User lethe - MathOverflowmost recent 30 from http://mathoverflow.net2013-05-26T02:14:52Zhttp://mathoverflow.net/feeds/user/10333http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/81811/contractible-noncompact-3-manifold-without-boundary-not-homeomorphic-to-bbb-r3Contractible noncompact 3-manifold without boundary not homeomorphic to $\Bbb R^3$lethe2011-11-24T15:19:11Z2011-11-25T01:18:48Z
<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 $V=W$?lethe2011-10-15T07:34:13Z2011-10-15T23:30:14Z
<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-claHow does all of the bundles over a certain manifold characterize the homotopy class of the base manifold?lethe2011-04-20T04:30:42Z2011-04-21T07:36:56Z
<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-pointCan a smooth, immersed loop in R^2 become not nullhomotopic by removing a point?lethe2011-03-11T09:40:22Z2011-03-12T10:33:15Z
<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-isCan we deduce that two rings R1 and R2 are isomorphic if their polynomial ring is iso.?lethe2010-12-08T13:39:58Z2010-12-08T14:11:11Z
<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-singularityComment by lethelethe2013-04-09T02:45:06Z2013-04-09T02:45:06ZIf $S=spec(k)$, $X$ 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-varietyComment by lethelethe2013-04-08T06:22:08Z2013-04-08T06:22:08ZSince 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#110541Comment by lethelethe2012-12-10T20:45:59Z2012-12-10T20:45:59ZShould $C_1, C_2$ 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-r3Comment by lethelethe2011-11-25T01:17:39Z2011-11-25T01:17:39ZSorry! I'm looking for a modern explanation about this example, but I wasn't so sure about whether the language is already "modern" in the original paper. Now it seems to be already settled...I didn't expect it can be just called "whitehead manifold"... Thanks all of you!http://mathoverflow.net/questions/81811/contractible-noncompact-3-manifold-without-boundary-not-homeomorphic-to-bbb-r3/81814#81814Comment by lethelethe2011-11-25T01:16:37Z2011-11-25T01:16:37ZI 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#78198Comment by lethelethe2011-10-15T11:51:02Z2011-10-15T11:51:02ZWOW, that's amazing!http://mathoverflow.net/questions/68732/what-to-write-for-an-undergraduate-thesisComment by lethelethe2011-06-24T13:57:35Z2011-06-24T13:57:35ZThanks!
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-thesisComment by lethelethe2011-06-24T13:29:32Z2011-06-24T13:29:32ZIndeed. I forgot that site. Then close it.http://mathoverflow.net/questions/66152/proof-that-higher-cech-cohomology-groups-vanish-for-fine-sheavesComment by lethelethe2011-05-27T06:27:13Z2011-05-27T06:27:13Zmiss a "$" in the third paragraph? It doesn't display correctly here...And I think in order to imitate the partition of unity, G&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#62411Comment by lethelethe2011-04-21T09:14:20Z2011-04-21T09:14:20ZI'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#62407Comment by lethelethe2011-04-21T01:52:53Z2011-04-21T01:52:53ZThanks 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#62411Comment by lethelethe2011-04-21T01:47:41Z2011-04-21T01:47:41ZWhy 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#62373Comment by lethelethe2011-04-20T11:00:51Z2011-04-20T11:00:51ZWhy 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-claComment by lethelethe2011-04-20T04:53:02Z2011-04-20T04:53:02ZTo 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-claComment by lethelethe2011-04-20T04:39:02Z2011-04-20T04:39:02ZAhh, indeed, my question was vague. Actually I'm merely curious about how to derive some information about bundles. Let me edit it.