so if charpoly(A) = f then charpoly(A^2) = f(x^(1/2)) f(-x^(1/2))

re Leaky; hi Fargle

Hey @Ted, long time no see.

Well, not that long. HNY.

You too!

I'm just stepping out at this second, but I should be back within an hour or so.

Okey dokey.

Maybe when I get back I'll stop forgetting that Taylor series are a thing

We can't be responsible for your faulty memory.

welcome back @TedShifrin

thanks

Ted, may you please tell me if my proof is correct: The closure of a set A is a closed set: Proof: Assume p is a limit point of the closure A and that p ∉ the closure so since p is a limit point then every neighborhood of p contains a point q in the closure such that p ≠ q, but since p∈ ℝn−closureofA then there exists a neighborhood of p completeley contained in ℝn−closureofA which is a contradiction, because p is a limit point of the closure

and may you please tell me what i can do to improve it?

I've been telling you for a few weeks you need to take the time to learn MathJax/LaTeX.

You have to start by saying what your definition of closure is. For me, it's automatically a closed set.

I know I know, I'm sorry, i'll try to learn it by tomorrow.

Closure is the union of the set and its limit points

it seems to me you assumed the closure was closed when you said there's a neighborhood of $p$ completely contained in the complement.

Hey Ted, you're back :D

Hi @CaptainAmerica.

Hope you had a nice trip

I did. I'm glad to be home, though.

what I was trying to show was really that the complement is open, which is the same thing, how can i modify my argument? i'm not really sure.

It seems to me your proof just asserted that the complement was open. How did you use that $p$ was a limit point of the closure?

@TedShifrin I'll bet, you've been gone since last year.

Because my definition of closed is that if p is a limit point of the set then the p is an element of the set

True, @CaptainAmerica.

Is this list of conditions enough for the existence of the envelope for the given family?

But you never used it, did you, @maths?

@N.Maneesh: Assuming the first two conditions (you have a typo) is sufficient to get the envelope locally ... I don't know a necessary condition, offhand.

@TedShifrin I think I did when I assumed that p is a limit point of the closure and for the sake of a contradiction that p $\notin$ closure

Yes, but you never actually used that assumption, did you?

@TedShifrin From that we can get only $x$ can be a function of $c$ and $y$ can be a function of $c$. What is the guarantee that we can find a relation between $x$ and $y$?

You don't need that. You parametrize the curve by $c$.

okay. Thank you

It's worthwhile to do some explicit computational examples, @N.Maneesh.

@TedShifrin but when I assumed that there exists a neighborhood of p completely contained in the complement, that doesn't satisfy the definition of a limit point because a limit point means that every neighborhood of that point contains another point q$\in$ closure such that p $\neq$ q.

@TedShifrin Okay

How did you get to assume the existence of that neighborhood?

@TedShifrin from the definition of a limit point

A point in $\mathbb{R}^n$ is a limit point of E if every neighborhood of the point contains a point q ∈ E such that p is not q.

oh

you mean the existence of the neighborhood of the point in the complement

Right.

because the complement is a subset of metric space, what I mean by neighborhood is an open ball

That's not my issue at all.

Did you write out carefully what it means to say that $p$ is in the complement of the closure without assuming that complement is an open set?

Oh oh, @Fargle returns.

Where?

Outer space?

@TedShifrin No, but i'm not sure what to say, the only thing I can think of is that since p $\in$ $\mathbb{R}^n$ and p$\notin$closure then it is not a limit point of the set or a point in the set

So you have to say precisely what it means to know that $p$ is not in the set and is also not a limit point. Why does that give you the neighborhood you claimed?

because of the negation of what it means to be a limit point

Hey everyone!

Hey @TedShifrin :)

A point in $\mathbb{R}^n$ is not a limit point of a set C if $\exists$ a nbhd of the point such that it does not contain any other point of C.

Have you also used the fact that $p$ doesn't belong to the set?

Hi @Perturb

@TedShifrin i'm not sure how I could justify that if p$\in$ $\mathbb{R}^n$ and p is a limit point of the closure of a set A but p is not in the closure then p is not a limit point of A, it makes intuitive sense, but i'm not sure how to justify it

Well, you need to use the fact that $p$ is not in the set, as well. Suppose your set was $\Bbb Z\subset\Bbb R$, with no limit points at all.

@TedShifrin is it true that if a point is in a set, then it must have a neighborhood that is completely contained in the set?

You answer that yourself?

No, it's not true

What sort of set must it be?

an open set

Right.

Lets say I have a spaces $X$ and $Y$ and I have the wedge product $X \vee Y$. Then I have a topological embedding $i : X \to X \sqcup Y$ and a quotient map $q : X \sqcup Y \to X \vee Y$. Is $q \circ i$ a topological embedding? It seems really really close to being one. Because $q \circ i$ is injective and continuous. All I'd have to check is that $q \circ i$ is an open map. So if I choose $U$ open in $X$, then $i[U]$ is open in $X \sqcup Y$.

Now if I can show that $i[U]$ is a saturated open set, then I'll be done since $q$ is a quotient map, but the only way to do that is to show $i[U] = q^{-1}[q[i[U]]]$ and this may not be true if $U$ contains the basepoint of $X$ that gets collapsed in the wedge product.

What's the problem if the basepoint is in there?

Okay so let $y \in Y$ be the basepoint for $Y$ and let $x$ be the basepoint for $X$, and let $i_Y : Y \to X \sqcup Y$ be the canonical embedding of $Y$ into the disjoint union. Then $q(i_Y(y)) = q(i(x)) \in q[i[U]]$, but $i_Y(y) \not\in i[U]$ and so $i[U]$ is a proper subset of $q^{-1}[q[i[U]]]$

Hey there everybody!

Hey! @Dami

Well, maybe you're right, @Perturb. Isn't $i(y)$ in that preimage? ... I don't like thinking about this stuff.

hi Demonark

@TedShifrin Yeah $i(y)$ would be in the preimage so we don't get equality of those two sets

If you have compactness and Hausdorff, then it's automatic.

Maybe you should just do it in more basic fashion.

Really it's automatic in that case?

I was actually trying to use this argument to show that if $K_1$ and $K_2$ are polyhedra, then for $n \geq 1$ we have $H_n(K_1 \vee K_2) \cong H_n(K_1) \oplus H_n(K_2)$ (where $H_n$ is through simplicial homology)

I have to use this argument to apply Mayer-Vietoris

Remember that a bijection from a compact space to a Hausdorff space is a homeomorphism if it is continuous.

Oh how could I forget

$q \circ i$ is clearly a bijection and continuous

So I'm done

Wait 'til you're my age!

Thanks! @TedShifrin

Sure.

Lol I remember the first time I had to use the fact was in manifolds to show that RP^3 is homeomorphic to SO(3)

One thing which I didn't get then (really didn't get it until I saw the analogous trick with group theory) was finding a map out of a space such that the relation f(x) = f(y) was the defining relation for the quotient space

So our TA said "Yeah this induces a continuous bijection out of RP^3" and I just shrugged. But yeah it's a fairly convenient fact

Wait, what's the analogous group theoretic trick?

Just a shot in the dark but I guess if you have a group $H$ and you wanna show it's isomorphic to something and you have another group $G$ along with an epmorphism $\phi : G \to H$, then you can show $H \cong G/\operatorname{ker}(\phi)$ through the first iso theorem

I guess I don't see the analogy clearly between that and what Ted formulated

if (X,d) is any metric space and S is an open and connected subset of it then it is also path-connected and polygonal-arc connected?

@famesyasd Consider the closed topologist's sine curve consisting of all points (x, sin (1/x)) with x > 0, and all points (0,y) with 0 <= y <= 1, as a subspace of the plane. This is a metric space, and it is an open, connected subset of itself, but it is not path connected.

What are the different kinds of mathematical thinking ?

To me , there is a clear difference between analytic and algebraic thinking , are there others?

Do you mean insofar as, like, different proof methods and stuff like that?

I could name combinatorical proof methods or categorical stuff off the top of my head.

i.e., respectively, "these numbers are equal because they count the same thing", and "x is true by y universal property"

I don't think these are sharp distinctions but you can definitely see different "flavors" of stuff going on there.

I mean in terms of actual ways of thinking @Fargle , like if I study ring theory vs general topology , my brain feels very different , it is easy for me to see how someone can be very good at one and terrible at the other. So am inclined to think they actually require different ways of thinking , one visual and one highly abstract and symbolic.

Let $(\Omega_X, \Sigma_X, \mu_X)$ and $(\Omega_Y, \Sigma_Y, \mu_Y)$ be normalized measure spaces (i.e. $\mu_X(\Omega_X) = \mu_Y(\Omega_Y) = 1$). Let $\mu$ be a measure on $(\Omega_X \times \Omega_Y, \Sigma_X \otimes \Sigma_Y)$ such that $\mu(A \times \Omega_Y) = \mu_X(A)$ for every $A \in \Sigma_X$. Conjecture: for integrable $f : \Omega_X \to \Bbb R$ we have $$\displaystyle \int_{\omega_X} f(\omega_X) \ \mathrm d\mu_X = \int_{(\omega_X,\omega_Y)} f(\omega_X) \ \mathrm d\mu$$

is the integral over $\Omega_X$?

the first one is over $\Omega_X$ and the second one is over $\Omega_X \times \Omega_Y$

Yeah I thought so, you have some small $\omega$ instead of $\Omega$

oh I meant $\displaystyle \int_{\omega_X \in \Omega_X} f(\omega_X) \ \mathrm d\mu_X$

anyway I believe it's true because it's true for simple $f$ because of the axiom I placed

so yeah that resolves the issue, thanks

Uhm you're welcome I guess, I did nothing :P

always approximate integrable functions by simple functions, I guess

Hello, why is $\lvert \log(1+z) \rvert \leq 2 \lvert z \rvert$ for $\lvert z\rvert \leq \frac{1}{2}$

I found the Taylor expansion of $log(1+z)=z-\frac{z^2}{2}+...$

I am confused by "edge-induced subgraph" when reading a paper.

Consider this graph:

What would be the subgraph induced by the edges 1-2 and 1-3? Is it the entire graph 1-2, 2-3, 3-1, or just 1-2, 1-3?

The paper makes sense if I assume the latter.

(But I would have assumed that an edge-induced subgraph is the same as the subgraph induced by the endpoints of those edges.)

An edge-induced subgraph contains all the inducing edges and every vertex incident with at least one of these edges. In this case, it would be 1-2,1-3.

@LeakyNun Uncountably many re s

Hello brethren (or not if this offends you) -

Surely this has been computed before? Ritsch fails me for some reason :(

wolframalpha.com/input/…

Plus I am lazy :(

My best guess now is $\sum_{t=0}^s \operatorname{Li}_{t+1}(x) \log^{s-t}(x) \frac{s!}{(s-t)!}$

Guess I should tack on a (-1)^t, and it's valid only for $s\geq2$. Can't find why though

@MatheinBoulomenos so for an extension $L/K$, we can take the separable elements to form a separable extension $L^s/K$, and then normal close it to get $N(L^s)/K$ which is galois

it will be canonically isomorphic with the extension $N(L)^s/K$

and also $\operatorname{Gal}(N(L^s)/K) = \operatorname{Aut}(L/K)$ canonically

you'll want $L$ algebraic for $N(L)$ to make sense, but yeah

yeah I forgot to say that

@LeakyNun wait that doesn't sound right

take $\Bbb{Q}(\sqrt[3]{2})/\Bbb{Q}$

oh no

you can say $\mathrm{Gal}(N(L^s)/K)=\mathrm{Aut}(N(L)/K)$ still

then how do I make Aut(L/K) a galois group

change the base to $L^{\mathrm{Aut}(L/K)}$

what if I don't want to change the base

oh and I proved that for $\operatorname{char}(K) \ne 3$ and $\mu_3 \subset K$ and $p, q \in K^\times/K^{\times3}$ we have $K(\sqrt[3]p) = K(\sqrt[3]q)$ iff $p=q \lor p=q^2$ and I'm wondering if there's a better result

oh $\mu_3 \subset K$ is just to ensure the extensions are normal

oh of course it's true, it's the special case of Kummer theory

was about to say that

I'm an idiot

no

what's your favourite proof of $\sqrt{p^\ast} \in \Bbb Q(\zeta_p)$?

where $p^\ast = \pm p$ so that $p^\ast \equiv 1 \pmod 4$

ramification

and how do you know that it ramifies?

you know that $p$ is the only prime that ramifies in $\Bbb{Q}(\zeta_p)$ because $\Phi_p$ is separable mod q iff $p \neq q$

aha nice

and the ring of integers is $\Bbb{Z}[\zeta_p]$

and then how do you get $p^\ast$?

let's assume $p$ is odd

sure

you look at what happens to $2$

oh

no, I mean you also want that $2$ doesn't ramify

if you disregard $2$ there are two possibilities: $\Bbb{Q}(\sqrt{p})$ and $\Bbb{Q}(\sqrt{-p})$

I keep saying nonsense don't I

have you worked out the ring of integers of a quadratic number field?

yeah sure

great, so that's monogenic, so we can read off the splitting/ramification on the reduction of the minimal polynomial

you mean read off

if $n$ is square-free and $n \not \equiv 1 \pmod{4}$, then the ring of integers in $\Bbb{Q}(\sqrt{n})$ is $\Bbb{Z}[\sqrt{n}]$ with minimal polynomial $x^2-n$, that's a square mod $2$

so $2$ is ramified

or you could also compute the discriminant

so that's where the condition $p^* \equiv 1 \pmod{4}$ comes from

I mean "read off ... from" actually, propositions are hard

splitting of rational primes in quadratic and cyclotomic extensions is something everyone interested in ANT should work out at some point

@Mathein apparently funktionalanalysis counts as an applied module at Heidelberg

lol, nice

Yeeeee hahah

Like.. it's a Veranstaltung in Grundmodul Angewandte Analysis und Modellierung

and Veranstaltungen in this Modul zählen als applied modules

idk why I'm speaking like this

es ist fine

lolol

es ist fun to vermisch German and Englisch

da have you right

nah that's not nice

@LeakyNun once you figured out the quadratic subfield of $\Bbb{Q}(\zeta_p)$, you're close to proving quadratic reciprocity

oh 2 isn't ramified simply because $x^p-1$ is separable mod 2 right?

@Leaky: You meant fein.

@LeakyNun right

Hi @Ted

hi, a @Balarka

I learnt this cool fact. If $G$ is a compact subgroup of the isometry group of a nonpositively curved complete simply-connected manifold, then the action of $G$ has a fixed point.

I think I once knew that ...

Not that I can see how to prove it now.

Guess I can use Cartan-Hadamard to lift to a vector space.

$\Bbb R^n$ with a weird non-positively curved metric, yeah

I guess the idea is to do an averaging trick.

Like you do with finite subgroup of isometries of standard R^n

Well, the averaging trick always works with any compact group; I just don't quite see it ...

Say $\mathcal{O}$ is an orbit. It's a compact subset of $\Bbb R^n$ with this different metric.

Take it's center of mass?

But a fixed point means the orbit is just a point.

I'm saying the center of mass, appropriately defined, will be a fixed point

Just starting with an arbitrary orbit $\mathcal{O}$

Hmm, very non-convex orbits ...

ah

But, yeah, it's clear that if you normalize the subgroup $G$ to have volume $1$, then for any $x$, $\bar x = \int_G g\cdot x\, dg$ is fixed by $G$. And this vector-valued integral makes sense.

Where am I using the fact that the group acts by isometries?

What does that integral mean?

$g \cdot x$ are points in $M$

No, I lifted to $\Bbb R^n$.

You can't sum them

But R^n with a very different metric

Vector-valued integral.

$G$ need not act linearly

Why do I care if it's linear?

Then $\bar{x}$ won't be fixed by $G$, no?

$g\cdot x$ is a vector-valued function.

Yeah, by (left-) invariance of $dg$. You do the usual thing of changing variables.

$g_0g = g'$ still runs over all of $G$.

I'm trying to understand. Pick any $h \in G$. Why is $h \cdot \overline{x} = \overline{x}$? Multiplication by $h$ need not interact well with $\int_G$, right?

That's what I was just doing. Bring $h$ into the integral and change variables, and use left-invariance of the measure.

@Ted He's saying that you can't bring $h$ into the integral because the action is not linear

How can you bring $h$ into the integral? Isn't that implicitly assuming linearity of the action of $G$?

YES

finally

Oh, I see.

Right

@BalarkaSen G cannot be finite, or else it would contradict nonexistence of finite dimensional models for BG. So it must be infinite and hence contains a circle subgroup. So this is equivalent to the case G a circle.

What? Lots of finite groups of isometries of R^n

Without fixed points you get a nonvanishing Killing field, but I don't know what to do with that

Oh, without fixed points.

That's a nice line of thought

You must be able to show that the circle action is impossible with nonpositive curvature. I think I see it if you said negative curvature

Closed geodesics are very rare in negatively curved manifolds

So maybe something with that

Are Jacobi fields nonexistent?

No, they exist, but they exponentially blow up if $K < -\epsilon < 0$

So if you move away from a closed geodesic you shouldn't ever get a closed geodesic

Can that be made into a proof?

I think so. Thanks.

Wait

Your question is false as stated

Nevermind sorry

Forgot simply connected

LOL ... howdy @MikeM.

Was thinking of the circle ktself

Yeah that's very necessary (otherwise torus :P)

Yes

Hi

[shows up, kills chat]

rip in ded

I disappeared for lunch. This is an interesting result, regardless.

@Ted Didn't we try to argue once some curvature constraint is forced by a foliations by geodesics?

I guess that can't be true, since that happens on a twice-punctured sphere.

You have to demand something more global

So if the metric is complete, maybe you can say something.

Balarka's idea to look at local changes near closed orbits seems nice to me

(I've given up on the foliations comment)

I still want to lift to $\Bbb R^n$ by Cartan-Hadamard, but maybe I shouldn't insist on it.

@MikeMiller Thoughts: Say $\rho : \pi_1(M) \to \text{Diffeo}(F)$ is a representation. Consider the suspension foliation on $E = M \times_{\pi_1(M)} F$ - this is a foliated $F$-bundle over $M$ with leaves parallel to $M$ which are covering spaces corresponding to $\ker \rho$, hitting the fibers transversely. If $M$ and $F$ are given metrics, they should collectively define a metric on $E$ under which this becomes a foliation by totally geodesic submanifolds.

Changing the metric on $F$ should change the curvature drastically but still leave this totally geodesic

Ha! Nice destruction of my idea

@TedShifrin I liked that idea but couldn't make anything of it.

Maybe I'm not so sure that what I said is a foliation by totally geodesic submanifolds. But whatever

I agree that you have constructed geodesic foliations with rather silly curvature.

Am I right in thinking that the suspension of $latex [0,1]$ is homeomorphic to $S^1$?

No.

When you suspend, you add a dimension, for starters.

suspension of "latex[0, 1]" is going to be quite ugly indeed

smacks Balarka

Hmm, what does that "latex" do, anyhow?

Some people probably like to be suspended in latex

kinky

I'll get the handcuffs.

Haha, no. You have to use it in wordpress.

Anyhow, @user193319, the suspension of an interval is a filled-in diamond.

Hey everyone!

Ah, course. I was thinking of $[0,1] \times [0,1]$ without its interior points...D'oh

hi @Perturb

Sort of relevant, I kept on pronouncing latex as "latex" instead of "lateck" until a PhD student told me basically I'm saying what they use in condoms not for math typesetting

Right, so now you get Mike's "joke." :P

Yeah :p

So, $[0,1]$ does not retract onto its endpoints because $[0,1]$ is connected while $\{0,1\}$ is disconnected...right?

So a (finite) Galois extension $L/K$ gives a representation of $\operatorname{Gal}(L/K)$ on $L$ the $K$-vector space

and there should be an interpretation of Hilbert 90

as some sort of a projection operator

Ok but what about Hilbert 360 noscope

ha ha

Given $\varphi \in H^1(G;L^\times)$, we have $\pi_\varphi : L \to L$ defined by $\displaystyle \pi_\varphi(x) = \sum_{\sigma \in G} \varphi(\sigma) \cdot \sigma(x)$

it should be a projection

hi @Daminark

Yo

@Daminark how are things

Yo Demonark

@user193319 Right.

wagwan yo

Problem: Let $\pi : E \to X$ be a covering map, and let $p : I \to X$ be a loop. Assume that $p$ has a lift $\tilde{p} : I \to E$ such that $\tilde{p}(0) \neq \tilde{p}(1)$. Show that $p$ is not nullhomotopic. Strategy: Assume $p$ is nullhomotopic, and then argue that $p$ is path homotopic to constant path. If that's true, then their lifts are path-homotopic and begin and end at the same point. But that would mean $\tilde{p}$ is a loop, but it isn't.

Does this seem like it would work? I've tried proving it, but I haven't had any luck.

Or is there some better strategy?

@user193319: Do you have a major theorem to use?

Your strategy is correct.

I was thinking of using the following: Let $\pi : E \to B$ be a covering map; let $p(e_0) = b_0$. Let $f$ and $g$ be two paths in $B$ from $b_0$ to $b_1$; let $\tilde{f}$ and $\tilde{g}$ be their respective lifts to paths in $E$ beginning at $e_0$. If $f$ and $g$ are path homotopic, then so are their lifts and they end at the same point.

I'm studying for a prelim, so pretty much any theorem in Munkres is available to me.

Things are going alright, classes are starting on Monday so that'll be fun

it's amazing how Fermat's last theorem generated a whole field of mathematics

At least an integral domain of mathematics :P

@user193319: I was referring to the homotopy lifting theorem/property.

But, yes, that theorem will do.

But yeah it's kinda weird to me that such a simple-sounding problem is that hard to solve