@Perturbative I just wanted to mention that Lectures on the h-Cobordism Theorem is slowly getting closer to 14 days of inactivity, after which it might get frozen. (I do not know whether you want to keep room or whether it is already abandoned.)
Quick question: I'm trying to show that a subgroup $\Gamma$ of a topological vector space $H$ is discrete. I have a compact set $X$ and $X \cap \Gamma$ is finite. Why can I conclude from the finiteness of $X \cap \Gamma$ that $\Gamma$ is discrete?
hi, anyone knows the actual name for the double cross product method to obtain orthogonal vectors? it's not gram schmidt right?
Anonymous
09:52
Is there a simple way to see that there are only $p^2$ solutions to the 2-cocycle identity $c(b_2,b_3) - c(b_1 b_2,b_3) + c(b_1,b_2 b_3) - c(b_1,b_2) = 0$ where $c: C_p \times C_p \to \Bbb Z/p\Bbb Z$?
If you replace our unit cube by a cube of side length $a$, what is the relationship between the area of its orthogonal projection onto the $x y$ -plane and the lenglh of its orthogonal projection onto the $z$ -axis? In other words, write the area as a function of the length.
:) and I guess what @Alessandro meant was that I can always shrink $X$ so that the intersection of $X$ and $\Gamma$ only contains $0$, and then just move this around by elements of $\Gamma$ ?
You can do it directly with your $X$, you want to use that $X$ has nonempty interior otherwise the Baire category theorem won't let you cover everything by translates of $X$
You do need some more argument than I thought earlier
I think that's the correct idea but it needs to be written down carefully
Yeah, the lecturer says "since $X \cap \Gamma$ is finite and $X$ is a neighbourhood of $0$, we can always shrink $X$ so that the intersection of $X$ and $\Gamma$ is $0$.. thus $\Gamma$ discrete"
Let f(x) = \begin{cases} x, & \text{if $x$ is rational} \\ 1-x, & \text{if $x$ is irrational}\\ \end{cases}. Does $\ f$ assume every value between $\ 0$ and $\ 1$?
@MaximilianJanisch what is orthogonal projection onto the xy -plane of cube is it area of one face of cube and what about orthogonal projhection on z-axis
@MaximilianJanisch so you mean if y is irrational, then y=1-(1-y)=f(1-y) and if y is rational then y=f(y) in either case there is a pre image of y and hence surjective?
$X := \lbrace x \in \Bbb R^n : \lvert x \rvert \leq c \rbrace$, and $X \cap \Gamma$ is finite for every $c > 0$, so in particular as $c \to 0$ we get $X \cap \Gamma = \lbrace 0 \rbrace$, and then $\gamma + X \cap \Gamma = \lbrace \gamma \rbrace$, so we isolate every point of $\Gamma$ .. ¯_(ツ)_/¯
I'm half-guessing my way through some of the topological arguments
\text { Let } P=\left\{\left(m^{2}+1\right)^{1 /(k+1)}: m \in \mathbb{Q}, k \in \mathbb{N}\right\} . \text { Let } g:[0,1] \rightarrow P \text { be a } function. Prove that there exist $x, y \in[0,1]$ such that $x \neq y$ and $g(x)=g(y)$
Let $P=\left\{\left(m^{2}+1\right)^{1 /(k+1)}: m \in \mathbb{Q}, k \in \mathbb{N}\right\} .$ Let $g:[0,1] \rightarrow P$ be a function. Prove that there exist $x, y \in[0,1]$ such that $x \neq y$ and $g(x)=g(y)$
or note that $\mathbb{Q}\times\mathbb{N}\rightarrow P,\,(m,k)\mapsto(m^2+1)^{1/(k+1)}$ is a surjection from a countable set onto $P$, hence $P$ is countable
@MaximilianJanisch Lol. You should have written something that was not linear (I wonder if the TA would have bothered to figure out what quadratic it was).
@feynhat I was able to figure it out. Devices per row is the square route of the closest perfect square that is greater than n devices. From there we can easily find rows. So basically this:
function getRows(n) {
dpr = Math.ceil(Math.sqrt(n));
return Math.ceil( n / dpr);
}
Let $\epsilon > 0$ then exists a $n \in \mathbb{N}$ such that $\epsilon > \frac{1}{n}$ this statement is usually justified by the archmedian axiom which states if $y>x>0$ then there exists $n \in \mathbb{N}$ such that $nx>y$. This might sound stupid. But could you please show me how this follows from the axiom? i just cant see it! i tried to put symbols and such but i just cant see how it follows from the axiom.
@feynhat In metric spaces sequentially compact, limit point compact and compact are all equivalent so that's usually a bad indicator for understanding these statements. Take $\Bbb N \times \{0, 1\}$ where $\Bbb N$ is equipped with discrete topology and $\{0, 1\}$ is equipped with indiscrete topology. This is first countable, limit point compact (because lol) but not sequentially compact. Then sequence $(n, 0)$ has no convergent subsequence.
The reason your argument using local base fails is because it's not $T_1$
What is true is that in first countable spaces sequential compactness is equivalent to countable compactness (every countable open cover has a finite subcover). In $T_1$ spaces countably compactness is equivalent to limit point compactness. Combining, you get in first countable $T_1$ spaces, sequential compactness is equivalent to limit point compactness.
I was implicitly assuming that any neighborhood of the limit point will intersect the set in infinitely many points. This may not be true for non-T1 spaces.
OK, let's see if I can solve a @Ted exercise on the fly. Let $M, M' \subset \Bbb R^3$ be surfaces and $f : M \to M'$ a bijective map such that the line segment joining $P$ and $f(P)$ is tangent to $M$ and $P$ and $M'$ and $f(P)$, has constant length $r$ and the angle between the normals $\mathbf{n}_P$ and $\mathbf{n}_{f(P)}$ of the corresponding surfaces is constant $\theta$.
Choose moving frames $(\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3)$ on $M$ and $(\mathbf{e}_1', \mathbf{e}_2', \mathbf{e}_3')$ on $M'$ such that $\mathbf{e}_1 = \mathbf{e}_1' = \vec{Pf(P)}$
I need a small hint: Suppose $\mathbf{g} : (a,b) \to \mathbb{R}^n$ is a differentiable parametrized curve and that for some point $\mathbf{p} \in \mathbb{R}^n$ we have $\lVert \mathbf{g}(t_0)-\mathbf{p} \rVert \le \lVert \mathbf{g}(t)-\mathbf{p} \rVert$ for all $t \in (a,b)$. Prove that $\mathbf{g}'(t_0) \cdot (\mathbf {t_0}-\mathbf{p})=0$.
@Semiclassical No, no, I think you can get wild stuff from easy surfaces. This gives the nonlinear superposition law for sine-Gordon solitons, as far as I am aware
@Simone I think you want $g^{\prime}(t_0)\cdot(g(t_0)-p)=0$. In this case, note that the assertion is trivial if $g(t_0)=p$, so we may assume this is not the case. Then, consider that $t_0$ is a minimum of the map $t\mapsto\lVert g(t)-p\rVert$.
The geometric image is fairly nice: The assertion says that the tangent line of the curve at $t_0$ is orthogonal to the direction from $g(t_0)$ to $p$. It has to be, for if it weren't, there would be points on the tangent closer to $p$ than $g(t_0)$ and after just moving a little bit on the curve, the curve should still be close enough to the tangent that there is a point on it closer to $p$ than $g(t_0)$, contrary to the hypothesis.
@BalarkaSen Oh I should I have gotten this immediately. If $\mathbf{x}$ and $\mathbf{x}'$ are local coordinates on $M$ and $M'$ then $\mathbf{x}' = \mathbf{x} + r \mathbf{e}_1$. So $d\mathbf{x}' = d\mathbf{x} + r d\mathbf{e}_1 = \omega_1 \mathbf{e}_1 + (\omega_2 + r\omega_{12}) \mathbf{e}_2 + r \omega_{13} \mathbf{e}_3$
Sounds like Balarka is delivering a soliloquy to me.
@Simone: I didn't give a hint. But I will point out that you'll always prefer to differentiate $\|x\|^2$ rather than $\|x\|$. I mean, sometimes you can't choose, but here you can.
@TedShifrin Because since $\omega_{12}' = d\mathbf{e}_1' \cdot \mathbf{e}_2'$, which if I plug in $\mathbf{e}_1' = \mathbf{e}_1$ and $\mathbf{e}_2' = \cos(\theta) \mathbf{e}_2 + \sin(\theta) \mathbf{e}_3$ I get $\omega_{12}' = \cos(\theta) \omega_{12} + \sin(\theta) \omega_{13}$, so if I have $\omega_{13}$ in terms of $\omega_2$ and $\omega_{12}$ I should be able to combine and compute Gaussian curvature by computing $d\omega_{12}'$ in terms of $d\omega_{12}$
And the above formula is precisely giving me $\omega_{13}$ explicitly aka computing the second fundamental form
@JBis: So you take the largest $a$ allowed by those two inequalities, @JBis. But, as I realized subsequently, maybe we need to turn the rectangle around (in case $x<y$) and take both $3a\le y$ and $3b=27a/16\le x$. Which case gives the larger value of $a$?
You wanted the largest rectangle. I'm just saying that you have to think about both ways of sticking the rectangle in the big one. $a$ lines up EITHER with $x$ OR with $y$. The answer will depend on the original ratio $x:y$.
If I have an differential equation $\dot r^2 + V(r) = E$, how can I show that if $\dot r(0) = 0$ and $r(0)$ is a critical point for $V$ then $r$ is constant ?
If I start at the critical point and stay there for a small $\epsilon$, I will have to remain there all the time. So the only interresting case is if there's a "take off", ie if $\dot r$ is nonzero
Well, let's think a little bit more. Since energy is constant, if the kinetic energy increases, then $V(r)$ must decrease, so $r(0)$ is a (local) max of $V$. That says $V''(r)\le 0$. Aren't we just supposed to argue by uniqueness of solutions to ODE? Why is this so complicated?
@TedShifrin I made a mistake. The last restriction is $b<=y$ not 3b. However, I was able to correct the equation you gave me. I tested it and it works thanks!!
If you replace our unit cube by a cube of side length $a$, what is the relationship between the area of its orthogonal projection onto the $x y$ -plane and the lenglh of its orthogonal projection onto the $z$ -axis? In other words, write the area as a function of the length.
