put a double product up the top and relabelled right

My $a_{ij}$ are the double products and are thus elements of $A$ (they're not the same as the $a_{ij}$ in the answer, but still elements of $A$)

I agree that that computation makes sense in terms of equivalence classes occurring in defining $M_\mathfrak{p}$ as an $A_\mathfrak{p}$-module, and I would also agree that this makes sense in $M_g$ if I knew that extending and then restricting yields an isomorphism

Like formally their computation occurs inside $M_\mathfrak{p}$

Like that equality apriori means equivalence in $M_\mathfrak{p}$

Or in other words I agree that this solves the problem if the unit of adjunction for $f_!\leftrightarrow f^*$ is a natural isomorphism

Which it might be for all I know

I see what you mean

Hi!

we only have $u(gb_i - \sum a_{ij}m_j) = 0$ for some $u\in A\setminus \mathfrak p$

Luckily we can take $ug$ instead of g and $ua_{ij}$ instead of $a_{ij}$ and all is good

@tigre

$C[a, b]$ dual is functions of bounded variations, right?

Or am I misremembering

How would I how would I find the minimum of ab/c + bc/a + ac/b if a^2+b^2+c^2=1?

@satan29 That is a doubt that I also have myself.

@BalarkaSen measures

The dual is huge

I think what I said is also true

@tigre pick an $A_\mathfrak p$-basis of $M_\mathfrak p$ that lives in $M$. This gives you an map $A^n \to M$ that is an isomorphism at $\mathfrak p$. Take the kernel and cokernel to be $K$ and $Q$ and form an exact sequence $0 \to K \to A^n \to M \to Q \to 0$ that becomes $0 \to 0 \to A_\mathfrak p^n \to M_\mathfrak p \to 0 \to 0$, so $K_\mathfrak p = 0$ and $Q_\mathfrak p = 0$, so there is $g_1$ and $g_2$ such that $K_{g_1} = 0$ and $Q_{g_2} = 0$. Now take $f = g_1 g_2$.

neat

@Alessandro Ok what's the dual of that? All measurable functions? Pairing is integration

The double dual inclusion $V \to V^{**}$ is $C[a, b] \to L^\infty[a, b]$?

The dual of $C(X)$ for compact $X$ is the space of Radon measures with bounded variation

that's the Riesz–Markov–Kakutani representation theorem

@LeakyNun You should put this up as an alternative answer, it seems much more clear

good idea

Can I use Muirhead without having to expand a huge expression?

@Alessandro It seems to be true that $C[0, 1]^*$ is all complex Borel measures on $[0, 1]$.

Thanks @Astyx @LeakyNun

I forgot R-M-K lol

This is too hard for me

It'd be better not to get ignored!

Hope what I said was helpful after all

@user83244835 If people ignore you, they probably don't have a good answer (or the effort to answer is too high for the quality of the proposed answer, etc)

@Astyx It was, thanks

@user83244835 Do you have an example. This is an extremely vague question

Alright uh.

I have to show that.

any1 there?

@anakhro Yes, absolutely. I bought the DVD. It's good. Lots of the obvious big name people talking about him, including Griffiths, Bryant, Singer.

@TedShifrin Hi

Howdy, Karim.

@BalarkaSen Is the notation in the smash product chosen because $\wedge = \times /\vee$

lol

thats brilliant

You could turn that into Arkiv

lol

why do we care about the behavior of f(z) at a possible natural boundary?

do we care?

Do we care about the behavior of f(z) at a possible natural boundary?

Possible?

"possible" as in one is not sure whether say, Re(z)=1 is a natural boundary or not

What is the definition of natural boundary?

If we have singularities so densely packed on Re(z)=1 that analytic continuation cannot be applied to extend the function across the boundary Re(z)=1

that's my current understanding

Well, yes, in general, you are asking for the largest possible domain the function can have (after you've analytically continued as much as you can). For example, if you have a power series with radius of convergence $1$, then you know that there is at least one point on the circle of radius $1$ where you cannot analytically continue. $f(z)=1/(1-z)$ is an example, but it's analytic everywhere else.

But there are famous power series with radius of convergence $1$ that cannot be analytically continued across the circle at all.

@Xnero: What do you think it is?

What is $f(2)$?

Cool. What is $f(y)$?

OK, so now what is $f(x+1)$?

Sure. It's $(x+1)^2+1$. If you want, you can expand it out.

Yup.

That's what a function does :)

Salutations @TedShifrin !.................just saying what's up.

Howdy, dc3rd.

Doing all this geometry has been fun and it is definitely adding a dimension to how I look at things......and as you'll love to hear......creating a picture of why the algebra makes sense.

@TedShifrin So if I want to probe $f(z)$ near $\Re(z)=1$ to try to identify a natural boundary, what should I try?

@dc3rd Maybe you can appreciate the first chapter more now :)

@geocalc You have to find, as you said, a dense set of points on that line across which you cannot analytically continue. That's usually very hard.

So what do I use tubular neighborhoods for. What type of problems

@TedShifrin: can I crash a tutoring session? ;-p

ah, I see I am 2 hours late to crash the party.

@robjohn: Stop your silliness. Grr. Someone just downvoted for no apparent reason.

@user2103480: Here's an example. Suppose you have an embedded submanifold $M\subset X$. You have a smooth function on $M$. You wish to extend it to be a smooth function on all of $X$.

@TedShifrin I get that all the time.

at least I think the answers are good, so I see no reason to downvote.

Another basic application is that a tubular neighborhood is diffeomorphic to a neighborhood of the zero section of the normal bundle of the submanifold. So you can transfer, for example, intersection theory questions in the normal bundle to intersection theory questions in the ambient manifold.

@robjohn: If you see something I should improve here, let me know.

@robjohn: Were you going to crash re natural boundary or evaluation of a function? :D

in the same spirit, tubular neighborhood thm tells you that the homology of the normal bundle relative complement of zero section is isomorphic to the homology X relative complement of M

@TedShifrin Just looking to subvert somewhere.

@TedShifrin Nothing looks bad there. I think someone is downvoting for other reasons. Not the way we ask people to vote, but it's hard to control since there is no requirement to comment when downvoting. I have a lot of unwarranted downvotes.

I should say seemingly unwarranted; it's hard to tell for sure without a comment.

The only thing that I can see that someone might disagree with is the statement: "You should always prefer to do a volume integral rather than a surface integral."

Other than that, there is nothing that seems opinion based.

Thanks both!

Hi everyone

@robjohn: I almost put "almost" in that sentence. But I added the phrase about satisfying hypotheses instead. It's sound pedagogical advice, not opinion :P

Hi, demonic Alessandro.

Hm. One thing we used was substitution of the tubular neighborhood into complements. Is the trick here that M\N is not necessarily a manifold for N embedded into M, but M\VN is?

so that we can use poincare duality

@TedShifrin and I did say "seems opinion based" :-p

It's not a closed manifold (with boundary).

@TedShifrin thanks!

There will be people who want to keep math pure and keep "physics" (e.g., centroid) out of it, I suppose, @robjohn. They should be summarily shot.

Grr I'm stuck on some point set topology that should be trivial

Your trivial point set topology is always beyond me.

@TedShifrin Oh, if they are complaining about the mention of a centroid, then, yes, ammunition could be employed.

I have a continuous surjection $f:X\to Y$ between compact spaces. For $B\subseteq X$ open let the fiber image $f_{fib}(B)$ be $\{y\in Y\mid f^{-1}(y)\subseteq B\}$. Apparently this fiber image is always open (possibly empty)

Sounds tube lemma-ish.

no hypotheses on $B$?

I have an argument thinking about convergence of the fibers in the hyperspaces of closed sets of $X$, but that's nonsense, it should be completely elementary

@Thorgott $B$ open, sorry I forgot to write it

I would know better what to do with a smooth submersion.

Has to be tube lemma.

@robjohn @TedShifrin can it not be integrated into the system that a reason (which involves a character minimum) be implemented in terms of down voting?...I guess it would have to go hand in hand with individuals asking for help to explicitly show the work they've done and have the attempt have substance...

@TedShifrin I always forget what that is, wikipedia to the rescue

Fundamental fact to prove product of (two) compact spaces is compact.

Ah yeah, it's immediate from the tube lemma I think

Time to read the proof of the tube lemma then! Thanks

This is something I never forget. I actually used it a number of times in teaching other subjects.

@dc3rd I am sure it could be implemented, but there are many who find that unacceptable.

@TedShifrin Hmm... I don't remember seeing that in the proof of Tychonoff's Theorem.

maybe it was but just not called that

Ah the proof is actually very easy (and it makes clear why compactness is important), nice, I'll try not to forget about it this time

@robjohn There's a bunch of different approaches, I had never seen it proved using the tube lemma

@AlessandroCodenotti I would be surprised if there weren't several approaches.

@robjohn: No, Tychonoff is based on the FIP. I said specifically two (parenthetically). :P

tube lemma can be avoided by arguing with nets

and arguing with nets generalizes the finite case to Tychonoff's theorem rather seamlessly

As I said, the tube lemma shows up numerous times in algebraic and differential topology, and things like dynamical systems, so I consider it fundamental.

@Thorgott but you need c h o i c e tho

what's choice man

I only need that transfinite induction is obviously possible

Agree that tube lemma is fundamental

Even the algebraic geometers know it, under the name "universally closed"

@AlessandroCodenotti Yes this is straightforward tube lemma which can equivalently be stated as: for all Y and compact X the projection X x Y -> Y is a closed map

Your set f_fib is best characterized by its complement: f_fib(B)^c = {y | exists x in B^c with f(x) = y} = p_2({(x, f(x)) | x in B^c}). Notice that {x, f(x) | x in B^c} is a closed subset of X x Y: the graph of f is closed because Y is Hausdorff and f is continuous, and this is its intersection with the basic closed set B^c x Y

Since p_2 is a closed map f_fib(B)^c is a closed set hence f_fib(B) is open

This is an abstract phrasing of a more concrete argument where you show f_fib(B) is locally open

All you use here is X compact and Y Hausdorff

hi @MikeMiller @TedShifrin

Do you guys know let us start with projective algebraic surface X do you guys know if spreading out X is linked with the geometric genus of X ?

I think the family determined by spread is connected with geometric genus

Is that true

?