User lethe - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T02:14:52Z http://mathoverflow.net/feeds/user/10333 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/81811/contractible-noncompact-3-manifold-without-boundary-not-homeomorphic-to-bbb-r3 Contractible noncompact 3-manifold without boundary not homeomorphic to $\Bbb R^3$ lethe 2011-11-24T15:19:11Z 2011-11-25T01:18:48Z <p>I heard this example was given in Whitehead's paper A CERTAIN OPEN MANIFOLD WHOSE GROUP IS UNITY.( <a href="http://qjmath.oxfordjournals.org/content/os-6/1/268.full.pdf" rel="nofollow">http://qjmath.oxfordjournals.org/content/os-6/1/268.full.pdf</a> ) But I was confused by his term. Thus I'm looking for an explanation in more standard terms about this example.</p> <p>But since my aim is to know about an example of this kind, any alternative will do either.</p> http://mathoverflow.net/questions/78194/v-w-are-varieties-does-v-times-mathbfp1w-times-mathbfp1-imply-v $V$, $W$ are varieties. Does $V\times \mathbf{P}^1=W\times \mathbf{P}^1$ imply $V=W$? lethe 2011-10-15T07:34:13Z 2011-10-15T23:30:14Z <p>If $\mathbf{P}^1$ is replaced by the affine line $\mathbf{A}^1$, this becomes the cancellation problem, and we have a pair of famous Danielewski surfaces ($xy=1-z^2$ and $x^2y=1-z^2$) as a counterexample (though I'm still seeking how to prove that..). I suppose in my case this counterexample might no longer work. </p> <p>Also one may replace $\mathbf{P}^1$ by $\mathbf{P}^n$ or other fixed varieties, or ask again after imposing some conditions on $V$ and $W$ (for example dimensions) if there are counterexamples for my question. And more wildly I may ask for what kind of family $X_n$, we will have the result that $V\times X_n=W\times X_n$ implies $V=W$. Any result of these kind of variations of the problem is also welcomed.</p> http://mathoverflow.net/questions/62368/how-does-all-of-the-bundles-over-a-certain-manifold-characterize-the-homotopy-cla How does all of the bundles over a certain manifold characterize the homotopy class of the base manifold? lethe 2011-04-20T04:30:42Z 2011-04-21T07:36:56Z <p>It is known that if $f:M\rightarrow N$ is a homotopy equivalent, then the the process of pullback gives a one-one correspondence between bundles over $N$ and $M$ up to isomorphism. Is the converse( that if $f$ gives a 1-1 correspondence between them, then $f$ is a homotopy equivalence)true? Or any counterexample?</p> <p>By the way, are bundles of a fixed rank over a compact manifold finite up to isomorphism? And is there any characterization of them like the holomorphic line bundles as a first cohomology group of a certain sheaf? If so, is there any way to calculate?</p> <hr> <p>$M,N$ are smooth manifolds, and I originally thought of real vector bundles. But since this is a wild guess, it might be better to think the most general bundles you could imagine that homotopy invariance holds, even though I only know the real case.</p> <p>It seems algori's answer apply to all kind of vector bundles I can imagine, though a bit beyond me. Thanx very much. And any reference for your arguments? </p> <p>And I'd like to know whether homotopy invariance holds in fiber bundles.</p> <hr> <p>Thanks all of you. But formally whose answer should I accept?</p> http://mathoverflow.net/questions/58137/can-a-smooth-immersed-loop-in-r2-become-not-nullhomotopic-by-removing-a-point Can a smooth, immersed loop in R^2 become not nullhomotopic by removing a point? lethe 2011-03-11T09:40:22Z 2011-03-12T10:33:15Z <p>ATT</p> <p>More precisely, let $\gamma :S^1\rightarrow R^2$ be a smooth immersed loop, the question is whether it is true that there is a point $p\in R^2-\gamma(S^1)$ such that $\gamma$ is not homotopic to constant map.</p> <p>Actually I'm not sure whether I choose the right tag. Tell me if I choose wrongly.</p> <p>I hope it won't turn out to be trivial.</p> <p>(Does the tex turn out all right? I don't seem to have the plug-in to display it.)</p> http://mathoverflow.net/questions/48643/can-we-deduce-that-two-rings-r1-and-r2-are-isomorphic-if-their-polynomial-ring-is Can we deduce that two rings R1 and R2 are isomorphic if their polynomial ring is iso.? lethe 2010-12-08T13:39:58Z 2010-12-08T14:11:11Z <p>I will state it again. Given two rings R_1 and R_2 (with or without identity. It's not specified.). If R_1[x] is isomorphic to R_2[y] (No such requirement that the isomorphism sends the constant terms to constant terms), can we deduce that R_1 iso. to R_2?</p> <p>I feel there might be a counterexample but it's quite hard to find one.</p> http://mathoverflow.net/questions/126928/relative-resolution-of-singularity Comment by lethe lethe 2013-04-09T02:45:06Z 2013-04-09T02:45:06Z If $S=spec(k)$, $X$ is a singular variety, I don't think you can use base change to resolve that. http://mathoverflow.net/questions/126814/constructing-a-projection-onto-a-variety Comment by lethe lethe 2013-04-08T06:22:08Z 2013-04-08T06:22:08Z Since you said projection, do you require the surjective map restricts to identity on $V$? http://mathoverflow.net/questions/110444/determining-homotopy-classes-t2-rp2/110541#110541 Comment by lethe lethe 2012-12-10T20:45:59Z 2012-12-10T20:45:59Z Should $C_1, C_2$ and $D_1, D_2$ be flipped in the commutative square? http://mathoverflow.net/questions/81811/contractible-noncompact-3-manifold-without-boundary-not-homeomorphic-to-bbb-r3 Comment by lethe lethe 2011-11-25T01:17:39Z 2011-11-25T01:17:39Z Sorry! I'm looking for a modern explanation about this example, but I wasn't so sure about whether the language is already &quot;modern&quot; in the original paper. Now it seems to be already settled...I didn't expect it can be just called &quot;whitehead manifold&quot;... Thanks all of you! http://mathoverflow.net/questions/81811/contractible-noncompact-3-manifold-without-boundary-not-homeomorphic-to-bbb-r3/81814#81814 Comment by lethe lethe 2011-11-25T01:16:37Z 2011-11-25T01:16:37Z I didn't know it is given such a name. Thanks! If I had been able to accept both of the answers. http://mathoverflow.net/questions/78194/v-w-are-varieties-does-v-times-mathbfp1w-times-mathbfp1-imply-v/78198#78198 Comment by lethe lethe 2011-10-15T11:51:02Z 2011-10-15T11:51:02Z WOW, that's amazing! http://mathoverflow.net/questions/68732/what-to-write-for-an-undergraduate-thesis Comment by lethe lethe 2011-06-24T13:57:35Z 2011-06-24T13:57:35Z Thanks! But I'm a little wonder what is meant by independent study. Ask my own questions and answer myself? That's a good idea but a bit hard. http://mathoverflow.net/questions/68732/what-to-write-for-an-undergraduate-thesis Comment by lethe lethe 2011-06-24T13:29:32Z 2011-06-24T13:29:32Z Indeed. I forgot that site. Then close it. http://mathoverflow.net/questions/66152/proof-that-higher-cech-cohomology-groups-vanish-for-fine-sheaves Comment by lethe lethe 2011-05-27T06:27:13Z 2011-05-27T06:27:13Z miss a &quot;$&quot; in the third paragraph? It doesn't display correctly here...And I think in order to imitate the partition of unity, G&amp;H abuses the language here. Probably you need a reference for a more detailed version about these. I learned these stuff from Voisin's hodge theory and complex algebraic geometry. http://mathoverflow.net/questions/62368/how-does-all-of-the-bundles-over-a-certain-manifold-characterize-the-homotopy-cla/62411#62411 Comment by lethe lethe 2011-04-21T09:14:20Z 2011-04-21T09:14:20Z I'm not able to judge. But if it is said to be right, I will accept this. I hope after finishing Hatcher and some references I would be able to read it. http://mathoverflow.net/questions/62368/how-does-all-of-the-bundles-over-a-certain-manifold-characterize-the-homotopy-cla/62407#62407 Comment by lethe lethe 2011-04-21T01:52:53Z 2011-04-21T01:52:53Z Thanks very much! I start to figure out why all of these answers work. Since all of the answers are of great help, I'm quite uncertain whose answer to accept, though it's merely formal. http://mathoverflow.net/questions/62368/how-does-all-of-the-bundles-over-a-certain-manifold-characterize-the-homotopy-cla/62411#62411 Comment by lethe lethe 2011-04-21T01:47:41Z 2011-04-21T01:47:41Z Why does a homology equivalence plus a$\pi_1$lead to weak homotopy equivalence? http://mathoverflow.net/questions/62368/how-does-all-of-the-bundles-over-a-certain-manifold-characterize-the-homotopy-cla/62373#62373 Comment by lethe lethe 2011-04-20T11:00:51Z 2011-04-20T11:00:51Z Why does these homotopy group calculation conclude all the bundles are trivial? Sorry I didn't even finish learning Hatcher's Algebraic Topology. And why$\pi_2(G_0)=0$? At the beginning I meant vector bundles of finite rank with structure group$GL(n,\mathbb R)\$ because I only know the homotopy invariance in this case. But when I realized I didn't specify what kind of bundles, I decided not to change it since such restriction seems unnatural. Thus, is it possible to extend the statement to fiber bundles? I mean, does homotopy invariance still hold in fiber bundles? http://mathoverflow.net/questions/62368/how-does-all-of-the-bundles-over-a-certain-manifold-characterize-the-homotopy-cla Comment by lethe lethe 2011-04-20T04:53:02Z 2011-04-20T04:53:02Z To budney, I mean the map induces the 1-1 correspondence between all bundles, not the induced map between a particular bundle and its pullback. Btw, why do all orientable 3 mfd have trivial tangent bundle? http://mathoverflow.net/questions/62368/how-does-all-of-the-bundles-over-a-certain-manifold-characterize-the-homotopy-cla Comment by lethe lethe 2011-04-20T04:39:02Z 2011-04-20T04:39:02Z Ahh, indeed, my question was vague. Actually I'm merely curious about how to derive some information about bundles. Let me edit it.