# Tag Info

28

Here is a proof that no sphere (of dimension $>0$) admits an averaging function. Suppose there is an averaging function $f:S^n\times S^n\to S^n$. Let $T(x_0,x_1,x_2,\dots,x_n)=(-x_0,-x_1,x_2,\dots,x_n)$; then $T:S^n\to S^n$ is homotopic to the identity and satisfies $T(T(x))=x$. Now consider the map $g:S^n\to S^n$ given by $g(x)=f(x,T(x))$. Since $T$ ...

22

Such functions are called means. The earliest paper on the subject that I’ve seen is G. Aumann, Über Räume mit Mittelbildungen, Mathematische Annalen (1943), Vol. 19, 210-215. He shows inter alia that no $S^k$ has a mean; that the only $2$-dimensional manifold with a mean is the open disk; and that if $X$ has a mean, then so does every retract and every ...

19

Edit: This post has been edited to take Eric Wofsey's comments into account. The end result is the following theorem: Suppose $X$ is a closed manifold of positive dimension. Then $X$ does NOT admit an averaging function. If an averaging $f:X\times X\rightarrow X$ exists, it induces a map $f_\ast: \pi_k(X)\oplus \pi_k(X)\rightarrow \pi_k(X)$. ...

6

(Let me answer the case of coefficients in $\mathbf Z$ (or $\mathbf Q$ or $\mathbf R$ or $\mathbf C$.) I think it should only take a bit of universal coefficient wizardry to deduce the answer in general, but it is too late here for that.) First, by the Lefschetz hyperplane theorem and Poincaré duality, we have an isomorphism $$H_i(\mathbf P^n, \mathbf Z) ... 5 By the Lefschetz Hyperplane theorem, the inclusion i : X \hookrightarrow \mathbb{CP}^n induces a map i^* : H^q(\mathbb{CP}^n, \mathbb{Z}) \to H^q(X, \mathbb{Z}) which is an isomorphism for q \leq n - 2 and injective for q = n - 1. Recall that$$b_q(\mathbb{CP}^n) = \dim H^q(\mathbb{CP}^n, \mathbb{Z}) = \begin{cases} 1 & q\ \text{even}, 0 \leq q ...

5

They are equivalent. Start with the left unknot. It has two upper strands: a left one and a right one. Grasp the right upper strand, and pull it towards the left, over the left upper strand. You will obtain a copy of the right unknot.

5

You can prove that they're not homeomorphic by employing the same strategy as in the answer here. If they were homeomorphic, so would be their one-point compactifications, and hence they would have the same cohomology. $TS^2$ has no embedded spheres with self-intersection number 1; the canonical line bundle does (the zero section!) Another way of phrasing ...

4

Yes. $S^1$ is path connected, so every element of $S^1$ acts by a map homotopic to the identity.

4

The NE and SW blocks both contain triangles with vertices 1,3,4. But these intersect in just the edge 13 and the vertex 4, while the intersection of two simplices in a triangulation must be a "face" of each (which might be the empty simplex), not a union of two or more simplices.

3

General idea: Take the set $U=U_{\alpha_0\ldots\alpha_p}$. By performing $\delta$ twice on some form on $U$, you will get the set $V=U_{\alpha_0\ldots\alpha_p\beta_0\beta_1}$ twice, and the two forms on $V$ will cancel each other out. This is due to the $(-1)^p$ in the definition of $\delta$. To get the right feel, you may calculate $\delta$ explicitly for ...

3

For $A\subset X$ you have long exact sequence $$\dots\to H_n(A)\to H_n(X)\to H_n(X,A)\to\dots$$ The composition $r_*\circ i_*:H_n(A)\to H_n(X)\to H_n(A)$ is identity, so we see that $i_*:H_n(A)\to H_n(X)$ is inclusion for all $n$. Thus, we can write $$0\to H_n(A)\to H_n(X)\to H_n(X,A)\to0,$$ and $r_*$ gives us splitting of this short sequence.

3

The answer is no in general, because if $X$ is any nonempty set with the indiscrete topology (only $\emptyset$ and itself are open) then $\mathcal{O}_X \cong \mathbb{C}$, while two indiscrete spaces of different cardinality are not homeomorphic. When you restrict to compact Hausdorff spaces, this becomes true. (See A theorem due to Gelfand and Kolmogorov ...

3

Here's another (probably unnecessarily elaborate) proof that they are not diffeomorphic using Eliashberg's amazing theorem about which open $2$-handlebodies admit Stein structures and the adjunction inequality for Stein surfaces. I am sorry if this is not really accessible for the asker. $M$ is the 4-manifold given by gluing an open $2$-handle along the ...

3

The second stiefel whitney class of your bundle $TS^2\oplus l$ is non-zero. For the chern classes you can see $$c(TS^2\oplus l)=(1+2a)(1-a)=1+a$$ where $a$ is the generator of $H^2(S^2;\mathbb{Z})$ (here I see $l$ with the complex structure still). Taking mod 2 reduction shows that the second stiefel whitney class is non-zero.

3

The contraction has to take place within $S^2$. As Hans Engler says in the comments, the continuous deformation is technically a continuous map $f:[0,1]\times S^2\to S^2$, specifically, one such that $f(0,x)=x$ for all $x\in S^2$, and there is a $p\in S^2$ such that $f(1,x)=p$ for all $x\in S^2$. For each $\alpha\in[0,1]$ define $f_\alpha:S^2\to S^2:x\mapsto ... 2 If$f: X \subseteq \mathbb{R}^n \to Y$is continuous it induces a homomorphism$f_*: \pi_1(X) \to \pi_1(Y)$. However if$F: \mathbb{R}^n \to Y$extends$f$then we have that$F_*: \{1\} \to \pi_1(Y)$is certainly trivial because$\mathbb{R}^n$is contractible. But this gives a nullhomotopy of$f(\gamma)$for every$\gamma: I \to X$where$H: I \times I \to ...

2

Here on this picture some simplexes are identified. When you give names to different vertices, it will look as follows

2

John has a good question. There is a tendency in algebraic topology to confuse a topological space and a topological space with a base point. Grothendieck wrote to me in 1983 in part: " both the choice of a base point, and the 0-connectedness assumption, however innocuous they may seem at first sight, seem to me of a very essential nature. To make an ...

2

The component $D^2$ is contractible, so $H_*(D^2\times S^1)=H_*(S^1)$, because homology are homotopy invariant. It is well-known that $H_1(S^1)=\mathbb Z$.

2

First of all, let me falsify $[\mathbb{CP}^1]+[\mathbb{CP}^1]=[\mathbb{CP}^1\#\mathbb{CP}^1]=[\mathbb{CP}^1]$ by a sledgehammer argument: The complex bordism ring $\Omega_*^U\cong \mathbb{Z}[x_{2i}|i\in\mathbb{N}]$ is an integral polynomial ring with one generator in each even dimension, so in particular has no torsion and $[X]+[X]=[X]$ can only hold for the ...

1

You've done half of the homotopy with the cone upside-down. Intuitively, the homotopy equivalence should shrink the cone to its vertex, so if your equivalence relation $\sim$ identifies $(x_1,s_1)$ with $(x_2,s_2)$ when $s_1=s_2=1$, then your homotopy should be between $\mathrm{id}_{\mathrm{Con}(X)}$ and $c_{[(x,1)]}$, not between ...

1


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