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02:22
@psie What kind of bug? And was it really a bug?
Are you sure it was from the order hemiptera?
@Jakobian yes, and also uniqueness
I mean, I need basis in $V$ given a basis in $V^*$ so that $f_i(\alpha_j)=\delta_{ij}$
As of yet i dont know what $V^{**}$ is
02:47
@nickbros123 I thought that this boils down to the same thing, so its a bit circular
what I said that is
$V^{**}$ are linear functionals on $V^*$
The map which sends $v$ to the linear functional on $V^*$ given by $f(v)(w) = w(v)$ is actually an isomorphism of vector spaces
since $V$ is finite-dimensional
and so we are guaranteed that we can find such element $v\in V$, but the issue is how
and I don't think I really know how, unless perhaps we set up some basis of $V$ and go from there
so what I think is best solution is to set up a basis of $V$ and then set up some linear system of equations to solve your problem
@Jakobian I was thinking the same. Assume some basis, write an arbitrary transformation (in terms of the given $f_i$ and do a change of basis by post multiplying an invertible matrix, and argue about that mattix
well, its just a matter of setting up the linear system of equations
the solution and the "argue about the matrix part" comes after
this is still somewhat complicated thing to do, since given some basis $\beta_j$ of $V$, you need to know what $f_k(\beta_j)$ are
but I think this is necessary, if you didn't know how $f_j$ act on $V$, then you wouldn't be able to solve for $\alpha_j$ most likely
I could attempt to make this "wouldn't be able to solve" more rigorous but I don't see a need to
 
1 hour later…
04:22
4
Q: Quinn's Uniqueness of Resolutions

user535484I am reading the proof of proposition 3.23, page 288 in https://www.maths.ed.ac.uk/~v1ranick/papers/ends1.pdf, concerning the uniqueness of resolutions. Following the notation in the paper, since $X$ has dimension at least 5 and is resolvable, then crossing with a 2-torus yields a finite dimensi...

 
6 hours later…
10:37
@XanderHenderson these guys, so apparently from the order Diptera :)
11:11
@Jakobian yeah I also think it should be true for a.e case, I feel like the proof should involve trying to guess the situation when this expression (the original expression I talked about involving the rational functions with double poles) is maximized at the origin (because we can always reduce to this case via a möbius transformation of the unit disk into itself), at the origin this expression is rotationally invariant, so maybe a good guess would be the symmetric configuration of z_j..
@Jakobian then it should be possible to show for specific “nice” choices of coefficients c_k ,c_k’ that when z_j are roots of unity, the absolute value of the expression at the origin cannot exceed the maximum boundary value. It’s just harder than it appears to actually make this concrete
Btw ignore the subsidiary questions I asked after the first one, they were misguided and don’t help solve the first one (for those subsidiary questions the answer is easy because we are asking if the zero set of an analytic function has measure zero, which it always does)
Basically the hardest part is showing that the symmetric configuration is extremal
Then finally we would have to perturb the “nice” choice of coefficients c_k to get the result for a.e such rational function, this part would also not be easy
I’m tempted to just to write some small script that just checks loads of random cases, does anyone know of a good tool to use to do this? I want to compute the maximum value of an n variable positive function of y_1,…y_n of the form f(x_1,…,x_n)(y_1,…,y_n) defined for (y_1,…,y_n) in [0,2\pi]^n, and the function is a polynomial in the parameters x_1,…,x_n , which are real and between 0 and 1. I’m guessing there are lots of software packages I can use to do this?
Actually it’s a polynomial in 1/x_1,…,1/x_n, but this isn’t important
So I want to be able to compute the maximum for loads of randomly selected x_i’s
With the constraint x_1 + … + x_n =2
11:27
@psie Crane flies are not bugs!
they're huge!
yeah probably not a bug then
@psie But not bugs.
12:11
When is a bug not a bug?
A: when it's a feature :P
@user20458579510081670432 if something is a bug, then it's a bug, if its not a bug, then its not a bug. how can a bug also not be a bug?
🐞👾🪲🐛
That was a play on the two meanings of the word "bug."
A "bug" can also occur in computer programs.
In engineering, a bug is a design defect in an engineered system that causes an undesired result. Although used exclusively to describe a technical issue, bug is a non-technical term; applicable without technical understanding of the system. The term bug applies exclusively to a system that is (human) designed; not to a natural system; and that the issue is within the influence of human control. For example, humans have faults but not bugs, and a server crash due to natural disaster is not a bug. In addition to or instead of defect, some use: error, flaw or fault. Engineered systems is a broad...
12:37
Right.
Mad
Mad
13:14
i am very confused about the usage of terminology.

We call elements of the dual space Covectors. And define a tensor as a multilinear map $T^s_r$ where s is the number of copies of $V^*$ and then we say s the "contravariant" part.
You would expect, since $V^*$ elements are called covectors, that $s$ is the covariant part?
Nevermind, i understand why they are calling it like that, very tricky
 
1 hour later…
Mad
Mad
14:33
i am also struggling to understand the concept of the order of the definition.

is there a fundemantal difference between $ V \times V^* \times V $ and $ V^* \times V \times V$ as domains of definition, i mean, these two spaces are isomorphic as product spaces.
@user20458579510081670432 There is also the ambiguity of the term "bug" in everyday English vs the more technical definition used in etymology. In everyday English, many people will call any insect a bug (I've even heard people call spiders and centipedes bugs).
15:39
I'm working an exercise on the completion of metric spaces. We say that a complete metric space $Y$ is the completion of $X$ if $X$ is a dense subspace of $Y$.
> Show that when $Y$ is a completion of $X$, then the inclusion map $X\to Y$ extends to an isometry of $\tilde X$ onto $Y$.
Here $\tilde X$ is the set of equivalence classes of the set of Cauchy sequences in $X$, under the relation $\tilde s=(s_n)\sim(t_n)=\tilde t$ to mean that $d(s_n,t_n)\to0$ as $n\to\infty$. The metric on $\tilde X$ is $\rho(\tilde{s},\tilde{t})=\lim_{n\to\infty} d(s_n,t_n)$.
Question: I don't understand the exercise. What does it mean for the inclusion map $X\to Y$ to extend to an isometry of $\tilde X$ onto $Y$?
Can anyone suggest literature for motivation of box spaces and asymptotic dimension?
(asymptotic dimension of groups)
@flowian as far as I can remember the main motivation for asymptotic dimension is that it is a quasi-isometry invariant, unlike classical notions of dimension
So it is "the correct" thing to consider in geometric group theory
@AlessandroCodenotti Makes sense, thank you.
I'm looking for some background to fill into my thesis.
Jam
Jam
16:25
Can you help me find all ideals of $\mathbb{z}[x]/<x^3-x>$
i can use either the correspondence theorem or the chinese remainder theorem for rings to proz my ring is isomorphic to $ZxZxZ$
16:36
How can I count the number of distinct weakly connected DAGs on 4 labelled nodes?
@Jakobian nvm, i realized what I hoped was true fails spectacularly badly. You can get a huge class of examples where the absolute value of the expression at 0 exceeds the maximum boundary value by taking the Schwarzian derivative of a Schwartz christoffel mapping onto a very long and thin rectangle , I.e with one dimension much larger than the other. This gives a positive measure class of rational functions which for that expression obtains a maximum value only in the interior
But it is likely that for Schwarzian derivatives of maps onto regular polygons, it doesn’t fail.
Any computer geeks around here? I was writing a code to find the root of a function, using a) bisection of interval method and b)Newton Raphson Method. What I noticed is that the bisection method takes 52 iterations to converge, for almost all but a few trivial intervals I give it. why is it like that?
newton method converges anywhere between 5-20 iterations as long as my input number is fairly "slopey" (i.e, its slope is not close to 0)
would it be better to ask this question in computer forum instead of math forum?
Jam
Jam
16:53
actually my isomorphism is wrong.
Jam
Jam
17:22
math.stackexchange.com/questions/4959674/… if any1 is up for some algebra.
17:50
is it possible to convert rotation to diffusion?
18:02
If I had to try to start solving that I would take some unique solution to a diffusion pde on a region in the plane, represent it with constant time slices (planar picture) and build a section (of a fiber bundle) of these slices, viewed as line bundle in R^3 with a prescribed geometry s.t. a rotation of the bundle over the base projects directly onto the diffusive pde solution
The question is what use would this be?
and is "projection" even a valid "conversion" technique here
perhaps its more kosher to say generalized diffusion and planar diffusion
@psie It means that there is a map $\tilde{X}\to Y$ which is an isometry and such that its restriction to $X$ (where we treat $X$ as the, up to equivalence, constant sequences) is the map which is the inclusion of $X$ into $Y$
18:52
@psie I threw a bug in boiling hot water once for daring to bite me
19:05
I can't find a simple question: how does one show a cardinality is infinite iff it is idempotent, not 0, and not 1?
If $\kappa$ is infinite then $\kappa^2 = \kappa$, this is standard and depends on axiom of choice
If $\kappa = 0$ or $\kappa = 1$ then $\kappa^2 = \kappa$ as well
@Jakobian thanks!
@冥王Hades oh no, poor bug...reminds me of a story once my grandmother told me of how she cooked her first and only lobster. She cooked it alive in boiling hot water and the lobster started giving away a screeching sound which she remembers to this day.
What about the other direction,@Jakobian? And thank you.
and if $2\leq \kappa < \aleph_0$ then $\kappa^2 \geq 2\kappa \geq \kappa+1 > \kappa$
Thanks! :)
19:13
now since either of these $3$ possibilities has to hold, since the order on cardinals is linear
the inequality $\kappa+1 > \kappa$ for finite $\kappa$ follows from finite ordinals and finite cardinals being the same
This is helpful, @Jakobian, but I was hoping for a question on here I could cite.
in In the search of a question, 27 secs ago, by Shaun
in Mathematics, 13 mins ago, by Shaun
I can't find a simple question: how does one show a cardinality is infinite iff it is idempotent, not 0, and not 1?
Ah. Find a question for reference. This is a different matter
well, would you be content with the proof that $\kappa^2 = \kappa$ for infinite cardinals alone?
also a reference from a book would be theorem 1.5.11 in Introduction to cardinal arithmetic by Holz, Steffens, Weitz
19:29
@SoumikMukherjee Thanks :)
@Jakobian do you know if the metric on the completion $Y$ is the same metric as on its dense subset $X$? I am unsure how to solve the exercise if it isn't. I believe the map that sends the equivalence class of $(s_k)$ to the limit of $(s_k)$ is an isometry from $\tilde{X}$ to $Y$, but to verify this, I need to know the metric on $Y$ I think.
The metric on $X$ and $\tilde{X}$ are known, but that on $Y$ I am unsure.
19:45
what is Y?
@Jam math.stackexchange.com/questions/3200710/… is a related exercise with R in place of Z. it uses (an appropriate generalization of) the chinese remainder theorem. does it help?
@psie what do you mean by the same
equivalence class $(s_k)$ to limit of $(s_k)$ in $Y$, yes
just assume $Y$ has some metric such which restricts to the metric on $X$
$d_0:Y\times Y\to [0, \infty)$, $d_0\restriction_{X\times X} = d$
the point is that the metric on $Y$ is somewhat of a mystery
and what you're proving is that $Y$ must essentially be $\tilde{X}$
@Jakobian ok, right, I am proving uniqueness up to isometry sort of (if I understand the exercise correctly), but how would one verify that this map is an isometry?
soumik: oh, i see. a degenerate case of the phoenician waw. many thanks
I guess we want to check $d_0(s,t)=\rho(\tilde{s},\tilde{t})$, where $\rho$ is the metric on $\tilde{X}$.
19:51
I am glad to help:)
@psie yes, uniqueness up to unique isometry
actually there is unique uniformly continuous such map, and it happens to be an isometry
the whole uniqueness talk is as if I'm attempting some kind of word salad
it's probably very confusing
well what do you know already about $\tilde{X}$
@Jakobian it is complete
did you check its a metric space, complete, $X$ is dense in it?
the map $x\mapsto\tilde x$ maps $X$ to a dense subset of $\tilde X$, indeed
@psie I hate bugs, I feel no empathy towards them.
20:06
okay so given a Cauchy sequence $(s_n)$ of $X$, you want to send it to $f((s_n))$, the limit in $Y$ of $s_n$
and now suppose $(s_n)\sim (t_n)$. Then check that $f((s_n)) = f((t_n))$
hence this defines a unique map $g:\tilde{X}\to Y$ given by $g([(s_n)]_\sim) = f((s_n))$
did you check that?
@冥王Hades and they probably don't feel any empathy for you too :P
@Jakobian is $f$ the map $x\mapsto\tilde x$, where $\tilde x$ is the equivalence class of the constant sequence $(x,x,\ldots)$?
@Jakobian Precisely, which is why it bit me. Why should I not fight fire with fire then, right?
@psie lobsters aren't bugs though
yeah no :) they're more noble
@psie no
$f$ is the map from the set of all Cauchy sequences to $Y$
20:11
I remember at a kitchen where a lobster was watching another lobster being cooked. Must've been a horrifying site.
the equivalence class of the constant sequence $(x, x, ...)$ is an element of $\tilde{X}$ so you're pretty far off about what $f$ is
@Jakobian I will check this. Give me some minutes and I'll write something.
If $(s_n)\sim (t_n)$, then $d(s_n,t_n)\to 0$ as $n\to\infty$. Intuitively I want to conclude they have the same limit, but I don't know if there's something more to motivate that $f((s_n)) = f((t_n))$.
@psie How about you move away from the area of intuition and into the area of formal proof
Let $s$ be the limit of $s_n$ in $Y$ for example. Can you show that $s$ is the limit of $t_n$ in $Y$?
This is the same thing but phrased differently
@Jakobian well, this is where my brain asks about the metric on $Y$, because we define convergence of a sequence to a limit in some space with respect to the metric of that space
what does it mean for $t_n$ to converge to $s$ in $Y$
$d_0$ is a metric on $Y$
you don't need to know "what it is"
how you should use intuition is to form a proof
allow it to direct you, but not to dictate what is true and what is not
20:35
I think I need a break and think some more about this.
yeah maybe it's hard to grasp all at once
@Jakobian regarding your comment here though, I get what you are saying, but is an equivalent way of putting it that there's an isometry $h:\tilde{X}\to Y$ such that $h \circ e = i$? Here $e:X\to \tilde{X}$ is the map $x\mapsto \tilde{x}$ and $i:X\to Y$ the inclusion map. Maybe that is what you said?
@psie yes
ok, cool 👍
this is basically what I said, although people usually don't pay much attention between distinguishing $X$ and copy of $X$ in $\tilde{X}$
and what is also to be checked that it really is a copy of $X$
but I imagine you did that already
20:43
yeah
Towards the end of this video there is a demonstration of the Game of life being undecidable because you can run the game of life "on" the game of life recursively. 29:30 onwards.
21:03
@MatsGranvik Note that Mat SE moderator Asaf Karagila consulted on that video.
 
1 hour later…
22:08
If the Cesaro means $\frac{1}{N}\sum_{k=1}^N a_k$ are bounded in $N$ with $a_k$ positive, must $a_k$ be bounded?
22:29
sorry I didn't see that $a_k$ are positive
We have $$\lim_{N\to\infty} \frac{\sum_{k=3}^N \log(\log(k))}{N} = \lim_{N\to\infty} (\log(\log(N+1))-\log(\log(N))) = \log\left(\lim_{N\to\infty} \frac{\log(N+1)}{\log(N)}\right) = 0$$
and so $\log(\log(k))$ for $k\geq 3$ consists of positive numbers, is increasing and converges to $0$ Cesaro, but is unbounded
 
1 hour later…
Jam
Jam
23:43
@leslietownes no cause Z is not a PID like R and also you cannot use chinese remainder theorem cause x+1 and x-1 are not coprime
23:53
jam: but can you at least use the CRT to get partial information? e.g. what about the ideals generated by x and (x+1)(x-1)?
jam: Z[i] is a PID, isn't it? math.stackexchange.com/questions/443842/…

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