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## Hot answers tagged compactifications

12

Neil has already given adequate reply; this answer is partly for Simon, and partly for those who do like category theory, and realize that its purpose is to make life simpler, not more complicated! First, IMHO that's not a very good definition in the wikipedia article. A better definition is given in Spivak's A Comprehensive Introduction to Differential ...

12

If $\widehat X$ is the 1-point compactification of $X$, then there is a homeomorphism (for, say, locally compact Hausdorff spaces) $$\widehat{X \times Y} \cong \widehat X \wedge \widehat Y$$ with the smash product. Moreover, the smash product preserves homotopy equivalences for well-pointed spaces, which $\widehat{\Bbb C^*}$ and $S^1 \vee S^2$ both are. ...

10

Let $A=\{0,2,4,\dots\}$ be the even numbers and let $B=\{1,3,5,\dots\}$ be the odd numbers. Topologize $A\cup \beta B$ so that $A$ is a sequence limiting to a unique point in $\beta B \setminus B$. This is a compactification of $\mathbb{N}$ that fails to be soft, since any homomorphism of the required form would give a non-trivial sequence in $\beta B$ ...

9

As you guessed, the answer is no. The following counterexample in dimension $2$ can be found in Vo Van Tan's paper On the compactification of strongly pseudoconvex surfaces III, Mathematische Zeitschrift 195 (1987), 259-267. The Stein surface $\mathbb{C}^* \times \mathbb{C}^*$ admits two algebraic compactifications $M_1$ and $M_2$ which are not ...

8

There are some differences on the categorical level. The compact Hausdorff spaces are a reflective subcategory of topological spaces, and the Stone-Cech compactification is left adjoint to the inclusion of compact Hausdorff spaces into topological spaces. This basically encodes the universality and is enough reasons for ubiquitous appearance. For the ...

8

If you consider only complements of points, which are particular dense $G_\delta$-sets, the answer is independent of $ZFC$. The following results can be found in van Mill´s article in the Handbook of Set-Theoretic Topology: Corollary 1.5.4: $(CH)$ If $x \in \omega^*$ then $\omega^* \setminus \{x\}$ is not $C^*$-embedded in $\omega^*$. and Corollary ...

8

The main idea is this: You can always find an $X$ such that $X\supseteq U$ and $X\setminus U\supseteq \Sigma := \mathrm{Sing} X$. Then in characteristic zero apply Hironaka's resolution theorem, which says that there exists a resolution of singularities $\pi:Y\to X$ such that $\pi$ is an isomorphism over $X\setminus \Sigma\supseteq U$. In particular, $\pi^{-... 8 In characteristic 0, yes, yes, no. Check out Kollar's paper: http://arxiv.org/abs/math/0508332 In characteristic p, unknown. 8 Let me denote$E_\pm(M)$the eigenspaces of$M$associated with the eigenvalues$\pm1$. Let me assume that the characteristic of the scalar field$k$is not$2$. Then your assumption is that $$k^n=E_+(A)\oplus E_-(A)=E_+(B)\oplus E_-(B).$$ In addition, the assumption that$A+B$is non-singular means$E_+(A)\cap E_-(B)=(0)$and$E_+(B)\cap E_-(A)=(0)$, from ... 8 You could have a look at https://arxiv.org/abs/1806.02470 and references therein. EDIT (taking into account the comment): compactification of the$\mathcal{N}=(2,0)$6d superconformal field theory on a 4-manifold (with an appropriate topological twist along the 4-manifold), produces a 2d (6=4+2)$\mathcal{N}=(2,0)$superconformal field theory. Taking its ... 7 I can show the following (which Anton was asking about in comments). Let$X$be locally compact and Hausdorff, and$U\subseteq X$open. Let$X_\infty$be the one-point compactication, so$U$is still open in$X_\infty$. By the universal property of the Stone-Cech compactification, there is a continuous map$\phi:\beta X\rightarrow X_\infty$which is the ... 7 Another example, different but diffeomorphic to Polizzi's, is given by$E \times \mathbb C$, where$E$is a fixed elliptic curve say$ \mathbb C^* / \lbrace z \mapsto 2z \rbrace$. It can be compactified as the projective surface$E \times \mathbb P^1$or as the (non-Kähler) Hopf surface$\mathbb C^2-\{0\} / \lbrace (z,w) \mapsto (2z,2w) \rbrace$. Notice ... 7 Let$M$be a punctured torus and$N$be a twice-punctured plane. Then$M$and$N$are homotopy equivalent, but their one-point compactifications are not (the first being a torus and the second having the homotopy type of$S^2\vee S^1\vee S^1$). In particular,$H_c^*(M)$has a nontrivial cup product but$H^*_c(N)$does not. 7 This desire fails on a much more basic level. A smooth morphism is flat, and you can't even have a flat morphism between two smooth compactifications. A flat morphism has equidimensional fibers. If it is also proper and birational, then it is finite and a finite birational morphism mapping onto something normal is automatically an isomorphism. The main ... 7 No. First of all, take note that for compact zero-dimensional spaces$X$, the$\sigma$-algebra generated by all clopen sets is precisely the$\sigma$-algebra of Baire sets (recall that in a completely regular space, the$\sigma$-algebra of Baire sets is the$\sigma$-algebra generated by collection of all zero sets). A space$X$is said to be perfectly ... 6 Here's a counterexample. Consider a non-abelian finite, 2-step nilpotent group$F$. Let$Z$be its derived subgroup. Define$G$as the quotient of the direct sum$F^{(\mathbf{N})}=\bigoplus F_n$of copies of$F$indexed by$\mathbf{N}$by identification of all copies of$Z$. Thus, writing$H=F/Z$, the group$G$lies in a central extension, with central ... 6 I am just writing my comment above as an answer, just in the special case that the parabolic is a Borel subgroup$B$. In this case, the reductive part$M$is a maximal torus$T$in$G$. Denote by$W\subset N_G(T)$a Weyl group. There is a conjugation action of$G$on itself, and there is an induced conjugation action of$W$on$T$. Denote by$q_T:T\to T/...

6

Here is a partial answer: the Continuum Hypothesis implies that all Parovichenko spaces are soft-Parovichenko; the proof is a bit long, so I put it in a PDF-file on my website. Also, I retract my claim in the comments that all compactifications with $\omega_1+1$ as a remainder are soft. It is true, in ZFC, that $\omega_1+1$ is soft-Parovichenko but "all ...

5

Firstly, I would suggest taking a look at the work of Peter Michor, who is actually a frequent contributor to MO. Regarding Riemannian metrics on spaces of curves, you might be interested in this nice recent paper. Also, for more on infinite-dimensional Riemannian geometry including some metric aspects, take a look at the work of Brian Clarke, Boris Khesin, ...

5

The main point is this. Let $(L_k)_{k=0}^\infty$ be another increasing sequence of compact sets whose interiors cover $X$. Each $X_n$ is compact and contained in the union of the sets $\text{int}(L_k)$, so it is contained in some finite union of these open sets. As the sets $\text{int}(L_k)$ are nested, it follows that $K_n\subseteq\text{int}(L_{k_n})\... 5 No that is not true. There is a much simpler example than in the MO answer above. I might have written the following example in one of my earlier MO answers. Let the proper target of the morphism be$\mathbb{P}^1_k = \text{Proj}\ k[S,T]$. The domain of the morphism will be an open subset of$$\mathbb{P}^1_k \times_{\text{Spec}\ k} \mathbb{P}^1_k = \text{... 4 The answer to the question asked is no, i.e. the Haar integral$I(f)$of a nonzero, nonnegative continuous function$f$is always positive. See Hewitt & Ross, Theorem (15.5)(i). The mistake in your argument is that$f(x)=1/(1+x^2)$is not almost periodic. Indeed, recall that$f$is almost periodic iff$\forall\varepsilon>0\exists L>0$such that ... 4 Yes, although the terminology is unusual this is standard. The compactifications of$X$correspond to quotients of$\beta X$which correspond to closed subalgebras of$C(\beta X) \cong BC(X)$. For example, see Theorem 3.55 of my book Measure Theory and Functional Analysis. If you don't care to look it up, the idea is to define an equivalence relation on$\...

4

The Stone-Čech compactification of $\mathbb{R}$ is not its one-point compactification. The former is the largest compactification of a space, while the latter, if it exists, is the smallest compactification, and in general there will be many compactifications in between. $\mathbb{R}$ has, for example, a two-point compactification, namely $[0, 1]$. The C*-...

4

Let $\Gamma=\mathbb Z$, and consider the sequence of all odd numbers, viewed as a filter. Pick an ultrafilter containing this filter. Then this ultrafilter is not idempotent. The reason is simple: odd + odd = even.

4

The following argument shows that the very best thing you could hope for does not work. I'm not sure whether I need all of my assumptions. Claim: There does not exist a compactification functor $(C,\eta)$ such that $\eta_{\mathbb A^2}$ is a [dense] open immersion into a smooth proper [hence projective] surface. Proof. Basically, $\mathbb A^2$ has ...

4

The answer is no: unless $G$ is compact, its subspace topology inside $bG$ (called its Bohr topology) is much weaker than its own. So $\iota:G\to bG$, while continuous with dense image, is not an embedding (homeomorphism onto $\iota(G)$), hence not a compactification in the sense of topologists. Rudin (1962, pp. 30-31) notes this without proof. Later ...

3

Here's a counterexample if you don't assume your algebra is self-adjoint. Let $X$ be the unit disk in $\mathbb{C}$ endowed with the discrete topology. Let $B$ be the algebra of complex-valued functions on $X$ that extend continuously to the one-point compactification. Note that every element of $B$ is constant on a cocountable set. Let $A$ be the closed ...

3

The closure of $X$ in $Z$ is compact , so there is no hope if $f$ is not bounded. If it is bounded then so is its extension to the closure of $X$ in $Y$ and this gives a bounded uniformly continuous function on the closure which can be extended to a bounded continuous function on $Y$ by the Tietze extension theorem. This, in turn, extends to a continuous ...

3

Here´s a ZFC answer to your question. All spaces are assumed to be Hausdorff. Recall that the tightness $t(x,X)$ of a point $x$ in the space $X$ is defined as the least cardinal $\kappa$ such that for every set $A \subset X$ such that $x \in \overline{A}$ there exists a $\kappa$-sized subset $B$ of $A$ such that $x \in \overline{B}$. The tightness $t(X)$ of ...

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