most recent 30 from http://mathoverflow.net 2013-05-18T16:20:15Z http://mathoverflow.net/feeds/user/9187 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/121504/on-duality-on-finite-projective-planes Juan OS 2013-02-11T17:04:17Z 2013-02-12T08:44:05Z

<p>Hey Everyone! In nearly all (if not all) projective geometry texts I have bumped into the following theorem:</p> <p>"Principle of duality: If in a theorem in $\mathfrak{P}$ one switches the word point for the word line and the corresponding incidence relations once again one obtains a theorem of $\mathfrak{P}$."</p> <p>So far so good. Then I found this awesome list by G. Eric Moorhouse: <a href="http://www.uwyo.edu/moorhouse/pub/planes/" rel="nofollow">http://www.uwyo.edu/moorhouse/pub/planes/</a></p> <p>I noted the distinction between a Hall Plane and its dual. So looking a bit into the matter I kept running into the claim "Hall Planes are non-Desarguian and non-self-dual" and the modified version of the principle of duality, which claims the dualized theorem is true <em>on the dual plane</em> (this makes much more sense in my mind). </p> <p>My question is twofold:</p> <ol> <li>How does one prove that Hall Planes are not self dual? I haven't managed to find a proof of this fact!</li> <li>What would be the true duality principle? If duality holds in the dual plane it should not hold in Hall Planes (as they're not self dual), yet all texts I've read claim that duality holds in any projective plane.</li> </ol> <p>Thanks in advance!</p> <p>P.S. Any good references on the concept of duality in projective geometry from a geometrical point of view would be much appreciated!</p>

http://mathoverflow.net/questions/97506/rational-homology-spheres-and-knots Juan OS 2012-05-20T20:45:50Z 2012-05-21T02:15:50Z

<p>It is known that the boundary of a branched double cover of the four ball branched over a surface bounded by a knot in is a rational homology $3$-sphere. Two questions come to my mind simply out of curiosity:</p> <ol> <li>Which rational homology $3$-spheres arise this way? By this I mean, is this set large or small (in the most vague terms, nothing formal) in the set of R.H.3S.'s? Is there an invariant capable of detecting when a RH3S arises this way?</li> <li>If we now substitute usual knots for embeddings $\mathbb{S}^2 \hookrightarrow \mathbb{S}^4$ (knotted spheres), and look at the branched double covers of the $5$ ball branched over a $3$-manifold bounded by the knotted sphere, is their boundary a RH4S? </li> </ol> <p>Many thanks! </p>

http://mathoverflow.net/questions/85212/why-categorify-relating-to-link-knot-homologies Juan OS 2012-01-08T21:10:05Z 2012-01-10T21:19:11Z

<p>Hey Everyone! So I am new blood in the topic of Khovanov Homology and related topics. According to my basic reading the idea is to get the Jones polynomial as the Euler Characteristic of a certain homology theory. My question is, why do this? I mean this in the sense that why is it important/interesting to have this "quantum invariants" $\rightarrow$ "homology theories" transformation? If this question is perhaps to basic and /or unsuited for the site I would appreciate some bibliography on the matter as Khovanov's "A categorification of the Jones polynomial" article doesn't seem to answer this question for me. Thanks!</p>

http://mathoverflow.net/questions/54278/orientation-of-a-glued-manifold Juan OS 2011-02-04T02:04:22Z 2011-02-04T06:13:18Z

<p>Im wondering if there's a short way to prove that when two manifolds with diffeomorphic boundaries are glued together along the boundaries the orientations of these must be inverse to each other. That is to say, suppose you have $M$ and $N$ oriented $n$-dimensional manifolds with $\partial M \cong \partial N$ under a diffeomorphism $\phi: \partial M \to \partial N$, you form $ C = M \cup_\phi N$, in order to do this you need that the orientation of $\partial M$ be opposite to that of $\partial N$, why is that? By homological means...</p> <p>I understand the reason via the orientation of the tangent spaces and the outward-first orientation of the boundaries, but how can i prove it with fundamental classes? I know the homology of the pair $(C,\partial M) \cong (M,\partial M) \oplus (N,\partial N)$ (relative Mayer-Vietoris) and the inclusion $j: (C, \emptyset) \to (C,\partial M)$ induces a monomorphism in the top homology because of the exact sequence of the pair $(C,\partial M)$, and I came up with a "proof" using this, but it is way to lengthy, maybe there is a "quick way" to do this?</p> <p>Thanks</p>

http://mathoverflow.net/questions/42553/morse-theory-and-exotic-spheres Juan OS 2010-10-17T22:55:14Z 2010-10-20T20:00:01Z

<p>Hey everyone! Im finally at the end of Milnor's "On manifolds homeomorphic to the 7-sphere", and i stumbled upon something i cant figure out...</p> <p>For those with the refference im talking about "lemma 5", it goes something like this, you have two $\mathbb{S}^3$ bundles over $\mathbb{S}^4$, we want to obtain the total space of this bundle, so you glue them via the transition function, one can think of this as having a pair of copies of $(\mathbb{R}^4 \setminus {0}) \times \mathbb{S}^3$ and gluing them by identifiying $(u,v) \mapsto (u',v')=(u / \|u\|^2, u^hvu^j/\|u\|)$ where $u$ and $v$ are quaternions, so far so good, now Milnor states that if $h+j =1$ then this manifold is a $7$-sphere, his reason is that the function $f(x) = \mathfrak{R}(v)/(1+\|u\|^2)^{1/2}$ is a morse function, this with the "first" coordinate chart, for the second he defines $u'' = u'(v')^{-1}$ and substitudes $(u',v')$ for $(u'',v')$ stating that the function $f$ is now given by $\mathfrak{R}(u'')/(1+\|u''\|^2)^{1/2}$. He then says "It is easily verified that f has only two critical points (namely $(u,v) = \pm (0,1)$) and that these are nondegenerate".</p> <p>Thats where i get lost, i dont understand his change of coordinates $(u',v') \mapsto (u'',v')$, nor why he states the function is now the one stated... I tried developing the algebra but i cant get it to work out, i tought maybe he was using the involution $v \mapsto v^{-1}$ somehow but it doesnt add up either... Any help is much appreciated! Thanks in advance!</p>

http://mathoverflow.net/questions/42235/rings-with-right-inverses Juan OS 2010-10-15T00:15:53Z 2010-10-15T05:17:06Z

<p>Hey everyone!</p> <p>Lately I remembered an exercise from an algebra class from Jacobson's book: Prove that if an element has more than one right inverse then it has infinitely many, Jacobson attributes this excercise to Kaplansky. Regardless of the solution I began to wonder:</p> <p>Does anybody know any explicit examples of rings that have this property of having elements with infinitely many (or, thanks to Kaplansky, multiple) right inverses? Is the same true for left inverses? I came across an article from the AMS Bulletin that studied this topic but skimming through it I could not find an explicit example, sorry I cant remember the author. Anyways, thanks and good luck!</p>

http://mathoverflow.net/questions/38497/obstruction-cocycles Juan OS 2010-09-12T18:13:32Z 2010-09-13T06:27:32Z

<p>Hey everyone, I was reading about obstruction theory, here's a bit of a summary. Take a cellular space $X$ and a fibre bundle $p:E \to X$ with fiber $F$; consider the problem of extending a section $s$, defined on the $(n-1)$-skeleton over to the $n$-skeleton. We work cell by cell pulling back the bundle via the characteristic map and the section via the restriction of the Char. Map to the boundary of our $n$-cell, since the cell is contractible, the bundle is isomorphic to $D^n \times F$ so the section defines a map from $S^{n-1} \to D^n \times F$ i.e. an element of $\pi_{n-1}(D^n \times F) \cong \pi_{n-1}(F)$. Define the obstruction cochain as the element in $C^n(X^n,\pi_{n-1}(F))$ taking each $n$-cell to the element in the $(n-1)$-homotopy group constructed before.</p> <p>Here's what's bothering me, in Steenrod's book (The topology of Fibre Bundles) he proves that this cochain is actually a cocycle in a really weird way, it looks to me as if he makes no distinction between the homological boundary and the topological boundary of a cell. Roughly he writes the following composition: $ C_{q+1}(X) \stackrel{\partial_*}\to Z_q(X) \stackrel{hurewicz}\to \pi_q(X^q) \stackrel{f=(p_2\circ \sigma)_*}\to \pi_q(F) $</p> <p>And claims that this composition is the value of the obstruction cochain in an $n+1$ cell, how might one verify this?</p> <p>Not being happy with this proof i went and looked at the one Kirk and Davis' book (Lecture notes on algebraic topology) and found it too complex (I know, I dont like anything sorry).</p> <p>What I was wondering is if there was a way to prove this affirmation (the obstruction cochain is a cocycle) directly, i.e. denoting the cochain by $\Theta$ doing something like:</p> <p>$\delta \Theta (e) = \Theta( \partial e) = \Theta (\sum [w_i;e]w_i) = \sum [w_i;e]\Theta(w_i) = \dots = 0$ (where $e$ is a $(n+1)$-cell, $w_i$ is a $n$-cell and $[w_i;e]$ is their incidence number, so that the third term is $\Theta$ evaluated on the cellular (homological) boundary).</p> <p>Any help on the subject or a good refference is very very much appreciated! Thanks and have a great week.</p>

http://mathoverflow.net/questions/121504/on-duality-on-finite-projective-planes/121533#121533 Juan OS 2013-02-11T23:22:40Z 2013-02-11T23:22:40Z

Thanks Chris! I will look into your reference, hopefully I won't get confused all over again. Regarding my statement of the principle of duality I don't think it's wrong (though, granted, I didn't add the meaning of $\mathfrak{P}$, rather, as Andreas said, I mistook the model for the theory. Thanks for your help!

http://mathoverflow.net/questions/121504/on-duality-on-finite-projective-planes/121518#121518 Juan OS 2013-02-11T23:19:29Z 2013-02-11T23:19:29Z

You are completely right Andreas, the symbol refers to the theory not the model. I foolishly interpreted duality as &quot;a projective plane is isomorphic to its dual&quot; which is quite naive in retrospect. Thanks for your help!

http://mathoverflow.net/questions/97506/rational-homology-spheres-and-knots/97519#97519 Juan OS 2012-05-21T10:09:16Z 2012-05-21T10:09:16Z

Thanks a lot, Daniel! Do you have any idea on a reference (or any thoughts!) for the situation in 4 dimensions?

http://mathoverflow.net/questions/97506/rational-homology-spheres-and-knots Juan OS 2012-05-21T10:07:22Z 2012-05-21T10:07:22Z

Hi Agol, I phrased the question that way because I was trying to connect this somehow to the &quot;differential&quot; part of the topology. From this point of view the DBCs on the four ball are relevant as from them arises a bound to the slice genus by the knot signature. I am not sure if this is good enough reason for you, if not I can always edit the question.

http://mathoverflow.net/questions/85212/why-categorify-relating-to-link-knot-homologies/85215#85215 Juan OS 2012-01-08T22:13:13Z 2012-01-08T22:13:13Z

Thanks for your answer Francesco! Do you know if this was the original motivation for the categorification or a side product? Can you please give me a reference regarding knots with same Jones poly different Khovanov homologies, please?

http://mathoverflow.net/questions/54278/orientation-of-a-glued-manifold Juan OS 2011-02-04T02:24:04Z 2011-02-04T02:24:04Z

Thank you Ryan, you're correct and that is precisely what I want to know, I guess I wrote everything wrongly. How can you see that the orientation of $C = M \cup_\phi N$ induces opposite orientations on $M$ and $N$?

http://mathoverflow.net/questions/42553/morse-theory-and-exotic-spheres/42930#42930 Juan OS 2010-10-21T06:04:14Z 2010-10-21T06:04:14Z

In fact, if $h+j \neq \pm 1$ then the total space is not homeomorphic to the sphere, so the Morse function cannot have only two critical points! Thanks again!

http://mathoverflow.net/questions/42553/morse-theory-and-exotic-spheres Juan OS 2010-10-21T06:02:57Z 2010-10-21T06:02:57Z

I believe the answer to my question is well outlined by Greg; Milnor is defining the function via the coordinate charts, and $u''$ is not a coordinate change but rather a way to simplify notation. Adressing the remarks Greg makes about how Milnor came up with this function, I believe you can get a more general picture by reading the latter article &quot;Differentiable structures on spheres&quot;, where Milnor uses rotation groups to construct higher dimensional examples.

http://mathoverflow.net/questions/42553/morse-theory-and-exotic-spheres/42930#42930 Juan OS 2010-10-21T05:59:47Z 2010-10-21T05:59:47Z

Thanks Greg, I realized this a couple of hours ago, conjugation is the key to it all, however, the Morse function doesnt piece together in quite the same way for all bundles, the fact that $h+j=1$ is crucial for in order for the function to be well defined, since otherwise the real part of $v$ would not be preserved!

http://mathoverflow.net/questions/42235/rings-with-right-inverses/42246#42246 Juan OS 2010-10-17T02:19:20Z 2010-10-17T02:19:20Z

Thanx Pete! Your example is very concrete.

http://mathoverflow.net/questions/42235/rings-with-right-inverses/42237#42237 Juan OS 2010-10-17T02:18:51Z 2010-10-17T02:18:51Z

True, that is why I chose your answer, it's more general thanks!

http://mathoverflow.net/questions/38497/obstruction-cocycles Juan OS 2010-09-13T02:54:39Z 2010-09-13T02:54:39Z

@Tim Oops, now I get your comments, you're right, I wrote &quot;vector bundle&quot; instead of fibre bundle! Sorry, my bad! Still the answers have helped me greatly.

http://mathoverflow.net/questions/38497/obstruction-cocycles/38513#38513 Juan OS 2010-09-12T21:58:38Z 2010-09-12T21:58:38Z

You are correct about the definition of the cell complex, however my problem is that Steenrod appears to be stating that the composite of the maps in the sequence I wrote is the value of the obs. cochain, and I feel in doing so he is stating something like &quot;the value of the map in the (topological) boundary is the same as its value in the (homological) boundary&quot;, I'll try to reread the proof from the first definition you gave of the cell complex. Thanks.

http://mathoverflow.net/questions/38497/obstruction-cocycles/38510#38510 Juan OS 2010-09-12T21:54:58Z 2010-09-12T21:54:58Z

Thank you for the links, I will look into them.

http://mathoverflow.net/questions/38497/obstruction-cocycles Juan OS 2010-09-12T21:03:22Z 2010-09-12T21:03:22Z

@Ryan, no I have not looked at Whiteheads book, I will try and find it, can you give me the title please? @Tim, I dont see where do the homotopy groups or $\mathbb{R}^n$ come in, theyre always 0 so there are no obstructions there... What i meant by &quot;roughly&quot; is that I wasn't going to write the whole of Steenrod's proof, if you read his book you'll see why I wrote the Hurewicz map that way (I may be bad at AT but Im not that bad) and about q and n, I did mess up there, but the meaning gets through I believe. Sorry for my bad english...