Tag Info

New answers tagged

3

First of all notice that $1-$dimensional manifolds generically embedded in $n-$manifolds are unlinked for $n\geq4$. As you already noticed your space $A := S^5 \setminus (S_1 \cup S_2 \cup S_3)$ is homeomorphic to $\mathbb{R}^5 \setminus (\mathbb{R} \cup S_2 \cup S_3)$. This space is homotopically equivalent to $(\mathbb{R}^5 \setminus \mathbb{R}) \vee ...


1

There is a long strand of literature relating hyperbolicity of groups and of group actions to second bounded cohomology of groups. The first result in this theory is Brooks' proof that free groups of positive rank have second bounded cohomology of uncountably infinite dimension. After further progress eventually Epstein and Fujiwara proved the same ...


0

The below is the content of Leon's comments above, which give a full answer to this question. A sketch of the proof for the first equality is given in 4-Manifolds and Kirby Calculus, Gompf, Stipsicz, 1999 AMS GSM 20, page 125, Proposition 4.5.11. The last question: if we cut out the unknot $S^1\times B^2$ = thickened z-axis from ...


2

A closed orientable surface embeds into $\Bbb R^3$; a closed non-orientable surface does not (codimension 1 connected closed manifolds in $\Bbb R^n$ separate it into two pieces; one is compact, with boundary the closed manifold; the boundary of an oriented manifold is oriented). Because $X \# Y$ is oriented iff $X$ and $Y$ are, $X \# X$ embeds into $\Bbb ...


2

Since stable and unstable subspaces complement each other (I mean, $\mathbb{R}^2 = E^{+} \oplus E^{-}$ and any vector $v = \pi_{E^{+}} v + \pi_{E^{-}} v$), the two dimensional case could be interpreted this way: There exists a (1) $C^r$-function $\psi\, \colon E^{+} \rightarrow E^{-}$ such that (2) $\psi(0) = 0$, (3) $D\psi = 0$ and set (4) $x + \psi ( ...


3

I'm not sure what might be the earliest spot where it's mentioned, but you could refer Example 1.4 from Saveliev's book "Invariants of Homology 3-Spheres".


0

If you require each boundary component to be fixed pointwise, the homeomorphism does not exist. Since $x$ separates the boundaries $\{b_1$, $b_2\}$ to the boundaries $\{b_3$, $b_4\}$, the images of $x$ under homeomorphisms that preserving the boundaries still hold this property. But $y$ does not have such a property.


2

\begin{align} A + B &= \{ a + b \mid a \in A, b \in B\} \\ &= \bigcup_{b \in B} \{ a + b \mid a \in A\} \\ &= \bigcup_{b \in B} (A+b)\,. \end{align} If one of the sets $A,B$ is finite, then $A+B$ is closed, because $A+b$ is closed (I think this holds in all normed spaces). But since infinite unions of closed sets are not closed in general, it ...


4

Take $n = 2$, take $A$ to be the $y$-axis and take $B$ to be the positive quadrant of the hyperbola $y = \frac{1}x$. Then $A$ and $B$ are both closed, but $A + B$ is the set of $(x, y)$ such that $x > 0$, which is not closed. If $A$ and $B$ are both closed and one of them is compact, then $A + B$ is closed. See Closed sum of sets for a proof.


6

Not always. Consider in $\mathbb{R}$: $$A = \mathbb{Z}, \quad B = \left\{ n + \frac{1}{n} : n \geqslant 2 \right\}$$ so $\frac{1}{n} \in A + B$ but $0 \not \in A+B$.


0

For the existence of this product you can use classification of the surfaces: if you cut $S_{0,4}$ along $x$, you obtain a pair of pants $P_{x}$, similarly if you cut $S_{0,4}$ along $y$ then you obtain another pair of pants $P_{y}$. There exist a homeomorphism which maps $P_{x}$ to $P_{y}$ and $x$ to $y$. So isotopy class of this homeomorphism can be ...


3

I really liked this question! The wikipedia page for configurations of skew lines has a reference to this beautiful paper by Viro & Viro that -among other things- answers the first question positively! Here is how the idea worsk: Consider two parallel planes in general position (that is, not paralled to any of the lines). In the first plane fix the ...


0

alternative approach given covering space actions of $G_1$ on $X_1$ and $G_2$ on $X_2$ the action of $G_1\times G_2$ on $X_1 \times X_2$ defined by $(g_1,g_2)(x_1,x_2)=(g_1(x_1),g_2(x_2))$ is a covering space action and $X_1\times X_2/{G_1\times G_2}$ is homeomorphic to $X_1/G_1 \times X_2/G_2$. so for this problem if we take $G_1 = \mathbb{Z} , G_2= ...


4

By Alexander's Lemma, knot complements are irreducible $3$-manifolds, hence also prime manifolds, see here.



Top 50 recent answers are included