User david cohen - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T00:22:31Z http://mathoverflow.net/feeds/user/9455 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/131356/what-does-a-singular-simplex-with-real-coefficient-mean/131379#131379 Answer by David Cohen for What does a singular simplex with real coefficient mean David Cohen 2013-05-21T19:07:43Z 2013-05-21T19:07:43Z <p>As Lee suggests, you should look at Allen Hatcher's book. In particular, chapter 2 and section 3.3.</p> <p>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 $M$.</p> <p>The prototypical examples are as follows:</p> <p>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 $0$.)</p> <p>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.</p> http://mathoverflow.net/questions/122104/area-of-triangles-vs-comparison-triangles Area of triangles vs. comparison triangles. David Cohen 2013-02-17T22:36:13Z 2013-02-17T23:59:52Z <p>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 $\leq C_{0}\ell^{2}+C_{0}$.)</p> <p>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 $d(c,a)$.</p> <p>Does there exist a constant $C$ such that for all $a,b,c\in X$ we have:</p> <p>$Area_{X}(a,b,c)\leq C Area_{comp}(a,b,c)+C$?</p> <p>If the answer is no, then are there counterexamples when $X$ is homogeneous?</p> http://mathoverflow.net/questions/112538/does-every-orientable-surface-embed-in-mathbbr3 Does every orientable surface embed in $\mathbb{R}^{3}$ David Cohen 2012-11-16T01:15:41Z 2012-11-18T01:38:13Z <p>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 $\mathbb{R}^{3}$?</p> http://mathoverflow.net/questions/100392/simply-connected-simplicial-complexes Simply connected simplicial complexes David Cohen 2012-06-22T20:29:52Z 2012-06-22T22:53:36Z <p>Let $Z$ be a simply connected, two dimensional simplicial complex.</p> <p>Let $X\subset Z$ be a finite subcomplex with nontrivial $\pi_{1}$.</p> <p>Must there exist a finite, simply connected subcomplex $Y\subset Z$ such that $Y\supset X$?</p> <p>(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.)</p> http://mathoverflow.net/questions/100187/is-it-true-that-the-sum-of-a-specific-floor-function-of-a-prime-1/100189#100189 Answer by David Cohen for Is it true that the sum of a specific floor function of a prime = 1? David Cohen 2012-06-20T23:08:57Z 2012-06-20T23:08:57Z <p>Assuming that $p\text{#}$ means the product over primes $\prod_{q\leq p}q$, then it is clear that $$\sum_{i\leq p} \mu(i)[\frac{p}{i}] = \sum_{i | p\text{#}} \mu(i)[\frac{p}{i}].$$</p> <p>But this formula is very well known: $$\sum_{d\leq n} \mu(d)[\frac{n}{d}]=\sum_{k\leq n}\sum_{d | k} \mu(d)=1+0+0+\ldots$$</p> http://mathoverflow.net/questions/58341/is-a-space-with-no-covering-spaces-simply-connected Is a space with no covering spaces simply connected? David Cohen 2011-03-13T16:00:33Z 2011-03-13T16:54:35Z <p>Suppose $X$ is a path connected space such that every connected covering space of $X$ is trivial (1-fold.) Must $X$ be simply connected?</p> <p>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.</p> http://mathoverflow.net/questions/48910/smooth-functions-for-which-fx-is-rational-if-and-only-if-x-is-rational/48918#48918 Answer by David Cohen for Smooth functions for which $f(x)$ is rational if and only if $x$ is rational David Cohen 2010-12-10T13:15:00Z 2010-12-10T13:15:00Z <p>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.</p> <p>I am fairly certain that some variation on this should work for your problem.</p> http://mathoverflow.net/questions/42580/is-this-quotient-space-of-q-p-contractible Is this quotient space of Q_p contractible? David Cohen 2010-10-18T04:36:03Z 2010-10-18T09:27:03Z <p>Let $X_{p} = \mathbb{Q}_{p} / \sim$, where $\sim$ is defined by:</p> <p>$x\sim 0 \Leftrightarrow x\in \mathbb{Q}$</p> <p>$X_{p}$ 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.</p> <p>Is $X_p$ contractible?</p> http://mathoverflow.net/questions/39798/in-an-inductive-family-of-groups-does-the-probability-that-a-particular-word-is/39819#39819 Answer by David Cohen for In an inductive family of groups, does the probability that a particular word is satisfied converge? David Cohen 2010-09-24T04:31:33Z 2010-09-24T05:19:32Z <p><strong>Edit</strong>: As John points out below, this doesn't work.</p> <p>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 $g\in G$.)</p> <p>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 $1/5$.</p> <p>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 $G_{2i+1} = G_{2i}\times\mathbb{Z}/5\mathbb{Z}$.)</p> http://mathoverflow.net/questions/131528/how-closed-form-conjectures-are-made Comment by David Cohen David Cohen 2013-05-22T23:21:49Z 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 &quot;Occam's razor&quot; and &quot;Kolmogorov prior&quot;. http://mathoverflow.net/questions/130780/power-series-whose-partial-sums-attain-only-finitely-many-values Comment by David Cohen David Cohen 2013-05-16T01:27:28Z 2013-05-16T01:27:28Z Hopefully this link works: <a href="http://en.wikipedia.org/wiki/Ces%C3%A0ro_mean" rel="nofollow">en.wikipedia.org/wiki/Ces%C3%A0ro_mean</a> If not, google &quot;cesaro mean&quot; http://mathoverflow.net/questions/128152/which-hard-mathematical-problems-do-you-have-to-solve-to-earn-bitcoins Comment by David Cohen David Cohen 2013-04-20T03:50:25Z 2013-04-20T03:50:25Z The problems involved are purely computational. http://mathoverflow.net/questions/124854/graph-drawing-maximizing-the-volume-of-the-convex-hull Comment by David Cohen David Cohen 2013-03-18T16:09:36Z 2013-03-18T16:09:36Z A volume maximizing embedding always exists because the space of embeddings is compact. http://mathoverflow.net/questions/105480/are-there-lower-bounds-on-the-quality-of-a-rational-approximation Comment by David Cohen David Cohen 2012-08-25T20:56:19Z 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?) http://mathoverflow.net/questions/100187/is-it-true-that-the-sum-of-a-specific-floor-function-of-a-prime-1/100189#100189 Comment by David Cohen David Cohen 2012-06-21T02:49:40Z 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}. http://mathoverflow.net/questions/100187/is-it-true-that-the-sum-of-a-specific-floor-function-of-a-prime-1 Comment by David Cohen David Cohen 2012-06-20T22:59:46Z 2012-06-20T22:59:46Z In your example, you seem to be summing mu(i)*floor(p/i) rather than floor(mu*p/i). http://mathoverflow.net/questions/98013/is-there-an-algorithm-that-can-reverse-engineer-a-regular-expression Comment by David Cohen David Cohen 2012-05-26T05:13:39Z 2012-05-26T05:13:39Z I feel like there ought to be a Bayesian approach to this problem. http://mathoverflow.net/questions/92111/a-sequence-of-finite-groups Comment by David Cohen David Cohen 2012-03-25T03:45:00Z 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. http://mathoverflow.net/questions/83349/a-p-form-taking-discrete-values-on-p-chains-must-be-0 Comment by David Cohen David Cohen 2011-12-13T17:56:03Z 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 $\omega(v_{1},...,v_{p})=0$. http://mathoverflow.net/questions/78526/maps-preserving-algebraic-numbers Comment by David Cohen David Cohen 2011-10-19T03:28:01Z 2011-10-19T03:28:01Z The same sorts of arguments used to answer this question: <a href="http://mathoverflow.net/questions/48910/smooth-functions-for-which-fx-is-rational-if-and-only-if-x-is-rational" rel="nofollow" title="smooth functions for which fx is rational if and only if x is rational">mathoverflow.net/questions/48910/&hellip;</a> also answer yours. Basically, given dense, countable sets $A,B\subset \mathbb{C}$, there is an entire function which takes $A$ to $B$. http://mathoverflow.net/questions/77883/can-nonabelian-groups-be-detected-locally Comment by David Cohen David Cohen 2011-10-12T03:31:03Z 2011-10-12T03:31:03Z Dihedral groups look like an interesting special case. http://mathoverflow.net/questions/77878/which-are-the-groups-that-still-havent-been-proven-as-fundamental-groups-of-a-to Comment by David Cohen David Cohen 2011-10-12T00:01:34Z 2011-10-12T00:01:34Z Perhaps the question is asking which topological groups arise as fundamental groups? http://mathoverflow.net/questions/73792/continuous-extensions-reals-and-to-p-adic-numbers Comment by David Cohen David Cohen 2011-08-26T19:18:24Z 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 $p$. http://mathoverflow.net/questions/59115/a-set-for-which-it-is-hard-to-determine-whether-or-not-it-is-countable/59121#59121 Comment by David Cohen David Cohen 2011-03-22T00:21:22Z 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.