User jinhyun park - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T03:00:54Z http://mathoverflow.net/feeds/user/3168 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/116975/are-filtered-colimits-of-weak-equivalences-of-spectra-again-weak-equivalences Are filtered colimits of weak-equivalences of spectra again weak-equivalences? Jinhyun Park 2012-12-21T14:56:20Z 2012-12-21T19:15:12Z <p>Hi, I have a question on weak-equivalences of spectra.</p> <p>More precisely, I wonder whether filtered colimits of weak-equivalences of spectra are again weak-equivalences of spectra. Here, spectra are in the sense of Bousfield-Friedlander, i.e. a sequence of pointed simplicial sets $(E_0, E_1, \cdots, )$ with the morphisms $\sigma_n : S^1 \wedge E_n \to E_{n+1}$ for all $n \geq 0$. Of course, weak-equivalences of spectra are by definition the stable weak-equivalences.</p> <p>After some efforts, I was able to find from a webpage that, if the spectra in question are all Kan spectra (spectra whose all levels are Kan simplicial sets), then this is indeed true. But, I have no idea how to proceed in general, partly because I do not think spectra are cofibrant in general.</p> <p>I was informally told by someone that, in general, if we are working with a combinatorial'' model category, then such questions are likely be true. But, is it true that the category of spectra is combinatorial, and if that is the case, then does the main question hold?</p> <p>I am fine with working with only Kan spectra if this question turns out to be negative, but since I didn't do my basic graduate school studies in algebraic topology, I hoped to know the answers for this question first for I thought the question might be easy for experts in the field. Thank you.</p> http://mathoverflow.net/questions/96067/equalizers-for-morphisms-of-connected-varieties-with-marked-points Equalizers for morphisms of connected varieties with marked points Jinhyun Park 2012-05-05T13:24:06Z 2012-05-05T13:24:06Z <p>I recently began to study some aspects of varieties with marked points, and I tried to understand their categorical collective behaviors.</p> <p>Here is one issue that I was not quite able to see immediately, and I wonder if someone could give me appropriate advice. For simplicity, let $(U, p,q), (V, r,s)$ be two connected varieties with two marked points over a field (nice enough, say). Suppose $p \not = q$, $r \not = s$.</p> <p>Question : Let $f, g : (U, p, q) \to (V, r,s)$ be two morphisms of varieties $U \to V$ that respect the marked points. Then, can we find a <em>connected</em> variety with two marketed points $(W, t, u)$ with $h : (W, t, u) \to (U, p,q)$ that equalizes $f, g$, in other words, $f \circ h = g \circ h$?</p> <p>Here, connectedness for $W$ is very important: if not, then I can simply take $(W, t, u) = (p \coprod q, p, q)$, and the inclusion map from $W$ to $U$ will does the job. </p> <p>If there is only one marked point for each variety, say $f, g : (U, p) \to (V, r)$, then the corresponding question is fairly easy. We can take $(W, t) = (p, p)$.</p> <p>So, this question looks nontrivial when one has more than one marked point. </p> <p>I guess when the number of marked points increases, there will be fewer morphisms (even none I guess) so that maybe one can have a nice way to answer it, but I do not know what attempts can be made to it.</p> <p>Does someone have nice suggestions to try?</p> http://mathoverflow.net/questions/4802/what-do-higher-chow-groups-mean/12764#12764 Answer by Jinhyun Park for What do higher Chow groups mean? Jinhyun Park 2010-01-23T19:49:27Z 2010-09-07T14:19:32Z <p>Benjamin Antineau's answer needs a minor correction. For $X= spec (k)$, we have $CH^n (X, n) = K^M _n (k)$, not $CH^{2n} (X, n)$. Indeed not too many is known, but it is a very interesting subject (at least for me) to pursue. A good start would be Burt Totaro's paper 'Milnor K-theory is the simplest part of K-theory' or something similarly titled, where you can find the cubical version of it. Using $\mathbb{A}^1$-invariance with some spectral sequence arguments, one can prove that the above "simplicial version" and "cubical version" are isomorphic, thus equivalent.</p> <p>Going back to Peter Arndt's question about 'intuition', the easiest one would be to look it as an algebro-geometric version of singular homology theory.</p> <p>For instance, when $X$ is a topological space, a singular $n$-simplex is given by a continuous map $s: \Delta ^n \to X$. We collect their formal finite sums over the integers, and apply some simplicial formalisms. That's how we get the singular complex.</p> <p>When $X$ is a variety, the problem is bad, even if we take $\Delta^n$ to be the algebraic n-simplex. One problem would be that there aren't enough morphisms of varieties $s: \Delta^n \to X$ to begin with. So, a way out is to take all "correspondences", i.e. closed subvarieties in the product space $\Delta^n \times X$. One problem that still persists here is that, to be able to apply the simplicial formalism, one has to have a good intersection property of correspondences with the faces of $\Delta^n$, but by taking all algebraic cycles, one may not get it. Consequently, we put conditions such as proper intersection with all faces. </p> <p>That's why we define things in this way.</p> <p>Regarding the question of what kernel/image does: it is difficult to explain everything, but the easiest case might worth paying attention: for instance, $z^i (X, 0)$ is the codimension i algebraic cycles on $X$, and the boundary map $z^i (X, 1) \to z^i (X, 0)$ by definition gives the rational equivalence of cycles on $X$. In this way, from the cokernel for instance, we recover the Chow group. </p> http://mathoverflow.net/questions/21816/proving-existence-of-non-special-divisors-of-a-given-degree-d-on-compact-riemann Proving existence of non-special divisors of a given degree d on compact Riemann surfaces Jinhyun Park 2010-04-19T05:24:36Z 2010-04-19T05:51:33Z <p>I have a simple question. Let $C$ be a compact Riemann surface of genus, say $g >= 2$, to avoid silly cases.</p> <p>I think it should be true, but I want to prove the following concretely: </p> <p>"there exists a divisor $D$ on $C$ of degree $g-1$, that is non-special."</p> <p>(For those who do not know what special divisors are: a divisor is called special if it has $h^0 (D) >0$ and $h^1 (D) >0$.)</p> <p>Notice that by the Riemann-Roch, for this degree $g-1$ case we immediately have $h^0 (D) = h^1 (D) = 0$. This is, in fact, equivalent to $D$ being non-special, when $\deg D = g-1$.</p> <p>Is there an interesting (or any) way to prove this? I believe it should be fairly easy, and maybe I am very dumb so that I can't immediately produce a proof.</p> <p>More generally, if this is possible, if the degree is a given $d$, when do we see that there exists a non-special or special divisor of given degree $d$ on a given compact Riemann surface?</p> http://mathoverflow.net/questions/21018/what-is-known-about-finite-morphisms-from-x-to-the-projective-line/21085#21085 Answer by Jinhyun Park for What is known about finite morphisms from X to the projective line Jinhyun Park 2010-04-12T09:18:08Z 2010-04-12T09:18:08Z <p>The question seems a bit vague, so let me also add one in addition to answers by Maharana, Kevin Buzzard and Qing Liu. </p> <p>One simple but the most useful result for this kind of situation is the Riemann-Hurwitz formula, that relates the genus of X, genus of P^1 and the ramification points. See Hartshorne's chapter 4, or probably R. Narasimhan's book on Compact Riemann surfaces.</p> <p>I encourage Ariyan to look at the general Riemann-Hurwitz formula for any nonconstant holomorphic maps from X to Y, where X and Y are both compact Riemann surfaces, and look at various consequences of this theorem.</p> http://mathoverflow.net/questions/12767/when-two-k-varieties-with-the-same-underlying-topological-spaces-isomorphic When two k-varieties with the same underlying topological spaces isomorphic? Jinhyun Park 2010-01-23T20:18:44Z 2010-01-24T15:08:16Z <p>I have a little problem. I'm probably being just so careless..... Here k-varieties are all integral separated k-schemes of finite type over k, where k is a field.</p> <p>Suppose $X, Y$ are $k$-varieties, and let $f :X \to Y$ be a morphism of $k$-varieties that is one to one and onto. Then, when can we say this $f$ is an isomorphism of $k$-varieties?</p> <p>If this is too vague, let me add that the case I would like to see is when each fibre of $f$ (which is a singleton) is reduced. Under this assumption, would this give an isomorphism?</p> http://mathoverflow.net/questions/12762/proving-if-fibres-are-reduced-or-not Proving if fibres are reduced or not. Jinhyun Park 2010-01-23T19:23:54Z 2010-01-23T23:46:11Z <p>Suppose X, Y, Z are k-varieties and $f: X \to Z$ factors through $f': X \to Y$ and $g: Y \to Z$. Suppose all of f, f', g are surjective. Assume that for $z \in Z$, the fibre $f^{-1} (z)$ is reduced. Then, is the fibre $g^{-1} (z)$ always reduced? </p> <p>If not, when will it be true?</p> http://mathoverflow.net/questions/116975/are-filtered-colimits-of-weak-equivalences-of-spectra-again-weak-equivalences/116987#116987 Comment by Jinhyun Park Jinhyun Park 2012-12-30T17:40:56Z 2012-12-30T17:40:56Z Oh Thanks. This is very helpful. http://mathoverflow.net/questions/116975/are-filtered-colimits-of-weak-equivalences-of-spectra-again-weak-equivalences/116987#116987 Comment by Jinhyun Park Jinhyun Park 2012-12-22T03:58:55Z 2012-12-22T03:58:55Z The above $\alpha_0$ should be $\aleph_0$. I don't know how to correct a comment.. Sorry. http://mathoverflow.net/questions/116975/are-filtered-colimits-of-weak-equivalences-of-spectra-again-weak-equivalences/116987#116987 Comment by Jinhyun Park Jinhyun Park 2012-12-22T03:57:26Z 2012-12-22T03:57:26Z I have one more question. From the references you mentioned, I found all informations I needed, except that the category of Bousfield-Friedlander spectra is locally $\alpha_0$-presentable, etc. Is this a standard fact, or an easy to prove statement? Or if there is a reference would you mind letting me know? http://mathoverflow.net/questions/116975/are-filtered-colimits-of-weak-equivalences-of-spectra-again-weak-equivalences/116987#116987 Comment by Jinhyun Park Jinhyun Park 2012-12-22T03:15:25Z 2012-12-22T03:15:25Z Thank you for your answer. http://mathoverflow.net/questions/96067/equalizers-for-morphisms-of-connected-varieties-with-marked-points Comment by Jinhyun Park Jinhyun Park 2012-05-06T14:18:36Z 2012-05-06T14:18:36Z Thank you. I think I was a bit careless in posting the question. Indeed, I was expecting too much here, and there seem to be easy counter examples. http://mathoverflow.net/questions/33556/do-homotopy-colimits-always-commute-with-homotopy-colimits/33559#33559 Comment by Jinhyun Park Jinhyun Park 2012-03-02T09:34:10Z 2012-03-02T09:34:10Z Ah, it was a few years ago when the answer was posted, but it turns out this is exactly what I was looking for right now. Thank you! For those who want to use this in their papers, let me just leave a remark here that it was published as part of Memoir of AMS, Vol 155, No. 736 (2002). http://mathoverflow.net/questions/21816/proving-existence-of-non-special-divisors-of-a-given-degree-d-on-compact-riemann/21817#21817 Comment by Jinhyun Park Jinhyun Park 2010-04-19T11:36:05Z 2010-04-19T11:36:05Z Ah, thank you very much! There was a very easy way! http://mathoverflow.net/questions/12762/proving-if-fibres-are-reduced-or-not Comment by Jinhyun Park Jinhyun Park 2010-01-23T19:53:06Z 2010-01-23T19:53:06Z Sorry for being imprecise. Here $k$-varieties mean, integral $k$-schemes of finite type over $k$, where the middle scheme isn't probably fitting into the case. Do you know any other example where X, Y, Z are all k-varieties?