Second day working construction, and biking to/from work site. I already feel lighter and detoxified.

I have a goal in mind: the Cybertruck (by Tesla). Either that or a Fisker Ocean, but I'm liking the performance specifications of the Cybertruck. I think with my student loan debt, I should be behind the wheel of one in at most 5 years, but I may get higher pay work on the way there, so who knows...

Because I also can code, and do math lol. I will also take some CC courses for math & physics, and maybe Spanish, but I want to save up on my own and pay myself, since they're reasonably priced (since they're non-university credits, but they are transferable to NAU).

I think my work on the twin prime problem will still be needed when I finally go to Grad school maybe in 8 years. In other words, I think that at that time in the future, it will still likely be an open problem.

I think I could go there in less time though, maybe in 5 years as well. 3 years of community college then 2 years to finish off a CS or Math degree (hopefully, if they can make use of ~10 year old credits from the same university (NAU)).

But I might be more interested in physics problems by then, who knows...

We need gravitic drives, after all, and portable fusion devices. Those are a must in order to make an efficient spacecraft, or even flying car.

We also need, which is probably closer on the horizon, 70-80% efficient solar cells.

When they reach 60%, and commercial availability, I'm putting them on my car.

"what do you mean by "the same"? Certainly, when resistance is added across the same voltage, current will drop. However, the current on both sides of the resistor will be the same." @robjohn

how do you know both sides will have same current ?

assuming resistor acts like friction

@robjohn you can ignore my question I figured it out.

Given two equivalent metrics say $d_1$ and $d_2$ on X, it is to be shown that convergent sequences in $(X,d_1)$ are same as $(X,d_2)$.

Does it mean that $S\subset (X,d_1)$ is open if and only if $S\subset (X,d_2)$ is open?

If so, then if a sequence $x_n$ has the limit $x$ in $(X,d_1)$ then $x_n$ could have a limit $x'\ne x$ in $(X,d_2)$? Or, $x_n$ must have limit x in $(X,d_2)$? Because the definition does not seem clear to me :(.

koro: yes, the thing you wrote about open subsets

also the thing you wrote about if x_n is convergent in (X,d_1) then it is convergent in (X,d_2)

buried into this, and maybe blended into it? is the uniqueness of the limit for a convergent sequence in a metric space

anyway it's enough to show if x_n converges to x in (X,d_1) it also converges to x in (X,d_2)

So the question asks me to show that $\lim x_n=x$ (in (X,d_1)) is same as showing $\lim x_n=x$ (in (X,d_2)).

if so then suppose that $x_n$ converges to x in $(X,d_1)$. Given $\epsilon\gt 0, \exists N$ such that $n\ge N\implies d_1(x_n,x)<\epsilon$. It follows that $x_n\in B_{d_1}(x, \epsilon)$ for all $n\ge N$. By equivalence of $d_1$ and $d_2$, $B_{d_2}(x,\epsilon)$ is open. So there exists $0<r<\epsilon$ such that $B_{d_2}(x,r)\subset B_{d_2}(x,\epsilon)$.

From here, how can it be shown that $x_n$'s (for $n\ge N$) also lie in $B_{d_2}(x,r)$?

I think by same open set, it means that $B_{d_1}(x,\epsilon)$ is open iff and only if $B_{d_2}(x,\epsilon)$ is open and $B_{d_1}(x,\epsilon)=B_{d_2}(x,\epsilon)$.

with that, the statement is proved.

not quite. knowing that E = B_{d_1}(x,\epsilon) is open in (X,d_2) just means that for any y in E, there's d > 0 with B_{d_2}(y,d) subset E.

they don't necessarily have the same open balls. think back to your linear algebra stuff, if we did that. all norms on a finite dimensional vector space are equivalent, but the euclidean ball of radius 1 about (0,0) in R^2 is not a max-norm ball of any radius about (0,0).

The question says 'same open subsets' so same would give the equality also. That is if S is $d_1-$open then S is $d_2-$open also.Is my understanding correct?

i forget if we actually did that, or if it was all a distraction from axler.

knowing that B_{d_1}(x,epsilon) is open in (X,d_2) doesn't tell you that B_{d_1}(x,epsilon) is a ball in (X,d_2), let alone one of the same 'radius'.

Ah, leslie: the 'equivalent' definition done in LA, is not equivalent to the definition being used in the question.

yes, the one in the question is even weaker.

which should show you that it won't have stronger conclusions.

yes, I'm not so familiar with using equivalent norm related definition in metric spaces :(.

you can characterize equivalence of metric spaces in terms of open balls instead of open sets, but it isn't that two spaces are equivalent iff they have the same open balls.

Let $E=B_{d_1}(x,\epsilon)$. $E$ is $d_1$-open so by equivalence of $d_1$ and $d_2$, $E$ is $d_2$ open as well. $x_n\in E$ for all $n\ge N$ in $(X,d_2)$ so $d_2(x_n,x)<\epsilon$. It follows that $x_n$ converges to $x$ in $(X,d_2)$.

this makes the presentation more clear.

why does x_n in E imply d_2(x_n,x) < epsilon ?

I'm thinking about that. It seems complicated.

normally in these epsilon-N proofs, epsilon has something to do with proximity in the space where you're trying to prove convergence. so if you assume convergence in (X,d_1), maybe start with an epsilon-ball in (X,d_2)...

@leslietownes Assume that $x_n\to x$ in (X,d_1). Let's consider $E=B_{d_2}(x,\epsilon)$. Since $E$ is $d_2$- open, it must be $d_1$ -open. $x\in E\implies $ there is an $r<\epsilon$ such that $B_{d_1}(x,r)\subset E$. By convergence, there exists N such that $n\ge N\implies x_n\in B_{d_1}(x,r)\subset E$. It follows that $x_n\in B_{d_2}(x,\epsilon)$ for all $n\ge N$.

that is, $d_2(x_n, x)<\epsilon$ for all $n\ge N$. It follows that $x_n\to x$ in (X,d_2).

hooray

A zonohedron is a finite Minkowski sum of line segments (the Minkowski sum of two sets $A$ and $B$ is $A+B:=\{a+b\mid a\in A,b\in B\}$). Prove that every zonohedron has a parallelogram face

no it isn't, and so what if it was, and you do it

@leslietownes thanks a lot :-).

This is a puzzle

an oscar anti-slap face mask?

soft question: is it normal to want to familiarize yourself well with cohomology theories for smooth manifolds, before really digging into cohomology theories currently being taught to you for a lecture course in complex manifolds?

or rather, is this good practice? or is it better practice to just dive into whatever is being taught

(assuming you can pull off learning both to personal satisfaction for the former and hopefully well enough exam-wise for the latter before things like exams)

i think im missing something obvious, why is it the case that for a closed manifold $M$, with path-components $M_{\alpha}$, the deRham cohomology groups are a direct sum for each path-component? For say singular homology I can see this because the image of simplices are connected and chains are finite formal sums of singular simplices, so the chain complex splits and the homology groups accordingly, but why is this true for $p$-forms?

if I have infinitely many path components, and my manifold is orientable, wouldn't any volume form be non-zero in every component? So how could I write a volume form as a finite $\mathbb{R}$-linear combination of top-forms on each component?

oh nvm, I guess it really should be a direct product, what I read seems to just be incorrect

for compactly supported cohomology groups, a direct sum makes sense

@robjohn That's problematic consciousness. Even it's omicron, it hurts.

people want to believe so badly that it's over, but there are two new variants out.

This game is over after all people get infected. I've been vaccinated three times but still I got covid.

So I always thought getting covid is an ultimate vaccine for covid.

@love_sodam the vaccine doesn't prevent one from getting the virus, it just aids the immune system in fighting it so that it doesn't get as bad as it would have.

@love_sodam even that won't prevent it. I know of some people who've had it three times.

Viruses are complicated.

@robjohn Ah, right right. I always mistake vaccines for treatments.

@robjohn So the end of the world is coming?

The end of the world as we knew it has already arrived.

Today I've been reading about fixed points of affine maps & non-expansive maps in Hilbert spaces

Why $\Bbb R$-dimensions of $SO(n)$ and $O(n)$ are the same?

@love_sodam: how are you now?

@Koro Still in pain. My throat hurts the most.

when I had it, I had fever so high that I couldn’t even stand up and I was feeling so cold.

@Koro Did you get a vaccine?

morning, @Ted

Hi @robjohn …. Where’s all the rain you promised?

We had about 0.85", but were expecting a half an inch more.

It poured here for an hour yesterday morning. No rain there?

Just a smidge.

Late afternoon.

The important part is whether the mountains got snow. That is important to the water supply.

I'm hoping that is where our missing half an inch went

That's a pretty decent rain

I’m thinking it’s fake news.

like i was saying. government block designs.

@love_sodam yes, I’d got two doses. And I still got covid.

@love_sodam Two components of $O(n)$.

sup

how many mathematicians in the chat

not a ton

how are you doing

OK

how are you?

We’re overweight, but expecting a ton is excessive.,

what a great joke ted

I'm doing good recently

Do we know you by another name?

Got to my ideal BMI in 2 months

only drinking chamomille tea

like Alex G

:)

Another name?

tons of other names

non memorable at least imho

You’re acting like you know us, but I have no clue who you are.

Asionmas heretorelax yorch emperoroficecream chronomas dream

Oh, I'm sorry I acted like I know you

I don't know you

I just have read a ton of your comments

But I'm friendly in general

I don't even know what " acting like you know us means"

I refer younto our attorney… paging @leslie

leslie is your attorney?

I heard that most lesliecoin nodes are compromised rn

that's false. you clearly don't know us if you are disparaging lesliecoin.

twiddles thumbs

yeah right

I'm shorting lc hard af

well, have fun in the cesspits behind the poorhouse.

I will, leslie coin will crumble sooner or later

lesliecoin will outlast the heat death of the universe. it is that immutable.

funds are safu?

is it also non-fungible ?

yes. everything is recorded forever.

I'm pretty sure I could funge tf out of lesliecoin with a couple of beers in me

i should stop aggressively promoting lesliecoin, i'm afraid of getting caught by some bot designed for fishing out scammers.

Chamomile tea leads quickly to beers.

chamomille tea is also a depressant

3 chamomille teas are equivalent to a beer under the chamomileometer

so if u mark 12 on the chamomileometer gl

@Asinomás leslie means that lesliecoin will be worth just as much at the heat death of the universe as it is now.

If $V_1,V_2$ are irreducible representations of a lie algebra $\mathcal{g}$ then $V_1\oplus V_2$ is irreducible too, right?

What does irreducible mean?

the only subrepresentations are 0 and itself

What are the subrepresentations of a direct sum?

$0$, subreps of $V_1$, subreps of $V_2$ and $V_1\oplus V_2$?

So, did you answer your own question?

okay, this was based purely on intuition, so I just wanted to make sure my intuition was correct, before trying anything

It's sort of like asking if the product of two prime numbers is prime.

now do tensor products.

Tensor product of two prime numbers?

yes.

thats subreps of symmetric product, alternating product, 0 and the tensor product, right?

@leslietownes

not sure how to interpret that if V_1 and V_2 are different, but, i was asking about prime numbers.

You was?

like i said. yes. and i can't seem to get a straight answer.

Primes aren't str8.

Yeah, I mean't $V_1=V_2$. Since $V\otimes V= SymV \oplus Alt(V)$

That's false in general, btw.

when is it true then? Here, $SymV$ should be $Sym^2V$ and $Alt(V)$ should be $Alt^2(V)$

For $2$ it's correct. Why? And why not for $k>2$?

i assume counting dimensions you see dimensions not equal

Do you have a more conceptual understanding than that?

if ted were here he'd say, work on some examples.

not sure what you're looking for, you can't write $v\otimes w =\frac{1}{2} (something k alternating + something k symmetric)$

for $v,w\in V$

Where do you think I will find a proof of the statements about the Lyapunov equation found here ee.ic.ac.uk/hp/staff/dmb/matrix/….

Also, looking at the Lyapunov equation, $AX+XA^*=-Q$. If $A$ is stable and $Q$ is pos-def, then $X$ is pos-def. So I can consider the matrix $XA^*X^{-1}$. Is this matrix skew-symmetric by any chance

What does it mean that the functor from Set to Grp sending a set to the free group generated by it commutes with pushouts? what do I need to show? that given a pushout (P,i,j) with i,j the pushout morphisms, we have F(i)F(j)=F(j)F(i)?

This makes no sense.

monty for the first thing about solvability you might consider the operators L and R on matrices given by L(X) = AX and R(X) = XA^H. L and R commute even if A and A* do not. so there is some basis where L and R are simultaneously upper triangularizable. the eigenvalue condition guarantees that the diagonal entries (eigenvalues) of L+R are nonzero so you have solvability. i think that works.

that's only the "if" part though, not the "only if."

to get a handle on that i guess you need to know/prove that the eigenvalues of L are eigenvalues of A and similarly for R and A^H but i think it is all pretty simple.

i.e. it does not go too far beyond definitions.

the other direction might be harder.

algebros

if A is an operator and not a matrix, that idea still works, you just consider L and R inside the C* algebra generated by L, R, and 1 and use the gelfand transform to identify the spectrum of L+R, instead of upper triangularization.

jesus, you know a lot of math @leslie

I don't know what you're saying myself :) Haven't gotten that far

or banach algebra generated by L, R, and 1, i guess.

I guess the area you're speaking of applies to Physics problems

We both used "I guess" strange weirdness

:D

Perhaps some electrons in our brains got entangled from the sunlight

lyapunov did not have the gelfand transform. there must be some concrete way of doing it. or maybe he only cared about matrices.

i wonder if lyapunov even had matrices.

Poor guy

obviously they had them but a lot of that old matrix theory stuff, particularly once you start solving DEs with it, got very ugly back then. at least some of the abstract theory did clean that up.

I'm glad they organized it better, for posterity

gelfand was still around when i was in grad school although i don't think he traveled much. i never met him.

lyapunov had a pretty epic hairstyle. it would have been unworthy of comment at the time, but i think it would make an impression today.

died in odessa.

well, that's enough wikipedia for one day.

I wonder if Gelfand had a girlfrand.

lol alliteration

Wikipedia is impossible to learn math from. I've tried it many a time. Good reference though

That's how I do my hair, he was ahead of his time in the hairstyle department, and mathematically as well

*if I combed it

@leslietownes super thanks for the one side of that proof. I am only dealing with matrices luckily. Any idea if $AX+XA^*=-Q$ is symmetric?

sorry skew-symmetric

is that physics terminology? i might refer to each of A, X, Q, or maybe operators involving the left or right hand side, as potentially being skew-symmetric, but not to the equation itself.

but rhetorical moves like that are pretty common in physics

ooops

I meant the matrix $XA^*X^{-1}$.

sorry

oh, back from the original question. i somehow ignored that :D i don't see why it would be without more information.

what does it mean to say that A is 'stable'? i never used that.

@Myprofileissaved you have to show that the free group on a pushout of sets is the pushout of the free groups of the sets involved

So if you have a pushout P of A<-G->B, then you need to show that F(P) is the pushout in Grp of F(A)<-F(G)->F(B)

(the proof for this is most likely a special case of the proof that a left adjoint preserves colimits, unless you want to get weirdly explicit and work with the actual constructions of pushouts in Set and Grp)