I've recently learned about projective geometry. The question i have is regarding finding the equation of asymototes. The procedure I understand for fiding the equation of the asymptote of a curve $F(x,y) = 0$ when the point at infinity is real is as follows: 1. Homogenezie the equation $F(x, y) = 0$ by substiting $x = X/W$ and $y = Y/W$; simplify the expression 2. Find points at infinity by setting W = 0 and solving the equation for Y; let's say one of the points at infinity have the homogenous coordinates of [a, b, 0]
This procedure seems to still work when the points at infinity are complex - for instance, if i find the equations for the asymptotes of a circle $x^2+y^2=R^2$ using the procedure above i seem to get the correct equations of $x+iy=0$ and $x-iy=0$
But isnt the sitatuon when the points at infinity are complex entirely different as we have to treat the function as a complex function right? Is there an intuitive explanation why this procedure still works?
From thinking about it myself, i couldn't really understand what a complex point at infinity really is as well; on the real projective plane, ppl talk about the points at infinity as representing the slope of the line
@TedShifrin hmmm do you mean the circle goes off to on the complex projective plane? I understand the notion of a number going off into infinity - its value grows and grows - but what does it mean for a closed geometrical object like a circle to go off into infinity?
Spent my plane ride reading about enumeration combinatorics
Look at this bonkers result
I guess I can copy/paste, actually
> A domino tower is a stack of horizontal 2 × 1 bricks in a brickwork pattern, so that no brick is directly above another brick, such that the bricks on the bottom level are contiguous, and every higher brick is (half) supported on at least one brick in the row below it. Let the size of a domino tower be the number of bricks. See Figure 1.7 for an illustration.
> [image]
> Remarkably, there are $3^{n-1}$ domino towers consisting of $n$ bricks. Equally remarkably, no simple bijection is known. The nicest argument known is as follows.
@BalarkaSen My first solution didn't have a spiral at all, the bad point was the rightmost point (so the lines didn't surround it), but in retrospect it's similar to this version sort-of "folded onto itself"
The uniform convergence for the question would follow from: $\sup_{x\in \mathbb R}|(1+x^{2n})^{\frac 1{2n}}-f(x)|\le \sup_{|x|\le 1} |(1+x^{2n})^{\frac 1{2n}}-1|+\sup_{|x|>1}|(1+x^{2n})^{\frac 1{2n}}-|x||$
So here is the situation for my Gauss map ponderings. Suppose $f : M \to N$ is a fixed smooth embedding of a closed manifold $M^m$ inside a closed manifold $N^n$, and let $G(df) : M \to \mathrm{Gr}_m(N)$ be the (tangential) Gauss map. Suppose $G_t : M \to \mathrm{Gr}_m(N)$ is a homotopy of the Gauss map starting at $G_0 = G(df)$, by rotating the planes of the Gauss map.
In general it's $G_t$ is not going to be the Gauss maps of anything. The planes have been disturbed too badly to integrate
@BalarkaSen It's also neat that it's not an accident of how I chose the frame / how I chose to crop it; any zoom will do basically the same thing because all that really matters is what happens near the center
However, if you allow mild singularities, it suddenly becomes true. There is a homotopy $f_t : M \to N$ of topological embeddings with so-called "wrinkle singularities" starting at $f_0 = f$ such that $G(df_t)$ is $C^0$-close to $G_t$ for all $t$.
Here is what wrinkle singularities look like:
The top row is a representation of the wrinkle as a movie of birth and death of a pair of semicubicle cusps
The caveat is that a wrinkle is only a neighborhood of the sphere (circle here, in dimension 2) traced out by the pair of cusp points.
@TedShifrin You're right on the money. This is an example of an $h$-principle; there is a classical variant of this which is solved exactly by trying to approximate the formal 1-jet given by the disturbed plane distribution by holonomic 1-jets.
This one's slightly weird because the proof of this is some hands-on trickery. The usual machine doesn't apply
I want to understand this in a better framework
Oh, by the way, I forgot to mention, but note that the special thing about wrinkles is that -- just like cusps -- even though they aren't smooth the Gauss map is defined
Somehow the point is that a wrinkle allows for an abrupt change in the tangential Gauss map. You are in a totally flat, constant tangential gauss map, world, and suddenly you hit the wrinkle and do a little whoop with the tangent plane
And since you can wrinkle inside a wrinkle, by doing this many times, one should be able to approximate any plane field
But it's a fidgety idea and I'd like to write it down in a better way.
I think there is some serious analogy here with what I am reading, actually.
The loose charts I told you about allows you to have certain h-principles for Legendrian submanifolds. Namely, it gives you a purely algebraic topology criterion for when two Legendrians are isotopic through Legendrians if one of them has a loose chart
The main procedure feels like the one in this simple sphere eversion video. Take a smooth isotopy between the Legendrians (this has to exist first), then you take the {zig-zag} x I, helicopter it to somewhere whenever you encounter non-Legendrian-ness like the little loop of self-intersections in this video, then shrink it down so that it looks like a wrinkle
Use those wrinkles to avoid problems like tangencies
@leslietownes I assumed you were referencing Chabad because you seemed knowledgeable about Jewish things from that conversation we had a while ago but I realize there's no reason for that to be true
Speaking of continued fractions, cool fact: if you take an unknot in $S^3$ and do a $p/q$-surgery on it, that's the same thing as a Hopf linkage where you do integer surgery on each component, which turn out to be the coefficients of the continued fraction
By the way did you know if you take a triangle of Hopf links (three unknots, every pair is a Hopf link, no further Borromean structure), the knot complement of that covers the knot complement of the trefoil knot
The following definition is in "Multivariable Mathematics" by Theodore Shifrin.
Definition Let $U\subset\mathbb{R}^n$ be an open subset containing a neighborhood of $\mathbf{a}\in\mathbb{R}^n$, and let $\mathbf{f}:U\to\mathbb{R}^m$. We say $\mathbf{f}$ is continuous at $\mathbf{a}$ if $$\lim_{\m...
Should be obvious from taking three 1/1 circles on the torus, and try to make it into the toral embedding of the trefoil somehow
three 1/1 circles on the torus is that triangle of Hopf links
Take the embedding of the triangle of Hopf links on $T^2$, and then act by $S_3 = \langle a, b \rangle$ where $a$ is a transposition and $b$ is a $3$-cycle, the action of $b$ being rotating by $2\pi/3$ along the axis that goes through the donut-hole of the torus and $\pi$ along the center circle of the torus, simultaneously.
You can do the same thing where the axis of rotation, instead of being the one which goes through the donut hole, is inside the solid torus -- namely the center circle.
@AkivaWeinberger Wait I'm not sure I understand. $(\pi, \pi)$ rotation does act freely on the complement of the links, why does it need to permute the circles? The 3-cycle already does that
For what value of $c$ does the spiral $r=e^{c\theta}$ have the property that, the points with vertical tangent are on the same y-coordinate as a point with horizontal tangent?
i am reading a proof of the "root" of semi definite hermetic endomorphisim. and in the proof he says, we will choose a base of eigen vectors. Why does this base exist? what attriute of the Semi definite hermetic endomorphism makes it possible for the existence of such base of Eigenvektoren
nvm found the solutiion
the matrix is n x n, so if we want to use spectral theorem, we need to show that the complex conjugate is equal to the matrix it self, since positive definit, this implies that the eigenvalues are real.
Given $B(x)$ then doing $B'(x)$ do you lose some information about $B(x)$? That is to say the properties $B(x)$ has, may not necessarily translate to that of $B'(x).$ Clearly in differential equations if you differentiate the terms it's a completely different equation.
you're operating on $B(x)$ with the differential operator
question to sequences usually i see its written something like $ 0 \rightarrow V \stackrel{\phi}{\rightarrow} V_2 $ what is meant by the first arrow? is it the zero function? or any function ? usually nothing is written above it
