most recent 30 from http://mathoverflow.net 2013-05-25T22:38:09Z http://mathoverflow.net/feeds/user/4517 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/129541/does-the-paper-on-the-cobordism-ring-omega-and-a-complex-analogue-ii-exist Glen M Wilson 2013-05-03T14:48:40Z 2013-05-03T16:16:57Z

<p>I've been investigating the Milnor hypersurfaces, and every reference seems to point to the paper by Milnor, "On the cobordism ring $\Omega_*$ and a complex analogue II". Despite my best efforts, I cannot seem to find it. Was this paper ever published? If not, is there a draft that is available? </p>

http://mathoverflow.net/questions/113969/computation-of-stable-homotopy-groups-of-rp2 Glen M Wilson 2012-11-20T18:21:18Z 2013-01-21T22:10:59Z

<p>I would like to compute the first few stable homotopy groups of $RP^2$. </p> <p>I first thought to use the Atiyah-Hirzebruch Spectral Sequence, (see Davis &amp; Kirk, pg. 242). Here is what I computed for the $E^2$ term of the spectral sequence: <code>$$E^2_{p,q}=\begin{array}{|ccc} \mathbb{Z}_2 &amp; \mathbb{Z}_2 &amp; \mathbb{Z}_2 \\ \mathbb{Z}_2 &amp; \mathbb{Z}_2 &amp; \mathbb{Z}_2 \\ \mathbb{Z} &amp; \mathbb{Z}_2 &amp; 0 \\\hline \end{array}$$</code></p> <p>From this, I compute that the associated graded complex to $\pi_1^s(RP^2)$ is $\mathbb{Z}_2\oplus\mathbb{Z}_2$. (I think I made a mistake here with the local coefficients. I believe I showed the local coefficients act trivially, so it should just reduce to ordinary homology with coefficients in $\pi_q^s(S^0)$.) So either $\pi_1^s(RP^2)$ is $\mathbb{Z}_4$ or $\mathbb{Z}_2\oplus\mathbb{Z}_2$. </p> <p>On the other hand, we know that $\pi_1^s(RP^2)=\pi_2(\Sigma RP^2)$ by the Freudenthal suspension theorem. Using the evident cell structure on $\Sigma RP^2$ consisting of a single 0-cell, a single 2-cell, and a single 3-cell, we see that $\pi_1(\Sigma RP^2)=0$ by cellular approximation. So by the Hurewicz theorem $\pi_2(\Sigma RP^2)\cong H_2(\Sigma RP^2) \cong H_1(RP^2) \cong \mathbb{Z_2}$. </p> <p>Where am I going wrong using the AHSS? How does one compute $\pi_2^s(RP^2)$?</p>

http://mathoverflow.net/questions/117494/math-for-a-cake/117566#117566 Glen M Wilson 2012-12-29T22:15:47Z 2012-12-29T22:15:47Z

<p>How about the snake lemma? It's not a formula, but it could still look great on a cake! Plenty of excellent .tex diagrams here: <a href="http://tex.stackexchange.com/questions/3892/how-do-you-draw-the-snake-arrow-for-the-connecting-homomorphism-in-the-snake-l" rel="nofollow">http://tex.stackexchange.com/questions/3892/how-do-you-draw-the-snake-arrow-for-the-connecting-homomorphism-in-the-snake-l</a></p>

http://mathoverflow.net/questions/111310/universal-covering-space-for-non-semilocally-simply-connected-spaces Glen M Wilson 2012-11-02T21:34:44Z 2012-11-03T17:02:35Z

<p>Consider a topological space $X$. Let us consider a universal covering space to be a covering $ p : \tilde{X} \rightarrow X$ which is a covering of all other covering spaces. (Perhaps I should call this an initial covering space). That is, for any other covering space $ q : Y \rightarrow X$, there is a map $f : \tilde{X} \rightarrow Y$ such that $p = q \circ f$. </p> <p><strong>Question:</strong> Does there exist a space $X$ which is not semilocally simply connected with a universal covering space in this sense? Such a covering space is necessarily not simply connected. </p> <p>I've seen some papers by Brazas, Biss, Cannon and Conner, but I couldn't see how to especially use their results in finding such an example. I'd expect such an example to be in the literature somewhere. Could somebody give me a reference in this case? </p> <p>One idea is to find a non semilocally simply connected space $X$ with finite fundamental group, e.g. $\mathbb{Z}_2$. In this case, it might be possible that $X$ itself acts as a universal covering space. </p>

http://mathoverflow.net/questions/46444/natural-numbers-n-which-satisfy-gnunn Glen M Wilson 2010-11-18T01:37:12Z 2010-11-18T04:56:21Z

<p>Are there any natural numbers $n$ (other than 1) for which $gnu(n)=n$? We define $gnu(n)$ to be the number of isomorphism classes of groups of order $n$. This question popped into my head today, and I couldn't come up with a proof one way or another. </p> <p>In the paper by Conway, et al. entitled <a href="http://www.math.auckland.ac.nz/~obrien/research/gnu.pdf%20%22Counting%20Groups%3A%20Gnus,%20Moas,%20and%20other%20Exotica%22" rel="nofollow">"Counting Groups: Gnus, Moas, and other Exotica</a>, Conjecture 10.1 implies that there should be no such natural number $n$ which satisfies $gnu(n)=n$. </p> <p>Does anybody have an argument to show that there is no natural number $n$ (other than 1) for which $gnu(n)=n$? If it is really straightforward, just say so and I'll work it out for myself when I find time. Thanks! </p>

http://mathoverflow.net/questions/25355/are-the-only-universal-co-universal-conditions-co-limits Glen M Wilson 2010-05-20T12:54:55Z 2010-06-07T14:28:26Z

<p>Reading this title, you may have thought there was a typo, but there isn't (well I don't think there is at least!). This question arises from a definition I formulated recently, and would like to understand better. Perhaps a more seasoned category theorist could take a stab at it, or point me in the right direction. </p> <p>$\Large{Background}$</p> <p>${\large Definition:}$ A category consists of a class of morphisms and a class of objects, which satisfy the usual properties. (I just want to stress the class part of it.)</p> <p>${\large Definition:}$ Consider the metacategory\footnote{this follows Mac Lane's definitions, or is just a category in the modern parlance.} $\underline{Cat}$ of categories. Consider a (meta)functor $\mathfrak{i} : \underline{Cat} \rightarrow \underline{Cat}$ which preserves adjoints, i.e. if $F \dashv G$, then $\mathfrak{i} F \dashv \mathfrak{i} G$; consider a natural transformation $P_{-}^{\mathfrak{i}} : id_{\underline{Cat}} \rightarrow \mathfrak{i}$. For any category $\mathfrak{i}$, we define a left adjoint to ${P_{\mathfrak{C}}^{\mathfrak{i}}} : \mathfrak{C} \rightarrow \mathfrak{i} \mathfrak{C}$ to be a universal universal construction, and a right adjoint to $P^{\mathfrak{i}}_{\mathfrak{C}}$ to be a universal couniversal construction. We typically will abbreviate these as UUCs and UCCs.</p> <p>Why this definition? I formulated this definition, because it is exactly the property that makes the following proposition work. Also, you can talk about a given UUC or UCC in any category $\mathfrak{C}$, like you can with the limit of a diagram in an arbitrary category, regardless of whether it exists or not. Thus a universal construction that arises in this way seems more "universal" to me, at least. I can discuss this in greater detail if anybody wants me to, but it isn't all that essential to the problem. </p> <p>${\large Proposition:}$ Consider $F:\mathfrak{C}\rightarrow\mathfrak{D} \in \underline{Cat}$. If $L\dashv F$ and a given UCC $(\mathfrak{i}, P_{-}^{\mathfrak{i}})$ exists in both $\mathfrak{C}$ and $\mathfrak{D}$, then the UCC commutes in $\mathfrak{D}$. That is, if <code>$P_{\mathfrak{C}}^{\mathfrak{i}} \dashv {\mathcal{R}}_{\mathfrak{C}}^{\mathfrak{i}}$</code> and <code>$P_{\mathfrak{D}}^{\mathfrak{i}} \dashv {\mathcal{R}}_{\mathfrak{D}}^{\mathfrak{i}}$</code>, then <code>$\mathcal{R}_{\mathfrak{D}}^{\mathfrak{i}}\circ\mathfrak{i} F \cong F \circ \mathcal{R}_{\mathfrak{C}}^{\mathfrak{i}}$</code>. A similar result holds for all manner of permutations of left and right adjoints.</p> <p>(I have an elaborate xymatrix diagram depicting the simple proof, but I don't think it works with jsmath...)</p> <p>${\Large Question}$ </p> <p>Are the only UUCs and UCCs just colimits and limits respectively, if we take the initial and terminal object constructions as a special case of a (co-)limit? </p> <p>${\large Remark:}$ I wasn't able to come up with any other examples, and the various adjoint functor theorems suggest that limits and colimits are the essential constructions that are needed for there to be an adjoint, which led me to believe the answer to my question is that limits and colimits are the only UCCs and UUCs. I must admit that I don't have a great understanding of the proofs of Freyd's Adjoint Functor Theorem or the SAFT, so perhaps the essential tool is stuck in there and I am just being too lazy. =)</p>

http://mathoverflow.net/questions/24288/sufficient-conditions-for-graph-non-isomorphisms/24316#24316 Glen M Wilson 2010-05-12T01:29:07Z 2010-05-12T01:29:07Z

<p>[Sorry to nitpick, but perhaps it will be helpful to Adam. And I also couldn't comment on Pete's answer, so sorry again!]</p> <p>"Intrinsic properties" are always relative to some kind of idea of "sameness" (or something like that). The fact that the edges are bridges in the one picture is not intrinsic with respect to the picture representing a graph. It is indeed an intrinsic property if we were trying to distinguish paintings though. In a more mathematical context, the situation could be something like dealing with the geometry and topology of an object: an ellipsoid and sphere are homeomorphic, but have different geometry. So Gaußian curvature is geometrically intrinsic but not topologically intrinsic.</p>

http://mathoverflow.net/questions/22659/does-the-forgetful-functor-hopf-algebrasalgebras-have-a-right-adjoint/22685#22685 Glen M Wilson 2010-04-27T05:30:03Z 2010-04-27T05:30:03Z

<p>The article <a href="http://arxiv4.library.cornell.edu/pdf/0905.2613" rel="nofollow">here</a> proves that the forgetful functor from $k$-Hopf algebras to $k$-algebras has a right adjoint. The main tool they use is the Special Adjoint Functor Theorem. </p>

http://mathoverflow.net/questions/22188/introductory-book-on-spectral-sequences/22225#22225 Glen M Wilson 2010-04-22T17:23:06Z 2010-04-22T17:23:06Z

<p>I'm going to have to agree with everyone who recommends Bott &amp; Tu. That provided me with a good understanding of the basic setup. After I was comfortable with that, I moved on to Hilton &amp; Stammbach's book "A Course in Homological Algebra" that did a good job of showing how the general idea works for Abelian categories. </p>

http://mathoverflow.net/questions/22189/what-is-your-favorite-strange-function/22223#22223 Glen M Wilson 2010-04-22T17:19:14Z 2010-04-22T17:19:14Z

<p>I like the Cantor function. A continuous, increasing function $f:[0,1]\rightarrow[0,1]$ with derivative $0$ almost everywhere. See wiki article <a href="http://en.wikipedia.org/wiki/Cantor_function" rel="nofollow">here</a>. </p>

http://mathoverflow.net/questions/21614/ideals-in-the-ring-of-smooth-endomorphisms-of-the-real-line Glen M Wilson 2010-04-16T20:31:36Z 2010-04-18T18:35:57Z

<p>My question is coming from the method Reid and Chris suggested in solving the problem <a href="http://mathoverflow.net/questions/19116/colimits-in-the-category-of-smooth-manifolds" rel="nofollow">here</a>. Help on any point is greatly appreciated! </p> <p>Question 1. For a real manifold $M$, consider $C^{\infty}(M,\mathbb{R})$. For a point $p\in M$, consider the ideal $I_p=\{ f : f(p)=0 \}$. Is $I_p^n$ equal to the set of smooth maps $f$ which have $n-1$st order contact (ref. Golubitsky, Guillemin pg. 37) with the $0$ function? </p> <p>Question 2. Consider $C(\mathbb{R},\mathbb{R})$ and the ideal $I_0= \{ f : f(0)=0 \}$. Is it the case that $I_0^2 = \{ f : f(0)=f'(0)=0\}$? Is it the case that $I_0^n=\{ f : f(0)=f'(0)=...=f^{(n-1)}(0)=0\}$? </p> <p>To question 2, the one inclusion is immediate. However the inclusion $ \{ f : f(0)=f'(0)=0\} \subseteq I_0^2 $ doesn't seem obvious to me right now. It seems like one could perhaps find a function which doesn't agree with it's Taylor series that satisfies the derivative condition, but not be in $I^2_0$. </p> <p>I hope this isn't too elementary for MO. Thanks for your help! </p>

http://mathoverflow.net/questions/21614/ideals-in-the-ring-of-smooth-endomorphisms-of-the-real-line/21759#21759 Glen M Wilson 2010-04-18T17:17:49Z 2010-04-18T17:17:49Z

<p>I thought I would add a little more elaboration to the answers given above. There is a general result in the book “Stable Mappings and Their Singularities” by Golubitsky and Guillemin in §II.6 that provides an answer to my Q2. Restricted to my special case, one first observes that if $f$ is smooth and $f(0)=f'(0)=0$ that $g(x):=\frac{f(x)}{x}=\int_{0}^{1}f'(tx)dt$. The right hand side is easily seen to be smooth as we can differentiate under the integral sign. Also $g(0)=f'(0)=0$. Induction then provides the general result (for $\mathbb{R}$). The case when $f:\mathbb{R}^n\rightarrow \mathbb{R}$ is handled analogously. </p> <p>As for the general statement, it looks like Robin has that taken care of! Thanks everyone! </p>

http://mathoverflow.net/questions/19116/colimits-in-the-category-of-smooth-manifolds Glen M Wilson 2010-03-23T15:48:40Z 2010-03-26T22:54:06Z

<p>In the category of smooth real manifolds, do all small colimits exist? In other words, is this category small-cocomplete? I can see that computing push-outs in the category of topological spaces of smooth manifolds need not no be manifolds, but this is not a proof. </p>

http://mathoverflow.net/questions/18988/if-f-is-left-adjoint-to-g-when-does-fg-preserve-limits-when-do-counits-intercha/19120#19120 Glen M Wilson 2010-03-23T16:27:08Z 2010-03-23T16:27:08Z

<p>Have you looked at the paper by B. Eckmann and P. J. Hilton entitled "Commuting Limits with Colimits" in the "Journal of Algebra", 11, 116-144 (1969)? </p>

http://mathoverflow.net/questions/129541/does-the-paper-on-the-cobordism-ring-omega-and-a-complex-analogue-ii-exist/129542#129542 Glen M Wilson 2013-05-03T17:51:45Z 2013-05-03T17:51:45Z

Awesome! Thank you very much.

http://mathoverflow.net/questions/128178/examples-of-applications-of-the-freyd-mitchell-embedding-theorem/128191#128191 Glen M Wilson 2013-04-21T15:27:35Z 2013-04-21T15:27:35Z

Take $\alpha \in \pi_i^s$, $\beta \in \pi_j^s$. THey can be represented by maps $\alph : S^{i+j+k} \rightarrow S^{j+k}$ and $\beta : S^{j+k} \rightarrow S^k$ so that the composition is defined $\alpha \circ \beta : S^{i+j+k} \rightarrow S^k$ which then determines an element of $\pi_{i+j}^s$. From a more general perspective $\pi_*^s = \mathbb{S}^*(pt)$---the cohomology of a point where $\mathbb{S}$ is the sphere spectrum. The sphere spectrum is a ring spectrum so cohomology groups have a product structure.

http://mathoverflow.net/questions/114834/gysin-sequence-for-s0-bundles Glen M Wilson 2012-11-29T02:41:37Z 2012-11-29T02:41:37Z

The comment asking what is meant by the Gysin sequence is not an answer as far as I am concerned.

http://mathoverflow.net/questions/114528/are-homeomorphic-open-subsets-of-mathbbrn-also-diffeomorphic/114532#114532 Glen M Wilson 2012-11-26T14:53:59Z 2012-11-26T14:53:59Z

Oops, missed that.

http://mathoverflow.net/questions/113969/computation-of-stable-homotopy-groups-of-rp2/113972#113972 Glen M Wilson 2012-11-20T22:09:47Z 2012-11-20T22:09:47Z

Even better! This seems to completely describe $\pi_k^s(RP^2)$ in terms of $\pi_*^s$. That is very nice indeed.

http://mathoverflow.net/questions/113969/computation-of-stable-homotopy-groups-of-rp2/113972#113972 Glen M Wilson 2012-11-20T19:31:33Z 2012-11-20T19:31:33Z

Oh I see now. So the generalized homology theories for the spectral sequence are to be unreduced ones. Since $\pi_*^s$ is a reduced homology theory, you make it into an unreduced one by adding the basepoint. Everything works out quite nicely now. Thank you for correcting my error! I will see if I can get this to work for $\pi_2^s(RP^2)$.

http://mathoverflow.net/questions/111310/universal-covering-space-for-non-semilocally-simply-connected-spaces Glen M Wilson 2012-11-07T19:24:52Z 2012-11-07T19:24:52Z

Thanks Lee! I haven't looked at Spanier much. I will check it out.

http://mathoverflow.net/questions/111310/universal-covering-space-for-non-semilocally-simply-connected-spaces Glen M Wilson 2012-11-06T18:39:35Z 2012-11-06T18:39:35Z

Yes, that was what I felt like it should be. However, it seemed Hatcher took a different approach by (sort of) defining simply connected covering spaces to be universal covers for nice enough spaces. At any rate, I agree with you entirely and am glad to now better understand things.

http://mathoverflow.net/questions/111310/universal-covering-space-for-non-semilocally-simply-connected-spaces/111322#111322 Glen M Wilson 2012-11-03T23:27:51Z 2012-11-03T23:27:51Z

Thank you so much for your answer Tom! You are spot on about my assumptions that I did not mention. The topologist's sine curve is a good example I will keep in mind.

http://mathoverflow.net/questions/111310/universal-covering-space-for-non-semilocally-simply-connected-spaces/111388#111388 Glen M Wilson 2012-11-03T23:17:15Z 2012-11-03T23:17:15Z

Jeremy, this is perfect! Thank you so much for the answer.

http://mathoverflow.net/questions/51056/are-there-any-good-websites-for-hosting-discussions-of-mathematical-papers Glen M Wilson 2011-01-13T18:17:09Z 2011-01-13T18:17:09Z

As a graduate student still just finding my way around the research community, I think such a forum would be a great way to interact with other people interested in the same material I am in, that I don't readilly get on MO or even at my graduate school. I can imagine that there are lots of advanced undergraduates and graduate students who become interested in a topic, and it turns out that the school that they are at does not have too many people that do research in that field! I was rather alone in my interest in category theory as an undergraduate; this is one reason I joined MO.

http://mathoverflow.net/questions/46444/natural-numbers-n-which-satisfy-gnunn Glen M Wilson 2010-11-18T02:36:02Z 2010-11-18T02:36:02Z

OK, I guess I should probably delve a bit deeper into that paper. Sorry for not doing my homework well enough!

http://mathoverflow.net/questions/46444/natural-numbers-n-which-satisfy-gnunn Glen M Wilson 2010-11-18T01:44:45Z 2010-11-18T01:44:45Z

Oops! Thanks for catching that.

http://mathoverflow.net/questions/19116/colimits-in-the-category-of-smooth-manifolds Glen M Wilson 2010-08-12T21:20:40Z 2010-08-12T21:20:40Z

Colin, see the answers below!

http://mathoverflow.net/questions/25355/are-the-only-universal-co-universal-conditions-co-limits Glen M Wilson 2010-06-09T11:18:02Z 2010-06-09T11:18:02Z

I'm in the process of thinking about how I should formalize when two UUCs or UCCs are the same. I'm a bit busy though, so it might take me a while.