Tag Info

New answers tagged

3

To complement Christian Remling's nice concrete explanation, let me just add that more is known. Your 2nd question can be rephrased as asking: does the function $x\mapsto |x|$ ``operate in'' the Fourier algebra? This kind of question used to be of interest to people working on Banach algebras of functions. For the Fourier algebra of a LCA group the answer ...


1

This won't work. I want to show that we can't take square roots in $A(\mathbb R)$. My function will be of the type $$ f(x) = \sum h_n \varphi\left( \frac{x-a_n}{L_n}\right) ,\quad\quad\quad\quad (1) $$ and here the individual summands will have disjoint supports. I will take $h_n\in\ell^2$, $h_n\notin\ell^1$. Since the $L^1$ norm of the Fourier transform of ...


1

I'm not sure about the proof I gave, But as I checked, it didn't use full AC, but as Asaf mentioned in a comment it uses DC. The following theorem is due to Pontrjagin. See Book by Montgomery and Zippin(Page 29). I will give a sketch of proof. Note: Maybe it's needed to add some separation axiom to the following theorem. please edit it, if needed. Theorem: ...


4

No, contrary to what you said, it is not always possible, even if $G$ is a semidirect product. Let us fix some $g \in N$, so $\sigma(gx) = \sigma(x)$ for all $x$. Then $\sigma(x)^{-1} g \sigma(x) = \omega(g,x)$ is a bounded function of $x$. It is easy to construct a counterexample to this, by arranging for the conjugacy class of $g$ to be unbounded. For ...


4

Actually, Santi Spadaro almost gave the answer in a comment to your previous question: Let $X$ be the $\Sigma$-product of $2^{\omega_1}$ (where $2=\{0,1\}$), that is, the subset of $2^{\omega_1}$ of all points with at most countably many coordinates different from $0$. Of course $X$ is a subgroup of $2^{\omega_1}$. Then, as explained in Henno Brandsma's ...


3

I am a little surprised that you think it's "intuitive" that Haar measure must be used. In section 9.13 of Jacobson's "Basic Algebra II" this result is proved for totally disconnected division algebras. In section 27 in Chapter IV of Warner's "Topological Fields" Theorem 27.2 covers the connected case. Neither book uses Haar measure.


6

The following is perhaps more of an extended comment than an answer. The sequence of sheaves is exact iff the quotient map $G\to H$ has a section over a neighborhood of every point (in fact, because of the group structure, it suffices to have a section over any single nonempty open set). In particular, for instance, this means the sequence of sheaves is ...


0

Here is a positive answer when $G$ is the universal covering group of $H$ (so then $F=\pi_1H$ which is discrete, and $\pi_1G=0$), such as $\mathbb{Z}\to\mathbb{R}\to S^1$. Denote by $M(U,G)$ the $G$-valued continuous functions on open $U\subset X$. Then covering space theory says the following induced sequence is exact: $$0\to M(U,F)\to M(U,G)\to ...


3

The paper that you mentioned is referring to a theorem proved by D. Remus in his dissertation: Theorem: If $G$ is abelian and $\tau$ is a minimal group topology on $G$, then there exists a subgroup $N$ such that $G/N \cong \mathbb{Z}/p\mathbb{Z}$ ($p$ prime) and $\{N\}$ is a fundamental system of neighborhoods of $e$ in $\tau$. Note that the only group ...


3

Pre-compact group topologies on $G$ are described by isomorphism classes of epimorphisms $G \to K$, where $K$ is a compact abelian group. (This follows from the observation that a homomorphism is an epimorphism if and only if it has dense range.) By Pontryagin duality, such isomorphism classes of maps $G \to K$ are in bijection with isomorphism classes of ...


1

I didn't think about the precompact group topologies, but the statement $|\hat G|=2^{2^{|G|}}$ is incorrect. A character of $G$ is a map from $G$ to the complex numbers. Hence there are no more than $(2^{\aleph_0})^{|G|}$ characters. If $G$ is infinite, then $(2^{\aleph_0})^{|G|}=2^{|G|}$ and not $2^{2^{|G|}}$. But I don't understand how you get the ...


10

This is more of a comment than an answer, but it's too long for a comment, so I'm putting it here. It sounds as though you are asking what sorts of groups of diffeomorphisms there are acting transitively on spheres, so that for each such subgroup $\Gamma\subset \mathrm{Diff}(S^n)$, one can define a notion of $\Gamma$-bundles over manifolds, where a ...


2

Any compact group with no nontrivial normal closed subgroups has this property, since there can be no coarser Hausdorff topology and in any coarser non-Hausdorff topology the closure of the identity would be a normal closed subgroup in the original topology. For instance, this includes all (centerless) compact simple Lie groups.


5

Yes. Since $G$ is $\sigma$-compact it is enough to show that $X_K= \lbrace f\in C(G):$ supp$(f)\subseteq K\rbrace$ is meager (of first category) in $L^2(G)$. Otherwise, the inclusion map $i_K: X_K \to L^2(G)$ between Banach spaces (of course, $X_K$ endowed with the sup-norm) would have a range of second category and the open mapping theorem (in the version ...



Top 50 recent answers are included