mathoverflow.net/q/473844/323920 This is a paper that quoted more than 600 times but super hard to read. I suspected nobody except the author really understood the proof but it seems it's true lol

Hi!

Hello 👋

In a filtered module we have that if the general term of a series converges to zero then the series itself converges to zero. The proof goes something like this: Let $M$ be a filtered module with filtration $(M_n)_{n\in\mathbb{Z}}$. Suppose $(x_n)\to 0$ where $x_n\in M$ then for any integer $p$ there exists $n(p)\in\mathbb{Z}$ such that for all $n\geq n(p)$ we have $x_n\in M_p$ Then for such $n$ and $k\geq 0$ we have $x_n+x_{n+1}+\cdots+x_{n+k}\in M_p$. Then the sequence of partial sums of

the series $\sum x_n$ is Cauchy and the Cauchy criterion applies which completes the proof.

Sorry I forgot to mention that the module was complete (and Hausdorff)

Well I was about to ask why this proof doesn't work in real numbers for example but now that I've typed it out I notice that we can't get much of a topology from a filtration in reals since it is a field...

The proposition "Let $f:X \to Y$ be a non surjective function. Then $Y \ne \varnothing$." can be logically stated as $\forall f \in \{g \ | \ g: X \to Y \ \text{function}\}, f \ \text{surjective} \implies Y \ne \varnothing$? If so, to prove this by contrapositive I must prove that $Y = \varnothing \implies \exists f:X \to Y \wedge f \ \text{surjective}$?

@ZaWarudo no

$Y = \emptyset \implies f$ surjective

Or $f$ not surjective $\implies Y\neq \emptyset$

Sorry I made a typo: the original claim was $\forall f \in \{g \ | \ g: X \to Y \ \text{function}\}, f \ \text{not surjective} \implies Y \ne \varnothing$

@Pizza It seems impossible to me

I did it by first doing the integral in dy, but then carrying it out when you have to integrate it for the second time in dx results in an expression that is a bit difficult to calculate

@Jakobian Can I ask you if you would do it this way, or another way pls?

@BinkyMcSquigglebottom: Why hard to integrate? In your first attempt $dydx$, after substituting the extrema $\sqrt{3}$ and $\sqrt{4-x^2}$ did you try the substitution $x=2\sin \theta$? That's a standard technique of calculus I

@ZaWarudo yes, I tried to use it but how would you do it starting from $$\iint_D x^2+y^2dxdy, \quad D=\{1\leq x^2+y^2 \leq 4, y \geq \sqrt{3 }\}$$ . I mean would you do the same things as me?

There has to be an easy shortcut

Why you think there "must be an easy shortcut"? Double integrals are usually full of calculations. The only shortcut I see (but of course wait for other points of view) is that you can notice that $x^2+y^2=(-x)^2+y^2$ and $1 \le x^2+y^2 \le 4$ can be also be written as $1 \le (-x)^2+y^2 \le 4$. So, by symmetry, you can evaluate $2I$ (where $I$ is the double integral) and assume $x \ge 0$ in $D$.

This follows from the fact that if $g$ is an even function, then $\int_{-a}^a g(t)dt=2\int_0^a g(t)dt$.

@ZaWarudo can you check my post pls

Is it better to do like binky or like me ? Or as the user who replied to my post

@BinkyMcSquigglebottom Anyway thanks for the help

Let $a_1, a_2,\ldots, a_n$ be ideals of a commutative ring $A$ and let $m$ be an integer. Then am I correct in saying that $(a_1\cap a_2\cap \cdots a_n)^m=\bigcap\limits_{i=1}^n a_i^m$? I have shown this but it seems too good to be true.

@Pizza the answer or heropup seems the fastest solution

@SineoftheTime He advised me to write the bounds of r in terms of theta instead of writing the bounds of theta in terms of r , right?

yes

@SineoftheTime how would you do it?

Pls

see heropup's answer

@Pizza here as you already understood, you can't just plug $r=2$ since the radius is not fixed in the region

@SineoftheTime By chance it becomes: $\int_{\frac{\pi}{3}}^{\frac{2\pi}{3}} \int_{\frac{\sqrt{3}}{\sin(\theta) }}^{2} r^3 \, dr \, d\theta$

?

Okay, I have to look at the answer better, now I'm without a pen, I don't know if I did the right thing above

@Jakobian I hope you don't leave chat though... You're obviously values here

Valued*

@ephe This is not true, though I'm unsure right now if there is an easy counter-example

@Thorgott I know that it is true for pairwise distinct maximal ideals but then I am missing something in my proof. Where would I need to use those even?

Not sure if there is a result for this, but if $S$ is a submanifold of $M$ with codimension 2 or more, is there an isomorphism $H^n(M)\cong H^n(M,S)$ of the top-degree cohomology groups?

@ephe I don't know what your proof attempt looks like. There is one inclusion that is always true, and the other one not necessarily. It is also useful to observe that $\mathfrak{a}\cap\mathfrak{b}=\mathfrak{a}\mathfrak{b}$ implies your result, and being pairwise distinct and maximal is sufficient (albeit far from necessary) for that.

@anak yes, this follows from the LES

@Thorgott Like $\to H^{n-1}(S)\to H^n(M,S)\to H^n(M) \to H^n(S)\to 0$?

yeah

@Thorgott Oh, the map $H^{n-1}(S)\to H^n(M,S)$ is zero because the codimension forces $H^{n-1}(S) \cong 0$! That's what you are seeing?

yeah, the terms involving just $S$ vanish for dimension reasons

Oh my I am stupid, I totally missed that.

Thanks, Thorgott!

I suppose we can also appeal to Poincaré duality, but that's murkier in the non-orientable case

no problem :)

You mentioned last time that you were learning about $\infty$-categories, @Thorgott? Is that for research?

How to solve $\int \frac{9}{(sin(x))^4}$ pls

@BinkyMcSquigglebottom What trig identities have you tried?

@BinkyMcSquigglebottom First off, a nit to pick: you don't "solve" an integral expression; you "solve" equations (and inequalities). Presumably, you want to evaluate this expression, i.e. you want to find some "nice" closed-form expression for the given anti-derivative.

You might find it easier to try doing 1/sin(x), 1/sin^2(x), ...

I've seen Weierstrass substitution used

@XanderHenderson Oh, ok thanks

@BinkyMcSquigglebottom Have you tried it? If you are familiar with the Weierstrass substitution, it is a natural thing to try in such a case. So... get a pencil and compute.

weierstrass substitution is totally not necessary i can just write csc^4 as csc^2 (1 + cot^2) and substitute u = cot

@anak for my masters thesis, perhaps research afterwards

@BinkyMcSquigglebottom Okay. That's fine, too. Again, same advice a gave before: just get a pencil, and compute.

@Thorgott Nice, what drew you to $\infty$-category theory?

@Thorgott I can see now why my direct proof attempt was completely wrong but I couldn't get anywhere with your intersection equals product suggestion. Would you care to give another hint?

@anak I always liked category theory and it sort of comes naturally upon studying things like triangulated categories and homotopy (co-)limits

@ephe Ah, perhaps I missed a step. My point is that the claim you want is easily true if we can replace the intersections with products. A sufficient condition for the product and intersection of two ideals to agree is that they are comaximal. Now, if you have two distinct maximal ideals, then any two powers of them will be comaximal.

@Thorgott Nice! Does it appear naturally in many other fields than foundations and category theory itself?

I imagine algebraic geometry might have some use for them as it often goes full out with the CT.

it is essential in modern interpretations of homotopy theory and also has relations to algebraic topology (TQFTs, spectra, etc.). derived algebraic geometry is also one of the main applications, though I know little about it

@nickbros123 I don't think I will although I won't be (substantially) helping anyone in chat this week

I'm on mobile so its hard to read, and write posts

Plus I'm on mini-vacations so I shouldn't bother about other people's problems

@Thorgott Yeah I've been trying to work it out since I've read your response but I really cannot figure out how to get that inclusion even after replacing the intersections of powers with the product of powers. I've only managed to get the intersection equals product part by showing that the radical of sums of powers of maximals is $(1)$ and hence they are comaximal.

@Thorgott oooh I was only replacing one side with the product. It really is obvious when you replace both sides with products. Thank you!

yeah, perhaps I wasn't clear enough about that

@Pizza right

@SineoftheTime i solved it!

Can I show you something for a moment?

If I did this, is it still right?

go ahead

(i write wait)

@SineoftheTime

why $r\in [1,2]$?

@SineoftheTime do you mean why I integrated from 1 to 2?

yes

you should integrate from $\sqrt 3 \csc \theta$ to $2$

because I switched to polar coordinates and r^2 varied between 1 and 4, so r varies between 1 and 2

did you read heropup's answer to your post?

@SineoftheTime yes, but it's not very clear to me

since $y\ge \sqrt 3$, you have $r\ge \frac{\sqrt 3}{\sin \theta}$ so that $\frac{\sqrt 3}{\sin \theta}\le r \le 2$

so cant be 1 because y >= sqrt(3)?

yes

but shouldn't the polar coordinates also be replaced in the conditions?

@SineoftheTime

yes

@SineoftheTime thanks!!

is it clear now?

@SineoftheTime Yes, so the error is just there, right?

@robjohn could you help me to set up mathjax on my phone? I don't know how to add the javascript to a bookmark

yes

and $r$ should multiply all the expression

Oh never mind I got it

Anyone familiar with transition probabilities/transition kernels/Markov kernels/regular conditional probabilities? I have a very elementary question. Let's denote it by $\nu: E\times\mathcal F\to[0,1]$. Such a thing is a probability measure for a fixed point $x\in E$, i.e. $A\mapsto\nu(x,A)$ and a measurable function for a fixed set $A\in \mathcal F$, i.e. $x\mapsto \nu(x,A)$. Now, the notation $\nu(X,A)$, where $X$ is a r.v., what does it mean?

Is it simply the composition of the function $x\mapsto \nu(x,A)$ with $X$?

So $\omega\mapsto \nu(X(\omega),A)$, with $\Omega\to[0,1]$.

@Jakobian how did u do it

cuz I have been trying for months

couldnt get it to work

@robjohn Kudos on the apt avatar!

🇺🇲

What do you actually call $gr(M)=M_n/M_{n+1}$ where $M$ is a filtered module. I tried to look up the name of this $gr$ but failed. I know that $gr(A)$ is the graded ring associated to the filtered ring $A$ so would $gr(M)$ then be graded module associated to the filtered module $M$?

@Jakobian what kind of phone do you have?

@ephe the individual $M_n/M_{n+1}$ are the filtration quotients (though I'm personally of the opinion that if your filtration is decreasing, the indices should be super- and not subscripts), the associated graded module is their direct sum over all $n$

@Thorgott Oh sorry! I thought I wrote the sum but I think I deleted it accidentally while messing about. Thank you for confirming the name!

I have chatjax on my phone, if anyone needs help I know how to add it

@Jakobian nice!

@robjohn an android with google chrome

that's the hardest. Have you read the section about this on the installation webpage?

@nickbros123 you copy the script, and you try to add a bookmark, but then you edit the bookmark to copy paste the javascript rather than an actual site

Because it doesn't allow you to add bookmarks directly

@robjohn yes, sorry, I've made it work soon after I pinged.

Ah, so you have it working?

Yep

great!

are there good references for whitehead group?

@Jakobian By chance, to activate LaTeX, you also have to write the name of your bookmark in the search bar then click on It, right?

that name can refer to multiple different things

@Pizza yes

@Jakobian ah okay, same

@Thorgott Wh(G) where G is a group