Mathematics - 2020-10-20 (page 3 of 3)

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11:01 PM

if it were a manifold, there would be an open subset homeomorphic to an open subset of some $\mathbb{R}^n$. That would contain a basic open set homeomorphic to an open subset of $\mathbb{R}^n$ and that basic open set contains a homeomorphic copy of $\mathbb{R}^{\mathbb{N}}$, so it would in fact embed in some $\mathbb{R}^n$.

This sounds highly impossible, but I don't have a good argument for why

It follows from invariance of dimension, but that's a big gun

Ted Shifrin

But no matter what $n$ you pick, I can find a subset that's $\Bbb R^m$ with $m>n$ and you can't stick $\Bbb R^m$ in $\Bbb R^n$.

Yes, well, you need a big gun if you're not using smooth maps.

Thorgott

I was hoping that perhaps embedding $\mathbb{R}^{\mathbb{N}}$ into $\mathbb{R}^n$ is more easily seen to be impossible than embedding $\mathbb{R}^m,m>n$ into $\mathbb{R}^n$

orientablesurface

11:14 PM

@TedShifrin with regards to your answer on, math.stackexchange.com/questions/3872942/…. , the last sentence implies that $X_ph=0?$

Ted Shifrin

What do you mean?

orientablesurface

@TedShifrin why does $dh_p|_{T_pf^{-1}(0)}=0$ imply $X_ph=0$?

Ted Shifrin

11:32 PM

$X_p h = dh_p(X_p)$.

orientablesurface

But $X_p\in T_p\mathbb{R}^3$

Ted Shifrin

No, it's tangent to the level surface.

orientablesurface

$T_pf^{-1}(0)=\{$ $v\in T_p\mathbb{R}^3$ $:$ $vF=0$ whenever $F\in C^{\infty}(\mathbb{R}^3)$ and $F|_{f^{-1}(0)}=0$ $\}$

geocalc33

I need to show that four integrals are equal

Thorgott

Let $I$ be an infinite set and $M_i,i\in I$ be manifolds, all of which have positive dimension. Assume $(P,(\pi_i)_{i\in I})$ is the product of the $M_i,i\in I$ in the category of manifolds. For each finite subset $J\subseteq I$, let $p_j\colon\prod_{j\in J}M_j\rightarrow M_j$, $j\in J$ be the projections and $p_i\colon\prod_{j\in J}M_j\rightarrow M_i$ for $i\not\in J$ be an arbitrary morphism (constant one does the job). This factors as a map $f\colon\prod_{j\in J}M_j\rightarrow P$.
Let $C$ be an object and $r,s\colon C\rightarrow\prod_{j\in J}M_j$ be two morphisms such that $f\circ r=f\ci…

Ted Shifrin

11:41 PM

That's what you asked to prove, isn’t it, @orientablesurface? I'm totally lost.

Thorgott

now on to the subtler question whether infinite products of 0-dim manifolds exist

orientablesurface

I know that $T_pf^{-1}(0)$ is defined as above. I am asked to show that the above is is equal to $\{$ $X_p\in T_p\mathbb{R}^3$ $:$ $X_pf=0$ $\}$

not for arbitrary $f$

Clearly the first set is contained in the second

I suppose I don't see why $X_p\in T_pf^{-1}(0)$ , if $X_pf=0$

Rithaniel

I've got more Cauchy-Schwarz stuff

I'm pretty sure the following is correct, but I wanna somehow get rid of the second term when I know that $\|f(x)\|\leq 1$
$$\int_a^b\big\vert k(x)f(x)\big\vert^2dx\leq\int_a^b\big\vert k(x)^2\big\vert^2dx\int_a^b\big\vert f(x)^2\big\vert^2dx$$