The reduction of the structute group of $\mathbb{R}^n$ -fiber bundles to a special subgroup of $Homeo(\mathbb{R}^n)$

Let $ G$ be the group of all homeomorphisms $ f$ of $ \mathbb{R}^n$ which satisfy $ $ f(x+m)=f(x)+m,\quad \forall m\in \mathbb{Z}^n$ $

In the other words, $ G$ is the group of all equivariant homeomorphisms of Euclidean space with respect to the standard action of $ \mathbb{Z}^n$ on $ \mathbb{R}^n$ .

Is there a fiber bundle whose fibers are homeomorphic to$ \mathbb{R}^n$ but the structure group of the bundle can not be reduced to $ G$ ? Is there a manifold $ M$ such that the structure group of $ TM\to M$ , which is counted as a fiber bundle, can not be reduced to $ G$ ?

Tiling $\mathbb{R}^n$ with a single shape of positive measure

Let $ n>1$ be an integer. For $ A\subseteq \mathbb{R}^n$ and $ x\in\mathbb{R}^n$ we set $ x+A = \{x+a: a\in A\}$ .

Is there a measurable subset $ A\subseteq \mathbb{R}^n$ with positive $ n$ -dimensional Lebesgue measure and the following properties?

  1. $ \mathbb{R}^n = \bigcup\{z+A: z\in \mathbb{Z}^n\}$ , and
  2. if $ z_1\neq z_2\in \mathbb{Z}^n$ then $ (z_1+A)\cap (z_2+A) = \emptyset$ .

Covering dimension of boundary of compact subset of $\mathbb{R}^n$

Let $ X$ be a compact subset of $ \mathbb{R}^n$ , with the inherited Euclidean topology. Does it follow that $ dim_{cov}(\partial X)\leq dim_{cov}(X)-1$ ? Or, at least, $ dim_{cov}(\partial X)\leq n-1$ ?

I would be happy to have a reference for that, if it is true. Otherwise, I would be happy to get a counter example, or sufficient conditions under which it holds.

Thanks

Formally proving that a metric is not induced by any norm in $\mathbb{R}^n$

What is the procedure to formally prove that no norm exists in $ \mathbb{R}^n$ , that induces a metric $ d$ ?

My first instinctive idea would be to show that $ d$ is a metric in $ \mathbb{R}^n$ , but after this I don’t know any further. What could I achieve by following this road?

The specific problem I am working on is to prove that no norm $ ||\cdot||$ in $ \mathbb{R}^2$ exists, that induces the metric $ d$ :

$ d((x_\text{1},y_\text{1}),(x_\text{2},y_\text{2})) := \begin{cases} |y_\text{1}-y_\text{2}|, & \text{if }x_\text{1} = x_\text{2},\ |y_\text{1}| + |x_\text{1}-x_\text{2}| + |y_\text{2}|, & \text{if }x_\text{1} \not= x_\text{2}. \end{cases}$

I do not intend to get solutions for my specific problem, but maybe a similar problem exists elsewhere, which I could study. Of course, I have searched MathOverflow, and other sources.

Double line integral of $1/|\mathbf{x}-\mathbf{y}|$ in $\mathbb{R}^n$

I was wondering how to express the closed-form solution of the following double line integral in $ \mathbb{R}^n$ :

$ $ \int_{\mathbf{x}_1}^{\mathbf{x}_2} \int_{\mathbf{y}_1}^{\mathbf{y}_2} \frac{1}{|\mathbf{x} – \mathbf{y}|} d\mathbf{x} d\mathbf{y} $ $

where

$ $ \mathbf{x} = [x_1, x_2, \dots, x_n]^T $ $ $ $ \mathbf{y} = [y_1, y_2, \dots, y_n]^T $ $ $ $ |\mathbf{x}| = \sqrt{\sum_{k=1}^n|x_k|^2} $ $ $ $ |\mathbf{y}| = \sqrt{\sum_{k=1}^n|y_k|^2} $ $

given that $ \mathbf{x}_1, \mathbf{x}_2, \mathbf{y}_1, \mathbf{y}_2$ are all distinct from each other and the lines $ \mathbf{x}_1, \mathbf{x}_2$ and $ \mathbf{y}_1, \mathbf{y}_2$ do not cross each other. That is, it is granted that $ |\mathbf{x} – \mathbf{y}| > 0$ when evaluating the integral.


I thought on using coordinate translation to do the integration along the line $ \mathbf{z}_1, \mathbf{z}_2$ in relation to the origin

$ $ \int_{\mathbf{z}_1}^{\mathbf{z}_2} \frac{1}{|\mathbf{z}|} d\mathbf{z} = \ln{\frac{ \sum_{k=1}^n \mathbf{z}_{2,k}(\mathbf{z}_{2,k} – \mathbf{z}_{1,k}) + \sqrt{ (\sum_{k=1}^n \mathbf{z}_{2,k}) (\sum_{k=1}^n (\mathbf{z}_{2,k} – \mathbf{z}_{1,k})) } }{ \sum_{k=1}^n \mathbf{z}_{1,k}(\mathbf{z}_{2,k} – \mathbf{z}_{1,k}) + \sqrt{ (\sum_{k=1}^n \mathbf{z}_{1,k}) (\sum_{k=1}^n (\mathbf{z}_{2,k} – \mathbf{z}_{1,k})) } }} $ $

where $ \mathbf{z}_{1,k}$ is the $ k$ -th coordinate of the point $ \mathbf{z}_{1}$ .

But then I run out of ideias on how to proceed further and do the second integral taking advantage of that result…

The set $D_v=\{ x\in \mathbb{R}^N: D_vf(x) \text{ exists } \}$ is measurable.

Let $ f:\mathbb{R}^n\to \mathbb{R}$ be Lipsticz function. The directional derivative in the direction of $ \mathbf{v}$ is given by $ $ D_\mathbf{v}f(\mathbf{x})=\lim_{t\to 0}\dfrac{f(\mathbf{x}+t\mathbf{v})-f(\mathbf{x})}{t}. $ $ Define a set $ $ D_\mathbf{v}=\{ \mathbf{x}\in \mathbb{R}^n:D_\mathbf{v}f(\mathbf{x})\text{ exists } \} .$ $ Show that $ D_\mathbf{v}$ is measurable.

Please help me in proving that.

Volume of manifolds embedded in $\mathbb{R}^n$

Let $ N$ be a closed, connected, oriented hypersurface of $ \mathbb{R}^n$ . Such a manifold inherits a volume form from the usual volume from on $ \mathbb{R}^n$ and has an associated volume given by integration of the inherited volume form over $ N$ .

I would like to know if there is a proof of the following intuivitely obvious (I think) fact: let $ X$ be a vector field on $ \mathbb{R}^n$ everywhere transverse to the hypersurface. Let $ \psi_\epsilon^X$ be the flow of such a vector field for a small time $ \epsilon$ (we can choose $ \epsilon$ small enough so that $ \psi_\epsilon^X(N)$ is diffeomorphic to $ N$ ). Then the volume of $ \psi_\epsilon^X(N)$ is not equal to the volume of $ N$ .

Does such a result exist? I assume, if so, it would not generalise to other manifolds/volumes other than the canonical one?

Classifying Radon partitions in $\mathbb{R}^n$ whose affine hull is $\mathbb{R}^n$

Specifically, I want to determine all distinct “types” of Radon partitions of $ n+2$ points in $ \mathbb{R}^n$ for which the affine hull is all of $ \mathbb{R}^n$ . This is a homework question, so I’m primarily looking for advice on getting started.

Radon’s theorem states that any set of $ n+2$ points in $ \mathbb{R}^n$ can be partitioned into two disjoint sets with intersecting convex hulls.

The affine hull of a set $ S$ is given by $ $ \mbox{aff}(S)=\left\{\sum_{i=1}^k\alpha_ix_i:k>0,x_i\in S,\alpha_i\in\mathbb{R},\sum_{i=1}^k\alpha_i=1\right\}.$ $

So, the simplest case is when $ n=1$ , and clearly the affine hull of any Radon partition of $ n+2=3$ points in $ \mathbb{R}$ is all of $ \mathbb{R}$ . When $ n=2$ , the $ n+2=4$ points can be partitioned as a triple and a singleton or two pairs of points. In the case of the former, the convex hull of the triple must contain the singleton. In the case of the latter, the pairs of points must form intersecting line segments. But the former is the only “type” of partition I want to consider, as it’s affine hull is in fact $ \mathbb{R}^2$ , but the affine hull of the partitions in the latter is not all of $ \mathbb{R}^2$ . So I might conjecture that the full classification requires that the convex hull of one of the partitions must be fully contained in the convex hull of the other partition, but I’m not sure if I even believe this intuitively.

Thanks in advance for any advice.

Extendind a a function defined in $\mathbb{R}^n\setminus\{0\}$ to a continuous function defined in $\mathbb{R}^{n}$.


Let $ g: \mathbb{R}^n\setminus\{0\} \to \mathbb{R}$ be a function of class $ C^{1}$ and suppose that there is $ M > 0$ such that $ $ \left|\frac{\partial}{\partial x_{i}}g(x)\right| \leq M.$ $ Prove that if $ n \geq 2$ then $ g$ can be extended to a continuous function defined in $ \mathbb{R}^n$ . Show that if $ n = 1$ the statement is false.

My attempt.

I want define the extension $ \bar{g}: \mathbb{R}^n \to \mathbb{R}$ by $ \bar{g}(x) = g(x)$ if $ x \in \mathbb{R}^{n}\setminus\{0$ } and $ \displaystyle \bar{g}(0) = \lim_{x \to 0}g(x)$ . Thus $ $ \lim_{x \to 0} \bar{g}(x) = \lim_{x \to 0}g(x) = \bar{g}(0)$ $ and so, $ \bar{g}$ is continuous. So the question is reduced to prove that $ \displaystyle \lim_{x \to 0}g(x)$ exists. Thus, I must show that $ $ \forall \epsilon > 0, \exists \delta > 0\text{ s.t. } \Vert X \Vert < \delta \Longrightarrow |g(x)|<\epsilon.$ $

The hypothesis $ $ \left|\frac{\partial}{\partial x_{i}}g(x)\right| \leq M$ $ seems necessary for get $ $ |g(x)-g(y)| \leq M|x-y|$ $ using the Means Value Inequality. This almost solves the problem, because if $ g(0) = 0$ we can write $ $ |g(x)| \leq M|x|.$ $ But $ g(0) = 0$ doesnt make sense. I’m stuck here.

Also, I cannot see the whe is necessary $ n \geq 2$ , where I use this in the demonstration and why it fails when $ n=1$ .

Reference for compact embedding between (weighted) Holder space on $\mathbb{R}^n$

Suppose $ 0<\alpha<\beta<1$ , and $ \Omega$ is a bounded subset of $ \mathbb{R}^n$ . Then the Holder space $ C^{\beta}(\Omega)$ is compactly embedded into $ C^{\alpha}(\Omega)$ . But if $ \Omega=\mathbb{R}^n$ , then the compact embedding is not true.

However, if we consider the weaker weighted Holder space $ C^{\alpha, -\delta}(\mathbb{R}^n)$ (for any $ \delta>0$ ) instead of $ C^{\alpha}(\mathbb{R}^n)$ . Then is $ C^{\beta}(\mathbb{R}^n)$ compactly embedded to $ C^{\alpha, -\delta}(\mathbb{R}^n)$ ?

Here $ $ \|f\|_{C^{\alpha, -\delta}}=\|(1+|\cdot|^2)^{-\frac{\delta}{2}}f\|_{C^{\alpha}}. $ $

I could not find a precise reference from some books on functional analysis. Any comment is welcome.