User david cohen - MathOverflow

Answer by David Cohen for What does a singular simplex with real coefficient mean

2013-05-21T19:07:43Z

As Lee suggests, you should look at Allen Hatcher's book. In particular, chapter 2 and section 3.3.

In practice, you can often think of $\sum a_{i}\sigma_{i}$ as coming from the following procedure. Let $\phi:X\rightarrow M$ be a degree $d$ map of $n$-manifolds. Then any singular cocycle (with integer coefficients) $\alpha$ representing the fundamental class of $X$ will push forward to a singular cocycle $\phi_{\ast}\alpha$ representing $d$ times the fundamental class of $M$. Thus, $\frac{1}{d}\phi_{\ast}\alpha$ will be a cocycle (with rational coefficients) representing the fundamental class of $M$.

The prototypical examples are as follows:

If $M$ is a torus, then there are covering maps $\phi: M\rightarrow M$ of arbitrarily high degree, hence $M$ has no simplicial volume. (We can iterate the above procedure, starting with any cocycle $\alpha$ representing the fundamental class of $M$, getting representatives $\alpha,\frac{1}{d}\phi_{\ast}(\alpha),\frac{1}{d^{2}}\phi_{\ast}^{2}(\alpha),\ldots$ with volume going to $0$.)

On the other hand, if $M$ is a higher genus surface, then this strategy is obstructed by Euler characteristic. Namely, if $\phi:Y\rightarrow M$ is a degree $d$ covering map, then $\chi(Y)=d\chi(X)$, so the genus of $Y$ grows linearly with $d$. In fact, you can show that $X$ has nontrivial simplicial volume using hyperbolic geometry.

Area of triangles vs. comparison triangles.

2013-02-17T22:36:13Z

Let $X$ be a complete, simply connected Riemannian manifold satisfying a quadratic (coarse) isoperimetric inequality. (I.e., there is a constant $C_{0}$ such that every loop of length $\ell$ has a filling disk of area $\leq C_{0}\ell^{2}+C_{0}$.)

For points $a,b,c\in X$, define $Area_{X}(a,b,c)$ to be the minimal area of any geodesic triangle in $X$ with vertices $a,b,c$. Define $Area_{comp}(a,b,c)$ to be the area of a Euclidean triangle with side lengths $d(a,b)$, $d(b,c)$ and $d(c,a)$.

Does there exist a constant $C$ such that for all $a,b,c\in X$ we have:

$Area_{X}(a,b,c)\leq C Area_{comp}(a,b,c)+C$?

If the answer is no, then are there counterexamples when $X$ is homogeneous?

Does every orientable surface embed in $\mathbb{R}^{3}$

2012-11-16T01:15:41Z

A topological surface can be pretty strange (consider, for instance, covers of $S^{2}-K$, where $K$ is a Cantor set.) Can every orientable topological surface be topologically embedded in $\mathbb{R}^{3}$?

Simply connected simplicial complexes

2012-06-22T20:29:52Z

Let $Z$ be a simply connected, two dimensional simplicial complex.

Let $X\subset Z$ be a finite subcomplex with nontrivial $\pi_{1}$.

Must there exist a finite, simply connected subcomplex $Y\subset Z$ such that $Y\supset X$?

(Motivation: the fact that Whitehead conjecture remains unproven indicates that there are probably some very weird things that can happen in two dimensional complexes. This question attempts to locate some of these pathologies.)

Answer by David Cohen for Is it true that the sum of a specific floor function of a prime = 1?

2012-06-20T23:08:57Z

Assuming that $p\text{#}$ means the product over primes $\prod_{q\leq p}q$, then it is clear that $$\sum_{i\leq p} \mu(i)[\frac{p}{i}] = \sum_{i | p\text{#}} \mu(i)[\frac{p}{i}].$$

But this formula is very well known: $$\sum_{d\leq n} \mu(d)[\frac{n}{d}]=\sum_{k\leq n}\sum_{d | k} \mu(d)=1+0+0+\ldots$$

Is a space with no covering spaces simply connected?

2011-03-13T16:00:33Z

Suppose $X$ is a path connected space such that every connected covering space of $X$ is trivial (1-fold.) Must $X$ be simply connected?

Intuitively, the answer seems to be no (imagine taking a disk, cutting out a square, and gluing in $T\times T$ where $T$ is the topologist's sine curve.) But this is a rather weak intuition.

Answer by David Cohen for Smooth functions for which $f(x)$ is rational if and only if $x$ is rational

2010-12-10T13:15:00Z

It is true that given any two dense subsets $A,B\subset(0,1)$, there is an absolutely monotone function $[0,1]\rightarrow[0,1]$ which carries $A$ to $B$. I don't know the exact proof of this, but the idea is very simple: you enumerate the elements of $A$ and $B$, then take the first element of $A$, and associate it with an appropriate element of $B$, then the first element of $B$ and associate it with an appropriate element of $A$, then associate the second element of $A$ to an appropriate element of $B$, etc.

I am fairly certain that some variation on this should work for your problem.

Is this quotient space of Q_p contractible?

2010-10-18T04:36:03Z

Let $X_{p} = \mathbb{Q}_{p} / \sim $, where $\sim$ is defined by:

$x\sim 0 \Leftrightarrow x\in \mathbb{Q}$

$X_{p}$ is path-connected, because (unless I'm making some horrible mistake,) for any $x\in X_{p} \backslash \mathbb{Q}$, we have that $\lbrace x,0\rbrace$ under the subspace topology is path-connected.

Is $X_p$ contractible?

Answer by David Cohen for In an inductive family of groups, does the probability that a particular word is satisfied converge?

2010-09-24T04:31:33Z

Edit: As John points out below, this doesn't work.

Let's look at the example where $w=x_{1}^{2}$. I'm pretty sure that $p_{w}(G\rtimes \mathbb{Z}/2\mathbb{Z}) \geq 1/2$ for any group $G$ (where we take the non-abelian choice for the semi direct product,) since if $g\in G$ and $z$ is the generator of $\mathbb{Z}/2\mathbb{Z}$, then $(gz)^{2}=gzgz=gg^{-1}zz=1$ (exactly half of the elements of $G\rtimes \mathbb{Z}/2\mathbb{Z}$ are of the form $gz$ where $g\in G$.)

On the other hand $p_{w}(G\times \mathbb{Z}/5\mathbb{Z}) \leq 1/5$, since if $g\in G$ and $z$ is the generator of $\mathbb{Z}/5\mathbb{Z}$, we know that $(gz^{a})^{2}=1$ can only happen if $z^{2a}=1$, which happens with probability $1/5$.

We can combine these two facts to get a sequence where $p_{w}$ won't converge. (E.g., $G_1 = \mathbb{Z}/2\mathbb{Z}$ and $\forall i\geq 1$, $G_{2i}=G_{2i-1}\rtimes \mathbb{Z}/2\mathbb{Z}$ and $G_{2i+1} = G_{2i}\times\mathbb{Z}/5\mathbb{Z}$.)

Comment by David Cohen

2013-05-22T23:21:49Z

Imagine an a priori probability distribution over all closed form expressions. For any $n$, there are only finitely many closed form expressions with at most $n$ characters. Hence, the probability of a closed form expression goes to zero as its length increases, thus, it makes at least some sense to weight closed forms according to their simplicity. If you are curious about the philosophical questions involved, you should look up "Occam's razor" and "Kolmogorov prior".

Comment by David Cohen

2013-05-16T01:27:28Z

Hopefully this link works: en.wikipedia.org/wiki/Ces%C3%A0ro_mean If not, google "cesaro mean"

Comment by David Cohen

2013-04-20T03:50:25Z

The problems involved are purely computational.

Comment by David Cohen

2013-03-18T16:09:36Z

A volume maximizing embedding always exists because the space of embeddings is compact.

Comment by David Cohen

2012-08-25T20:56:19Z

Doesn't the article on Liouville numbers answer your question? (I.e., for a Liouville number, such an A cannot possible exist?)

Comment by David Cohen

2012-06-21T02:49:40Z

I am using [] to denote floor. The first equality in the second equation comes from reversing the order of summation on the right hand side, since d divides exactly [n/d] elements of {1,...,n}.

Comment by David Cohen

2012-06-20T22:59:46Z

In your example, you seem to be summing mu(i)*floor(p/i) rather than floor(mu*p/i).

Comment by David Cohen

2012-05-26T05:13:39Z

I feel like there ought to be a Bayesian approach to this problem.

Comment by David Cohen

2012-03-25T03:45:00Z

I think in that case, $G_{i}=S_{i}$ and $a_{i}=(i i+1)\in G_{i+1}$ would work.

Comment by David Cohen

2011-12-13T17:56:03Z

It suffices to do this in $\mathbb{R}^{n}$, in which case it is just the fundamental theorem of calculus: choose vectors $v_{1},...,v_{p}$, and now integrate $\omega$ over the parallelpiped boxed in by $\epsilon v_{1},...,\epsilon v_{p}$: this will give you $\omega(v_{1},...,v_{p})\epsilon + o(\epsilon)$ by the FTC. So by taking $\epsilon$ small, we conclude that $\omega(v_{1},...,v_{p})=0$.

Comment by David Cohen

2011-10-19T03:28:01Z

The same sorts of arguments used to answer this question: mathoverflow.net/questions/48910/… also answer yours. Basically, given dense, countable sets $A,B\subset \mathbb{C}$, there is an entire function which takes $A$ to $B$.

Comment by David Cohen

2011-10-12T03:31:03Z

Dihedral groups look like an interesting special case.

Comment by David Cohen

2011-10-12T00:01:34Z

Perhaps the question is asking which topological groups arise as fundamental groups?

Comment by David Cohen

2011-08-26T19:18:24Z

Given two countable dense subsets $A,B\subset \mathbb{R}$, there is a continuous homeomorphism $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $f(A)=B$. You can construct this function by just carefully choosing, one-by-one, the images of points of A (and the preimages of points of B). I believe this problem is exactly the same: just enumerate the rationals $q_{0},q_{1},...$, then choose $f(q_{i})$ so that it is near where you would expect it in $\mathbb{R}$, and $\mathbb{Q_{p})$ for the first $i$ primes $p$.

Comment by David Cohen

2011-03-22T00:21:22Z

Surely there are uncountably many. Just remove countably many tame knots from $\mathbb{R}^{3}$, there are uncountably many ways to do this.