User daniel miller - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T07:40:47Z http://mathoverflow.net/feeds/user/6856 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/127045/fixed-point-theorems/127349#127349 Answer by Daniel Miller for Fixed point theorems Daniel Miller 2013-04-12T11:47:50Z 2013-04-12T14:04:37Z <p>Allow me to mention another version of the <a href="http://en.wikipedia.org/wiki/Lefschetz_fixed_point_theorem" rel="nofollow">Lefschetz fixed-point theorem</a>. If $X$ is a (say smooth projective, though this works in greater generality) variety over $\mathbb F_q$ of dimension $d$, then $$\left|X\left(\mathbb F_{q^n}\right)\right| = q^d \sum_i (-1)^i \mathrm{tr}\left(\Phi_{q^n} : H_{et}^i(\bar X,\mathbb Q_\ell)\right)$$ where $\ell$ is prime to $q$ and $\Phi_{q^n}$ is the geometric Frobenius.</p> <p>As a corollary one gets the rationality of the zeta-function of $X$.</p> <p>(Note that this actually is a fixed-point theorem. $X(\mathbb F_{q^n})$ is just the set of fixed points under $\Phi_{q^n}$ applied to $X$.)</p> http://mathoverflow.net/questions/124959/smallest-class-of-rings-closed-under-familiar-operations Smallest class of rings closed under familiar operations Daniel Miller 2013-03-19T13:19:02Z 2013-03-19T19:43:27Z <p>Suppose I start out with the ring $\mathbb{Z}$, and call $\mathcal{C}$ the smallest collection of (commutative, unital) rings closed under the following list of operations (which I am aware has some redundancy):</p> <ol> <li>Localization</li> <li>Taking quotients</li> <li>Adding a variable: $A\mapsto A[X]$</li> <li>Completion with respect to an ideal</li> <li><strike>Tensor products</li> <li>More generally: finite limits and colimits</strike></li> </ol> <p>Since the tensor product of noetherian rings can be non-noetherian, I'm not even sure if everything in $\mathcal{C}$ is noetherian. </p> <blockquote> <ul> <li>Is there an easy-to-state characterization of which rings are in $\mathcal{C}$? </li> <li>Can much be said about what "nice" properties rings in $\mathcal{C}$ have? (I'm interpreting "nice" pretty loosely here). </li> </ul> </blockquote> <p>I apologize in advance if this is well-known: even an answer of the form "this is all worked out in [X]" would be appreciated. </p> <p><b>Edit:</b> I would also be quite happy if anything could be said about $\mathcal{C}$ if we drop condition 6. </p> <p><b>Edit:</b> It looks like my naive hope that the objects of $\mathcal{C}$ might be "nice" is rather hopeless. If we drop conditions 5 and 6, then certainly everything in $\mathcal{C}$ is noetherian. Here is one way in which such rings might be called "nice":</p> <blockquote> <p>If we define $\mathcal{C}$ using only 1-4, is the isomorphism problem for $\mathcal{C}$ decidable? </p> </blockquote> http://mathoverflow.net/questions/92206/what-properties-make-0-1-a-good-candidate-for-defining-fundamental-groups What properties make $[0,1]$ a good candidate for defining fundamental groups? Daniel Miller 2012-03-25T22:21:08Z 2013-03-07T05:25:26Z <p>The title essentially says it all. Consider the category $\mathfrak{Top}_2$ of triples $(J,e_0,e_1)$ where $J$ is a topological space, and $e_i \in J$. There is an obvious generalization of the definition of homotopic maps. Suppose we have selected $(J,e_0,e_1)\in \mathfrak{Top}_2$. We could say that two continuous maps $f,g:X\to Y$ are "$J$-homotopic" if there is a continuous map $h:X\times J\to Y$ such that $h(x,e_0) = f(x)$ and $h(x,e_1) = g(x)$. We could then define $\pi_1 (X,x)$ to be the set of continuous maps $f:J\to X$ satisfying $f(e_0)=f(e_1)=x$, with $J$-homotopic maps identified. Here in order to define composition of paths in the naive way, we need to have picked some <strike>homeomorphism</strike> continuous map from $J$ to $({0}\times J\cup {1}\times J)/(({0},e_1) = (1,e_0))$, taking $e_0$ to $e_0\times 0$ and $e_1$ to $e_1\times 1$. I have two questions:</p> <ol> <li><p>Can $([0,1],0,1)$ be characterized as an object in $\mathfrak{Top}_2$ in a purely categorical manner?</p></li> <li><p>When is $\pi_1 (X,x)$ a group? For that matter, when is $\pi_1 (X,x)$ associative?</p></li> </ol> <p>Essentially, the question comes down to: what properties of $[0,1]$ are needed in order to do homotopy theory?</p> http://mathoverflow.net/questions/37792/a-possible-generalization-of-the-homotopy-groups A possible generalization of the homotopy groups. Daniel Miller 2010-09-05T13:33:45Z 2013-02-16T11:37:22Z <p>The homotopy groups $\pi_{n}(X)$ arise from considering equivalence classes of based maps from the $n$-sphere $S^{n}$ to the space $X$. As is well known, these maps can be composed, giving arise to a group operation. The resulting group contains a great deal of information about the given space. My question is: is there any extra information about a space that can be discovered by considering equivalence classes of based maps from the $n$-tori $T^{n}=S^{1}\times S^{1}\times \cdots \times S^{1}$. In the case of $T^{2}$, it would seem that since any path $S^{1}\to X$ can be "thickened" to create a path $T^{2}\to X$ if $X$ is three-dimensional, the group arising from based paths $T^{2}\to X$ would contain $\pi_{1}(X)$. Perhaps more generally, can useful information be gained by examining equivalence classes of based maps from some arbitrary space $Y$ to a given space $X$.</p> http://mathoverflow.net/questions/120534/basic-questions-on-the-homotopy-category Basic questions on the homotopy category Daniel Miller 2013-02-01T18:50:47Z 2013-02-02T20:27:30Z <p>I apologize in advance if this the answer to this question is standard or well-known. I am not in any way an algebraic topologist.</p> <p>$\newcommand{\s}{\mathscr}$Let $\s T$ be the category of topological spaces, $\s T_*$ the category of pointed topological spaces. We can construct the homotopy category $\s H$ and the "pointed homotopy category" $\s H_{*}$ by letting $\hom_{\s H}(X,Y)$ be $\hom_{\s T}(X,Y)$ modulo homotopy and $\hom_{\s H_*}(X,Y)$ be $\hom_{\s T_*}(X,Y)$ modulo homotopies preserving basepoints. There are obvious functors $F:\s T\to \s H$, $F_{*}:\s T_{*}\to \s H_{*}$. </p> <p>My questions are pretty basic: do $F,F_*$ preserve products? arbitrary limits? only arbitrary finite limits? coproducts? arbitrary colimits? It not, how does one compute colimits in $\s H$ and $\s H_{*}$?</p> <p>If this can be found in a standard reference, just write an answer saying "this is a stupid question - this is all worked out in...," and I'll upvote and accept.</p> http://mathoverflow.net/questions/111950/when-does-pontryagin-duality-generalize When does Pontryagin duality generalize? Daniel Miller 2012-11-10T00:21:52Z 2012-11-12T10:42:59Z <p>Let $T$ be a locally compact abelian (LCA) group. For any other LCA group $G$, let $\hom(G,T)$ be the set of continuous homomorphisms $G\to T$. With the compact-open topology, $\hom(G,T)$ is certainly a topological group, but is not in general locally compact, even if $T$ is compact. In any case there is an obvious homomorphism $$\alpha_G : G \to \hom(\hom(G,T),T)$$ sending $x\in G$ to the functional $\chi\mapsto \chi(x)$. </p> <p>I have three questions: </p> <ol> <li>For what LCA groups $T$ is $\hom(G,T)$ locally compact for all locally compact $G$?</li> <li>For what groups $T$ is $\alpha_G$ always a (topological) isomorphism? <li>More generally, is there any full subcategory $\mathcal L$ of the category of LCA groups for which $\hom(G,H)$ is in $\mathcal L$ whenever $G$ and $H$ are in $\mathcal L$? </ol> <p>(<b>Edit</b>: I'm most interested in 1 and 2. If there is some $\mathcal L$ satisfying 3 where the objects in $\mathcal L$ can be characterized by some interesting topological criterion, that would be fantastic. My question is not whether there are "ad-hoc" ways of constructing $\mathcal L$.)</p> http://mathoverflow.net/questions/26848/looking-for-an-interesting-problem-riddle-involving-triple-integrals/30812#30812 Answer by Daniel Miller for Looking for an interesting problem/riddle involving triple integrals. Daniel Miller 2010-07-06T19:58:06Z 2012-10-18T21:19:06Z <p>A good example that my Honors Calc prof gave me is the following: $$\zeta(n) = \int_{[0,1]^n}\frac{d\boldsymbol x}{1-x_1\dotsm x_n}$$ The proof is an easy induction on $n$.</p> http://mathoverflow.net/questions/38219/intuition-on-finite-homotopy-groups Intuition on finite homotopy groups Daniel Miller 2010-09-09T19:24:10Z 2012-08-23T01:58:01Z <p>As I have been studying algebraic topology, something that I found puzzling was the existence of finite homotopy groups. For instance, $\pi_{4}(S^{2})\cong\pi_{5}(S^{4})\cong\mathbb{Z}/2\mathbb{Z}$. I was wondering if there was any kind of intuitive reason for why this might be true, and if there are spaces $X$ such that $\pi_{1}(X)$ is finite. Speaking very roughly, it would seem that a finite, nontrivial fundamental group means that if you repeat a closed path enough times, it can be contracted to a point, something which I find rather hard to visualize. So the question is: Is there any intuitive reason for the existence of finite homotopy groups?</p> http://mathoverflow.net/questions/93124/automorphisms-of-mathbbc/93170#93170 Answer by Daniel Miller for Automorphisms of $\mathbb{C}$ Daniel Miller 2012-04-04T21:46:14Z 2012-04-04T22:41:29Z <p>There is the more general fact that any automorphism of any subfield of $\mathbb{C}$ can be extended to an automorphism of $\mathbb{C}$. For a proof, see the paper <em>Automorphisms of the Complex Numbers</em> by Paul Yale of Pomona College. Here is a <a href="http://www.jstor.org/discover/10.2307/2689301?uid=3739792&amp;uid=2129&amp;uid=2&amp;uid=70&amp;uid=4&amp;uid=3739256&amp;sid=55994902273" rel="nofollow">JSTOR link</a>. In general, if $k$ is an arbitrary (EDIT: algebraically closed) field, my guess would be Yale' argument could be easily extended to show that any automorphism of a subfield $h\subset k$ can be extended to an automorphism of $k$.</p> http://mathoverflow.net/questions/61409/using-higher-order-bring-radicals-to-solve-arbitrary-polynomials Using higher-order Bring radicals to solve arbitrary polynomials Daniel Miller 2011-04-12T15:07:14Z 2011-10-06T13:30:34Z <p>It is well known that there is no general formula for the solution of the quintic. Of course, what this really means is that there is no general formula that only involves addition, subtraction, multiplication, division, and the extraction of $n$-th roots. Indeed, if one is allowed to use the Bring radical, that is, solutions of the equation $x^5+x+a=0$, then it is indeed possible to solve any quintic. It would seem that if one introduced higher order Bring radicals, it would be possible to solve polynomials of higher degree. More precisely, define a Bring radical of order $n$ to be a continuous function $B_n(t)$ such that $B_n(t)$ is a solution to an $n$-th degree polynomial, one of whose coefficients is $t$. (Of course, I am being rather vague, most of these Bring radicals are only continuously definable on some proper subset of $\mathbb{C}$) It is trivial that any $n$-th degree polynomial may be solved by means of some $n$-th order Bring radical. However, it is not at all apparent that for some fixed $n$, there exist a finite collection $B_n^1,B_n^2,\cdots,B_n^k$ such that <em>any</em> $n$-th degree polynomial may be solved using the $B_n^i$. So my question is: </p> <blockquote> <p>Is it the case that for any $n$, there is a finite collection of Bring radicals that may be used to solve any $n$-th degree polynomial? </p> </blockquote> <p>Another question, to which the answer is most likely negative, is whether there exists a finite collection $B_{r_1}^1,B_{r_2}^2,\cdots$ of Bring radicals such that any polynomial of any degree is solvable using the $B_{r_i}^j$.</p> <p><strong>Edit</strong>: My definition of a higher-order Bring radical was rather narrow. I'd also be interested in any answer that involved Bring radicals $B_n(t)$ that were solutions to a polynomial of the form $x^n+p_{n-1}(t)x^{n-1}+\cdots+p_{1}(t)x+p_{0}(t)$. The general idea concerns whether or not one can solve <em>all</em> $n$-th degree polynomials just be adjoining the roots of some finite family of them polynomials, whose coefficients depend smoothly on $t$.</p> http://mathoverflow.net/questions/73102/self-satisfying-properties Self-Satisfying Properties Daniel Miller 2011-08-17T21:41:21Z 2011-08-18T02:09:19Z <p>When one is first learning about topologies or $\sigma$-algebras, a common (trivial) exercise is showing that they are closed under intersections, but not in general under unions, i.e., any intersection of topologies is a topology, but the union of two topologies may not be a topology. When I was first learning topology, I found this to be rather disappointing, for otherwise one could form the "topology of topologies" on a given set $X$.</p> <p>Suppose we have a property $P$, such as "$\forall y,z\in x \:(y\cap z\in x)$." Now for any set $x$, we may form the set $K_P (x)$ of all $t\subset x$ such that $P(t)$ holds. Can we say anything about $P$ if, for all $x$, $P(K_P (x))$ holds? Of course, a trivial example of such a $P$ is the property of being closed under intersections. Is that the only such $P$ ?</p> http://mathoverflow.net/questions/68208/elementary-number-theory-text-from-a-categorical-perspective Elementary Number Theory Text from a Categorical Perspective Daniel Miller 2011-06-19T12:56:18Z 2011-06-27T23:54:39Z <p>My question is somewhat similar to <a href="http://mathoverflow.net/questions/8097/number-theory-textbook-with-an-algebraic-perspective" rel="nofollow">this previous question</a>, but from a slightly different perspective. Is there any textbook on elementary number theory that develops the properties of $\mathbb{Z}$ as, say, the initial object in the category of commutative rings with identity? I am looking for something that presupposes a knowledge of category theory at the level of Categories for the Working Mathematician.</p> <p><strong>Edit:</strong> I had no idea that this question would provoke the storm of criticism that is has. My intention was not to imply that number theory is best learned from a categorical perspective, or that number theory should be subsumed by category theory. I was simply wondering what sort of interesting things one could say about $\mathbb{Z}$ from a category-theoretic perspective. So, I'll narrow the question: "Are there any good sources for learning about the properties of a natural numbers object in an arbitrary topos (possibly well-pointed and satisfying the axiom of choice)?"</p> http://mathoverflow.net/questions/64151/categorical-invariants Categorical Invariants Daniel Miller 2011-05-06T20:17:30Z 2011-05-06T21:16:55Z <p>I apologize in advance if this question seems too vague. </p> <p>In many topology courses, concepts like the fundamental group and homology groups are introduced as a means of distinguishing non-homeomorphic spaces - for instance, $\mathbb{T}^2$ and $S^2$. Similarly, things like the rank of an abelian group, and the Krull dimension of a ring are (relatively) interesting ways of taking an object and capturing useful information in a number. Of course, the most interesting invariants are those that are functorial in some way or another. In <a href="http://mathoverflow.net/questions/40562/proving-the-impossibility-of-an-embedding-of-categories" rel="nofollow">my earlier question</a>, I asked for a reason that the category of topological spaces cannot be embedded in the category of groups. Now it turns out that one nice reason is that the category of groups is not cartesian closed, while the category of (compactly generated weakly Hausdorff) spaces is. I found this to be rather nice, as cartesian closedness is a rather global property. On the other hand, lots of categories are cartesian closed, and it would be nice if there were some kind of categorical invariant capable of distinguishing them. So my question is: Are there any nice categorical invariants? Preferably a categorical invariant would take the form of a functor $\mathfrak{C}\mathfrak{A}\mathfrak{T}\to\mathscr{C}$, where $\mathfrak{C}\mathfrak{A}\mathfrak{T}$ is the (meta) category of all categories and $\mathscr{C}$ is some nice category (abelian groups, sets, etc). But I'd be interested in any interesting way of capturing global information about a category.</p> http://mathoverflow.net/questions/60457/elementaryshortuseful/60671#60671 Answer by Daniel Miller for Elementary+Short+Useful Daniel Miller 2011-04-05T11:20:40Z 2011-04-05T11:20:40Z <p>At the risk of incurring the wrath of some here, I would propose the <a href="http://en.wikipedia.org/wiki/Yoneda_lemma" rel="nofollow">Yoneda Lemma</a>, along with the minimum of necessary category theory. Like it or not, category theory is hugely useful to algebraists, and early exposure can be very helpful. (It was to me!)</p> http://mathoverflow.net/questions/44705/cardinalities-larger-than-the-continuum-in-areas-besides-set-theory Cardinalities larger than the continuum in areas besides set theory Daniel Miller 2010-11-03T17:27:44Z 2011-03-11T20:17:14Z <p>It seems that in most theorems outside of set theory where the size of some set is used in the proof, there are three possibilities: either the set is finite, countably infinite, or uncountably infinite. Are there any well known results within say, algebra or analysis that require some given set to be of cardinality strictly greater than $2^{\aleph_{0}}$? Perhaps in a similar vein, are any objects encountered that must have size larger than $2^{\aleph_{0}}$ in order for certain properties to hold?</p> http://mathoverflow.net/questions/58148/polynomial-group-laws-on-mathbbr2 Polynomial group Laws on $\mathbb{R}^2$ Daniel Miller 2011-03-11T12:53:15Z 2011-03-11T15:24:36Z <p>When students are first learning about groups, a classic example of a group that is <em>not</em> defined as a set of functions is the group whose underlying set is $\mathbb{R}\setminus-1$, and whose operation is $x*y=x+y+xy$. This naturally leads one to wonder about what other polynomials in two variables give rise to a group law on $\mathbb{R}$. Is there any nice criteria for such polynomials, or, in the case that there is not, are there any nice classes of polynomials that are group laws? </p> http://mathoverflow.net/questions/56938/what-does-the-adjective-natural-actually-mean/56943#56943 Answer by Daniel Miller for What does the adjective "natural" actually mean? Daniel Miller 2011-03-01T00:01:49Z 2011-03-02T19:53:51Z <p>I'm not sure if this is what you are looking for, but the word "natural" is often used to imply that the object being discussed is in fact a <a href="http://en.wikipedia.org/wiki/Natural_transformation" rel="nofollow">Natural Transformation</a>. For example, the isomorphism $V\cong V^{**}$ between a finite dimensional vector space and its double dual is natural in the sense that it is a natural isomorphism between functors. The same holds for the isomorphism $G/\mbox{Ker}\:\phi\cong H$ when $\phi:G\to H$ is a surjective homomorphism and $G,H$ are groups.</p> http://mathoverflow.net/questions/35408/naturally-occuring-groups-with-cardinality-greater-than-the-reals Naturally occuring groups with cardinality greater than the reals. Daniel Miller 2010-08-13T00:26:38Z 2011-03-01T18:26:51Z <p>In group theory, the single most important piece of information about a group is its cardinality, which is of course either finite, countably infinite, or uncountably infinite. Usually, however, uncountably infinite simply means a cardinality of $\aleph_{1}$, the same as $\mathbb{R}$. My question is: is there anywhere that groups with cardinality strictly greater than $\aleph_{1}$ arise naturally? Of course, it is easy enough to construct groups with arbitrarily large cardinality, but I cannot recall ever seeing them used.</p> http://mathoverflow.net/questions/54877/countable-fields-with-no-countable-extension Countable Fields with No Countable Extension Daniel Miller 2011-02-09T12:45:00Z 2011-02-09T14:02:35Z <p>Let $\mathscr{S}$ be the set of all countable subfields of $\mathbb{C}$. Clearly, $\mathscr{S}$ is a partially ordered set under inclusion, and if $K_1\subseteq K_2 \subseteq \cdots$ is an ascending chain of countable subfields, then $\bigcup_{i=1}^{\infty}K_i$ is a countable union of countable fields, and is hence an upper bound for $K_1\subseteq K_2 \subseteq \cdots$ which is in $\mathscr{S}$. But then $\mathscr{S}$ satisfies the conditions of Zorn's lemma, so there is some maximal element $K$. It would then seem that $K$ is a countable field such that whenever $K\subsetneq L\subseteq \mathbb{C}$ is a pair of field extensions, we have that $L$ is uncountable. This seems quite unintuitive to me. Has anyone exhibited such a subfield of $\mathbb{C}$ and proved that it has the properties required? It would seem that for any countable subfield $K\subseteq \mathbb{C}$, there will be some complex number $\alpha\notin K$, in which case $K\subsetneq K(\alpha)\subset\mathbb{C}$ and $K(\alpha)$ is also countable.</p> http://mathoverflow.net/questions/40562/proving-the-impossibility-of-an-embedding-of-categories Proving the impossibility of an embedding of categories Daniel Miller 2010-09-30T01:07:16Z 2011-01-16T22:14:28Z <p>A number of topological invariants take the form of functors $\mathscr{T}\to\mathscr{G}$, where $\mathscr{T}$ is the category of all topological spaces and continuous functions, and $\mathscr{G}$ is the category of all groups and homomorphisms. For examples, consider the homology groups $H_{n}(X)$ or the homotopy groups $\pi_{n}(X)$. Of course, a problem with these invariants is that they are not fully faithful functors, i.e., $H_{n}(X)\cong H_{n}(Y)$ does not imply that $X$ and $Y$ are homeomorphic. The existence of a fully faithful functor $F:\mathscr{T}\to\mathscr{G}$ would imply that $\mathscr{G}$ has a subcategory $F\mathscr{T}$ equivalent to $\mathscr{T}$. This would be both rather disturbing and extremely interesting. First, it would mean that in a sense, all of topology is just a subset of group theory, which would be rather disturbing to topologists, but it would also reveal a fundamental connection between two seemingly disparate disciplines. My question is: is it possible to prove that no such functor exists? In other words, could one exhibit some categorical property that $\mathscr{T}$ posesses that $\mathscr{G}$ does not. This question can naturally be extended to other important categories, like $\mathscr{M}$, the category of all modules, or $\mathscr{R}$, the category of all rings. So in general, given arbitrary categories $\mathscr{C}$, $\mathscr{D}$, is there any natural way of showing that no fully faithful functor $F:\mathscr{C}\to\mathscr{D}$ exists, i.e., are there any nice "categorical invariants?"</p> <p>EDIT: A couple people pointed out that I really ought to be discussing fully faithful functors, rather than just faithful functors. Also, I have changed the title in accordance with Martin Brandenburg's recommendation.</p> http://mathoverflow.net/questions/51068/open-questions-in-riemannian-geometry/51072#51072 Answer by Daniel Miller for Open Questions in Riemannian Geometry Daniel Miller 2011-01-03T22:58:42Z 2011-01-03T22:58:42Z <p>The book "A Panoramic View of Riemannian Geometry" by Marcel Berger includes a number of open problems.</p> http://mathoverflow.net/questions/45308/choice-function-on-the-powerset-of-the-reals Choice Function on the Powerset of the Reals Daniel Miller 2010-11-08T14:24:57Z 2010-11-08T22:56:06Z <p>I'm not sure if this question is appropriate for mathoverflow, but I can't help but think that other people have wondered about it as well. When anyone first learns about the axiom of choice, the standard example used to convince the listener as to its necessity is the problem of finding a choice function on <code>$\mathscr{P}(\mathbb{R})\backslash\{\emptyset\}$</code>, the powerset of the reals, without the emptyset. I have always wondered: Is the axiom of choice really necessary to construct the function? In other words, is it possible to prove that without the axiom of choice, no such choice function exists. Or, if it is possible to directly construct a choice function on <code>$\mathscr{P}(\mathbb{R})\backslash\{\emptyset\}$</code>, are there any good examples?</p> http://mathoverflow.net/questions/40005/generalizing-a-problem-to-make-it-easier/40016#40016 Answer by Daniel Miller for Generalizing a problem to make it easier Daniel Miller 2010-09-26T12:41:44Z 2010-09-26T12:41:44Z <p>This may be a little trivial, but there are a number of identities for the Fibonacci numbers that are most easily proved by generalizing them. For example, proving</p> <p>$f_{2n-1}=f_{n+1}f_{n}-f_{n-1}f_{n-2}$</p> <p>requires a rather convoluted process involving a couple lemmas unless one realizes that it is far easier to prove</p> <p>$f_{m+n}=f_{m}f_{n+2}-f_{m-2}f_{n}$</p> <p>and then substitude $m=n,n=n-1$.</p> <p>I think a classic example of generalizing in order to prove a simple result is Galois Theory. Ruffini's attempted proofs of the unsolvability of the quintic were enormously long and tremendously complicated. However, once the machinery of Galois Theory is developed, which is rather easy, it is almost trivial to demonstrate that there exist quintic equations that are not solvable by radicals. </p> http://mathoverflow.net/questions/39901/set-theory-within-the-framework-of-category-theory Set theory within the framework of category theory Daniel Miller 2010-09-24T20:58:49Z 2010-09-24T21:45:02Z <p>I started studying the basics of category theory recently, and after seeing how a great deal of group theory could be described categorically, I began to wonder if it were possible to describe set theory, or set-theoretic concepts, without reference to elements, i.e., by only using sets and functions. For example, instead of saying $x\in S$, one could say that there is a map $f:0\to S$, where $0$ is a singleton. Similarly, one could describe the power set $\mathscr{P}(X)$ by saying that for any function $f:Y\to X$, there are functions $g:0\to \mathscr{P}(X)$ and $h:\mathscr{P}(X)\to X$ such that $f=hg$. Disjoint unions can be described as coproducts, and cartesian products can be described by means of a universal property. I was wondering if it were possible to describe all of naive set theory in this way, and if so, whether any attempts have been made to do so.</p> http://mathoverflow.net/questions/39490/dualizing-the-definition-of-a-free-group Dualizing the definition of a free group Daniel Miller 2010-09-21T12:48:02Z 2010-09-21T18:13:33Z <p>In most basic abstract algebra courses, the free group is directly constructed, a process that I find rather unwieldy. An alternate method of characterizing the free group is by means of its universal property: for any function $f:S\to G$, an arbitrary group, there is a function $g:S\to F_{S}$ and a unique homomorphism $\varphi: F_{S}\to G$ such that $f=\varphi g$. Of course, a direct construction of the free group is necessary to show that any group actually satisfies this definition. I was wondering what happened when the definition was dualized. In other words, let $P_{S}$ be the group such that for any function $f:G\to S$, there is a function $g:P_{S}\to S$ and a unique homomorphism $\varphi:G\to P_{S}$ such that $f=g\varphi$. It would seem, in light of Cayley's theorem, that $P_{S}$ is just the set of permutations on $S$, but I'm not sure of this. Does anyone know what $P_{S}$ is?</p> http://mathoverflow.net/questions/39042/infinite-field-theory-and-category-theory Infinite Field Theory and Category Theory Daniel Miller 2010-09-17T00:53:05Z 2010-09-17T13:24:54Z <p>I should start by saying that I have not studied field theory in depth, so if this question is totally off base, I apologize. Something I noticed as I studied group theory is many concepts that were very difficult to define directly had simple and elegant categorical definitions. For example, the direct definition of the free group is rather long and arduous, whereas the categorical definition, i.e. any function $S\to G$, where $G$ is a group factors through a homomorphism from the free group generated by $S$ to $G$, is quite simple. However, for the most part, it seems to me that categorical methods are most easily used on infinite groups, and in particular, infinite abelian groups. Despite this limitation, categorical methods seemed so natural that I couldn't help but wonder if they can be applied to field theory with similar results. So my question is: (1) is it beneficial to study infinite field theory in the generality that category theory necessitates, and (2) are there any good books that use this approach.</p> http://mathoverflow.net/questions/18271/what-out-of-print-books-would-you-like-to-see-re-printed/39039#39039 Answer by Daniel Miller for What out-of-print books would you like to see re-printed? Daniel Miller 2010-09-17T00:23:38Z 2010-09-17T00:23:38Z <p>The whole Academic Press series on pure and applied mathematics contains a number of gems, including Mordell's work on Diophantine equations and Fuchs' work on infinite abelian groups. Unfortunately, it is out of print and used editions are usually horribly expensive. </p> http://mathoverflow.net/questions/38752/analysis-from-a-categorical-perspective Analysis from a categorical perspective Daniel Miller 2010-09-15T00:42:10Z 2010-09-15T04:39:44Z <p>I have not studied category theory in extreme depth, so perhaps this question is a little naive, but I have always wondered if analysis could be taught naturally using categories. I ask this because it seems like a quite a lot of topological and group theoretic concepts can be defined most succinctly using categorical concepts, and the categorical definitions are more revealing. So my question is: (1) Is it possible/beneficial to teach analysis using category theory? and (2) Are there any good textbooks that use this method?</p> http://mathoverflow.net/questions/18847/homotopy-first-courses-in-algebraic-topology/37986#37986 Answer by Daniel Miller for "Homotopy-first" courses in algebraic topology Daniel Miller 2010-09-07T16:14:40Z 2010-09-08T00:31:11Z <p>J. P. May's superb book, "A Concise Course in Algebraic Topology," starts with a great deal on homotopy theory, and doesn't really get to homology until nearly half way through. I learned a great deal from this approach, and think that it is the best way to teach algebraic topology. But May's book is probably too difficult for a "first course" in algebraic topology.</p> http://mathoverflow.net/questions/36218/criteria-for-autg-to-be-simple Criteria for Aut(G) to be simple Daniel Miller 2010-08-20T20:18:47Z 2010-08-26T21:42:42Z <p>It is well known that the automorphisms of a group $G$ form a group under composition, and that the group of inner automorphisms $\phi (x)=gxg^{-1}$ forms a normal subgroup of $\mbox{Aut}(G)$. Thus, $\mbox{Aut}(G)$ is simple if and only if either $\mbox{Inn}(G)=\mbox{Aut}(G)$ or $\mbox{Inn}(G)$ is trivial. In the second case, since $G/Z(G)=\mbox{Inn}(G)$, $G$ must be abelian. My question is, when does $\mbox{Inn}(G)=\mbox{Aut}(G)$? Or, as it is unlikely that the general case is not fully understood, are there nice classes of groups for which there are a nice set of criteria for $\mbox{Inn}(G)=\mbox{Aut}(G)$.</p> http://mathoverflow.net/questions/131435/why-dont-more-mathematicians-improve-wikipedia-articles Comment by Daniel Miller Daniel Miller 2013-05-22T20:19:30Z 2013-05-22T20:19:30Z For what it's worth, seeing this question made me feel guilty enough to go create a Wikipedia article on a topic not yet covered. http://mathoverflow.net/questions/7439/algebraic-varieties-which-are-also-manifolds/7459#7459 Comment by Daniel Miller Daniel Miller 2013-05-22T02:24:04Z 2013-05-22T02:24:04Z +1 for considering more general fields that e.g. number theorists care about http://mathoverflow.net/questions/128490/ring-with-prescribed-k-group/128495#128495 Comment by Daniel Miller Daniel Miller 2013-04-23T15:33:23Z 2013-04-23T15:33:23Z Very nice! What about getting prescribed non-reduced $K$-theory via non-commutative rings? http://mathoverflow.net/questions/106560/philosophy-behind-mochizukis-work-on-the-abc-conjecture/107279#107279 Comment by Daniel Miller Daniel Miller 2013-04-14T23:57:26Z 2013-04-14T23:57:26Z I'd just like to point out that Mochizuki has released revised versions of IUTT I,III and IV. http://mathoverflow.net/questions/852/what-is-inter-universal-geometry Comment by Daniel Miller Daniel Miller 2013-03-19T12:53:39Z 2013-03-19T12:53:39Z For those interested in Mochizuki's proof (or claimed proof) of ABC, it's worth noting that he has recently posted revised versions of IUTT I,II,III [here](<a href="http://www.kurims.kyoto-u.ac.jp/~motizuki/papers-english.html" rel="nofollow">kurims.kyoto-u.ac.jp/~motizuki/&hellip;</a>). http://mathoverflow.net/questions/120534/basic-questions-on-the-homotopy-category/120559#120559 Comment by Daniel Miller Daniel Miller 2013-02-02T13:04:48Z 2013-02-02T13:04:48Z Great answer! This is exactly the sort of perspective I was looking for. http://mathoverflow.net/questions/111950/when-does-pontryagin-duality-generalize Comment by Daniel Miller Daniel Miller 2012-11-10T20:48:58Z 2012-11-10T20:48:58Z @Benjamin: thanks! Armacost's book looks like exactly the sort of reference I was hoping existed! http://mathoverflow.net/questions/111950/when-does-pontryagin-duality-generalize Comment by Daniel Miller Daniel Miller 2012-11-10T13:24:01Z 2012-11-10T13:24:01Z @Todd of course, but I'm hoping there are more interesting examples. http://mathoverflow.net/questions/22189/what-is-your-favorite-strange-function/36311#36311 Comment by Daniel Miller Daniel Miller 2012-07-17T11:56:18Z 2012-07-17T11:56:18Z The proof of their existence comes from the structure theorem for divisible abelian groups. Both are isomorphic to $\mathbb{Z}/\mathbb{Q}\times \bigoplus_\omega \mathbb{Q}$. http://mathoverflow.net/questions/92206/what-properties-make-0-1-a-good-candidate-for-defining-fundamental-groups/92223#92223 Comment by Daniel Miller Daniel Miller 2012-03-26T11:22:56Z 2012-03-26T11:22:56Z Thanks! This is exactly the sort of thing I was looking for. http://mathoverflow.net/questions/92206/what-properties-make-0-1-a-good-candidate-for-defining-fundamental-groups/92216#92216 Comment by Daniel Miller Daniel Miller 2012-03-26T01:26:32Z 2012-03-26T01:26:32Z Nice answer, but I don't feel that you are getting to the heart of my question. Showing that the unit interval behaves well &quot;up to homotopy&quot; is presupposing the fact that we have defined homotopy using the unit interval. If we, so to speak, we were not already using $[0,1]$ to do homotopy theory, then how could we identify the relevant categorical properties of $[0,1]$? By categorical here, I mean expressible in the category of topological spaces and continuous maps (not up to homotopy). http://mathoverflow.net/questions/92206/what-properties-make-0-1-a-good-candidate-for-defining-fundamental-groups Comment by Daniel Miller Daniel Miller 2012-03-25T22:40:00Z 2012-03-25T22:40:00Z You're quite right. Edited to reflect the change. http://mathoverflow.net/questions/73102/self-satisfying-properties/73107#73107 Comment by Daniel Miller Daniel Miller 2011-08-18T02:28:14Z 2011-08-18T02:28:14Z Sorry, I was a little bit vague about whether or not I wanted $P(x)$ to hold. I am interested in when $P(K_P (X))$ holds for arbitrary $x$. http://mathoverflow.net/questions/73102/self-satisfying-properties Comment by Daniel Miller Daniel Miller 2011-08-17T23:15:58Z 2011-08-17T23:15:58Z @Oliver: Edited to clarify http://mathoverflow.net/questions/68208/elementary-number-theory-text-from-a-categorical-perspective Comment by Daniel Miller Daniel Miller 2011-06-19T17:21:11Z 2011-06-19T17:21:11Z @Qiaochu: I find that I understand things much better when I am able to look at them from a categorical perspective. So, the only real application I have in mind is an improved understanding ot the topic.