User alan wilder - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T01:41:35Z http://mathoverflow.net/feeds/user/11300 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/68470/mapping-into-a-geometric-realization Mapping into a geometric realization. Alan Wilder 2011-06-22T02:28:20Z 2011-06-22T02:28:20Z <p>Suppose $S$ is a simplicial set, $X$ is a space, and we are given a map <code>$f: \text{Sing}\,X\to S.$</code> When is is possible to produce a map $X\to |S|$?</p> <p>We can take the realization of $f$, to get $|f|:|\text{Sing}\;X|\to |S|$, and then the question becomes: when does $|f|$ factor through the counit map $|\text{Sing}\;X|\to X$ (at least up to homotopy)?</p> <p>It does if $X$ is a CW-complex, which is not the case for my example. What about if $S$ is a Kan complex? Does that help, or are the properties of $X$ really the key here?</p> http://mathoverflow.net/questions/68027/embeddings-of-vector-spaces Embeddings of vector spaces Alan Wilder 2011-06-17T04:06:24Z 2011-06-17T08:23:09Z <p>Let $V$ be an $n$-dimensional vector space. Is the space of embeddings <code>$\coprod_1^{k} V \to V$</code> path connected for large enough $n$? Clearly $n=1$ is not enough, but I feel like $n=2$ is enough for $1$-connected. Does the space become highly connected as $n\to \infty$? This feels like it is equivalent to a question about the little disks operads, but I don't know how to frame it as such.</p> http://mathoverflow.net/questions/67083/reference-request-lax-ends Reference Request: Lax Ends Alan Wilder 2011-06-06T20:19:01Z 2011-06-07T20:38:55Z <p>I've read in a few different places that the standard fact <code>$\text{Nat}\,(F,G) \cong \int_x \text{Hom}\,(Fx,Gx)$</code> can be upgraded to <code>$\textbf{LaxNat}\,(F,G) \cong \oint_x\textbf{Hom}\,(Fx,Gx)$</code> Where the left hand side is the category of lax natural transformations and modifications, and the right hand side is a lax end.</p> <p>I am looking for a reference that gives the definition of lax end and proves this equivalence. I do know of the reference</p> <p>S. Bozapalides, Th<code>\'{e}</code>orie formelle des bicat<code>\'{e}</code>gories</p> <p>but I can't read French and I also can't find a copy. If someone can link me to the Bozapalides reference would be great. Or even better would be if there is a reference in English. Thanks!</p> http://mathoverflow.net/questions/66939/pseudofunctors-and-pseudonatural-transformations pseudofunctors and pseudonatural transformations Alan Wilder 2011-06-05T05:13:58Z 2011-06-05T19:49:07Z <p>Based on the discussion <a href="http://ncatlab.org/nlab/show/transfor" rel="nofollow">here</a> I feel like there should be a bijection between pseudonatural transformations of pseudofunctors $J\to\mathcal{C}$ and pseudofunctors $J\times [1]\to\mathcal{C}$, at least morally ($[1]$ denotes the poset 0&lt;1). </p> <p>The map from 'homotopies' $J\times [1]\to\mathcal{C}$ to pseudonatural transformations works out nicely, but the other direction seems problematic. In particular, given $\alpha:F\Rightarrow G$, if we try to form a pseudofunctor $\tilde{\alpha}:J\times [1]\to\mathcal{C}$ it's clear that we should set $$\tilde{\alpha}(\text{id}_j,0\to 1) = \alpha (j),\; \text{and}\quad \tilde{\alpha}(j'\to j,\text{id}_0)=F(j'\to j),\tilde{\alpha}(j'\to j,\text{id}_1) = G(g)$$ but what about $\tilde{\alpha}(j'\to j,0\to 1)$? The problem is pseudonatural transformations only tell how to fill squares, not the triangles individually (i.p. they don't give any diagonal 1-morphism in the middle). One could just choose either $\alpha(j)\circ F(g)$ or $G(g) \circ \alpha(j)$ but this won't result in a bijection.</p> <p>Is the best one can hope for an equivalence modulo modification?</p> http://mathoverflow.net/questions/59357/why-chain-homotopy-when-there-is-no-topology-in-the-background/59366#59366 Answer by Alan Wilder for Why chain homotopy when there is no topology in the background? Alan Wilder 2011-03-23T22:54:19Z 2011-03-25T01:14:54Z <p>There is an inner-hom in $\mathbf{Chain}$, and the 1-chains are chain homotopies. The definition is $$\underline{\mathbf{Chain}}(C_\bullet,D_\bullet)_k = \Pi_n \textrm{Hom} (C_{n-k},D_n)$$ so a 0-chain is just a map $f_n:C_n\rightarrow D_n$. The differential of this complex is given by $$df(c) = d_D(f(c)) - (-1)^{|f|}\left(f(d_C(c)\right)$$ So a 0-cycle is just a chain map. A 1-chain whose boundary is $f-g$ is exactly a chain homotopy from $g$ to $f$.</p> <p>Alternatively, there is a model structure on $\mathbf{Chain}$ where the weak equivalences are quasi-isomorphisms, and you can make sense of cylinder as a cylinder object for a chain complex, and then your topological motivation should all still make sense. I don't know all the details though so I won't try...</p> <p>EDIT: crosspost...</p> http://mathoverflow.net/questions/59499/end-of-a-weak-equivalence end of a weak equivalence Alan Wilder 2011-03-24T22:27:47Z 2011-03-24T22:27:47Z <p>I would like to get a concrete description of sufficient conditions for the end of a morphism in $\mathcal{C}^{J^{op}\times J}$ (which is a point-wise weak equivalence) to be a weak equivalence.</p> <p>In thinking about this problem, I've come to sufficient conditions that seem to be very rarely satisfied. Here's what I have:</p> <p>An end is the same as a limit over the subdivison category, which I'll denote with $'$s. Subdivision categories are always inverse categories, and in particular Reedy, so we can put the Reedy model structure on $\mathcal{C}^{J'}$. When the index category is inverse, limit preserves trivial fibrations, so by Ken Brown's lemma, a limit of a point-wise weak equivalence between fibrant objects is a weak equivalence. So we need to figure out what fibrant means in $\mathcal{C}^{J'}$.</p> <p>Let $X\in\mathcal{C}^{J^{op}\times J}$, and let $X'\in\mathcal{C}^{J'}$ be the associated 'subdivison'. If $f$ is a morphism of $J$, $M_f X' = \ast$ is the terminal object, because there are no non-identity morphisms with source $f$ in $J'$. That implies that in order for $X$ to be s.t. $X'$ is fibrant, for $f:s\rightarrow t$, we need $$X(s,t) = X'(f) \rightarrow \ast\times_{M_f\ast}M_fX \cong \ast$$ to be a fibration. No surprises here--in order to be fibrant it needs to be point-wise fibrant (at least for objects that have a morphism between them).</p> <p>When we look at objects $i\in J'$ coming from objects of $J$ though, the matching space becomes a limit over the discrete category of morphisms $f$ with source or target $i$. So the "matching space" condition here becomes $$X(i,i) = X'(i) \rightarrow \ast\times_\ast M_i X' \cong \prod_{f:c\rightarrow i} X(c,i) \times \prod_{g:i\rightarrow c} X(i,c)$$ This seems like it's asking too much. A map into a product being a fibration would require something like the map to each factor being a fibration and all the lifts have to agree.</p> <p>So, in short, the question is just: </p> <p>Are there more reasonable sufficient conditions for the end of a weak-equivalence to be weak-equivalence?</p> <p>Alternatively, if I made a mistake in the above, pointing it out would be great too!</p> http://mathoverflow.net/questions/59252/geometric-realization-on-mathbfstop geometric realization on $\mathbf{sTop}$ Alan Wilder 2011-03-22T23:47:16Z 2011-03-23T10:57:50Z <p>Is geometric realization $|\cdot|:\mathbf{Top}^{\mathbf{\Delta}^{\textrm{op}}}\rightarrow \mathbf{Top}$ a left Quillen functor? If so, under what model structure on $\mathbf{Top}^{\mathbf{\Delta}^{\textrm{op}}}$? I would guess the Reedy model structure.</p> <p>A reference would be ideal.</p> <p>Thanks</p> http://mathoverflow.net/questions/55669/nerve-groupoids-kan-complexes-nerve-bicategories-w-adjoints Nerve: Groupoids-> Kan Complexes. Nerve: Bicategories w. adjoints -> ? Alan Wilder 2011-02-16T21:57:17Z 2011-02-23T12:44:01Z <p>If you take the nerve of a groupoid, you get a Kan complex.</p> <p>Question:</p> <p>Take a bicategory that has adjoints for 1-morphisms, which is one notion of 'weak' groupoid (if all 2-morphisms are isomorphisms, then such a bicategory is a 2-groupoid), and take its nerve.</p> <p>Is there a name for a bisimplicial set arising in this way? Does it have some nice properties? For example, is there a model structure on $\mathbf{ssSet}$ such that these are fibrant?</p> http://mathoverflow.net/questions/55807/inner-hom-and-geometric-realization Inner hom and geometric realization. Alan Wilder 2011-02-18T00:44:55Z 2011-02-18T04:20:14Z <p>I would like to prove the following fact, which I learned from a previous MO question.</p> <p>Let <code>$S_\cdot,T_\cdot\in\mathbf{sSET}$</code> be simplicial sets, and assume that <code>$T_\cdot$</code> is Kan. Then there is a weak equivalence <code>$$|\underline{\mathbf{sSET}} (S_\cdot,T_\cdot)|\simeq \underline{\mathbf{TOP}} (|S_\cdot|,|T_\cdot|)$$</code></p> <p>Here is what I have so far: By a Quillen adjunction, <code>$$\mathbf{TOP} (|\underline{\mathbf{sSET}} (S_\cdot,T_\cdot)|, \underline{\mathbf{TOP}} (|S_\cdot|,|T_\cdot|)) \cong \mathbf{sSET}(\underline{\mathbf{sSET}} (S_\cdot,T_\cdot),\textrm{Sing}\ \underline{\mathbf{TOP}} (|S_\cdot|,|T_\cdot|))$$</code> So we need to find a weak equivalence in the second set. Notice <code>$$\textrm{Sing}\ \underline{\mathbf{TOP}} (|S_\cdot|,|T_\cdot|))_k \cong \mathbf{TOP}(\Delta^k\times |S_\cdot|,|T_\cdot|)$$</code> And by the universal property of coends, there's a map (*) <code>$$|\Delta(k)|\rightarrow \Delta^k$$</code><br> ($\Delta(k)$ is the simplicial set corepresenting $[k]\in\mathbb{\Delta}$), so we get a map to $\mathbf{TOP}(|\Delta(k) \times S_\cdot|,|T_\cdot|)$, which is in turn in bijection with $\mathbf{sSET}(\Delta(k) \times S_\cdot,\textrm{Sing} |T_\cdot|)$, and that is the $k^{\textrm{th}}$ space of <code>$\underline{\mathbf{sSET}}(S_\cdot,\textrm{Sing}\ |T_\cdot|)$</code>. So unraveling, we need to find a weak equivalence <code>$$\underline{\mathbf{sSET}} (S_\cdot,T_\cdot) \rightarrow \underline{\mathbf{sSET}} (S_\cdot, \textrm{Sing} |T_\cdot|)$$</code> We do have the unit map $T_\cdot\rightarrow \textrm{Sing} |T_\cdot|$ of the adjunction in the target, and since all simplicial sets are cofibrant, the result would follow if this unit map is a trivial fibration when <code>$T_\cdot$</code> is fibrant (by compatibility of inner-hom with the model structure in $\mathbf{sSET}$). Here's where I'm stuck; it seems like I'm missing a key ingredient to finish.</p> <p>(*) also here I need to show that these set maps <code>$$\mathbf{TOP}(\Delta^k\times |S_\cdot|,|T_\cdot|) \rightarrow \mathbf{TOP}(|\Delta(k) \times S_\cdot|,|T_\cdot|)$$</code> assemble to a weak equivalence of simplicial sets.</p> http://mathoverflow.net/questions/48237/compatibility-of-classifying-space-with-inner-hom Compatibility of classifying space with inner-hom? Alan Wilder 2010-12-04T00:32:00Z 2010-12-04T03:32:11Z <p>Let $\mathbf{sTop}$ be the functor category $\mathbf{Top}^{{\mathbf{\Delta}}^{\textit{op}}}$, and let $\mathbf{sCat}$ be the functor category $\mathbf{Cat}^{{\mathbf{\Delta}}^{\textit{op}}}$, and let $B:\mathbf{Cat}\rightarrow\mathbf{Top}$ be the classifying space functor (take nerve then realize). How do $B\underline{\mathbf{sCat}}(\mathcal{C},\mathcal{D})$ and $\underline{\mathbf{sTop}}(B\mathcal{C},B\mathcal{D})$ compare? </p> <p>I think they are weakly equivalent (in the Reedy model structure), and I'm hoping that there might be a trivial cofibration between them. Anyone know a reference for something like this? </p> http://mathoverflow.net/questions/68470/mapping-into-a-geometric-realization Comment by Alan Wilder Alan Wilder 2011-06-22T06:50:48Z 2011-06-22T06:50:48Z Yeah that is more or less what I expected. Thanks, Tom. http://mathoverflow.net/questions/68027/embeddings-of-vector-spaces/68030#68030 Comment by Alan Wilder Alan Wilder 2011-06-18T00:38:05Z 2011-06-18T00:38:05Z Thanks for the clarification. To take the last statement further, if I demand that the embeddings preserve a framing of $V$ do I get a homotopy equivalence to $F_k$ with unlabeled points, and so the highly connected as $n\to\infty$ result? http://mathoverflow.net/questions/68027/embeddings-of-vector-spaces Comment by Alan Wilder Alan Wilder 2011-06-17T06:01:12Z 2011-06-17T06:01:12Z No not proper, just smooth. I think the answer below is what I need. http://mathoverflow.net/questions/68027/embeddings-of-vector-spaces Comment by Alan Wilder Alan Wilder 2011-06-17T05:46:56Z 2011-06-17T05:46:56Z Smooth embeddings as manifolds. Proper I'm not sure... http://mathoverflow.net/questions/67083/reference-request-lax-ends Comment by Alan Wilder Alan Wilder 2011-06-08T01:49:21Z 2011-06-08T01:49:21Z Yes as long as one sticks to strict 2-categories and strict functors, the details in proving the lax transformation identity are not too terrible using the obvious definition of lax wedge/lax end. I'll take that as evidence that obvious is right in this case. Still, a reference would be nice. http://mathoverflow.net/questions/67083/reference-request-lax-ends Comment by Alan Wilder Alan Wilder 2011-06-08T00:00:00Z 2011-06-08T00:00:00Z I guess there's an obvious candidate for what a &quot;lax wedge&quot; $c\Rightarrow F$ should be. Maybe the rest is straightforward too... http://mathoverflow.net/questions/67083/reference-request-lax-ends/67092#67092 Comment by Alan Wilder Alan Wilder 2011-06-06T22:55:37Z 2011-06-06T22:55:37Z Thanks, Finn. I'm still holding out hope for a reference in English, but all of the above is helpful. http://mathoverflow.net/questions/66939/pseudofunctors-and-pseudonatural-transformations Comment by Alan Wilder Alan Wilder 2011-06-06T19:57:09Z 2011-06-06T19:57:09Z @David: The Gray tensor product has more data than the cartesian product, which makes the problem worse, i.e. there are &quot;more&quot; p.functors $J\otimes [1]\to\mathcal{C}$ than p.functors $J\times [1]\to\mathcal{C}$, because $J\times [1]$ embeds in $J\otimes [1]$. http://mathoverflow.net/questions/66939/pseudofunctors-and-pseudonatural-transformations/66970#66970 Comment by Alan Wilder Alan Wilder 2011-06-05T19:49:57Z 2011-06-05T19:49:57Z @Martin: I'm not sure what you mean by the middle paragraph, but assuming you meant a(j)o F(g)=&gt;G(g)o a(j), you actually do have this data on both sides. Clearly it is part of the natural transformation data, and for a p.functor $h:J\times [1]\to C$, this isomorphism is produced by applying the compositors of $h$ to the equal factorizations $(j'\to j,\text{id}_1)\circ (\text{id}_j, 0\to 1) = (\text{id}_{j'},0\to 1)\circ (j'\to j, \text{id}_0)$ in the middle we have $h(j'\to j,0\to 1)$, which is the datum that has no counterpart on the p.natural transformation side http://mathoverflow.net/questions/66939/pseudofunctors-and-pseudonatural-transformations/66970#66970 Comment by Alan Wilder Alan Wilder 2011-06-05T18:11:40Z 2011-06-05T18:11:40Z Thanks, Tim, can you go into more detail please? http://mathoverflow.net/questions/61409/using-higher-order-bring-radicals-to-solve-arbitrary-polynomials Comment by Alan Wilder Alan Wilder 2011-04-12T17:59:54Z 2011-04-12T17:59:54Z @JC define the Bring radical to be the polynomial itself. http://mathoverflow.net/questions/59499/end-of-a-weak-equivalence Comment by Alan Wilder Alan Wilder 2011-03-25T21:26:42Z 2011-03-25T21:26:42Z some motivation: Segal space seems to refer interchangeably to a simplicial space with extra conditions or bisimplicial set with extra conditions. This makes me think that the two models are at least weakly equivalent in every way that matters, and so there should be weak equivalences $$|ssSET(S,T)| \sim sTOP(|S|,|T|)$$ (these are inner homs) for $S$, $T$ Segal spaces (of the bisimplicial flavor). Trying to produce this weak equivalence led me to the above question. http://mathoverflow.net/questions/59496/direct-sum-decomposition-of-abelian-groups Comment by Alan Wilder Alan Wilder 2011-03-24T22:34:35Z 2011-03-24T22:34:35Z <a href="http://en.wikipedia.org/wiki/Chinese_remainder_theorem" rel="nofollow">en.wikipedia.org/wiki/Chinese_remainder_theorem</a> http://mathoverflow.net/questions/59252/geometric-realization-on-mathbfstop Comment by Alan Wilder Alan Wilder 2011-03-23T05:51:55Z 2011-03-23T05:51:55Z yeah left. oops, thanks! http://mathoverflow.net/questions/55807/inner-hom-and-geometric-realization/55816#55816 Comment by Alan Wilder Alan Wilder 2011-02-22T23:30:30Z 2011-02-22T23:30:30Z Ahh, that is the piece I was missing. Thank you!