User joseph victor - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T12:41:28Z http://mathoverflow.net/feeds/user/24021 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/128976/can-one-make-the-category-of-pairs-of-topological-spaces-a-model-category Can one make the category of pairs of topological spaces a model category? Joseph Victor 2013-04-28T04:09:16Z 2013-04-29T18:12:26Z <p>Can you make the category whose objects are pairs of spaces $(X,A)$, and morphisms the obvious diagrams, into a model category? Of course I want this to be done in a meaningful way, that is, agreeing with the adjoint functors $X\mapsto (X,\emptyset)$ and $(X,A)\mapsto X$?</p> <p>There might be some intuitive reason that it is wrong to expect this, but I don't see it yet. </p> <p>Thanks!</p> http://mathoverflow.net/questions/127881/is-there-a-picture-i-should-have-in-my-head-of-rational-homotopy-equivalence Is there a picture I should have in my head of rational homotopy equivalence? Joseph Victor 2013-04-17T19:37:36Z 2013-04-17T19:37:36Z <p>My understanding is that one thinks to rational homotopy theory for computational advantage. However, thinking about things in terms of localizations still lacks some amount of intuition for me.</p> <p>In particular, if $f,g$ are continuous functions and $\gamma$ is localization functor by rational homotopy equivalence and $\gamma(f)=\gamma(g)$, is there something I can say analogous to continuously ("rationally"?) transforming $f$ to $g$". </p> <p>Thanks!</p> http://mathoverflow.net/questions/124360/when-is-a-differential-abelian-group-a-chain-complex-and-when-can-you-extend-map When is a differential abelian group a chain complex, and when can you extend maps from a projective resolution? Joseph Victor 2013-03-13T00:04:31Z 2013-03-13T01:49:15Z <p>By differential abelian group I mean an abelian group $A$ with a self-map $\partial:A\to A$ with $\partial^2=0$ (or equivalently, a $\mathbb{Z}[\partial]/\partial^2$-module).<br> Under what circumstances can you put a grading on $A$ such that $\partial$ lowers dimension? I don't require the grading be non-negative, so raising dimension is equivalent. Is the grading introduced unique (up to dimension shifts)? </p> <p>In particular, can you always grade a differential abelian group? I don't necessarily believe the answer is yes (I suspect the answer is no), but I am finding it surprisingly difficult to find a counter-example. </p> <p>Finally, if $F_*$ is a positively graded free resolution, then then a map from $f:F_0\to A$, can we extend this to a "chain map" $F_*\to A$ (the formula $\partial f = f \partial$ still makes sense!), even if $A$ is not actually a chain complex. Note that if $A$ is a chain complex but the image of $f$ is not in $A_0$, we can still extend $f$ to a direct sum of chain maps in different degrees. </p> <p>Thanks, -Joseph</p> http://mathoverflow.net/questions/117928/finding-a-subspace-disjoint-from-a-union-of-subspaces Finding a subspace disjoint from a union of subspaces Joseph Victor 2013-01-03T06:23:39Z 2013-01-03T18:19:21Z <p>Let $k$ be a finite field (I care about $\mathbb F_p$, especially $\mathbb F_2$) and let $V_1,...,V_N\subset k^n$ be subspaces.<br> I want to find a subspace $S\subset k^n$ such that $S\cap V_i=0$ for each $i$, with $dim(S)$ as large as possible, and I want to do this in polynomial time (in exponential time we can try every possibility, as this is a finite vector space). </p> <p>In $\mathbb{R}^n$ this is easy since if we pick $S$ randomly of dimension $n-\max dim(V_i)$ it will almost always be disjoint from the $V_i$. Of course, in a finite field you can fill up all of $k^n$ with just 1-dimensional $V_i$, so this approach fails. </p> <p>Thanks</p> <p>EDIT: What if $N\lt \lt n$, though $dim(V_i)$ can be around $n/2$ or so. </p> http://mathoverflow.net/questions/114452/computing-slim-extensions-representing-ext Computing Slim Extensions representing Ext Joseph Victor 2012-11-25T21:30:10Z 2012-11-26T18:11:50Z <p>Hey Everyone</p> <p>Let $A$ be an algebra over a field (group rings $k[G]$ for group cohomology, the Steenrod Algebra). We want to compute, say, $Ext_A(k,k)$, so let $F_*\to k$ be an $A$-free resolution. Elements of $Ext^j_A(k,k)$ can be represented as Yoneda extensions, essentially exact sequence $k\to M_1 \to ... \to M_{j} \to k$, which we denote $\mathcal{X}$, and a chain map map $F_*\to \mathcal{X}$ with the cocycle, $F_{j+1}\to k$, on one side and the identity map $k\to k$ on the other. Using pushouts one can, given a cocycle, generate $\mathcal{X}$, but the modules in $\mathcal{X}$ can be rather large. </p> <p>My question is then this. Is there a way to either generate a "small" extension (measured however you'd like, perhaps the size of an extension should be the largest dimension over $k$ in any bi-digree) representing a cocycle, or given the a Yoneda extension, is there a way to make some sort of local change to it that can make it "smaller" (in which case we can search for a "local optimum"). </p> <p>This question is motivated as a workaround to the problem presented by Bob Bruner's in <a href="http://www.math.wayne.edu/~rrb/papers/yoneda.pdf" rel="nofollow">http://www.math.wayne.edu/~rrb/papers/yoneda.pdf</a>. </p> <p>Thanks</p> http://mathoverflow.net/questions/112554/do-people-still-use-massey-products-for-computations-in-the-adams-spectral-sequen Do people still use Massey Products for computations in the Adams Spectral Sequence Joseph Victor 2012-11-16T05:59:29Z 2012-11-16T17:54:20Z <p>Hey everyone,</p> <p>It seems to me like in the literature of the Adams Spectral Sequence, older publications (Toda, May, Tengora+Mahowald) make heavy and explicit use of Massey Products for computations.<br> More "recent" sources (Kahn, Milgram, Ravenel, May (again), Bruner) seem to like to make references to the Massey products, even just for the sake of naming, but use Steenrod Squares for actual computations. The few computations I know of using Massey products can be more easily done with Steenrod Squares, but this may just be my ignorance about Massey products. My question then is do these things still play an important computational role?</p> <p>Thanks</p> http://mathoverflow.net/questions/106269/which-algebraic-integers-in-a-cyclotomic-field-give-you-integer-absolute-value/106275#106275 Answer by Joseph Victor for which algebraic integers in a cyclotomic field give you integer absolute value? Joseph Victor 2012-09-03T20:55:20Z 2012-09-04T00:24:36Z <p>There is a primitive way to do it, given $n$.<br> You can of course an element of such a ring of integers can be written (non-uniquely, but this does not matter) as $\sum a_i\zeta_n^i$, and the absolute value then is $\sum_{i,j} a_ia_jcos(2\pi(i-j)/n)$. So just figure the relations in these cosine values over $\mathbb{Z}$ (for a specific value of $n$), and then write down some relations you need between the $a_i$'s and $a_j$'s to make sure you get an integer in the end.</p> <p>Of course, using a basis instead of a spanning-set simplifies these relations, but it doesn't matter: you can just write the relations and then decompose any element this way and just check the relations. </p> <p>Hope this helps. </p> http://mathoverflow.net/questions/106268/image-of-j-in-the-classical-adams-spectral-sequence Image of J in the classical Adams Spectral Sequence Joseph Victor 2012-09-03T19:29:55Z 2012-09-03T19:50:57Z <p>Hey all,</p> <p>I know that in some versions of the Adams Spectral Sequence you can easily identify the image of $J$, and I was wondering if there was a way to identify the image of $J$ in the $E_2$ page of the classical mod-2 version, especially for $t-s=3$ (mod 4). Since the order of this image is known and it is known that the image is a direct summand, it isn't so hard to find it in $E_\infty$. Of course, if you can identify $Im(J)$ in an earlier page, then you learn a huge amount about the differentials in that column. This might imply that identifying the image of $J$ is almost as hard as calculating the differentials, so maybe this is too much to hope for, but maybe just maybe there's a trick. </p> <p>Thanks</p> http://mathoverflow.net/questions/105753/universal-property-of-the-smash-product-of-pointed-spaces Universal Property of the Smash Product (of pointed spaces) Joseph Victor 2012-08-28T19:26:52Z 2012-08-29T13:39:35Z <p>Hey</p> <p>Is there a universal property for the smash product (of pointed spaces or pointed CW-complexes or something of that ilk)? I've seen the smash product of spectra defined with a universal property in terms of the smash product of pointed-spaces, but I was wondering if there was just some simple universal property you could put on these somewhat mysterious (to me) space-level operations.</p> <p>EDIT: I was hoping for something more satisfying than the internal Hom adjoint. The tensor product can be defined similarly, but I find the universal product in terms of bilinear maps more intuitive (although, when unraveled, they are the same thing). I was hoping for something similar for smash. </p> <p>Thanks :)</p> http://mathoverflow.net/questions/104877/what-do-empty-parens-symbol-mean what do empty parens symbol mean? Joseph Victor 2012-08-16T23:23:32Z 2012-08-17T06:07:27Z <p>Quick easy question: what is the meaning of the symbol $(\space\space )$. I've seen it now in two papers, one of which is Milgram's Group Representations and the Adams Spectral Sequence, avaiable at <a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.pjm/1102968426" rel="nofollow">here</a>. </p> <p>On the top of page 170, Milgram writes $\partial(\space\space) \to S^k\vee S^k\to S^k\to v$</p> <p>I figure if I knew what was meant by mathematical colloquialism "$\partial(\space \space)$", interpereting this part of the paper would be easy enough, I'm just not sure wht it means. </p> <p>Thanks</p> http://mathoverflow.net/questions/104380/why-are-cup-i-products-and-steenrod-squares-often-always-unary Why are cup-i products and Steenrod Squares often (always?) unary? Joseph Victor 2012-08-09T23:39:32Z 2012-08-11T14:42:34Z <p>One way to define the Steenrod Operations is to use the cup-i product, as in Mosher and Tangora's book. It basically says, given the chain complex from mod-2 homology $C_*$, define</p> <p>$D_0 : C_*\to C_*\otimes C_*$, </p> <p>so that the cup product is given (on cocycles)</p> <p>$(u\cup v)(\sigma) = (u\otimes v)(D_0\sigma)$</p> <p>and then for higher $i$, define $D_i$ so that </p> <p>$D_{i-1}+\rho D_{i-1} = D_i\partial + \partial D_i$</p> <p>where $\rho$ is the flipping map. Then the $\cup_i$ product is just</p> <p>$(u\cup_i u)(\sigma) = (u\otimes u)(D_i\sigma)$</p> <p>And then define for $[u]\in H^n$ $Sq^{2n-i}([u]) = [u\cup_{i}u]$</p> <p>This definition seems perfectly well-defined as a binary operation, and yet wherever I've seen it done it has only even been used as a unary operation. </p> <p>Is there a reason why this is the case, why either the product is undefined as a binary product or not useful as a binary product or just too hard to use?<br> Is this a dumb question?</p> <p>Thanks, -Joseph</p> http://mathoverflow.net/questions/104048/defining-the-cup-product-in-ext-using-a-kunneth-formula Defining the cup product in Ext using a Kunneth formula Joseph Victor 2012-08-05T20:25:31Z 2012-08-05T20:25:31Z <p>I want to make a Kunneth product of sorts on Ext. In particular, letting $C_*$ be a $R$-free resolution for $k$ over a $k$-hopf algebra $R$, elements in $Ext_R(k,k)$ are represented by maps in $Hom_R(C_*,k)$. Of course, we can consider $Hom_R(C_* \otimes C_*,k)$, the cohomology of this being the same since $k\otimes k \cong k$. </p> <p>What I want then is a sort of Kunneth product:</p> <p>$\times :Hom_R(C_*,k)\otimes Hom_R(C_*,k)\to Hom_R(C_*\otimes C_*,k)$</p> <p>I want it to be the case that this defines the cup product (aka Yoneda product). If we get a magical map </p> <p>$\Delta : C_*\to C_*\otimes C_*$,</p> <p>then </p> <p>$\cup : Ext_R(k,k)\otimes Ext_R(k,k)\to Ext_R(k,k)$</p> <p>can be given, on the co-chain level,</p> <p>$(a\cup b)(\sigma) = (a\times b)(\Delta \sigma)$</p> <p>Now, $\Delta$ is possibly defined up to homotopy by the requirement that it commutes with the augmentation in $R$. </p> <p>My question is this: is there a way to describe this Kunneth product using any $C_*$, or do I need to use a Bar resolution of sorts like in group cohomology. Does this method even make sense? Can you help flesh this out? </p> <p>Thanks!</p> http://mathoverflow.net/questions/103056/when-is-an-acylic-chain-complex-contractible When is an acylic chain complex contractible Joseph Victor 2012-07-25T00:07:23Z 2012-07-25T05:56:21Z <p>When is an acyclic chain complex contractible? </p> <p>I know an acyclic chain complex of free modules over a PID (or field) are always contractible, but what about over a more complicated ring, like a graded algebra over Z/p (for instance the mod p Steenrod Algebra)? </p> <p>EDIT: I want to assume the chain complex is bounded below, but not necessarily that the ring is commutative. I suspect that it isn't always true for a non-commutative ring, but I don't really have a counter example. </p> <p>Thanks everybody! </p> http://mathoverflow.net/questions/102316/differentials-in-the-adams-spectral-sequence-for-spheres-p2 Differentials in the Adams Spectral Sequence for spheres p=2 Joseph Victor 2012-07-15T22:37:32Z 2012-07-16T01:05:09Z <p>Hey everybody,</p> <p>How does one compute the differentials in the Adams Spectral Sequence for spheres at prime 2 in the range $13\le t-s\le 20$? There seem to be 6 nonzero differentials, and at this point I only understand $d_2(h_4)=h^2_3h_0$. </p> <p>There seem to be two methods that are used or referenced in various texts, but I haven't figured out exactly how to apply either in this context. The first is the Massey Product/Toda Product (apparently they are the same, but Massey is algebraic and works in $E_2$ and Today is topological and works in $\pi_*^s$). The second is by building a cofiber sequences $S^0\to S^0\cup_f e^i\to S^i$ which gives a long exact sequence in both the $\pi^s_*$ and the spectral sequence itself. </p> <p>If possible, could somebody point me to a resource where they use these methods in this range, or give me a hint on how I can try to do this. </p> <p>Thanks a bunch -Joseph</p> http://mathoverflow.net/questions/101540/how-do-you-know-when-something-must-die-in-the-adams-spectral-sequence-for-pi How do you know when something must die in the Adams Spectral Sequence for $\pi_*^s$ Joseph Victor 2012-07-06T23:21:06Z 2012-07-07T04:42:34Z <p>Hey everybody,</p> <p>I think this question might be just a simple oversight on my part, but this has been bugging me a few days. </p> <p>I am reading Hatcher's <a href="http://www.math.cornell.edu/~hatcher/SSAT/SSch2.pdf" rel="nofollow">Spectral Sequences book</a>, and trying to understand his example where he computes $\pi_*^s$ for $p=2$ (page 21-23), and I'm a bit confused about a certain step. He claims that the element corresponding to $h_3^2$ must have order 2 in $\pi_{14}^s$, because of "the commutativity property of the composition product, since $h_3$ has odd degree". Now, I see why $h_3^2$ can have order at most 4, because $h_3^2h_0^2=0 \in E_2$, but why must it have order 2 exactly? What does the odd degree have to do with it? If I am not mistaken, the Yoneda product on $Ext_A(Z/2,Z/2)$ induces the composition product on $\pi_*^s$, which, mod 2, is commutative, but the Yoneda product has $h_3h_0=h_0h_3$ in the $E_2$ page, so I can't from that derive the induced composition product is 0. Do I need to use a fact about $\pi_s^*$ that doesn't come from this spectral sequence? </p> <p>Thanks for the help everybody! -Joseph Victor</p> http://mathoverflow.net/questions/99050/4-polytope-with-vertices-at-the-binary-octahedral-group 4-polytope with vertices at the binary octahedral group Joseph Victor 2012-06-07T18:28:15Z 2012-06-07T19:17:31Z <p>Hey everybody,</p> <p>Does anybody know if there is a convex polytope in $R^4$ with vertices at the <a href="http://en.wikipedia.org/wiki/Binary_octahedral_group" rel="nofollow">binary octahedral group</a> (identitfying $H$ with $R^4$). </p> <p>The binary tetrahedral group lies at the vertices of the so-called 24-cell, and the binary octahedral group is just a direct some of two binary tetrahedral groups, but it is not clear how to interpret that geometrically. </p> <p>Experimentally, I have found that, for each octahedron in the 24-cell, each vertex in that octahedron is equidistance from exactly one point in binoct not in bintet. I don't know if this is relevant at all. </p> <p>Thanks so much!</p> <p>-Joseph</p> http://mathoverflow.net/questions/128976/can-one-make-the-category-of-pairs-of-topological-spaces-a-model-category/128982#128982 Comment by Joseph Victor Joseph Victor 2013-04-29T20:19:37Z 2013-04-29T20:19:37Z I just picked u the Hovey book. Thanks for the reference! http://mathoverflow.net/questions/128976/can-one-make-the-category-of-pairs-of-topological-spaces-a-model-category/129137#129137 Comment by Joseph Victor Joseph Victor 2013-04-29T20:19:01Z 2013-04-29T20:19:01Z I feared as much, ah well. http://mathoverflow.net/questions/128976/can-one-make-the-category-of-pairs-of-topological-spaces-a-model-category Comment by Joseph Victor Joseph Victor 2013-04-28T06:16:35Z 2013-04-28T06:16:35Z That sounds interesting! Is there a good reference for that? Do you know what goes wrong if I restrict to subspaces or $A\to X$ a cofibration? http://mathoverflow.net/questions/102138/a-toolbox-for-algebraic-topology/102146#102146 Comment by Joseph Victor Joseph Victor 2013-04-28T03:42:30Z 2013-04-28T03:42:30Z I think an AT wiki would be a wonderful thing, especially if it were more readable than nlab! http://mathoverflow.net/questions/124360/when-is-a-differential-abelian-group-a-chain-complex-and-when-can-you-extend-map Comment by Joseph Victor Joseph Victor 2013-03-13T01:48:44Z 2013-03-13T01:48:44Z oops... yeah. made an edit. http://mathoverflow.net/questions/117928/finding-a-subspace-disjoint-from-a-union-of-subspaces/117950#117950 Comment by Joseph Victor Joseph Victor 2013-01-03T18:32:28Z 2013-01-03T18:32:28Z I'm afraid I don't know as many combinatorial algorithms as I probably should. How would you use inclusion-exclusion to calculate the subspace? http://mathoverflow.net/questions/117928/finding-a-subspace-disjoint-from-a-union-of-subspaces/117950#117950 Comment by Joseph Victor Joseph Victor 2013-01-03T18:19:32Z 2013-01-03T18:19:32Z How sad. Can we do better if $N$ is small. http://mathoverflow.net/questions/117928/finding-a-subspace-disjoint-from-a-union-of-subspaces Comment by Joseph Victor Joseph Victor 2013-01-03T18:19:16Z 2013-01-03T18:19:16Z I want small $N$, and I added an edit to this effect. http://mathoverflow.net/questions/117928/finding-a-subspace-disjoint-from-a-union-of-subspaces Comment by Joseph Victor Joseph Victor 2013-01-03T09:33:44Z 2013-01-03T09:33:44Z 3Sat is probably a pretty easy way to do this, especially for Z/2, but boy is that unsatisfying. I'd rather have a nice lil algorithm. http://mathoverflow.net/questions/114452/computing-slim-extensions-representing-ext/114506#114506 Comment by Joseph Victor Joseph Victor 2012-11-27T18:37:59Z 2012-11-27T18:37:59Z Good point, mt. These are clearly exact. http://mathoverflow.net/questions/114452/computing-slim-extensions-representing-ext/114506#114506 Comment by Joseph Victor Joseph Victor 2012-11-27T03:26:12Z 2012-11-27T03:26:12Z I tried something like this but game up. I don't think these are still exact. Pick some 1-d $N$ which maps onto $k$, then $j^{-1}(N)=0$ but exactness, but $k\to 0$ is not injective. Am I missing something? http://mathoverflow.net/questions/114452/computing-slim-extensions-representing-ext Comment by Joseph Victor Joseph Victor 2012-11-26T18:47:51Z 2012-11-26T18:47:51Z To be honest, I didn't mean to write &quot;graded algebra of finite type&quot;. I fixed the finite type part. http://mathoverflow.net/questions/114452/computing-slim-extensions-representing-ext Comment by Joseph Victor Joseph Victor 2012-11-26T18:12:45Z 2012-11-26T18:12:45Z trivial, unless $G$ itself is graded. http://mathoverflow.net/questions/114452/computing-slim-extensions-representing-ext/114464#114464 Comment by Joseph Victor Joseph Victor 2012-11-26T07:05:05Z 2012-11-26T07:05:05Z Thanks Ralph. In the case of group cohomology, in playing around I was able to make $\mathcal{X}$ much smaller than the bar resolution, which obviously is not minimal. However, I was hoping to make \mathcal{X} smaller than the resolution being used even if that resolution was itself minimal (in the sense of the algorithm from the last paragraph). Is this something that can be done? http://mathoverflow.net/questions/114452/computing-slim-extensions-representing-ext Comment by Joseph Victor Joseph Victor 2012-11-26T06:58:03Z 2012-11-26T06:58:03Z Oops, Why on earth did I write finite type... I must be typing without thinking.