00:00 - 20:0020:00 - 00:00

12:29 AM
@AlessandroCodenotti I think $(-1)^n \dbinom{-r}{n}$ is more beautiful
less arbitrary

@Rithaniel Where some things get sent is already determined by where you've sent the others.
@Semiclassic: Yes, I did all that pretty much in the last chapter of my algebra book. I don't remember details now.

on a different note: Uber and Lyft are usually about the same for me. Right now, though, Lyft has my desired route as about 25 bucks
Uber is $47 :3 Uber seems to be complaining a lot about being broke. @Semiclassic: Do you find my comment here completely rude? Wrong? that question makes my eye twitch LOL what about your brain? 12:35 AM it makes my brain go "what" I cannot begin to fathom. But was I too rude? Maybe it's OK if I was.$du'^2+dv'^2=1$riiiight I didn't even start listing things that were crap. The list would have been everything. so @Leaky: What's your report after a week of classes? 2 hours ago, by Mathphile infinite tetration of x converges for$e^{-e} \lt x \le e^{1/e} $2 hours ago, by Mathphile does infinite pentation converge in any range too? can someone help me with this? 1:05 AM @TedShifrin I mainly know the material for now so I think it's ok; might have more challenge after a month 2 hours later… 2:40 AM What's the simplest EXPTIME-complete game you can think of? 6 hours later… 8:45 AM @Semiclassical I think I might be an idiot You’re supposed to calculate the Jacobian of the residuals aren’t you? Not the Jacobian of the whole dang function. 3 hours later… 12:04 PM$n≤2^{k+1}−1$If I take log_{2} on both sides, what do I get on the right side? Can$log_{2}{2^{k+1}-1}$get simplified to something? 12:45 PM 1 Let$w = \{w(i,j)\}_{1 \leq i,j \leq m}$be an$m \times m$symmetric matrix with non-negative real entries such that$w(i,j)=0$if and only if$i=j$. Show that $$d(i,j)=\min\left\{\sum_{j=0}^{k-1}w(i_j,i_{j+1}) \mid k \geq 1, i_0=i, i_k=j,i_j \in \{1,2,...,m\}\right\}$$ is a metric on$\{1,...,m...

I have a request for verification.

1:03 PM

This has been bugging me for a while.

Hi everyone. Still experimenting with Mandelbrot rendering, just in case anyone would like a pretty picture :)

If my opinion matters at all: The picture is really pretty, but I can't appreciate/understand the mathematics involved in it.

Fair enough. I can't say I understood very much of your question either. Also, I don't really understand a lot of stuff related to Mandelbrot set or fractals. But rendering/coloring them is addictive, even without understanding everything

1:27 PM
@AlessandroCodenotti @Semiclassical @LeakyNun plis help math.stackexchange.com/q/3356270/389992
I am afraid I have gone wrong somewhere and not realizing it.

My picture is already pretty, don't you think? :)

1:59 PM
Any examples of non-expansive operators or metric maps? I know that max is a non-expansive operator, but I am still trying to understand why exactly.
For example, suppose that we have the numbers x=3 and y=4 and our metric is the absolute value. So, |x-y| = |3-4| = |-1| = 1 = |4-3|.
What confuses me is: if max is a non-expansive operator, how can we show it, given that max requires two elements?
The formula is $d_{Y}(f(x),f(y)) \leq d_{X}(x,y) . \!$
What is f(x), in case f=max?
max(x)=max(3)=3
In this case, we will have 1 <= 1.
Very nice! But I want a more useful example that is not trivial

multiplication by a constant smaller than $1$ as a map $\Bbb R\to \Bbb R$

Anyway, the absolute value is apparently not a metric, but d, in the above formula, should be a metric

The projection $\Bbb R^2\to \Bbb R$

The absolute value is a norm, so it induces a metric

2:09 PM
@AlessandroCodenotti Right!
@AlessandroCodenotti So, for example, the function f(x) = 1/2x is a non-expansion, because, given two numbers x1 and x2, then |1/2*x1 - 1/2*x2| <= |x1 - x2|. For instance, x1=6 and x2=9, then |3 - 4.5| <= |-3| <=> 1.5 <= 3

2:21 PM
For which of the subspaces $U$ and $V$ of $\textbf{R}^4$ is the sum $U+V$ direct? $U=\{\textbf{x}\in\textbf{R}^4: x_1+x_2+x_3+x_4=0\}$ and $V=\{\textbf{x}\in\textbf{R}^4: x_1-x_2+x_3-x_4=0\}$.

2:36 PM
Consider the following theorem: Let $G$ be a locally compact abelian group, and suppose that the Gelfand homomorphism $\Gamma : L^{1}(G) \to C_{0}(\hat{G})$ given by $f \mapsto \hat{f}$ (Fourier transform, I think) is surjective, then $G$ must be discrete.
My question is, is the converse true? If $G$ is discrete, does it follow that the Gelfand transform is surjective?
Should I ask this one the main?

3:17 PM
Do you want $L^1(G)$ or $C^\ast(G)$ as the domain?

According to the theorem, $L^{1}(G)$ is the domain.

3:40 PM
Alright, yesterday I posted that incorrect result, and my mistake was thinking that thinking about subgroups of order 2 would be the same as thinking about all subgroups. After reviewing, what I posted actually does hold when $n=2$, because $S_3\cong GL_2(\mathbb{F}_2)$.

That's true
@user193319 Which theorem? Usually the Gelfand transorm has a commutative $C^\ast$ algebra as its domain

The one I typed out above.

If I wanted to motivate the $n=3$ case by thinking about permutations on subgroups, I'm kind of coming up short. I know there are seven subgroups of order 4 in this group, and there are six automorphisms on the group those subgroups are isomorphic to, so multiplication gives 42 automorphisms which leaves at least one group of order 4 fixed, but there are 168 automorphisms, so I'm missing a factor of 4, and I don't know where to find that factor.

So how is the Gelfand homomorphism defined?

I think the case I'm interested in the Gelfand transform is just the Fourier transform.

4:10 PM
Hi, I would like to know how to calculate the following:
(a-cb)/(a-b) = 2-c?
\frac{a-cb}{a-b}=2-c?

4:21 PM
@SubhasisBiswas Please do not ping people like this. You did it to me earlier. You can post your question, of course, but do not annoy people with pings.
@YOUSEFY Did you try clearing out the denominator? I don't know what you mean by "calculate", though.

@TedShifrin sorry Ted
I mean prove that \frac{a-cb}{a-b} = 2 -c.

Prove that starting from what?

I don't know how \frac{a-cb}{a-b} is equal to 2 - c.

It isn't.
You would need $2a-cb$ in the numerator to get that.

Okay, I see the point, but I don't know why in the paper shows the following:
take a look
it is from: An approximation algorithm for the asymmetric Traveling salesman problem with distances one and two, page 3.
by Vishwanathan

4:30 PM
This is a much more complicated question. They're asking for the inf of a quantity as certain things vary.
They're asking for the largest the quantity $$\frac{2n-rW}{2n-W}$$ can be when $W\le n$.

I'm thinking about it

I don't want the paper.

If I want to ignore the condition, does this make the question easier

You can't ignore it.
You can do it by calculus or you can do it with some simple algebra to rewrite the fraction as a function of $W$.

Because n is the number of vertices in graph and W is the maximum weight of a tour in the graph (a tour is a walk from a vertex and ending the same vertex such that you visit all vertices exactly once) and the weight of each edge is either 1 or 2. Thus, W is maximum will be 2n.

4:37 PM
No, they specifically said the maximum of $W$ is $n$.
And then, on the interval $[0,n]$, the maximum of that function I displayed is, in fact, $2-r$.

How did you do it? Do I need to write some values of W from the domain [0,n] and see when function has its maximum value

Not write some values. Understand what the graph of that function of $W$ looks like.

Okay, I will do it Ted. Thank you so much!

You're welcome.

hi @ted

4:46 PM
Hi @Semiclassic

There's a computation I've been using a bunch lately, and I'm trying to make sure I actually understand it
If I want to project onto the 1D subspace in the direction of unit vector $u$, that's easy: The projector is $P_u=uu^\top$
More generally, if I have a subspace $U$with orthonormal basis $u_1,u_2,\ldots,u_n$, then the projector onto $U$ is just the sum of the projectors onto the basis vectors: $P_U=P_{u_1}+P_{u_2}+\cdots +P_{u_n}$
That's familiar enough to me.
What I wasn't familiar with was how to write the projector if the basis isn't orthonormal

That's the math that solves the normal equations.

Right.

Just do $A(A^\top A)^{-1}A^\top$, where any basis for your subspace goes into the columns of $A$.

yeah, I was about to say
What's the best way to understand that formula? I mean, it certainly is idempotent

4:51 PM
That middle inverse "unscrambles" the arithmetic when the basis isn't orthonormal.
Your second example is precisely of that form when $A^\top A = I$.

That's true.
(And if I assume the basis is orthogonal but not orthonormal, then $A^\top A$ is diagonal)

Right.
You know where this formula comes from?

I know that it shows up when doing linear least squares
that's how I got interested in it, in fact
But I don't know how one would derive it without knowing about it ahead of time

But least squares is precisely orthogonal projection onto the column space of $A$.
I could tell you to watch my video, but it's easy enough.
No, I messed it up. Sorry.
To be the orthogonal projection means that $x-p$ is orthogonal to the column space of $A$. That means that $A^\top (x-Ax) = 0$.

hmm, let me follow this forward myself a bit
ah, ok

4:56 PM
Hang on. We start with a vector $x$. We want to write its projection $p$ onto the column space. By definition, that means $p=Av$ for some $v$.
Then $A^\top(x-p) = A^\top(x-Av) = 0$. We want to solve for $Av=p$.
Note that this equation is $(A^\top A)v = A^\top x$.
OK, now you finish.

Okay. Then one has $v=(A^\top A)^{-1}A^\top x$
and therefore $p=A(A^\top A)^{-1}A^\top x$. Nice

Sorry about my variable confusion. smacks self

hah, no worries
so things are easy once one formulates the orthogonality condition

Yes, that's the whole point.

to be precise, I've been interested in the history of regression/correlation
and wanted to get a geometrical perspective on how Yule (one of the early workers in the subject) derived one of his results (specifically, an inequality on correlation coefficients)
The geometric story you get is simple enough, happily: $\|x-p\|^2\geq 0$
That's all he needs to derive the inequality. So that's cute.

5:02 PM
@TedShifrin. Sorry

What's up guys?

Hi, Demonark.

@Daminark The all seeing eye of the alien overlords the sky

@Semiclassic: I've commented before that a lot of the statistical stuff has very pretty linear algebraic (geometric) interpretations, but the statisticians obscure/hide those.

('-')

5:05 PM
Quite.

:0
Actually it would've worked better if I looked side to side
( '-')
3
('-' )

One exception to that is Fisher, though that case is a bit strange

For example, the fact that the algebraic sum of the errors when you do least squares is $0$, is immediate from the fact that one of the columns of your matrix is all $1$'s. @Semiclassic

But, for instance, one of his papers specifically interprets the correlation coefficient as the angle between two vectors of sample data

As it should :)

5:07 PM
and another interprets the partial correlation coefficient as the angle you get when you project two vectors onto a plane. (this one shows up in Jacobson's thesis, though I don't remember if he cited Fisher)

I certainly never looked at any of the stat literature. Some of the stuff we figured out ourselves, but I'm sure he looked at the literature.

that last one is especially fun because it's equivalent to the spherical law of cosines :>

No, Fisher is not in his biblio.

the Fisher bit, not the spherical bit
(though one of the late papers of Pearson they cite is how I got the idea for the spherical law of cosines)
"R.A. Fisher (1 890-1962) used geometry throughout his work on distribution theory but his treatment of correlation is of most interest here. In his paper on the exact distribution of the correlation coefficient he (1915, p. 509) wrote, “The five quantities [associated with the bivariate normal] have . . . an exceedingly
beautiful interpretation in generalised space.” The n observations on a pair of variables can be represented by two points P and Q in n-dimensional space and the correlation coefficient interpreted as the cosine of the angle between OP and OQ."

So I just pulled out my algebra book. The spherical law of cosines I did with a bunch of cross products and dot products (and considering the dual triangle).

5:13 PM
"Later Fisher (1924, p. 330) observed that when there is a third vector corresponding to point R, the partial correlation of the original vectors is the “cosine of the angle between the projections of OP and OQ upon the region perpendicular to OR”. This is a geometric interpretation of residuals and of (8.2) above."
So yeah, Fisher definitely had that
@TedShifrin One of the points of that paper, though, is how long it took that geometric perspective to make headway into statistical regression

When proving if a sum of two subspaces $U,V$ of a linear space $W$ is a direct sum, is it sufficient to prove that the intersection of $U$ with $V$ is the zero set (hence $U+V$ is a direct sum?), or does one also have to prove the equality $W=U+V$? If yes to the latter, how would one go about doing that?

Yup.

to the extent that it has, it's via projections and Hilbert space

You're asking two different questions, @schn.
@Semiclassic: Most of the stuff I've ever seen is finite-dimensional.
So I dunno about Hilbert spaces.

"The treatise on the trigonometry of correlation that Pearson (1916, p. 237) thought “greatly to be desired” never materialised. Kendall’s (1961) Course in the Geometry of n Dimensions is a partial offering but it appeared just as a new approach was taking off. This drew on the Hilbert space theory developed in the early part of the century and assembled by Stone (1932): Birkhoff & Kreysig (1984) and Dieudonne (1 98 1) discuss the development. The
“projections” that invaded least squares in the 1960s (see Section 13) owed more to Stone (pp. 70-5) and his idempotent operators than to Fisher."
I am trusting this author's reading on the situation, of course

5:23 PM
@TedShifrin What are the two questions? Is it $W=U+V$ $\textit{and}$ whether $U+V$ is a direct sum?

Right.

I have been trying to find an elegant proof of this mathb.in/36275 I was able to show that it is true if mathb.in/36276 is true. I can't figure out an elegant way to prove either. Does anyone have any hints?

For the conjugate gradient method, you use the hessian and gradient of f, right? Not of the residuals of f like you do with Gauss-Newton?

@WilliamOliver So, in other words: Show that every real number in [0,1] has a unique base-3 representation uusing only 0 and 2?

I suppose so, I didn't think of it like that.

5:34 PM
Sounds pretty wrong, then.

Maybe I have been trying to prove the wrong exercise this entire time O_o

If I knew the answer to this question, I would have no more problems

You can't force the math. The progression from one step to the other has to be natural

@WilliamOliver that makes two of us

@ShineOnYouCrazyDiamond the sad part is that I lack that natural flow of maths in my hands..

5:36 PM
@ankii the trick is to take a long break, then come back to it
That's harder than it sounds

only if professors extend the deadlines..
but I'll see

Hm... Maybe this book just has an erroneous exercise..,
I have spent the last like 3-4 days trying to figure this out

I'm not convinced it's wrong , myself
It'd certainly be false if $X_n=\{0,1,2\}$
But if you only can use 0 and 2 as your digits, I think it might be true

Well I have come very close to a proof, I guess I can post the entire thing in mathb.in I am not sure if its correct or not though.

At the least, you can choose digits in the following way. Multiply $X$ by 3 to get a number between $0$ and $3$. It'll either by closer to 0 or to 2; pick the closer one
and then repeat

5:41 PM
@Semiclassic: I haven't looked at the original question, but what you posted describes the Cantor set, not all of $[0,1]$.
2

oh, wait
not "whichever is closer"
but "If 3X>2, pick 2. Else, pick 0."
@TedShifrin true.

The book actually says the image of $f$ is the cantor set, but I am not sure if I believe it

LOL ... I don't know what $f$ is, but the ternary expansion @Semiclassic described is precisely giving the Cantor set.

So I guess the operative example to look at would be $X=1/2$

I am very confused

5:45 PM
Any number between $1/3$ and $2/3$ needs to start with $.1$.

^
in base 3, that is

I thought we were doing ternary.

yeah, we are
just wanted to make sure that was understood

Sure.

The reasoning being: If the base-3 expansion starts as 0.0..., then its largest possible value is 0.02222.... = 0.1=1/3
and if it starts as 0.2..., then the smallest possible value is 0.200... = 2/3

5:47 PM
@WilliamOliver That is correct

Hi, demonic @Alessandro.

actually, I think I'm mischaracterizing the question a bit.
it only asks for the function to be one-to-one, not onto

Well, I don't like having to read things in links, so I haven't looked.

$\{0,2\}^{\Bbb N}$ is a Cantor space, if you can prove that $f$ is continuous and injective then its image will also be a Cantor space since the domain is compact and the codomain is Hausdorff
(It is actually the standard ternary Cantor set)
Hi @Ted

mathb.in/36277 Here is my proof so far

5:49 PM
@Alessandro: I suspect you're using a cannon without permission.

"For each $n \in \mathbb{N}^+$, let $X_n = \{0, 2\}$. Let $X = \Pi_{n \in \mathbb{N}^+} X_n$. Define a function $f: X \to [0, 1]$ by setting $$f(x) = \sum_{n = 1}^{\infty} \frac{x_n}{3^n}.$$ Prove that $f$ is one-to-one."

Oh wait its a bit messy sorry

Luckily it's a very superfluous cannon in this case!

So yeah, my characterization of it was wrong

@Semiclassic: That's easy enough. But, yes, the image is the Cantor set.
Hint: Look at the first place $x$ and $y$ "disagree" and give a positive lower bound on $|f(x)-f(y)|$.

5:52 PM
It should instead be: Every $X\in[0,1]$ has at most one ternary representation using only 0 and 2.
(It may have no such representation.)

Anyone willing to help with a uniqueness proof? (Scalar projection in R^2)

Where are you stuck, @psa?
That is, show us what you've tried to do; don't just ask us to do it.

I'll give a link to the question I asked on the site: math.stackexchange.com/questions/3357064/…

There is a perfectly good geometric solution posted.

Can you help me understand it then? How does he know that the line passes through the point $a\frac{u}{||u||}$?

5:56 PM
@Semiclassical I think I am going to post this as a math stack exchange question. I felt like I was missing something quick and obvious which is why I posted it in the chat

Ted's hint makes it pretty simple imo

Thats what I tried to do essentially

Also, what is the difference between affine line and a linear line? How does a real number become an "affine line"?

(It also matches with how I first saw the Cantor set, in the context of iterating the map ...no, not that one)

The response went a bit above my background @TedShifrin - I'm in second year.

5:58 PM
@psa: Everything perpendicular to $u$ projects to $0$, so all the vectors that project to $p=cu$ will be the line perpendicular to $u$ passing through $p$.

@Semiclassical Idk how to find the positive lower bound. I came up with this mathb.in/36276

@psa: Can you do the problem when $u$ and $v$ are perpendicular to start with?

Right, simple cases...

That's not an argument, that's just symbol pushing

Lets support

6:02 PM
If $\mathbf{u}$ and $\mathbf{v}$ are perpendicular, then their dot product is 0 and the angle between them is $\pi / 2$ rad, for starters...

@Semiclassical Referring to me? I can see why you'd think that, but that is a direct consequence of "subtracting" and seeing where they differ as @TedShifrin suggested..

OK, go on, @psa.
@William: Note that I specified that the $x$ and $y$ disagreed for the first time in a certain place. I.e., $x_i = y_i$ for $1\le i\le N-1$ and $x_N\ne y_N$.
You have to use that.

@TedShifrin Are you suggesting the positive lower bound is the sum from $1 \leq i \leq N - 1$?

Um, no.
Why don't you try writing down an explicit example. Like $x=(2,0,2,0,2,2,2,0,0, \dots)$ and $y=(2,0,2,0,2,0,2,2,2,0,\dots)$?
Examples don't make a proof, but they often give useful insight.

Alright, I will try an explicit example, thanks

6:11 PM
Oh now that I've got a bit of time I should do the twisted cubic thing in projective space

Now that you reminded me to pester you, Demonark, yes!

So in affine space it's $t\mapsto (t,t^2,t^3)$, this goes $\mathbb{A}^1\to \mathbb{A}^3$

Ayup.
And let's agree to use $(x,y,z,w)$ as homogeneous coordinates ...

Yeah lmao

or $(w,x,y,z)$, whichever you are "used to."
I mean, do you want the $1$ at the beginning or at the end?

6:13 PM
In projective space, we're going $\mathbb{P}^1\to\mathbb{P}^3$ and we want it to be a cubic, so $[x:y] \mapsto [x^3:x^2y:xy^2:y^3]$?

Please change notation so that it fits what we agreed on.
I.e., use different letters in the domain.

Start over: let $\mathbf{u}$ and $\mathbf{v}$ be perpendicular unit vectors [with u = (1,0) and v = (0,1)], and $\mathbf{r}$ be the vector of interest. Then if the scalar projection of r on u is $a=|r|\cos{\theta}$ where $\theta$ is the angle between $\mathbf{r}$ and $\mathbf{u}$ and $\mathbf{r}$ is in the first quadrant, then the scalar projection of $\mathbf{r}$ on $\mathbf{v}$ is $b=|r|\cos(\pi/2 - \theta)$ where the argument of the cosines is measured in radians.

Oh do you mean we're gonna write the $\mathbb{P}^3$ equations in the $(x,y,z,w)$, now I get it. In that case $[s:t] \mapsto [s^3:s^2t:st^2:t^3]$

@psa: Do you want to do stuff geometrically (talking about lines) or algebraically (working with specific coordinates for $\mathbf r$?
OK, @Demonark, and so your affine line comes with $s=1$.

eh, geometrically is probably easier and then I was thinking I could come up with a set of linear equations which I could show only has one solution which would indicate uniqueness
geometrically would be easier just because I can see where I'm going

6:17 PM
Well, if we're being geometric, we don't need systems of equations.
Draw me all the vectors that have $(3,0)$ as their projection on $u$.

OK, well geometric sounds more elegant anyways

So we definitely want $yz = xw$

(@psa: I wrote a linear algebra book with "A Geometric Approach" as a subtitle, so I'm fine with that. :P)
OK, @Demonark. So you have a surface now.
I need to stop @ ing you.

Haha, yeah not quite getting pinged :P

Okay... I would keep trying this but I have been thinking about this for way too long... haha... Could you tell me what you had in mind for the lower bound @TedShifrin? I give up...

6:18 PM
With the $x$ and $y$ I gave you, @William, what is $N$?

Ha, this is actually from a multivariable calculus class but we're basically just doing introduction to planes/boundaries/neighbourhoods/vectors now, which is what this question is from.
OK, so...

That's ok, @psa. Some of my videos might still help you ...
Anyhow, answer the $(3,0)$ question.

@TedShifrin In that specific case N is 6

OK, so now let's write down $f(x)-f(y)$ ...
For as many terms as you know ...

So prob now it'd be good to find an equation which separates $y$ and $z$. I guess $yw = z^2$. And if we're being symmetric $xz = y^2$

6:21 PM
Demonark: OK, so do those three equations give you precisely the curve you want?

@TedShifrin Its $\frac{2}{3^6} - \frac{2}{3^9}$ I see in this case that it is non 0, the problem is that there can be infinite terms like this and I can't guarantee that one term won't equal all of the others...

What's the worst case scenario you could get for the terms with $i\ge 7$?

So, if we have a point $[x:y:z:w]$ satisfying those equations... Well, let's say $x=0$. Then $y^2 = xz = 0$, so $y=0$. But then $z^2 = yw = 0$. So that's just the point $[0:0:0:1]$
Otherwise $x=1$. Then $z=y^2$ and $w = wx = yz = y^3$
So yeah this should be our twisted cubic, it's a copy of the affine twisted cubic plus one point at infinity

So those three equations cut out precisely the twisted cubic. What happens if you use only two equations?

@TedShifrin Do you mean in this specific case where they are all equal except 6 and 9?

6:26 PM
By the way, my ultimate challenge to you will be to figure out the normal bundle of the twisted cubic in $\Bbb P^3$. One of my all-time favorite exercises.
@William: No, worst case scenario is that they'll be different for all $i\ge 7$.
And the signs all are the same to make you most worried.

You mean two among these three? Or two in general?

@TedShifrin Yes... I am very worried...

Demonark: Start with two among these three, of course. [And make note of the normal bundle question for the future.]
@William: Seriously, what infinite series do you write down if they are all different?
Starting with $i=7$.

Well, first, I am trying to imagine why that might be the worst case... I am not convinced that that will have maginuted closest to 0

So let's say you have $yw = z^2$ and $xz = y^2$. If $x=0$ we're still stuck with the point $[0:0:0:1]$, and if $x=1$, well $z=y^2$, and $yw = z^2 = y^4$. So the question here becomes whether $w$ could be something other than $y^3$

6:32 PM
Well, obviously terms where they agree are going to make $|f(x)-f(y)|$ smaller.

The answer is we would need $y=0$
But then $w$ could be anything

@TedShifrin Is that true?

So yeah we'd have a twisted cubic union another copy of $\mathbb{P}^1$?

I think thats where I am getting tripped up

If you leave out positive terms from a sum of nonnegative numbers, then the sum is smaller.

6:33 PM
And the two would intersect at $[1:0:0:1]$

Looks that way, Demonark.

@TedShifrin They aren't nonegative right? f(x) - f(y) is a sum of positive and negative numbers potentially

If you get cancellations, you get a smaller sum. So worst case scenario is for all the terms to have the same sign.
You can make this rigorous with the triangle inequality.

Oh I think I see what you are saying...
I think I just need to think about it more then...

But you should be able to explicitly sum that geometric series and the proof (of this part) will be done.

6:38 PM
Yeah.. okay thanks!

You're welcome.

Now let's try $yz = xw$ and $xz=y^2$. If $x=1$ we have $[1:y:y^2:y^3]$. If $x=0$, then $y=0$, but then $z$ and $w$ can do anything, so we have a copy of $\mathbb{P}^1$ at infinity

Yup.

In this case they don't intersect

What?
You were already doing the intersection.

6:41 PM
No I mean there are two components that don't intersect here
The affine twisted cubic at $x=1$ and the copy of $\mathbb{P}^1$ at infinity

The closure of the affine cubic includes one of those points at infinity. So the intersection of the two surfaces is the twisted cubic union the line at infinity.

Ah yeah that's true

7:26 PM
@TedShifrin I want to come up with an intuitive way of specifying a trivialization of the normal bundle of an orientable knot (in an oriented 3-manifold) via a non-vanishing section.
I originally thought of just using the tangent vector to complete the section to a basis of the tangent space of the ambient space.
That is, if $K\subset M$ is the knot in the 3-manifold, then at each point $p\in K$ a section $v\colon K\to NK$ specifies a 2-frame of $NK$ via completing $(K'(p),v_p)$ to a basis of $T_pM$, where $K'(p)$ is the oriented tangent vector of the knot.

@TedShifrin imgur.com/DYofsHV These would be all the vectors that have a projection of (3,0) on u, given that u is aligned on the vertical axis.

This left me dissatisfied, though.

Suppose I want to find the parameter $a$ that maximises q(s, a), that is, $\max_a q(s, a)$.
Now, suppose now that I multiply q(s, a) by a negative constant, then $\max_{a}q(s, a)$ should now become $-c \min_a q(s, a)$?
If I multiply q(s, a) by a constant c > 0, then $\max_a q(s, a) = \max_a c q(s, a)$.
But what if c < 0?
Then I think that $\max_a q(s, a) = \min_a c q(s, a)$, which should be equal to $-c\max_a q(s, a)$?
Back to the basics

7:42 PM
@anakhro: You have to tell me how to get a nowhere-vanishing section in the first place.

Anyway, the maximization of a function is equal to the minimization of its negation

@psa: Right: So that's what I said a while ago — you get the line perpendicular to $u$ passing through the desired projection point.

I did a mistake in the reasoning above

@TedShifrin I guess that's a good point to make, but there I would be only assuming the existence of v, then I want to show I get a 2-frame for NK.

You have some mistakes, @nbro.

7:43 PM
Does it still not work?

@TedShifrin Yes
If $c > 0$, then $c \max_a q(s, a) = \max_a c q(s, a)$ (and not what I wrote above)

@anakhro: Any time you have an oriented 2-dimensional vector space, you can always complete a non-zero vector to a basis. But finding a (smooth nonvanishing) section in the first place is the challenge.
Right, when $c>0$. And what if $c<0$? @nbro

@TedShifrin i.e., the line formed by joining the tips of all those vectors and passing through (3,0)?

Remember that $c<0$, so you need to write $c=-|c|$ if you want to think in terms of positive numbers.
@psa: Yes, the tips of all the vectors lie on the line I said.

@TedShifrin So the non-vanishing section does indeed imply triviality of NK, but isn't any easier to find than a trivialization of NK in the first place.

7:46 PM
So this is the "affine line orthogonal to u passing through the point $a\frac{u}{||u||}$"

I don't know how to trivialize it without finding a section.
OK @psa
@anakhro: Unless you make very abstract arguments, but there are nontrivial circle bundles over the circle, so you have to use something concrete.

If $c < 0$, then the maximum of the function is its minimum, so it should be $\max_a c q(s, a) = c \min_a q(s, a)$?

"The maximum of the function is its minimum"????

@TedShifrin I guess if I had a Seifert surface $S$ for $K$ which I knew were orientable, then I'd have a non-vanishing section for $NS$ which I could use as $v$.

@TedShifrin The maximum of $q(s, a)$ is equal to the minimum of $-q(s, a)$

7:49 PM
@nbro: No, that's not right.
@anakhro Well, yes, but that's a very nontrivial theorem.

@TedShifrin Why is that?

@TedShifrin So, my thoughts so far: by a similar argument, there should be a geometric argument applied to v so that we have a line orthogonal to v. Now all that's left to show is that these orthogonal lines cannot be parallel, which will lead to a necessary condition that there is (by some Euclidean axiom) one meeting place for these non-parallel straight lines. This means that the vector r we picked to get a scalar projection of r on u to be a and r on v to be b must be unique.

@anakhro: I think you can get it out of Sard's Theorem.
@nbro: You just told me that if $1\le q\le 5$, then $5=-5$.
@psa: Yes, that's the gist of the argument the answer to your question gave. But the two lines are parallel if and only if the original vectors are (nonzero) scalar multiples of one another.

@TedShifrin thanks I will think about that, then.

Which they aren't, so r is unique.

7:53 PM
Right.

: )

@TedShifrin I didn't say this (I think). I wanted to say that q is maximised at a if -q is minimized at a. In other words, if q has a maximum at a, then this corresponds to a minimum of -q

Question: Is the trivial group a field?

@nbro: Did you think about the example I just gave you?
@Rithaniel: In my definition of a field, $0\ne 1$.

@TedShifrin I don't understand what you mean by the notation $1 \leq q \leq 5$

7:56 PM
What?

Alright, thank you, that makes things easier. (Trying to prove that $\{x^i:i\in\mathbb{N}\}$ is not a basis for $K[[ x]]$)

How can you not understand that notation and be talking about maximizing and minimizing?

@TedShifrin how have you been in any case? Cooked anything good lately?

Yeah, I cooked a great meal last night for friends. A little bit of French and a lot of middle Eastern.

I love middle eastern. I moved to a different city and they don't have any good middle eastern grocery places around here. :(
The toum has egg in it. :((((

7:59 PM
I don't know what that is.

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