User chris gerig - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T18:21:23Z http://mathoverflow.net/feeds/user/12310 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109848/pathological-examples-of-dimension Pathological Examples of Dimension Chris Gerig 2012-10-16T20:42:13Z 2013-05-03T04:42:50Z <p>I am trying to wrap my head around all the different notions of dimension (and their equivalences). To get a sense of this, it would be nice to know the subtle difficulties that arise. I thus think it would be nice to have a list of such examples! (I dug through the internet without locating such a collection.)</p> <p>This question/request can be interpreted as either<br> 1) An example that obeys a particular definition of dimension but goes against our intuition. Said differently: an example that <em>should</em> obey a particular definition of dimension, but doesn't.<br> 3) An example that disagrees with two different definitions of dimension.<br> 4) An example which hinges on a hypothesis of the dimension. </p> <p>*This last one is what got me to start this post, because I came across an example involving the Krull dimension: If our ring $R$ is Noetherian then $\dim R[x]=1+\dim R$, but if $R$ is not Noetherian then we can have $\dim R[x]=2+\dim R$. Found at <a href="http://www.jstor.org/stable/2373549?origin=crossref" rel="nofollow">http://www.jstor.org/stable/2373549?origin=crossref</a> (The Dimension Sequence of a Commutative Ring, by Gilmer).<br> *I am not sure where our space-filling curves fit in here.</p> <p><strong>Some standard definitions of dimension</strong></p> <ul> <li>Lebesgue covering dimension (of a topological space)</li> <li>Cohomological dimension (of a topological space)</li> <li>Hausdorff dimension (of a metric space)</li> <li>Krull dimension (of a ring or module)</li> </ul> http://mathoverflow.net/questions/129174/vector-fields-on-4n1-spheres Vector fields on $(4n+1)$-spheres Chris Gerig 2013-04-30T02:17:02Z 2013-05-01T01:19:05Z <p>If $n$ is odd then $S^{n-1}$ doesn't admit a nowhere-vanishing vector field, and if $n$ is even then there does exist one (Hairy Ball Theorem). We can then ask, on $S^{n-1}$, <em>what is the maximum number $k(n)$ of linearly independent vector fields?</em> Rewriting $n=2^{4a+b}(2s+1)$, Adams computes $k(n)=2^b+8a-1$.</p> <p>In particular, on the $(4n+1)$-spheres, <em>there is only one nowhere-vanishing vector field up to linear-independence</em>, whereas in every other (odd) dimension there are more.<br> <strong>Example:</strong> on the circle $S^1$ there are the vector fields generated by (counter)clockwise rotation, but these are the same up to a scalar. This makes sense: I start flowing along this single dimension and then I have to continue flowing in that direction until I come back to my starting point.<br> I tried considering the difference between $S^3$ and $S^5$, which fiber over $\mathbb{C}P^1$ and $\mathbb{C}P^2$ respectively. A nowhere-vanishing vector field in both cases is given by taking the standard nowhere-vanishing vector field on the $S^1$-fiber. But for $S^3$ there are three linearly-independent fields (the $i,j,k$-directions when representing $S^3$ as the unit quaternions <em>-- is there a way to see this using the fibration picture?</em>), whereas for some reason $S^5$ can only admit the one.</p> <p><strong>What can be the differential/topological reasoning behind this? I.e. is there a down-to-earth way to deduce this result on $S^{4n+1}$, or for starters, $S^5$?</strong></p> <p>Could there possibly be an analogous index theorem going on here, in the same way that the Poincare-Hopf theorem provides us the Hairy Ball result?</p> http://mathoverflow.net/questions/129174/vector-fields-on-4n1-spheres/129274#129274 Answer by Chris Gerig for Vector fields on $(4n+1)$-spheres Chris Gerig 2013-05-01T01:19:05Z 2013-05-01T01:19:05Z <p><em>Thanks Misha for the reference! (Just rewriting it here to complete this thread).</em><br> It seems that an off-the-cuff calculation (what I refer to as "down-to-earth") probably isn't going to suffice; there is some intricate stuff going on in the proofs involving the homotopy groups of the rotation groups of the spheres and orthogonal frames.</p> <p><strong>Homotopy Properties of the Real Orthogonal Groups</strong> (Whitehead), 1941, which focuses precisely on our case in question. The proof comes down to the obstruction of certain maps induced from rotation groups and their fixed subspaces on spheres.</p> <p><strong>Vector Fields on the n-Sphere</strong> (Steenrod, Whitehead) 1950, which studies particular fibrations of Stiefel manifolds over the spheres. It proves that for $n=2^k(2m+1)-1$ we have $k(n)&lt;2^k$, which includes our case in question.</p> http://mathoverflow.net/questions/128153/a-basic-question-related-to-hamiltonian-isotopy-in-symplectic-geometry/128157#128157 Answer by Chris Gerig for A basic question related to Hamiltonian isotopy in symplectic geometry Chris Gerig 2013-04-20T07:42:09Z 2013-04-20T07:42:09Z <p>Why not look at it in specific examples? Say the height function $H$ on the sphere $S^2\subset\mathbb{R}^3$. Or how it doesn't work for the obvious rotations on the torus.</p> <p>The <em>canonical</em> context in which these are used is <em>Classical Mechanics</em> on cotangent bundles... Hamilton equations! In other words, the Hamiltonian is the energy of your system, and your symplectomorphisms describe the physical motion in phase space.</p> <p>In terms of visualization: Note that the vector field $X_t$ preserves $H_t$ (using Lie derivative, noting that $\omega$ is skew-commutative), and so the integral curves $\lbrace\phi_t(x)\rbrace_{t\in\mathbb{R}}$ are contained in level sets of $H$, i.e. $\phi^*_tH_t=H_t$.</p> http://mathoverflow.net/questions/105951/finite-dimensional-mountain-pass-lemma/128037#128037 Answer by Chris Gerig for Finite dimensional "Mountain Pass Lemma" Chris Gerig 2013-04-19T02:33:37Z 2013-04-19T02:33:37Z <p>I have stumbled across Richard Palais' (co-author Chuu-lian Terng) <strong>Critical Point Theory and Submanifold Geometry</strong> (Springer Lecture Notes in Math 1353). This is an awesome book!</p> <p>The "Mountain Pass Lemma" for finite-dimensional manifolds is presented as <strong>Theorem 9.2.7 (pg189)</strong>.</p> http://mathoverflow.net/questions/125671/sum-of-two-tangent-bundles-of-s2n/125674#125674 Answer by Chris Gerig for Sum of two tangent bundles of $S^{2n}$ Chris Gerig 2013-03-26T21:54:44Z 2013-03-26T22:26:06Z <p>[]: I hastily interpreted "sum" as the "direct product", although $\oplus$ is almost surely (or surely) referring to Whitney sum:</p> <p>No. $\chi(S^{2n}\times S^{2n})=\chi(S^{2n})^2=4$ and so the Euler class of $S^{2n}\times S^{2n}$ is nontrivial (it pairs to the Euler characteristic under Poincare-duality). But that means there cannot exist a nonvanishing section of $TS^{2n}\oplus TS^{2n}$ (such a section would split the bundle with a trivial summand and the Euler class of a trivial bundle is zero), i.e. it is not trivial.</p> http://mathoverflow.net/questions/90251/understanding-mastering-analysis-in-topology-necessary Understanding/Mastering Analysis in Topology, necessary? Chris Gerig 2012-03-05T04:43:21Z 2013-03-15T20:57:36Z <p>I have spoken to one professor so far about this, which of course was helpful, and so I am looking for additional opinions: <strong>To work with topological tools that were built via analysis, should I be a "master" at that analysis?</strong> By this I mean, for instance, to use Seiberg-Witten Theory and Floer Homologies.<br> As an "entering" graduate student I am "purely" a pure topologist, as in I have no real training in analysis but Algebraic Topology under my belt for $\approx 6$ years. Now learning Seiberg-Witten Floer Homology and other Floer homologies, I tend to put all/most of the analysis (ex: compactness of moduli spaces) in a black box, and then continue to "learn". As a result, I am unsure if I am kind of wasting my time, i.e. if I can still utilize the theories effectively (and of course, I would like to extend theories). <strong>Is there a "good" balance between</strong> 1) <strong>simply accepting the analysis and</strong> 2) <strong>being able to do the analysis with both hands tied behind your back (as Kronheimer-Mrowka seem to do in their</strong> <em>Monopoles and 3-Manifolds</em> <strong>book)?</strong></p> <p>I am unsure how to make this question less vague / more precise, but I feel that there is a good underlying question here that can have an informative response.</p> http://mathoverflow.net/questions/94066/yang-mills-and-chern-simons-functionals-as-morse-functions Yang-Mills and Chern-Simons functionals as Morse functions Chris Gerig 2012-04-14T21:40:21Z 2013-03-15T19:20:05Z <p><strong>Can the Yang-Mills or Chern-Simons action functionals be considered as [possibly <em>perfect</em>] Morse functions?</strong> I assume we would be in an equivariant scenario due to considering the configuration spaces with gauge-groups/transformations. Or at least how far away are they from a Morse-Bott function (and from being perfect)?</p> http://mathoverflow.net/questions/124028/what-goes-wrong-for-the-sobolev-embeddings-at-kn-p What goes wrong for the Sobolev embeddings at $k=n/p$? Chris Gerig 2013-03-08T22:09:26Z 2013-03-12T18:14:34Z <p>For $u\in W^{k,p}(U)$, where $U\subseteq\mathbb{R}^n$ is open and bounded with $C^1$-boundary, we have the <em>celebrated Sobolev inequalities</em>:<br> If $k &lt; n/p$ then $u\in L^q(U)$ for $q$ satisfying $\frac{1}{q}=\frac{1}{p}-\frac{k}{n}$,<br> If $k > n/p$ then $u$ lies in a particular Hölder space. </p> <p>It is also known that this <em>doesn't work</em> for the borderline case $k=n/p$ (which is related to the Sobolev conjugate $p^*\to \infty$ as $p\to n$), although one would expect/hope for $u\in L^\infty(U)$.<br> **An exception as Denis mentions: it works for $(k,p)=(n,1)$ via the fundamental theorem of calculus.<br> **As a counterexample, the function $u(x)=\log\log(1+\frac{1}{|x|})$ with domain $U=int(\mathbb{D}^n)$ lies in $W^{1,n}$ but not in $L^\infty$ (for $n>1$). Instead, apparently the result we end up with is that the functions lie in <em>"the space of functions with bounded mean oscillation".</em></p> <p>1) <strong>Is there an intuitive / deeper reason as to what goes wrong?</strong><br> 2) <strong>Is there some sort of geometric reason why the Sobolev embedding theorem has to fail for $k=n/p$?</strong> I've been told that this critical case seems to arise often in geometry/topology.</p> http://mathoverflow.net/questions/122022/computing-the-cardinality-of-cohomology-groups/122028#122028 Answer by Chris Gerig for Computing the cardinality of cohomology groups Chris Gerig 2013-02-16T23:28:58Z 2013-02-17T04:37:10Z <p>A particular paper that comes to mind is:<br> <strong>On orders and vanishing of integral cohomology groups</strong>, Angelina Chin<br> Here she obtains a recurrence relation involving the orders of the cohomology groups of a finite group $G$ whose quotient by a normal subgroup is cyclic.</p> <p>1) Using Hopf's formula or $H^2(G,\mathbb{Z})\cong Hom(G,\mathbb{Q}/\mathbb{Z})$ should get you the orders of groups for 2nd cohomology.<br> 2) You can check to see if your group has <em>periodic cohomology</em> (period $d$), and then $|H^n(G,\mathbb{Z})|=|H^{n+d}(G,\mathbb{Z})|=|G|$ for some $n$.<br> 3) (Trivially) You can check to see if your group is cohomologically trivial, and then its cohomology vanishes.<br> 4) For abelian groups you can probably get some information from the explicit calculation of $H_*(G,\mathbb{Z}_p)$, given in Theorem V.6.6 of Ken Brown's <em>Cohomology of Groups</em> textbook.</p> http://mathoverflow.net/questions/121130/how-many-proofs-that-pi-nsn-mathbbz-are-there/121131#121131 Answer by Chris Gerig for How many proofs that $\pi_n(S^n)=\mathbb{Z}$ are there? Chris Gerig 2013-02-08T00:38:02Z 2013-02-15T21:36:59Z <p>$\pi_n(S^n)=[S^n,S^n]=\lbrace$cobordism classes of framed 0-submanifolds$\rbrace$ by the Pontrjagin-Thom construction. These are collections of points (with sign) which add up to give the degree of the maps, so this set is precisely $\mathbb{Z}$.</p> http://mathoverflow.net/questions/99472/what-else-is-seiberg-witten-theory-equal-to What else is Seiberg-Witten Theory equal to? Chris Gerig 2012-06-13T16:41:21Z 2013-02-11T08:11:23Z <p>In low-dimensional topology there have been a bunch of invariants defined, and Seiberg-Witten Theory seems to make its appearance in [a lot of] them:<br> 1) <em>Heegaard Floer homology</em> = SW Floer homology (Kutluhan, Lee, Taubes)<br> 2) <em>Embedded Contact homology</em> = SW Floer homology (Taubes)<br> 3) <em>Gromov-Witten invariant</em> = 4-dimensional SW-invariant (Taubes)<br> 4) <em>Turaev torsion</em> = 3-dimensional SW-invariant (Turaev)<br> 5) <em>Milnor torsion</em> (hence <em>Alexander invariant</em>) = 3-dimensional SW-invariant (Meng, Taubes)<br> 6) <em>Donaldson-Smith standard surface count</em> = 4-dimensional SW-invariant (Usher)<br> 7) <em>Casson invariant</em> (hence <em>integral Theta divisor</em>) = 3-dimensional SW-invariant (Lim)</p> <p>Conjectured:<br> 8) <em>Heegaard Floer closed 4-manifold invariant</em> = SW-invariant (Ozsvath, Szabo)<br> *Analog of (1) above in dimension 4<br> 9) <em>Lagrangian matching invariant</em> = SW-invariant (Perutz)<br> *Analog of (6) above for broken Lefschetz fibrations<br> 10) <em>Near-symplectic Gromov-Witten count</em> = SW-invariant (Taubes)<br> *Analog of (4) above for near-symplectic manifolds, counting holomorphic curves in the complement of the degenerate circles of the near-symplectic form -- but this invariant hasn't really been defined yet</p> <p><strong>Does/should it stop there?</strong> Are there constructions out there that Seiberg-Witten Theory could possibly have a link with?</p> http://mathoverflow.net/questions/52552/nontrivial-finite-group-with-trivial-group-homologies Nontrivial finite group with trivial group homologies? Chris Gerig 2011-01-19T22:17:06Z 2013-02-10T22:44:43Z <p>I stumbled across this question in a seminar-paper a long time ago:</p> <p><em>Does there exist a positive integer</em> $N$ <em>such that if</em> $G$ <em>is a finite group with</em> $\bigoplus_{i=1}^NH_i(G)=0$ <em>then</em> $G=\lbrace 1\rbrace$?</p> <p>I believe this to still be an open problem. For $N=1$, any perfect group (ex: $A_5$) is a counterexample. For $N=2$, the binary icosahedral group $SL_2(F_5)$ suffices (perfect group with periodic Tate cohomology). And I found in one of Milgram's papers a result for $N=5$, the sporadic Mathieu group $M_{23}$. Note that this question is answered for infinite groups, because we can always construct a topological space (hence a $BG$ for some discrete group $G$) with prescribed homologies.</p> <p><strong>Is there another known group with a larger $N\ge 5$ before homology becomes nontrivial?<br> Are there any classifications of obstructions in higher homology groups?</strong></p> <p>[]: Another view. A group is $\textit{acyclic}$ if it has trivial integral homology. There are no nontrivial finite acyclic groups. Indeed, a result of Richard Swan says that a group with $p$-torsion has nontrivial mod-$p$ cohomology in infinitely many dimensions, hence nontrivial integral homology.</p> http://mathoverflow.net/questions/120799/manifolds-admitting-cw-structure-with-single-n-cell Manifolds admitting CW-structure with single n-cell Chris Gerig 2013-02-04T20:31:18Z 2013-02-08T20:21:46Z <p>Let $M$ be a topological $n$-manifold, closed and connected (not necessarily oriented):<br> <strong>When does $M$</strong> <em>not</em> <strong>admit (up to homotopy-type) a CW-structure with a single $n$-cell?</strong></p> <p>By classification of surfaces we assume $n>2$. By existence of smooth structures we assume $n>3$. In particular, if $M$ is smoothable then Morse theory provides us the desired structure.</p> <p>[]: To put this question into context, we have various ways of showing that $H_{n-1}(M)$ has either $0$ or $\mathbb{Z}_2$ as its torsion subgroup depending on orientability. One way, when $M$ is such a CW-complex, is to quickly look at the chain-complex differential $d:C_n(M)\cong\mathbb{Z}\to C_{n-1}(M)$ and note that $H_n(M)\cong\mathbb{Z}$ for $M$ orientable and $H_n(M;\mathbb{Z}_2)\cong\mathbb{Z}_2$ otherwise. So I would like to see how large of a class of manifolds this argument holds for.</p> <p>[[Addendum]]: After chatting with Allen Hatcher and Rob Kirby, who reaffirm the comments below, here are their resulting thoughts:<br> <strong>1)</strong> We should be careful with the Kirby theorem of $M$ being homotopy-equivalent to a finite complex, because this complex is obtained by first embedding $M$ into $\mathbb{R}^N$ and then wiggling the boundary of a tubular neighborhood ($M\times D^{N-n}$) of $M$ to be PL, and so the resulting complex could have $i$-cells with $i>n$.<br> <strong>2)</strong> When $\dim M\ne 4$ there is a handlebody-decomposition, and this can be arranged to have a single 0-handle (canceling the other 0-handles with available 1-handles -- we can do this because there are no smoothing obstructions in a neighborhood of the 3-skeleton). Taking the dual handlebody, we have a decomposition with a single n-handle. Passing from the handlebody-decomposition to the CW-decomposition (shrinking everything to their cores), we obtain the desired CW-complex with a single n-cell.<br> <strong>3)</strong> When $\dim M=4$ then a handlebody-decomposition exists if and only if $M$ is smoothable. So when $M$ is smoothable we can apply the argument in (2).<br> <strong>4)</strong> But even when $M$ is not smooth we get some positive results, in particular for the $E_8$ manifold. We build $E_8$ using Kirby calculus on an 8-link diagram, giving a decomposition of $E_8$ into a 0-handle plus eight 2-handles plus a contractible piece (without the contractible piece we get a space with boundary being a homology 3-sphere, namely the Poincare-sphere $S^3/G$ with $G=$ binary icosahedral group). In particular, flipping this structure over we see that $E_8$ is homotopy-equivalent to a CW-complex with a single 4-cell. <em>Furthermore, Lennart Meier's remark gets us all other simply-connected 4-manifolds.</em></p> <p><strong>We are thus left with the scenario that $M$ (up to homotopy) is a closed connected non-simply-connected non-smoothable 4-manifold.</strong> (which the comments below assert)</p> http://mathoverflow.net/questions/119803/relation-between-groups-and-classifying-spaces Relation between groups and classifying spaces Chris Gerig 2013-01-25T01:52:43Z 2013-01-30T02:34:38Z <p>Let $G$ be a nonabelian group, with classifying space $BG$.<br> <strong>Motivation:</strong> We can compute its homology, $H_\ast(BG)=H_\ast(G)$. It would be nice to see some equivariant computations, like $H_\ast^G(BG)$ where $G$ acts by conjugation, but I need a particular model of $BG$ to work with. In any case, if $G$ acted freely it would reduce to computing the homology of $BG/G$. Although this action won't be free, I still wonder what $BG/G$ looks like.</p> <p><em>Preferred Explicit model:</em> Let $EG$ be the weakly contractible space (constructed simplicially using the elements of $G$), and consider the action of $G\times G$ on $G$ by $(g_1,g_2)\cdot g=g_1gg_2^{-1}$. Then we recover the classifying space $BG$ as $EG/G$ (with $G=G\times\lbrace 1\rbrace$), and we get the space $BG/G$ as $EG/(G\times G)$.<br> [: As pointed out, the outcome will depend on the model. If this isn't a "good" model, then I'll settle for a better one! (Although I would want to understand this model).</p> <p>So I have this space $BG/G$, dividing out the classifying space by the conjugation $G$-action. Here is where I get some big help: By the Kan-Thurston theorem, there exists a group $K$ such that $BG/G$ and $BK$ have the same homologies.</p> <p><strong>What can $K$ be?</strong> (Note that if $G$ were abelian then we'd trivially have $K=G$).<br> <strong>Is there a deeper connection between these two spaces?</strong></p> http://mathoverflow.net/questions/119050/fixed-point-of-s1-action-using-roots-of-unity Fixed point of $S^1$-action using roots of unity Chris Gerig 2013-01-16T10:36:06Z 2013-01-18T10:14:11Z <p><strong>Fact:</strong> For any (continuous) $S^1$-action on the closed unit disk $\mathbb{D}^n$, there is a fixed point $x_0\in\mathbb{D}^n$.<br> I have thought of a possible argument that re-proves this, but am not sure how to complete it:</p> <p>Let $U_p\subset S^1$ be the subgroup of $p^\text{th}$-roots of unity ($p$ prime). An $S^1$-action on a compact contractible space $X$ will induce a $U_p$-action on $X$. Smith Theory then implies that $X^{U_p}$ is nonempty, i.e. there is a fixed point $x_p\in X$ under $U_p$, for any given prime $p$. <em>Now here is where I want to say:</em> Taking $p$ sufficiently large, we find a fixed point $x_\infty$ under $S^1$. (The intuition is that $\lim_{p\to\infty}U_p\approx S^1$, and denseness will be sufficient by continuity of the action.)</p> <p>1) <strong>Is it possible to fill this gap, i.e. can this 'proof' make sense?</strong> Not sure how to make sense of this limit/sequence of $U_p$'s, and whether the fixed points hop back and forth forever.</p> <p>2) <strong>Is such a sequence $\lbrace x_p\rbrace_{p=\text{prime}}$ Cauchy? Or, does there exist a prime $p_0$ where $x_p=x_{p_0}$ for all primes $p>p_0$?</strong></p> http://mathoverflow.net/questions/105370/when-does-a-hypersurface-have-contact-type When does a hypersurface have contact-type? Chris Gerig 2012-08-24T07:18:05Z 2013-01-01T17:04:29Z <p>In a symplectic manifold $(X^{2n},\omega)$, a hypersurface $Y\subset X$ has <em>contact-type</em> if there is a contact form $\lambda$ such that $d\lambda=\omega|_Y$. Recall that a contact form is a 1-form with $\lambda\wedge(d\lambda)^{n-1}>0$, i.e. the opposite of a foliation. For example, a starshaped hypersurface has contact-type in $\mathbb{R}^{2n}$, or more generally an $Y\subset\mathbb{R}^{2n}$ transverse to a Liouville vector field defined in a neighborhood of $Y$. In particular, any contact manifold $Y$ is a contact-type hypersurface in a symplectic manifold (the symplectization $\mathbb{R}\times Y$). </p> <p>Now it's nice and useful to consider symplectic manifolds where its boundary has contact-type. And this can usually be done given a compact symplectic manifold (cobordism between contact manifolds).</p> <p>This leads to the question of whether or not you can always build such a space. In other words:<br> <strong>Given a random hypersurface in $\mathbb{R}^{2n}$, is it of contact-type?</strong> How do you tell when it's not of contact-type?</p> <p><em>Edit:</em> In light of the posted responses, I think it would be appropriate to tweak one of the questions above. In particular, it was essentially pointed out twice that in the "set of hypersurfaces" there are open neighborhoods which contain no contact-type ones. But, <strong>are "most" hypersurfaces in $\mathbb{R}^{2n}$ of contact-type?</strong> i.e. Should I expect my hypersurface to be of contact-type?</p> http://mathoverflow.net/questions/117287/motivation-of-virasoro-algebra/117290#117290 Answer by Chris Gerig for Motivation of Virasoro algebra Chris Gerig 2012-12-27T00:18:12Z 2012-12-27T01:19:36Z <p>I will answer (2) quickly: You can refer to my response to this other question: <a href="http://mathoverflow.net/questions/99643/why-does-bosonic-string-theory-require-26-spacetime-dimensions/99650#99650" rel="nofollow">http://mathoverflow.net/questions/99643/why-does-bosonic-string-theory-require-26-spacetime-dimensions/99650#99650</a></p> <p>In String Theory (physics) there are "quantum operators", and the relation they satisfy are precisely this Virasoro relation. And not just that, but $c=D$, the <strong>space-time dimension</strong>! So it is at least extremely important in unifying the theories of physics via strings, because this relation helps us determine the dimension of our universe. You can view this term proportional to the central charge as a "quantum effect" (i.e. it only appears when you take your classical system and quantize it). </p> <p>Why $c=D$?: The propagation ("worldsheet") of a 1-dimensional string (fundamental physical object in the theory) in space-time (dimension $D$) is described by functions $X^\mu$, where the index $\mu$ ranges from 0 to $Dâˆ’2$. They decompose into modes $a^\mu_n$ (for satisfying the string wave-equation). These modes end up mixing and defining quantum operators $L_m$, and the commutator-relations amongst these modes spews out the Virasoro relation with $c=D$.</p> <p>As for some rigorous intuition which will help with question (1): $c$ can be regarded as multiplying the unit operator, and when adjoined to the Lie algebra generated by the $L_m$ it lies in the center of that extended algebra. (I picked this up when working through Becker-Becker-Schwarz String theory textbook).</p> http://mathoverflow.net/questions/117125/a-novice-question-on-quantum-mechanics/117137#117137 Answer by Chris Gerig for A novice question on Quantum Mechanics Chris Gerig 2012-12-24T10:44:46Z 2012-12-24T10:44:46Z <p>Try not to lose sight of the physics (the ultimate goal, really) in this mathematical jargon: </p> <p>This addition-operation is literally <em>superposition</em>, so let's understand what's going on in such terms. The coefficients represent the probability that your object (described by the superposed vector) is actually in one of those state-vectors or the other. That being said, note that if your object is only in $|A\rangle$, then the wavefunction is literally $1|A\rangle$, because you're 100% there... it doesn't make sense for it to only be $\frac{1}{3}|A\rangle$, but regardless that's fine because we normalize the wavefunction, so any coefficient will disappear (i.e. become "1")... so this pertains to Dirac's first statement. Now For $|R\rangle$ there is still normalization, that being $\sqrt{|c_1|^2+|c_2|^2}=1$. But we have freedom here in this condition, so the coefficients themselves don't just pass to 1. In particular, they are important numbers: such a pair $(c_1,c_2)$ is telling you with what likelihood your object lies in $|A\rangle$ versus $|B\rangle$, in such a way that the object is <em>definitely somewhere</em> (that normalization above). Now we see clearly that varying those coefficients are going to change your state.</p> <p>(If you want the pure math explanation, see the other posts; I just kept with the flow of the title of the post).</p> http://mathoverflow.net/questions/117058/rigorous-solution-to-ricci-flow-on-dumbbell-s3 Rigorous solution to Ricci Flow on dumbbell $S^3$ Chris Gerig 2012-12-23T00:35:31Z 2012-12-23T03:27:50Z <p>To begin a small interest in Ricci Flow and similar tools, I am starting with Hamilton's expository paper <em>The Formation of Singularities in the Ricci Flow</em>. This was posted in 1995, so I am wondering if some of his "intuitive pictures" can now be made rigorous:</p> <p>Consider a dumbbell metric on $S^3$, where the neck looks like $S^2\times B^1$ (something like <a href="http://www.springerreference.com/MediaObjectsUnp/featuredArticles/115/186912/978-0-387-92920-0_31_Fig4_HTML.png" rel="nofollow">this</a>, except that picture is for $S^2$). The positive curvature at the ends of the dumbbell will cause the metric to contract (rounding out those parts), whereas the neck area should shrink. In particular, the neck should pinch off at some time $t_{pinch}>0$ (topologically a wedge-sum $S^3\vee S^3$). <strong>Has this been put into experimental practice?</strong> More to the point, Hamilton remarks that <em>there may be a weak solution extending past the pinching moment when the sphere splits into two spheres, although weak solutions haven't been defined for the Ricci flow</em>. <strong>To what extent can we currently rigorous this in practice (if at all)?</strong></p> http://mathoverflow.net/questions/116894/cohomology-ring-of-bg/116904#116904 Answer by Chris Gerig for Cohomology ring of BG Chris Gerig 2012-12-20T20:15:07Z 2012-12-20T20:15:07Z <p>I don't remember where I heard the following proof/sketch:</p> <p>Using the fibering $G/T\to BT\to BG$ and the fact that the Euler class of $G/T$ is nonzero, we have that $H^\ast(BG)$ embeds into $H^\ast(BT)$ (it composes with the transfer map to be multiplication by the Euler class); and the desired isomorphism comes from the fact that $W$ acts on $H^*(G/T)$ as the regular representation. </p> <p>This is actually a special case of equivariant cohomology, where we instead use the Borel construction and the fibering $G/T\to M_T\to M_G$.</p> http://mathoverflow.net/questions/47836/generalized-geometry-and-spin-structures/115964#115964 Answer by Chris Gerig for Generalized geometry and spin structures Chris Gerig 2012-12-10T11:23:14Z 2012-12-14T18:54:07Z <p>OK, I am making an assumption: <em>I can re-interpret the problem (using the musical isomorphism) as $V$ being the diagonal embedding of $TM$ inside $TM\oplus TM$ and studying spin structures on them</em>.</p> <p>Actually, it turns out (see comments) somehow that this "re-interpretation" is slightly different from the original construction. But I still believe the answer is no, based on its resemblance to the "re-interpretation":</p> <p>I will restrict to $\dim M=3$, in which case our (closed oriented) manifold is always spinnable. A spin structure $\mathfrak{s}$ on $M$ induces a canonical spin structure $\mathfrak{S}_0=\mathfrak{s}\oplus\mathfrak{s}$ on $TM\oplus TM$, and this is actually independent of the choice of $\mathfrak{s}$ (these appear in the notion of a <em>2-framing</em> on 3-manifolds, which Atiyah and Witten have used for some of their QFT studies). As a result, the "restriction" $\mathfrak{S}_0|_V$ on $M$ is ill-defined.<br> <em>[proof of claim of canonical spin structure (learned from conversation with Rob Kirby): the spin structure fixes a trivialization over the 1-skeleton, and over circles there are two trivializations, so changing a trivialization of $\mathfrak{s}$ is doubled in $\mathfrak{s}\oplus\mathfrak{s}$ which modulo-2 is no change.]</em></p> <p>This also implies that the "restriction" to each $TM$-summand is ill-defined. (What I know that works: the collar-neighborhood theorem does allow an induced spin structure on $T(\partial X$) from a spin structure on $TX$ thanks to the splitting $TX|_\partial=T(\partial X)\oplus\underline{\mathbb{R}}$ near the boundary.)</p> http://mathoverflow.net/questions/116251/mathematician-trying-to-learn-string-theory/116286#116286 Answer by Chris Gerig for Mathematician trying to learn string theory Chris Gerig 2012-12-13T15:34:05Z 2012-12-13T17:21:04Z <p>I mean, if you are <em>really</em> trying to understand String Theory, then you're going to have to become fluent in Classical Mechanics, Quantum Mechanics, Quantum Field Theory, and General Relativity first... otherwise the papers are going to be unmotivated and you won't understand the linguistics and you won't know how the results connect to the universe (i.e. they're more than just a sequence of symbols which we call math).</p> <p>That being said, assuming CM/QM/QFT/GR are under the belt, the best place to <em>start</em> is Green-Witten-Schwarz's (GWS) <a href="http://www.amazon.com/Superstring-Theory-Introduction-Monographs-Mathematical/dp/0521357527" rel="nofollow"><strong>Superstring Theory</strong></a>, followed by skimming Polchinski's <a href="http://www.amazon.com/String-Cambridge-Monographs-Mathematical-Physics/dp/0521672279" rel="nofollow"><strong>String Theory</strong></a>. This is supported by my string theory professor when I took it a while ago, Petr Horava (discoverer of D-branes). From here you can supplement other notes and papers.</p> <p>In Vol.1 of GWS, chapter 2/3 will explain the bosonic string theory (i.e. ignoring fermions) and BRST quantization, which leads to a critical dimension $D=26$. Then chapter 4 will fix this with supersymmetry (i.e. putting back in the fermions), leading to the actual critical dimension $D=10$. After this, gauge anomalies and compactification and dualities and D-branes can start being assessed.</p> http://mathoverflow.net/questions/115340/concerning-strata-in-c-inftym Concerning strata in $C^\infty(M)$ Chris Gerig 2012-12-03T22:03:23Z 2012-12-04T08:08:36Z <p>The Morse functions are dense in $C^\infty(M)$, and you can ask if a 1-parameter family of smooth functions between two given Morse functions will be a homotopy through Morse functions. Well, Cerf Theory shows that there is a codimension-one stratum $\mathcal{F}^1$ in $C^\infty(M)$, and so any path between two Morse functions will transversely intersect $\mathcal{F}^1$ a finite number of times (in a few specific ways).</p> <p><strong>Can we somehow 'measure' how far away a given function is from $\mathcal{F}^1$?</strong></p> <p>(Cerf's $\mathcal{F}^1$ consists of two components: the set of smooth functions that have exactly one birth point and distinct critical values, and the set of Morse functions where exactly two distinct critical points have the same value. A birth/death point of a smooth function on an $n$-manifold is a point with local neighborhood $-x_1^2-\cdots-x^2_i+x_{i+1}^2+\cdots+x^2_n+x^3_{n+1}$. The codimension-zero stratum $\mathcal{F}^0$ is essentially the space of Morse functions, but only the ones with distinct critical values. This separation is slightly weird to me.)</p> http://mathoverflow.net/questions/115340/concerning-strata-in-c-inftym/115372#115372 Answer by Chris Gerig for Concerning strata in $C^\infty(M)$ Chris Gerig 2012-12-04T08:08:36Z 2012-12-04T08:08:36Z <p>For completeness I will write the comments here, to close the question.</p> <p>Given $f\in C^\infty(M)$, in relation to the given stratum $\mathcal{F}^1$ it might be hard to measure distance because we want to detect when the function has <em>precisely one</em> birth point or has <em>precisely one</em> pair of critical points with identical values (the strata is explained nicely in Hatcher-Wagoner's book on the subject). But if we ignore this "precisely one birth or pair" condition (by perturbing from higher codimension strata), then</p> <p>the distance can be $f\mapsto \min_{p\in crit(f)}|\det(Hess(f,p))|\cdot\inf_{p\ne q\in crit(f)}|f(p)-f(q)|$</p> <p>Here the first factor checks for birth points, and the second factor checks for distinct critical points with identical values. If we further want to consider the stratum which doesn't contain Morse functions (so that we only worry how far away $f$ is from giving birth to / killing a critical point), then we can drop the second factor.</p> http://mathoverflow.net/questions/110808/how-many-4-manifolds-are-symplectic How Many 4-Manifolds are Symplectic? Chris Gerig 2012-10-27T04:23:27Z 2012-11-28T07:25:21Z <p>As an honest question (probably with some subjectivity), <strong>how many smooth oriented 4-manifolds are actually symplectic?</strong> Can I say half (perhaps under some mild assumptions)? I ask this question because every compact smooth oriented 4-manifold with $b^2_+\ge 1$ admits a <em>near-symplectic</em> form, i.e. a closed 2-form which is symplectic away from a finite set of circles.</p> <p>Some results that might push the percentage one way or the other:<br> 1) Gompf has shown that any finitely presented group can be realized as the fundamental group of a compact symplectic 4-manifold.<br> 2) The Seiberg-Witten invariants are nonzero for symplectic 4-manifolds, and in a sense show that they are the "irreducible" basic forms of smooth 4-manifolds.<br> 3) Every compact symplectic 4-manifold is a branched cover of $\mathbb{C}P^2$.</p> <p>The responses/comments show that we can ask this question (on when can I expect my 4-manifold to be symplectic) in many different ways, each with different expectations. So I am interested in some further thoughts on Tim's and Dmitri's questions.</p> http://mathoverflow.net/questions/113993/infinite-suspension-functor-on-the-pontryagin-thom-construction (Infinite) Suspension Functor on the Pontryagin-Thom Construction Chris Gerig 2012-11-20T22:28:10Z 2012-11-27T10:02:08Z <p>This is a slightly revamped version of a <a href="http://math.stackexchange.com/questions/241159/suspension-operation-on-the-pontryagin-thom-construction" rel="nofollow">question</a> I asked on the stackexchange forum. That question was asking if the Pontryagin-Thom constructon respects the suspension operation, alluding to stable homotopy. I now would like to dig deeper on this allusion.</p> <p>For a compact simply-connected oriented manifold $M$, view the Pontryagin-Thom construction as the bijective correspondence between $[M,S^r]$ and the set of (appropriate equivalence classes of) framed submanifolds of codimension $r$ in $M$. A quick subquestion is: <strong>Is this in some way functorial?</strong><br> For applying suspension $\Sigma^1$, I would ideally like to keep our general $M$, but for now I would be happy understanding what occurs for $M\approx S^n$. I am actually unsure if this will work for general $M$ because suspension doesn't typically produce a manifold right? -- Maybe we can tweak it (smooth the corners, or homotope it) to produce a manifold.<br> <strong>In perhaps a non-rigorous sense, does $\Sigma^1$ represent a natural transformation for the above construction?</strong> In particular, under $\Sigma^1$ we pass from $[M,S^r]$ to $[\Sigma^1M,S^{r+1}]$. So I would expect to get a correspondence $\Sigma^1\lbrace\text{framed }(n-r)\text{-submanifolds in }M\rbrace\simeq\lbrace\text{framed }(n-r)\text{-submanifolds in }\Sigma^1M\rbrace$.</p> <p>Now if this all works out, what happens on repeated iterations $\Sigma^n$? <strong>Does the Pontryagin-Thom construction blossom into anything under the infinite-suspension $\Sigma^\infty$?</strong></p> http://mathoverflow.net/questions/113801/computation-of-homotopy-groups-of-spheres-via-pontryagin-thom/113806#113806 Answer by Chris Gerig for Computation of homotopy groups of spheres via Pontryagin-Thom Chris Gerig 2012-11-19T07:38:35Z 2012-11-19T19:49:34Z <p>The $k=0$ and $k=1$ case are drawn up nicely in the second appendix of Freed and Uhlenbeck's classic book <em>Instantons and Four-Manifolds</em>. It's entitled the Pontrjagin-Thom Construction, and is motivated by wanting to compute $[M,S^3]$ (for any compact simply-connected 4-manifold) whose nontriviality depends on the parity of the natural intersection form.<br> The best part: there is a cool picture of a dinosaur (that is, a framed cobordism) being cut open (that is, by two homotopy-equivalent framings).</p> http://mathoverflow.net/questions/112417/why-donaldsons-four-six-conjecture Why Donaldson's Four-Six Conjecture? Chris Gerig 2012-11-14T21:22:07Z 2012-11-15T20:55:01Z <p>Simon Donaldson apparently made the following conjecture: Two closed symplectic 4-manifolds $(X_1,\omega_1)$ and $(X_2,\omega_2)$ are diffeomorphic if and only if $(X_1\times S^2,\omega_1\oplus\omega)$ is deformation-equivalent to $(X_2\times S^2,\omega_2\oplus\omega)$. Here $\omega$ is a symplectic structure on $S^2$, and a <em>deformation-equivalence</em> is a diffeomorphism $\phi:X_1\times S^2\to X_2\times S^2$ such that $\omega_1\oplus\omega$ and $\phi^*(\omega_2\oplus\omega)$ can be joined by a path of symplectic forms.</p> <p>However, where I read this did not contain any background or the original source. <strong>Where did Donaldson make this claim? And why did he make this claim? What is the motivation / are there good examples where this holds?</strong> Ivan Smith showed (through examples) that this conjecture fails when we replace $S^2$ by $\mathbb{T}^2$, so the statement itself seems pretty rigid.</p> http://mathoverflow.net/questions/104642/floer-homology-and-invariants-for-einstein-field-equations Floer homology and Invariants for Einstein Field Equations? Chris Gerig 2012-08-13T20:49:56Z 2012-11-08T23:36:52Z <p><strong>Motivation:</strong> There have been the <em>instanton</em> (anti-self dual connection) solutions to the Yang-Mills equation $d_A^\ast F_A=0$ which extremize the YM energy $\int_M|F_A|^2$, leading to the Donaldson invariants and even a Floer homology. There have been the <em>monopole</em> (connection + spinor) solutions to the Seiberg-Witten equations $D_A\psi=0$ and $F_A^+=\psi\otimes\psi^\ast-\frac{1}{2}|\psi|^2$ which extremize the Chern-Simons-Dirac functional, leading to the SW invariants and a nice Floer homology. These utilize the fundamental particles in the Standard-Model of physics... but not of General Relativity, where the <em>gravitons</em> arise.</p> <p>So I would be interested in a Floer homology and/or invariants arising from <em>gravitational instantons</em> (Riemannian metrics), i.e. solutions to the Einstein Field Equations $R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R-g_{\mu\nu}\Lambda=0$ in vacuum (no stress-energy term $T$). Here $\Lambda$ is the "cosmological constant", which we may or may not want to assume is zero. Surely these have been studied extensively. (This term 'gravitational instanton' is used first (I think) in Stephen Hawking's seminal 1977 paper "Gravitational Instantons", and basic examples are the Schwarzschild and Taub-NUT metrics.)</p> <p><strong>Should I expect something to arise? Are there immediate obstacles?</strong> Otherwise this would have been done by now, right?</p> <p><strong>Downfall?</strong>: Perhaps the moduli space is too big, or boring, or unknown.<br> <strong>Progress?</strong>: Witten has even shown that (2+1)-dimensional gravity (with no cosmological constant) on $M=\Sigma\times \mathbb{R}$ (compact surface $\Sigma$) is an ISO(2,1) Chern-Simons theory, i.e. the equations of motion of the CS-action are precisely the field equations. And if there is a cosmological constant $\Lambda\ne 0$, then the same result holds when we replace the gauge group ISO(2,1) by SO(3,1) or SO(2,2) depending on the sign of $\Lambda$.<br> <strong>More</strong>: There is something to be said from Witten's recent paper <em>"Analytic Continuation Of Chern-Simons Theory"</em>, but I am not ready to understand it.</p> http://mathoverflow.net/questions/130117/trivial-action-on-the-second-cohomology-group Comment by Chris Gerig Chris Gerig 2013-05-08T21:26:42Z 2013-05-08T21:26:42Z Except here he isn't acting on the group $G$, he is acting solely on the coefficients module $A$... I am actually unsure this defines a $G$-action for chomology. http://mathoverflow.net/questions/129752/does-a-connected-manifold-with-vanishing-euler-characteristic-admit-a-nowhere-van Comment by Chris Gerig Chris Gerig 2013-05-05T19:36:12Z 2013-05-05T19:36:12Z (It's due to Poincare and Hopf and called their index theorem.) http://mathoverflow.net/questions/129174/vector-fields-on-4n1-spheres/129182#129182 Comment by Chris Gerig Chris Gerig 2013-04-30T05:36:32Z 2013-04-30T05:36:32Z Correct, but this is what sparked my questions. http://mathoverflow.net/questions/128556/isomorphisms-induced-on-h Comment by Chris Gerig Chris Gerig 2013-04-23T23:57:06Z 2013-04-23T23:57:06Z Homework questions are 1) not permitted here, and 2) not permitted <i>anywhere</i> in the way you wrote because you have to portray your attempt and where you are stuck. Once you fix this, try: <a href="http://math.stackexchange.com/" rel="nofollow">math.stackexchange.com</a> http://mathoverflow.net/questions/47836/generalized-geometry-and-spin-structures/115964#115964 Comment by Chris Gerig Chris Gerig 2013-04-18T03:21:33Z 2013-04-18T03:21:33Z @Pedro: Yes I agree (the 'doubling' disappearing over the 1-skeleton for the spin structures); this is the way to get around the noncanonical-ness of using $TM$ instead of $T^*M$. I hope this was of help! http://mathoverflow.net/questions/127576/what-is-data-involved-in-a-mathematical-construction Comment by Chris Gerig Chris Gerig 2013-04-15T04:06:11Z 2013-04-15T04:06:11Z It means what Webster's dictionary says it means. http://mathoverflow.net/questions/126588/the-second-homology-of-a-group-g-and-presentation-complex-of-g Comment by Chris Gerig Chris Gerig 2013-04-05T08:18:07Z 2013-04-05T08:18:07Z Group-cohomology encompasses group homology; the previous tag is more abundant. http://mathoverflow.net/questions/126508/geometric-meaning-of-ricci-flatness Comment by Chris Gerig Chris Gerig 2013-04-04T18:56:06Z 2013-04-04T18:56:06Z <a href="http://math.stackexchange.com/questions/339057/geometric-meaning-of-ricci-flatness?noredirect=1#comment731170_339057" rel="nofollow" title="geometric meaning of ricci flatness%3fnoredirect%3d1%23comment731170_339057">math.stackexchange.com/questions/339057/&hellip;</a> ... I suggest making an edit to your post on the other site, so that you can receive better help. http://mathoverflow.net/questions/126365/two-ways-of-getting-a-cohomology-class-from-an-extension-of-a-discrete-group-by Comment by Chris Gerig Chris Gerig 2013-04-04T00:14:13Z 2013-04-04T00:14:13Z Addendum: This fibration I think is a principal bundle, and so existence of a section would mean the bundle is trivial, which should mean that the cohomology class in $H^2$ is zero, so that its image in $H^3$ is also zero, giving agreement here. http://mathoverflow.net/questions/126365/two-ways-of-getting-a-cohomology-class-from-an-extension-of-a-discrete-group-by Comment by Chris Gerig Chris Gerig 2013-04-04T00:11:26Z 2013-04-04T00:11:26Z Also, your extension $\alpha$ corresponds to the $B\mathbb{C}^*$-fibration that you write, and so we want an explicit description, in terms of the underlying groups, of the failure of a set-theoretic section to be the desired map. Ideally this will have either a cocyle-description (and then check it with $\delta\alpha$) or a crossed module extension description (and then compare extensions). On afterthought, these comments are probably recasting your question into a harder one. http://mathoverflow.net/questions/126365/two-ways-of-getting-a-cohomology-class-from-an-extension-of-a-discrete-group-by Comment by Chris Gerig Chris Gerig 2013-04-04T00:06:31Z 2013-04-04T00:06:31Z Just in case you haven't already looked in this direction (I got stuck): $H^3(A,B)$ corresponds to <i>crossed module extensions</i>. So your extension $\alpha\in H^2(G,\mathbb{C}^*)$ maps under the connecting homomorphism to a particular $\mathbb{Z}\to N\to E\to G$, which ideally you can read off from the definitions (with the help of MacLane's paper on this notion). http://mathoverflow.net/questions/126458/example-of-riemannian-manifold-with-no-minimizing-geodesic-between-certain-points Comment by Chris Gerig Chris Gerig 2013-04-03T22:36:32Z 2013-04-03T22:36:32Z plane with origin removed: points $(1,0)$ and $(-1,0)$ http://mathoverflow.net/questions/126427/suspension-of-the-n-sphere Comment by Chris Gerig Chris Gerig 2013-04-03T19:18:01Z 2013-04-03T19:18:01Z <a href="http://math.stackexchange.com/" rel="nofollow">math.stackexchange.com</a> will help you (check out the FAQ) http://mathoverflow.net/questions/125671/sum-of-two-tangent-bundles-of-s2n/125674#125674 Comment by Chris Gerig Chris Gerig 2013-03-26T22:08:45Z 2013-03-26T22:08:45Z Ah I took &quot;sum&quot; as direct product, although $\oplus$ is standard notation for Whitney sum. http://mathoverflow.net/questions/124568/fluid-mechanics-and-topology Comment by Chris Gerig Chris Gerig 2013-03-15T01:18:58Z 2013-03-15T01:18:58Z Google search <b>Robert Ghrist</b> and your wishes may come true.