@Jakobian do you have any clue how to show $K_n\cap S=\emptyset$?
There is the following hint in the exercise:
For the first part, begin by showing that there exists a positive integer $\nu$ such that $B\subset K_n$ for all $n\geq \nu$. Then suppose that for each $n\geq \nu$ there exists $s_n\in K_n\cap S$. Show that there exists a subsequence $(s_{n_k})$ converging to $x$, and hence find $k$ such that $s_{n_k}\in K\cap S$, a contradiction.
I need to construct a faithful representation of the frobenius group $F_{20}$ of degree 4. I knew in particular I can realize $F_{20}$ to be the group generating by the relations $x^5=y^4=1$ and $xy=y^2x$. The representation is supposed to be over $F_5$. I can map $y$ to the diagonal matrix with entries $1,2,3,4$, but I not sure what $x$ should be...
ted: stackexchange seems to have unconstitutionally barred the middle finger emoji from its list of emojis that are supported in chat, but a pair of those would also work
For local coordinates on a manifold, one wants to show by using defintion, that it satisfy the leibniz rule, by x denote local coordinates, by u the standard on R^n the proof goes as follows for a chart phi.
D fg)
Partial derivative of (fg) at p in respect to xi= partial of (fg composed phi-1) of phi(p) in respect to ui = partial of ( f composed phi-1).(g composed phi-1) of phi(p) in respect to dui
Sorry i am on my phone.
How does the third equality arise? The composition of product is not defined like this
Also, I'd be very grateful if you could explain why the right-hand side of $d(z, K^c) \geq d(z+\frac{1}{n}(y-x), K^c) - d(z, z+\frac{1}{n}(y-x))$ is positive?
we have $d(z+\frac{1}{n}(y-x), K^c)\geq d(B, K^c)$, right? $d(B, K^c)$ considers all points in $B$ whereas $d(z+\frac{1}{n}(y-x), K^c)$ only considers the single point $z+\frac{1}{n}(y-x)$ in $B$, so potentially they could be equal
I also drew a picture of the two scenarios. I do not really understand why $y$ gets to be the center of the ball in one scenario and the intersection with $K$ in the other. This still remains a mystery to me.
It seems like the center of the ball $B$ becomes meaningless when $B$ intersects $K$.
@LuckyChouhan I teach at a community college. Most days are spent either preparing lectures, giving lectures, or grading student work. On Fridays, we don't teach---we have meetings instead. I am currently chair of the Instructional Council, so every other Friday I get to spend three hours leading that meeting.
Very rarely, I find the time to think about actual mathematics, and pretend to work on this stupid paper that I really need to get out before I die.
@LuckyChouhan I'm too old for sports. I used to fence, but that pretty much wrecked my knees. I do like to go hiking, and live in one of the best places in the world for it.
en.wikipedia.org/wiki/Median says that "...any probability distribution has at least 1 median...", given the definition: "let X be a random variable on $R$, and if $x_0$ obeys the following: $P(X<x_0)\leq \frac{1}{2}$ and $P(X \leq x_0) \geq \frac{1}{2}$ then $x_0$ is a median of $X$" im unable to show that the existence is guaranteed for a general probability distribution
@LuckyChouhan Did I "like" it? I don't know that I remember it all that well. My recollection is that it was a lot of uplifting (though not necessarily useful) advice given by a person with a terminal cancer diagnosis.
its just that in both cases, we consider two sets and we have the notion of proving there exists one number in both sets (in the clopen problem, its $S$ and $\mathbb{R}^n-S$.)
these are just some random thoughts dont take me too seriously
@nickbros123 maybe you don't know, but this clopen problem is essentially just that $\mathbb{R}^n$ is connected. This is what it means
to be connected is to not contain a non-empty proper clopen subset
well, regardless of how you proved it, it boils down to intervals and their completeness They are similar yes, we use that there exist suprema and infima in real numbers. But its just that
you're correct but the connection isn't deep
@nickbros123 I don't see why don't we discuss about it? It certainly can be made very formal
there is a precise reason why this appeals to your intuition here
@Jakobian I havent learnt about the concept of connectedness, but I did assume a function $f:[0,1]\to \mathbb{R}^n$ of the form $\vec x_0+t(\vec y_0 -\vec x_0)$ for two points $x_0, y_0$ in $\mathbb{R}^n$. my knowledge of basic topology of $\mathbb{R}^n$ comes from Hubbard and Hubbard, they didnt talk on this (atleast not in the 1st chapter)
@Jakobian they only really spoke about open n clsoed sets, sequential definition of closed sets, closure, interior (left a few proofs regarding closure n interior), and didnt speak on the clopen stuff.
@Jakobian im not sure if that was what the authors had in mind
ted you'll like this. munchkin earned a cookie for doing a week of reading practice at home. instead of waiting to be given the cookie after dinner she did this: when my wife got up mid-meal to get something from the kitchen for the baby, munchkin called out "i guess i'm ready for the cookie"
did i ever tell you about the time she was talking maybe a little too much to another kid's dad at the pool, and i sarcastically asked her if she could count to a hundred, and then she did that
Hi, I'm trying to calculate $$\frac{\sum_{d=4}^{98} d(99-d)(d-1)(d-2)(d-3)}{\sum_{d=4}^{98} (99-d)(d-1)(d-2)(d-3)}$$. It simplifies to a neat 200/3, but I fail to find a way to actually go about simplifying this. Any help will be appreciated. I tried to simplify the expression by writing binomial coefficients in place of the products but that doesn't not make it anymore obvious.
I am trying to construct representations of the frobenius group $F_{20}$, I suppose the representation is construct over the field $F_5$ since it is indicated in the questions that it has four distinct one dimensional representation. But I am not sure how to construct a faithful four dimensional one. Do anyone have a clue?
oscar: i am not sure that the average weird math nerd has any idea of what 'the frobenius group F_20' is. is it defined to be, or otherwise equal to, the group realized on slide 9 of sporadic.stanford.edu/Math122/lecture14.pdf (roughly speaking, the "ax+b" group of affine transformations of F_5)?
oscar: because if you look at page 7 of albany.edu/~mz498674/waffle_paper.pdf, "exercise 4.5" on numbered page 103, they give formulas for a 4d representation of that group ("epsilon_5" is a primitive 5th root of unity)
daniel bump's order 4 and order 5 generators "h" and "k" [first doc] correspond to the second doc's "b" and "a", respectively
so the order 4 thing goes to a "shift the coordinates" operator, and the order 5 thing goes to a multiplication operator, with appropriate powers of the root of unity on the diagonal
sahaj: you should hear what our math says about you
@Sahaj Mathematicians keep on expanding new lengths of knowledge which gets studied over and over again. We are way past level of high school and it shouldn't come as a surprise
@leslietownes Yes, the $F_{20}$ is the group of affine transformation $x\to ax+b$ for $x\in\mathbb{Z}_5^{\times}$ and $b\in\mathbb{Z}$, sorry about the confusion. Thank you for the reference, I am reading it now.
oscar: if you don't care where the rep comes from, you can probably skip all of the stuff about waffles (which seem to be just some graphical trick for organizing some, not all, irreps of those groups) and swipe the formulas
oscar: might be worth considering whether there's some general thing where in <a,b: a^m, b^n, aba^{-1} = b^p> on C^m, where a goes to cyclic permutation of the coordinates and b goes to some diagonal matrix with powers of an nth root of unity along the diagonal [perhaps appropriately chosen] where you get a rep/irrep that way
or if this is somehow specific to one or more of the values of m, n, and p for F_20
@TedShifrin If you take a basis $(v_1, ..., v_n)$ then skeleton of $v$ with respect to this is basis is the vector $(a_1, ..., a_n)^t$ such that $v = \sum a_iv_i$
There are two legitimate contexts in which the word "skeleton" can be used: (1) biology class, and (2) Dungeons and Dragons. All other uses are an abomination.
let $Q:=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ and $G:=\{A\in GL_2(\Bbb{R}): A^tQA=Q\}$. I want to find the Lie algebra of $G$.
I know that by definition $Lie(G):=\{X: \exp(tX)\in G ~~\forall t\in \Bbb{R}\}$. And I also know that one has the characterization that $Lie(G)=T_I(G)$ which is the ta...
