Mathematics - 2024-04-26

Dannyu NDos

00:28

Lemme check this: Under ZFC, regardless of CH, there is a sub-field of the reals with cardinality $\aleph_1$, right?

If CH holds, $\beth_1 = \aleph_1$, so the reals themselves are an example. If ￢CH holds, using Axiom of Choice, we choose $\aleph_1$ elements from the $\beth_1$ elements that extend $\mathbb{Q}$.

Jakobian

@DannyuNDos sure. Choose aleph_1 distinct elements of R and the field generated by them has size aleph_1

Dannyu NDos

01:08

I've just noticed something peculiar about group completion. For consider the ordinal $\omega^2$ as an additive monoid. Since $1 + \omega = \omega$, the group completion will cancel the $\omega$ out to yield $1 = 0$.

A following conclusion is that, group completion is not necessarily injective.

leslie townes

yeah, you need some kind of cancellation condition to hold in the original thing for there to be any embedding in a group

otherwise, the collapsing can be very dramatic

copper.hat

02:06

place is very quiet without Ted the troublemaker

3

Dannyu NDos

I'm playing with Cat.

Equalizers and coequalizers, pushouts and pullbacks, short exact sequences, and such things under Cat.

In particular, I'm testing whether the commutative diagram of forgetful functors between Ab, Grp, CMon, and Mon is a pushout square, or a pullback square, or both.

Lukas Heger

02:23

@DannyuNDos I think it's a pullback square. It should boil down to the fact that a monoid is an abelian group iff it's commutative and a group

Dannyu NDos

That was my thought as well, but a proof should explicitly give $g : \mathbf{C} \to \mathbf{Ab}$ from $f_1 : \mathbf{C} \to \mathbf{Grp}$ and $f_2 : \mathbf{C} \to \mathbf{CMon}$.

user10478

Can someone explain if this (stackoverflow.com/a/64974659/2364796) makes any sense at all Linear Algebraically? I don't get "The literal position of that vector maybe garbage because by checking it in the euclidean space, you will anchor it on the origin." I get vector addition geometry, but surely the only way Queen - Woman ≈ King - Man and Queen - King ≈ Woman - Man is if the vector Queen - Women is near Royal or some similar words and Woman - Man is near some gender toggling words?

Dannyu NDos

@LukasHeger I think I found the answer is negative. For let $\mathbf{C}$ be the category of commutative rings, let $f_1$ be to forget multiplicative structures, and let $f_2$ be to forget additive structures. There is certainly no $g$.

Lucky Chouhan

04:34

@Jakobian yeah,

@XanderHenderson so there is no transparency haha,

@copper.hat then bring him back, call him

@DannyuNDos Are you Schrodinger's fan?

Dannyu NDos

04:58

No, but Heisenberg's.

Sine of the Time

05:57

@Jakobian I don't know him 😅

Soumik Mukherjee

@user10478 I am not sure but it feels like they are saying that the 4 points( queen, king, woman, man) are the vertices of a parallelogram. So the opposite sides represent the same vector. And I guess by anchoring it on the origin, they mean affine transformation of the plane to slide one of the vertices to origin.

@SineoftheTime Hey! Long time no see

Dannyu NDos

Is it possible to justify which the Hopf fibration $\phi : S^3 \to S^2$ is, $1$ or $-1$?

I mean, as a member of $\pi_3(S^2) = \mathbb{Z}$?

Jakobian

10:00

@SineoftheTime oh

@DannyuNDos as a group there is no way to distinguish $1$ and $-1$

Dannyu NDos

@Jakobian Yeah, but as a ring, $1$ is the unity, while $-1$ is not.

So what I'm asking is the ability to dedicate the Hopf fibration as $1$ or $-1$.

For comparison, the identity map in $\pi_2(S^2) = \mathbb{Z}$ is definitely $1$ and not $-1$, justified by the notion of degree.

nickbros123

10:20

isnt this inaccurate?

shouldnt we be using stars and sticks method to count this

Thorgott

10:34

@DannyuNDos It is a pullback square for the reason Lukas stated. It is not a pushout square, because the functor taking a category to its underlying set (insert parenthetical remark about size issues here) is a left adjoint, hence preserves colimits, but not every monoid is commutative or a group

Dannyu NDos

@Thorgott Why doesn't my counterexample work then?

For pullback

Thorgott

cause your two functors don't agree as functors into the category of monoids

Dannyu NDos

Oh.

Thorgott

@DannyuNDos also, this question only makes sense if you specify how you identify $\pi_3(S^2)=\mathbb{Z}$ and that is done by specifying a Hopf fibration, so the question is ultimately circular

Dannyu NDos

I should've looked into definitions more carefully.

Thorgott

10:38

nor do I think there even is "the" Hopf fibration, in the literature oftentimes either one is called Hopf fibration

Jakobian

10:53

@Thorgott isn't this basically my point

@DannyuNDos like this is conventional also. And those aren't rings

No matter what isomorphism you choose $G\cong \mathbb{Z}$, you can compose it with $x\mapsto -x$

So question of the form "what does $g\in G$ corresponds to" can just as well be $n$ as $-n$. Only when comparing two elements one can more clearly tell their relative signs

Or when you "fix an isomorphism" by letting, for example, the identity to correspond to $1$

Those aren't rings, we can't distinguish 1 and -1

YourLordJoyBoy

11:21

Final today....

How's everyone today?

Thorgott

typically one would probably say "the" Hopf fibration is the sphere bundle $S^3\rightarrow\mathbb{CP}^1$ of the tautological bundle followed by an orientation-preserving homeomorphism $\mathbb{CP}^1\cong S^2$ (this is at least well-defined up to homotopy or even isotopy) and use this fibration $S^3\rightarrow S^2$ as generator of $\pi_3(S^2)$, so that it corresponds to $1$ once cyclicity is established, but there's nothing stopping you nor any dire consequences to reversing orientation instead

Pizza

12:07

Hello, can someone help me understand Lagrange's theorem (or mean value or finite increment). Suppose $f(x)$ is continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$. Then there exists a number $( c )$ in $(a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.Why must f(x) be continuous in $[a,b]$ , and why do we care that it is only differentiable in $(a,b)$?

YourLordJoyBoy

I am struggling with a problem that is to help with my prep for the final. The electric bill charge for a certain utility company is $$0.082$$ per kilowatt-hour plus a fixed monthly tax of $$11.63$$. The total cost, y, depends on the number of kilowatt-hours, x, according to the equation $$y=0.082+11.63, x≥0.$$

I do hope this makes sense. Can I have some assistance in understanding what to do?

Jakobian

12:43

@Pizza are you asking why there's a difference between where we assume its differentiable and where we assume its continuous?

If you want to understand the assumptions of this theorem, look at the proof of it

Soumik Mukherjee

13:02

@YourLordJoyBoy What is your question? also the equation you wrote is incorrect

Pizza

@Jakobian Suppose $f(x)$ is continuous on the closed interval $[a, b]$. Why does it have to be continuous?

Jakobian

@Pizza I don't understand your question

Pizza

@Jakobian why the function in the theorem must be continuous, what happens if it isn't?

Jakobian

But $f$ is still differentiable on $(a, b)$?

Pick some function for which the theorem is true, lets say $f(x) = x$ for $x\in [0, 1]$. And now replace it with $g(x) = f(x)$ for $x\in [0, 1)$ and $g(1) = s$

Theorem of Lagrange says that there should exists $c\in (0, 1)$ with $g'(c) = f'(c) = 1 = \frac{g(1)-g(0)}{1-0} = s$

but this can only happen if $s = 1$

so continuity on whole interval $[a, b]$ is necessary

Pizza

@Jakobian i dont know

Jakobian

13:12

the example I just gave you shows that you can't get rid of the conditions of this theorem completely

I'm sure there exist some generalizations of it e.g. we might not require differentiability on whole of $(a, b)$ and still manage for the theorem to hold

But to really make sure of why we are assuming those things, see the proof of this theorem and steps where its necessary for the function $f$ to be continuous/differentiable

Pizza

@Jakobian this pls

I don't understand why f'(c)=1

Soumik Mukherjee

@Pizza Derivative of $f(x)=x$ is 1 right?

Pizza

yes

Xander Henderson

13:27

@Pizza Look at the proof of the theorem. Is differentiability at the endpoints ever used?

If the answer is "no", why add in an extra assumption? particularly when that assumption requires us to define "differentiability at the endpoints", which can be a little dicey (most elementary texts define differentiability at a point $a$ which is contained in an open interval which is contained in the domain of the function of interest.

Pizza

@XanderHenderson the function could also be differentiable at the extremes, but it doesn't matter, right?

Xander Henderson

@Pizza Look at the proof of the theorem! Where is differentiability used?

Pizza

@XanderHenderson it is not used, because it is not important

Xander Henderson

@Pizza Go reread the proof. Differentiability is definitely used.

Pizza

I mean it doesn't matter if it's differentiable at the extremes

Xander Henderson

13:30

@Pizza I don't know what you mean by that.

Do you mean at the endpoints?

Pizza

yes

Xander Henderson

("Extremes" usually refers to extreme values, i.e. minima and maxima.)

Yes, differentiability at the endpoints is not used. So why would you make the additional, unnecessary assumption that the function is differentiable at the endpoints?

Pizza

@XanderHenderson and why is it not needed?

Xander Henderson

@Pizza Look at the proof.

Nowhere in the proof do you need differentiability at the endpoints. YOU agreed to that.

So why add in the extra assumption?

Pizza

okok

instead continuity is necessary as Jakobian told me, right?

Xander Henderson

13:37

Yes, because there are counterexamples if you remove continuity.

Soumik Mukherjee

@Pizza Also if you add that unnecessary assumption of differentiability at endpoints, you lose valid examples like sqrt x on [0,1]

Pizza

@SoumikMukherjee why?

Xander Henderson

@Pizza Because that function is not differentiable at $0$, but it otherwise has the "mean value property".

Pizza

I saw this example from Jakobian but I don't understand this: $g(1) = s$.
what is g(1) and what is s?

@XanderHenderson oh... yes

Xander Henderson

I.e. on any interval $[0,n]$, there exists a point $c \in (0,n)$ such that $n \sqrt{c} ' = \sqrt{n}$.

Pizza

13:48

nice

Xander Henderson

(by abuse of notation)

John Zimmerman

14:08

any graph theorists in the house?

a graph is a collection of vertices and edges. I'm casually wondering what if the vertices were edges and the edges were surfaces. Then the edges would have to be connected by a surface. And then a homotopy would get you from edge to edge, or sea to shining sea.

Xander Henderson

@JohnZimmerman That isn't quite right. A graph is a triple $(V,E,\varphi)$, where $V$ and $E$ sets (thought of as vertices and edges), and $\varphi$ is a function $E \to V\times V$ (for directed graphs).

Your vertices and edges can be whatever you like, but the structure of the graph comes from the incidence function $\varphi$, which tells you how the edges are "glued into" the graph.

It is not clear to me how your construction works in that setting.

Koro

27bslash6.com/strata.html

Xander Henderson

@Koro You might want to explain why you are linking that here, 'cause it looks real spammy to me at first blush.

Pizza

14:23

@XanderHenderson is this mean value theorem for derivatives?

Xander Henderson

Though reading past the first two paragraphs is a bit more enlightening.

Koro

I see. It's a funny email exchange between tenant and the owner.

(too late to delete it now)

Xander Henderson

No, it's fine. The first paragraph really threw me.

Koro

This link didn't post with a little preview.

(sometimes a preview is also added with the link but it didn't this time). Nvm

Lukas Heger

@JohnZimmerman I think maybe a simplicial complex or a simplicial set is what you're after

a 1-d simplicial complex is a graph

a 2-d simplicial complex also has vertices and edges, but it also has something like "surfaces between the edges" if you want

this continuous to any dimension

I'm not 100% sure this is what you want, but it seems the closest thing that came to mind

Jakobian

15:05

@Pizza $g(1)$ is $g$ evaluated at $1$, and $s\in\mathbb{R}$ is some unspecified parameter

Pizza

@Jakobian ok thanks!

Jakobian

For Lagrange theorem to be true we need to have $s = 1$

Pizza

Can I ask you if this Lagrange proof is ok?

Jakobian

sure. Don't ask to ask

Pizza

For a function $f(x)$ continuous on $[a,b]$ and differentiable on $(a,b)$, there exists a $c ∈ (a,b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$

Proof:
$g(x) = f(x) + kx$ (this function is such that it verifies all the hypotheses of Rolle's theorem)

$g'(x) = f'(x) + k$ the function g(x) is differentiable at the points (a,b)

For Rolle I have to impose this condition:
$g(a) = g (b) \Longrightarrow f(a) + ka = f(b) + kb$
$\Longrightarrow f(b) - f(a) = ka - kb$
$\Longrightarrow f(b) - f(a) = -k(b-a)$

Jakobian

15:08

@Pizza yep that's correct

Xander Henderson

The proof is fine, though I have some stylistic complaints.

First, I would just define $$g(x) = f(x) - kx$$ where $k = \frac{f(b)-f(a)}{b-a}$ from the start. You don't need to "discover" this value in the proof.

Pizza

the arrows that I put maybe?

I didn't know how to connect the steps

Xander Henderson

Then verify that $g$ satisfies the hypotheses of Rolle:

Pizza

@XanderHenderson ah ok

Xander Henderson

> Note that $g(a) = g(b) = 0$, that $g$ is the sum of continuous functions on $[a,b]$ and is therefore continuous, and $g$ is the sum of differentiable functions on $(a,b)$ and is therefore differentiable.

The proof I give to my students looks like the following:

Pizza

15:16

now i see

but are your proof public ?

if there is a pdf I mean

but there should be a point where the tangent is parallel to the secant right?

@XanderHenderson but what is that part written in light blue at the beginning?

John Zimmerman

@LukasHeger yes thank you.

Pizza

in theory this proof is connected to the proof of some Lagrange applications, right?

psie

15:54

I'm reading about the Lebesgue-Stieltjes outer measure on $\mathbb R$, $$\mu ^{\ast }\left(A\right)=\inf \left\{\sum _{i\in \mathbb{N}}^{ }\left(F\left(b_i\right)-F\left(a_i\right)\right):\ A\subset \bigcup _{i\in \mathbb{N}}\left(a_i{,}b_i\right]\right\},\tag1$$where the infimum is taken over all countable covers with half-open intervals. There is a well-known theorem that describes all finite measures on $(\mathbb R,\mathcal{B}(\mathbb R))$.

In particular, I'm reading a passage of the proof that every bounded, right-continuous and increasing function induces a unique finite measure $\mu$. The proof of the existence of this measure is clear to me, but what I don't understand is why $\mu$ is finite. I'm trying to plug in $\mathbb R$ into $(1)$, but without success. Any ideas?

Jakobian

@psie because the function is bounded

$\mu^*((a, b])\leq F(b)-F(a)$ should be clear from definition

psie

ok, I think I understand, thanks

Jakobian

In fact this should be equality, but for arguing that its bounded an inequality is enough

psie

so we simply take $(-\infty, b]$ and let $b\to\infty$

I should have added, $F(-\infty)=0$

Pizza

16:22

Let $f,g : (a,b) \to \Bbb R$ differentiable, with $g'(x) \neq 0 \ \forall x\in(a,b)$
If these 3 conditions are met:
1) $\lim_{x\to a^+} f(x)=0, \ \lim_{x\to a^+} g(x) = 0$
2) there is finite or infinite $\lim_{x\to a^+}\frac{f'(x)}{g'(x)}$
If these things happen then also the $\lim_{x\to a^+}\frac{f(x)}{g(x)} = \frac{f'(x)}{g'(x)}$

If $\lim_{x\to a^+}\frac{f'(x)}{g'(x)}$ does not exist, this does not mean that $\lim_{x\to a^+} \frac{f(x)}{g(x)}$ cannot exist.
What does it mean that $\lim_{x\to a^+}\frac{f'(x)}{g'(x)}$ doesn't exist?

I was thinking of a limit that cannot be resolved

but if someone can clarify this thing, it would be better

Jakobian

@psie sure

Jakobian

16:40

@Thorgott can you tell what is going on in this theorem?

Xander Henderson

@Pizza No, not really. Those are the notes created during lecture with my class.

@Pizza That is an explanation to the students that the function $\ell$ "just' the line through a point with a slope, so we are using the point-slope equation for a line to generate that function.

The intuition is that $\ell$ is the secant line through $(a, f(a))$ and $(b, f(b))$.

@Pizza Yes, that is the point given to you by Rolle's theorem. It is called $c$ in the argument presented.

Pizza

thanks!

Xander Henderson

@Pizza What does it mean, in general, to say that $\lim_{x\to a} f(x)$ does not exist?

Pizza

@XanderHenderson neither the definition of finite limit nor that of infinite limit is satisfied

Xander Henderson

Okay, so you've answered your question...

Pizza

16:55

@XanderHenderson Do you have any examples in mind of limits that don't exist?

Xander Henderson

@Pizza $x \sin(1/x)$ at zero?

(thought of as $\frac{\sin(1/x)}{1/x}$)?

Pizza

@XanderHenderson but if I don't use hopital I can solve it right?

maybe $\sin(\frac{1}{x})$

Xander Henderson

@Pizza That's what I meant.

Pizza

ah okok

so with hopital this limit cannot be resolved, but using Taylor's developments for example it is possible right?

Jakobian

I've figured out that they mean $\text{End}(X^2)$ and not $(\text{End}X)^2$

not sure where the tensor product comes from

I suspect the last thing should be $\text{End}(X)\otimes_\mathbb{R} \mathbb{R}(2)$ because dimension should increase four times

not sure how you'd interpret it as a subset of $\text{End}(X^2)$

oh my bad

$\text{End}(X^2)\cong \text{End}(X)\otimes_\mathbb{R} \mathbb{R}(2)$ so it should be some error in the text maybe

Thorgott

17:53

@Jakobian I don't even understand the terminology/notation. What is ${}^2K$? What is a subset of type $(p,q)$?

Jakobian

18:10

@Thorgott ${}^2K$ is $K^2$ with pointwise product iirc

and an orthonormal subset of type $(p, q)$ of an algebra is a set of points $e_1, ..., e_{p+q}$ such that $e_i^2 = 1$ for $1\leq i\leq p$ and $e_i^2 = -1$ for $p+1\leq i\leq p+q$ and $e_i, e_j$ anti-commute for $i\neq j$

because once you put a symmetric bilinear form $a\cdot b = -\frac{1}{2}(ab+ba)$ on the linear span of $\{e_1, ..., e_{p+1}\}$ then it becomes an orthonormal subset in the sense that $e_i\cdot e_i = \pm 1$ and $e_i\cdot e_j = 0$ for $i\neq j$

which is isomorphic to a specific vector space with bilinear form called $\mathbb{R}^{p, q}$

which is $\mathbb{R}^{p+q}$ with quadratic form $(a_1, ..., a_{p+q})\mapsto -\sum_{i=1}^p a_i^2 + \sum_{i=p+1}^{p+q} a_i^2$

this is part of my study of Clifford algebras

Jakobian

20:12

@Thorgott It seems to be that $\text{End}(X^2)$ is $A$-linear endomorphisms and so it seems to make sense that the author wrote $\text{End}(X)\otimes_\mathbb{R} \mathbb{R}(2)$ instead of $\text{End}(X^2)$

or wait... should it be $\text{End}(X^2)$ actually

no there shouldn't be any square

Thorgott

22:09

weird stuff

Jakobian

22:20

@Thorgott can you confirm to me that $A(n)\otimes_\mathbb{R} \mathbb{R}(2) = A(2n)$ for an $\mathbb{R}$-algebra $A$?

$K(n)$ meaning $n\times n$ matrices over $K$

honestly I thought it might come close to your interests with all the linear algebra

and how close it feels to matrix groups

Thorgott

@Jakobian yeah, you have (coordinate-free) $\mathrm{End}_K(A)\otimes_K\mathrm{End}_K(B)\cong\mathrm{End}_K(A\otimes_KB)$ whenever $A,B$ are finite-dimensional algebras over a field $K$

the isomorphism takes the formal tensor product of two maps to their actual tensor product

what's going on here is that $\mathrm{End}(X^2)=\mathrm{End}(X)(2)=\mathrm{End}(X)\otimes_{\mathbb{R}}\mathbb{R}(2)$ (so I believe there is a typo at the very end in your screenshot)

actually, I may be wrong

I don't see how the Prop would be true for $X=\mathbb{R}$

these diagonal matrices, reflection along the diagonal and a 90° rotation do not generate all 2x2 matrices

idk perhaps I'm tired

Jakobian

22:42

@Thorgott $\begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}+\begin{bmatrix} 1 & 0 \\ 0 & -1\end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 0 & 0\end{bmatrix}$ for example where $\begin{bmatrix} 0 & 1 \\ 1 & 0\end{bmatrix}^2 = \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}$ and $\begin{bmatrix} 0 & 1 \\ 1 & 0\end{bmatrix}\begin{bmatrix} 0 & -1 \\ 1 & 0\end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & -1\end{bmatrix}$

and from $\begin{bmatrix} 0 & -1 \\ 1 & 0\end{bmatrix}$ and $\begin{bmatrix} 0 & 1 \\ 1 & 0\end{bmatrix}$ you can get $\begin{bmatrix} 0 & 1 \\ 0 & 0\end{bmatrix}$ and $\begin{bmatrix} 0 & 0 \\ 1 & 0\end{bmatrix}$

this should then get you $\begin{bmatrix} a & 0 \\ 0 & 0\end{bmatrix}$ and so on for $a\in S$ so you'll get $\begin{bmatrix} x & 0 \\ 0 & 0\end{bmatrix}$ for $x\in\text{End}(X)$

and same for every coordinate

maybe you were thinking only about generation multiplication-wise

Thorgott

oh yeah, addition exists

I was indeed just thinking of the multiplicative monoid

tired indeed lol

John Zimmerman

23:23

Can a non-meromorphic function be meromorphic at infinity?

Lukas Heger

@JohnZimmerman sure

just define the function to be some non-meromorphic nonsense inside the unit ball, and zero outside of it

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