Mathematics - 2021-12-09 (page 2 of 2)

7:03 PM

If x is in D/S’ then x is not a limit point of S hence there’s a deleted nbd around x that doesn’t intersect S. Ohh and inside that nbd, I can’t go and say let V be a small nbd inside. Why? Because if there are two elements inside they will be separated by distance 1.

It still needs work

Nah, it’s fine. If a point y is such that d(x,y)=1 then y must lie in S.

leslie townes

yeah, OK. subsets of a discrete space can't have limit points

S' will be empty for any S, even if S is the whole space

Koro

You beat up my question to death again :)

Thanks a lot for that :-)

@Alex I think that needs more work. Unless I’m missing something, I understand that using Cesaro you can show that $\lim \frac {\sum x_n}n=\lim \frac {x_{n+1}}{n+1-n}\to 0 $. I’m not sure how you obtained $x_n\sim 1/n$.

robjohn

@amWhy Ted is so fun to provoke; how else can one get smacked?!

amWhy

@robjohn Good point! ;D

robjohn

7:21 PM

@Koro $x_n-x_{n+1}=x_n^2\implies\frac1{x_{n+1}}-\frac1{x_n}=\frac{x_n}{x_{n+1}}\sim1$

monoidaltransform

What is $[\nabla_{\alpha},\nabla_{\beta}]g_{\mu,\nu}$?

robjohn

Perhaps $\nabla_\alpha\nabla_\beta g_{\mu,\nu}-\nabla_\beta\nabla_\alpha g_{\mu,\nu}$? or were you looking for something more descriptive?

monoidaltransform

More descriptive, with christoffel symbols

that's true and i'm assuming connection is levi civita

Koro

@robjohn yes, thinking more on this made me understand :)

monoidaltransform

so it must be 0

but I want expression in terms of christoffel symbols

Or instead, I would like to know that $\nabla_{\alpha}(\partial_{\beta}g_{\mu,\nu})$ is

robjohn

7:27 PM

is $\nabla_\alpha=\alpha\cdot\nabla$?

monoidaltransform

$\nabla$ is connection on manifold

Alex

@robjohn Is there an alternative way to write the proof without using Cesaro? A more elemental way?

I would like to write a proof without Cesaro 🤔

robjohn

@Alex you could essentially reprove Cesaro

@WojciechKulma that image won't display

Alex

@robjohn I was thinking in the following way: we have $a_{n}\sim 0$

Wojciech Kulma

hey @robjohn, many thanks, I figured it out already!

Alex

7:38 PM

I will try to write again

monoidaltransform

$\partial_{\alpha}\partial_{\beta}g_{\mu,\nu}-\Gamma^{\sigma}_{\alpha \beta}\partial_{\sigma}g_{\mu\nu}-\Gamma^{\sigma}_{\alpha \nu}\partial_{\beta}g_{\sigma \nu}-\Gamma^{\sigma}_{\alpha \nu}\partial_{\beta}g_{\mu \sigma}$

is that it?

Alex

Does it work? We have $x_n \sim 0$. Suppose that $\sum x_n <\infty$ so wr have $\prod (1-x_n)\sim a>0$. But $x_{n+1}/x_{1}=\prod (1-x_n)\sim a>0$, contradiction to $a_n\sim 0$

$x_{n+1}/x_n=1-x_n$

Koro

It looks correct. But $x_n\sim 0$ is probably an abuse of notation? As by $a_n\sim b_n$, it is understood that $\frac {a_n}{b_n}\to 1$.

Alex

@Koro yes, but it's more easy to write that $\lim x_n=0$. I'm sorry :(

Koro

I see. But I like your argument to avoid Cesaro's theorem. :)

Alex

7:48 PM

Thank you so much for your corrections

Koro

You're welcome!

Alex

@Koro :3 thanks

For now I'll go play chess for a bit or maybe go for a run like @copper.hat :) Today seems to be a good day

copper.hat

@Alex unfortunately my 'running' is more fast walking. my competitive running days are over :-)

Alex

@copper.hat At some point in my life, I participated in a 5km race. I did not win, but I was in a position number 10 of a total of about 200 participants. Did you also do athletics?

copper.hat

@Alex just running, cross country.

i like outdoors stuff, but have some restrictions due to accumulated injuries :-).

Koro

8:01 PM

Copper, did you ever go to Hyderabad?

Thorgott

@monoidaltransform that's awful

copper.hat

@Koro Unfortunately the furthest south I went was Kolkata.

monoidaltransform

i know

i'm trying to get an expression for the commutator of the metric tensor

commutator of the connections on the metric tensor I mean

Thorgott

8:37 PM

are you sure you don't want curvature instead

monoidaltransform

what do you mean? The problem is : Calculate the left hand side of : $[\nabla_{\alpha},\nabla_{\beta}]g_{\mu \nu}=0$

@Thorgott

Thorgott

that expression does not make sense to me

monoidaltransform

$[\nabla_{\alpha},\nabla_{\beta}]g_{\mu \nu}=\nabla_{\alpha}\nabla_{\beta}g_{\mu \nu}-\nabla_{\beta}\nabla_{\alpha}g_{\mu \nu}$

@Thorgott

Thorgott

that still doesn't actually explain the symbols

there's a ton of implicit things going on when you write an expression like that and I want you to spell them out, both so that you get a better understanding of what you wanna do and so that I get an understanding of what you wanna do at all

monoidaltransform

So $\nabla_{\beta}g_{\mu \nu}=\partial_{\beta}g_{\mu \nu}-\Gamma^{\sigma}_{\beta \mu}g_{\sigma \nu}-\Gamma^{\sigma}_{\beta \nu}g_{\mu \sigma}$

which is 0

Thorgott

8:53 PM

writing down equations is not what explaining means

jay

9:07 PM

hi

people use the term robust quite loosely in mathematics right

Xander Henderson

@jay Context?

jay

for instance, say you have some method to solve ODEs $\dot{x}(t)=f(x(t))$, which only works ( or has only been proved for ) a small class of functions $f$, would it be ok to say such a particular method is not robust ?

I guess more specifically you would just say not yet generalised

i just hear people throw around the word robust and wonder if it has a more specific meaning than 'strong/healthy'

Ted Shifrin

Maybe stability of a numerical solution?

Xander Henderson

@jay I don't like that usage very much.

I think that @TedShifrin has the context moar gooder.

In my experience (which is far from exhaustive), "robust" tends to refer to numerical methods.

A numerical method is "robust" if, for example, small perturbations or errors don't cause problems (i.e. if the method is "stable").

Ted Shifrin

@monoidal if the covariant derivative of the metric is $0$ (which it is), then any further covariant derivative is the derivative of $0$.

jay

9:19 PM

@XanderHenderson hmm ok, I hear it outside numerics community too, anyhow thanks :)

Xander Henderson

@jay Which is why I asked for context.

jay

Yeah sorry I cant rememebr exactly

Xander Henderson

As I said, in my experience, "robustness" is an adjective which I associate with numerical methods.

jay

someone talking about martingales is all haha :)

yes I thought so too, cheers!

Xander Henderson

Numerical stability

In the mathematical subfield of numerical analysis, numerical stability is a generally desirable property of numerical algorithms. The precise definition of stability depends on the context. One is numerical linear algebra and the other is algorithms for solving ordinary and partial differential equations by discrete approximation. In numerical linear algebra the principal concern is instabilities caused by proximity to singularities of various kinds, such as very small or nearly colliding eigenvalues. On the other hand, in numerical algorithms for differential equations the concern is the growth...

> Some numerical algorithms may damp out the small fluctuations (errors) in the input data; others might magnify such errors. Calculations that can be proven not to magnify approximation errors are called numerically stable. One of the common tasks of numerical analysis is to try to select algorithms which are robust – that is to say, do not produce a wildly different result for very small change in the input data. [emph added]

copper.hat

9:27 PM

I would not say that the word is used loosely, I would say that it is qualitative.

monoidaltransform

@TedShifrin yes ofcourse

But he wants us to find another expression of the covariant derivative

Xander Henderson

@copper.hat Something something condition number something something.

maths researcher

@monoidaltransform More generally, could you find an expression for $[\nabla_{i},\nabla_{j}]T_{a b}?$ where $T_{a b }$ are the components of a (0,2)$ tensor?

copper.hat

@XanderHenderson I have seen the term used in a variety of situations, not just numerical methods, but always in the context of a sensitivity of sorts. I suppose that just replaces one word by another...

monoidaltransform

10:03 PM

$T^{\sigma}_{\beta}g_{\sigma \mu}$ evaluate to? $T_{\beta \mu}$? Note the metric is on the right now

robjohn

10:16 PM

I am forced to use the mobile interface for a while. I notice that the buttons at the top of the page to see responses and reputation don’t create a drop down menu like they used to. Is anyone else seeing this? On my laptop, they were working fine.

geocalc33

11:29 PM

Can the manifold $\Bbb R^2$ be thought of as the gluing of four components along boundaries?

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$Mathematics$ Mathematics