Mathematics - 2025-01-26

hbghlyj

00:58

To prove that "$\mathbb{R}^{n_1}$ and $\mathbb{R}^{n_2}$ are not homeomorphic for $n_1 \neq n_2$", I wonder if the following is a valid proof:
Suppose $\mathbb{R}^{n_1}$ and $\mathbb{R}^{n_2}$ are homeomorphic. Then $\mathbb{R}^{n_1} \backslash \{0\}$ and $\mathbb{R}^{n_2} \backslash \{0\}$ are homotopy equivalent. But $\mathbb{R}^{n_1} \backslash \{0\} \simeq S^{n_1-1}$ and $\mathbb{R}^{n_2} \backslash \{0\} \simeq S^{n_2-1}$ have different homology groups, so they are not homotopy equivalent. Contradiction.

Ben Steffan

That would in fact be the standard proof of that fact, pretty much :)

you're being a bit sloppy: how do you know $\mathbb{R}^{n_1} \setminus \{0\}$ and $\mathbb{R}^{n_2} \setminus \{0\}$ are homotopy equivalent?

hbghlyj

Because $f$ restricts to a homeomorphism between $\mathbb{R}^{n_1} \backslash \{0\}$ and $\mathbb{R}^{n_2} \backslash \{f(0)\}$.

Ben Steffan

well now you've magically pulled out some $f$ and changed the statement

but yes, fix the $f$ to begin with and work with $\mathbb{R}^{n_2} \setminus \{f(0)\}$ and all's well

also the two punctured spaces will be homeomorphic, not just homotopy equivalent

hbghlyj

Ok

The proof in ncatlab.org/nlab/show/topological+invariance+of+dimension also uses that $\mathbb{R}^{n} \backslash \{0\}$ is homotopy equivalent to $S^{n}$, but continues to compute the relative cohomology groups $H^k(\mathbb{R}^n, \mathbb{R}^n \backslash \{0\})$. It seems that the proof in ncatlab is more complicated than the above proof.

Ben Steffan

marginally so

using cohomology is a rather strange choice, but computing $H_k(\mathbb{R}^n, \mathbb{R}^n \setminus \{0\})$ is not hard, and a group you will either know already or have to know at some point anyways so

and if you know that and some basic facts about cohomology, it's rather immediate that this is the same as the cohomology group

hbghlyj

01:14

Ok

Thorgott

01:58

I wager they do that cause the relative group is actually the one witnessing the dimension

Ben Steffan

02:27

yeah, but why they choose to use cohomology is beyond me

It's kind of funny that they say "We discuss various proofs [...]" and then all they give is one using ordinary cohomology with too many details and a note to the effect of "you can also use adams operations to prove this :)"

nickbros123

06:34

reading ring theory from Dummit and Foote right now

reads kinda boring, not gonna lie

Soumik Mukherjee

How was the exam @nickbros123?

pie

07:27

@SineoftheTime we are dividing $\sqrt[n]{a_n}$ by $b_n$ which itself tends to infinity the $\inf_{n\ge N} a_{n+1}/a_n$ tends to ifinity but it could be something like $O(n)$ and $\sup_{n \ge N} a_{n+1}/a_n$ could be something like $O(n^2)$ and $b_n$ can be something like $O(n \ln(n))$ the value of $\sqrt[n]{a_n}/b_n$ can be anything btween $(0,\infty)$

at least that what I think

I couldn't come up with a counter example

Should I make a post about this?

nickbros123

08:26

@SoumikMukherjee exam was rather easy, but I made stupid blunders. still hoping to cross the cutoff. I couldnt rly focus during the exam for some reason

Sine of the Time

12:04

@pie I thought you were applying the root test to the whole expression, my bad

Let's see if this works

$a_n=e^{n^2}$ and $b_n=e^{2n}$

we have $\sqrt[n]{e^{n^2}}=e^n\to \infty$ and $b_n \to \infty$

$\frac{a_{n+1}}{a_nb_n}=\frac{e^{(n+1)^2}}{e^{n^2}e^{2n}}=\frac{e^{n^2}e^{2n}e}{e^{n^2}e^{2n}}\to e$

but $\frac{\sqrt[n]{e^{n^2}}}{e^{2n}}=\frac{e^n}{e^ne^n}=e^{-n}\to 0$

Soumik Mukherjee

12:49

@nickbros123 I see, all the best for the upcoming interviews.

Thorgott

13:30

@Ben I got some useful feedback. It comes down to the following question: Can you find a smooth degree $n$ map $f\colon S^3\rightarrow S^3$ such that $S^3\rightarrow S^3\times\mathbb{R}^2,\,(z,w)\mapsto(f(z,w),w)$ is a smooth embedding (or something similar)?

("comes down to" as in "that would be sufficient")

Ryder Rude

13:46

Newcomb's paradox

In philosophy and mathematics, Newcomb's paradox, also known as Newcomb's problem, is a thought experiment involving a game between two players, one of whom is able to predict the future. Newcomb's paradox was created by William Newcomb of the University of California's Lawrence Livermore Laboratory. However, it was first analyzed in a philosophy paper by Robert Nozick in 1969 and appeared in the March 1973 issue of Scientific American, in Martin Gardner's "Mathematical Games". Today it is a much debated problem in the philosophical branch of decision theory. == The problem == There is a reliable...

> To almost everyone, it is perfectly clear and obvious what should be done. The difficulty is that these people seem to divide almost evenly on the problem, with large numbers thinking that the opposing half is just being silly

this paradox divides people 39% vs 31% and each side thinks the answer is obvious

Ben Steffan

@Thorgott that makes sense, but it looks rather unlikely to me

Thorgott

interesting, I would believe it's true (though evidently annoying to write down an example)

then again, we know it cannot be true if you replace $\mathbb{R}^2$ with $\mathbb{R}$, so it's subtle

however, it's clear that it works if we have sufficiently many $\mathbb{R}$ factors (certainly, $4$ will be enough), so a similar construction does yield an explicit manifold model of $K(\mathbb{Q},3)$ without relying on the folklore-ish result that every countable, finite-dimensional CW-complex has the homotopy type of a manifold (I'm in the process of bothering Moishe to properly source that argument)

Ben Steffan

14:14

@Thorgott you could ask another question...

Thorgott

nooo

though perhaps I should make an edit to make reference to the strategy clearer

Ryder Rude

15:07

Kyu Sakamoto ~ Sukiyaki (Official Music Video) [上を向いて歩こう坂本九]

►

Ker

15:52

Hey,

I have to prepare an exam on on Perturbation Theory and spectral theory for unbounded operators and I feel kinda stuck because I lost motivation to keep studying. I am looking for a study buddy to stay motivated and study together these topics, if you are interested please dm me.

References: notes from my course, Reed-Simon vol 1 and 2; A comprhenesive course in analysis vol 4, Spectral theory by borthwick, Quantum theory for Mathematicians by B.C. Hall and others.

Language: English or Italian.

psie

16:56

I've been skimming through LADR by Axler. I was wondering, does he ever prove anywhere that an invertible linear transformation (on $\mathbb R^n$ say) can be written as a finite product of elementary transformations? I'm just curious if this is something he included in his text. On the other hand, the linear algebra book by FIS, does anyone know if they ever prove the uniqueness of a positive square root of a positive operator?

According to my own research, it looks like Axler omits elementary matrices, whereas FIS omit the uniqueness of a positive square root of a positive operator.

Ben Steffan

@psie surely the text discusses gaussian elimination?

I cannot imagine a linear algebra text not discussing it

psie

@BenSteffan Maybe I'm completely overlooking something, but it doesn't look like LADR discuss Gaussian elimination. It is open access and navigating to the index, where you do find 'Gaussian elimination' it just redirects you to pages where they mention that word, but not really give a treatment of Gaussian elimination. FIS (4th edition), on the other hand, have an exercise on the square root of a unitary operator, but it merely is about existence, not uniqueness.

Uniqueness allows you to prove uniqueness of the polar decomposition (although I'm sure there are other ways too, maybe easier ways).

Ben Steffan

17:15

looks like the book assume Gaussian elimination known

that's a very strange choice to make

psie

@BenSteffan it does look like that

Ben Steffan

but perhaps it fits with the US education system

Xander Henderson

@BenSteffan Gaussian elimination is often taught in algebra / precalculus classes in the US. I know that it is on our curriculum form here for our "advanced algebra" class.

It is usually the last thing I teach each semester and I tend not to put it on the final, so I don't know how well students learn it. But it is on the syllabus.

Ben Steffan

makes sense

psie

For example, FIS has a lot about elementary matrices, and it seems to be on an advanced level.

Xander Henderson

17:20

I also think that Axler is aiming for his text to be for a second course in linear algebra, so certain stuff is assumed.

Ben Steffan

yes, I think I recall something like that

psie

18:15

I'm reading about the construction of the reals from Dedekind cuts in Rudin's PMA. We have $R$, the set of cuts which is an ordered set by proper inclusion. After proving the axioms of addition, he proves the axioms of multiplication. He does this only for $R^+$, the set of cuts $\alpha>0^\ast$.

Then he defines $\alpha 0^\ast=0^\ast\alpha=0^\ast$ and defines $$\alpha\beta=\begin{cases} (-\alpha)(-\beta),\text{ if }\alpha<0^\ast,\beta <0^\ast\\ -[(-\alpha)\beta],\text{ if }\alpha < 0^\ast,\beta > 0^\ast\\ -[\alpha(-\beta)],\text{ if }\alpha > 0^\ast,\beta < 0^\ast .\end{cases}\tag1$$

Then he says that the axioms of multiplication of $R$ follow now from repeated application of the identity $\gamma=-(-\gamma)$. What does he mean by repeated application of that identity? For me, $(1)$ looks perfectly fine already, since he has defined the product $\alpha\beta$ for $\alpha,\beta>0^\ast$ and such products appear in $(1)$.

Thorgott

it's well-defined, but he's talking about verifying the axioms

psie

yes, right

pie

19:09

@SineoftheTime I wonder what extra condition should $a_n, b_n$ satisfy so that one can replace $\sqrt[n]{a_n}$ by $a_{n+1}/a_n$

Maxime Jaccon

this site is kind of addicting :)

Soham Saha

19:26

Noticed something strange today: 2yrs ago, Desmos used to give overflow error for $(k!)!$ with $k>5.19862193175811$. But I checked it again today, and it seems that now the upper limit is somewhere around $5.20411$

Looks like the upper bound changed from $2^1012$ to $2^1024$…

Maxime Jaccon

19:38

any reason that the number was 1012?

ILikeMathematics

Let $$U_1 = \langle X^2 + 1, X^3 + 1\rangle_{\mathbb Q}$$ and $$U_2 = \langle -1 + X, -1 + X^2, -1 + X^3 \rangle_{\mathbb Q}.$$ Determine a basis of $U_1 \cap U_2$.

So, the idea is that we form a spanning set of $U_1 + U_2$ first by unioning the bases and whatever we throw out, that is in $U_1 \cap U_2$. So: \begin{align*} U_1 + U_2 &= \langle X^2 + 1, X^3 + 1, -1 + X, -1 + X^2, - 1 + X^3 \rangle_{\mathbb Q} \\&= \langle 2X^2, 2X^3, -1 + X, -1 + X^2, - 1 + X^3 \rangle_{\mathbb Q} \\ &= \langle X^2, X^3, -1 + X, -1, - 1 \rangle_{\mathbb Q} \\ &= \langle X^2, X^3, -1 + X, -1 \rangle_{\mathbb Q}\end{align*}

So, we threw out the $-1$. But obviously that is not in the intersection. So what went wrong?

Sine of the Time

20:13

@pie your question is basically: when $\lim_n c_n \sqrt[n]{a_n}=\lim_n c_n \frac{a_{n+1}}{a_n}$. If $\lim_n c_n \sqrt[n]{a_n}=\lim_nc_n \lim_n \sqrt[n]{a_n}$ then $ \lim_n \sqrt[n]{a_n}=\lim_n \frac{a_{n+1}}{a_n}$ (assuming the lim exists).

So I assume you need $c_n$ and $a_n$ such that the property of the limit of the product holds

I don't know if there's a weaker version

psie

22:05

I'm working an exercise in Rudin's PMA on the expression $b^r$ for $b>1$ and $r$ rational. Nowhere in this chapter has he stated computational rules for $b$ real (except possibly $0$) and $r\in\mathbb N$. I'm wondering, does Rudin assume the reader should know about say $b^{rq}=(b^r)^q=(b^q)^r$ for $r,q\in\mathbb N$ and $b$ real (except possibly $0$)? I think these rules are needed to solve that exercise.

I have an awkward omission of 'not' above. I meant to write "except possibly not $0$".

Frieren

I was wondering: is $\{x \in A \ | \ x \in B\}$ just a different way to write $A \cap B$?

Ben Steffan

Yes.

psie

I second that.

Sine of the Time

(removed)

Ben Steffan

Putting kumquats in your salad is a good idea

Frieren

22:19

So, basically, 'such that' is just a natural-language-way to say 'and'? For instance, when we say "Let $n$ be a even number such that $n \ge 4$" we mean $(\exists k \in \mathbb{Z}, n = 2k) \wedge (n \ge 4)$?

Ben Steffan

no

"There is an $n$ such that $n$ is a natural number" is $\exists n\colon n \in \mathbb{N}$

no "and" in sight

@Frieren the pedantic natural language way to spell that is "there exists a $k \in \mathbb{Z}$ such that $n = 2k$ and $n \geq 4$"

the "such that" notationally corresponds to the ,

I also understand now what the deal is with making your own beans

they're just that much better

I also have two kakis softening as we speak @SineoftheTime

psie

22:39

@BenSteffan are you mixing fruit with other stuff that is not fruit? Gross... :)

Sine of the Time

@BenSteffan nice. I no longer can find kakis at a reasonable price

Ben Steffan

@psie what, have you never made a vinaigrette with some lemon juice?

this is not so far off :)

Sine of the Time

Last time 4€/kg :(

Ben Steffan

:(

Frieren

@BenSteffan Sorry if I'm slow, but I don't completely get this. Isn't saying that an even number is such that the number itself is greater than $4$ the same as saying for it to be an even number and being greater than $4$?

Ben Steffan

22:42

one of the great benefits of living in germany is the abundance of turkish grocery stores and they carry them at reasonable prices still

@Frieren yes, but you're playing a dangerous game of trying to set up a correspondence between natural language and propositional logic

Sine of the Time

Turkish people carrying the economy :D

Ben Steffan

you can do that reasonably well, but only at the cost of being pedantic with your language

Frieren

@BenSteffan: Indeed, I was about to say that maybe I am getting too much philosophical :D

Ben Steffan

it's a perfectly fine thing to ask and/or worry about, but yeah, it requires pedantry

translating from the ordinary mathematical language we use to formal expressions like that is a much less precise business: you should always be able to do it, but usually you'll have to introduce more quantifiers, notation, etc. than the ordinary sentence might contain prima facie

I'm essentially saying that that's what's going on in your example

if you're being precise then "such that" should correspond to a ':' after a quantifier, but since we rarely bother it often looks like it corresponds to a $\wedge$

In particular, $\forall x \in M\colon P(x)$ is already a shorthand for $\forall x\colon (x \in M) \wedge P(x)$

@SineoftheTime the national dish of germany is the döner kebap :)

Thorgott

22:59

I hate the German equivalent of "such that" cause I never know where to put the comma

good thing I write math in English now

Ben Steffan

Not sure why I wrote that $\forall x \in M\colon P(x)$ is shorthand for $\forall x\colon (x \in M) \wedge P(x)$ there but that should obviously by $\forall x\colon (x \in M \implies P(x))$

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