User manuel rivera - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T22:55:59Z http://mathoverflow.net/feeds/user/5450 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/103620/geometric-information-on-transferred-structure Geometric information on transferred structure Manuel Rivera 2012-07-31T15:08:48Z 2012-12-18T08:22:00Z <p>Let $(M,g)$ be a Riemannian manifold, let <code>$\Omega^*(M)$</code> denote the cochain complex of differential forms on $M$ and <code>$H^*(M)$</code> its cohomology considered as a chain complex with trivial differential. We have a map <code>$f:\Omega^*(M) \to H^*(M)$</code> given by projecting a form to its harmonic part and taking its cohomology class and we have a map <code>$h: H^*(M) \to \Omega^*(M)$</code> which sends a cohomology class to its harmonic representative. Then $fh$ is the identity and $hf$ is chain homotopic to the identity. Hence, we have the right setting to transfer the $C_{\infty}$-algebra structure of $\Omega^*(M)$ with structure maps $m_1=d$, $m_2=\wedge$, and $0=m_3=m_4=...$ to obtain an $C_{\infty}$-algebra structure on <code>$H^*(M)$</code>. We can get formulas for the structure maps of the transferred structure on $H^*(M)$ by taking sums over trees and putting the chain homotopies in the right internal edges, etc...</p> <p>The topological information information obtained from this transferred structure is understood: up to homotopy, the transferred structure contains rational homotopy information. </p> <p>However, my question is the following: What kind of <em>geometric</em> information is contained in the transferred structure (which involves many choices as for example the metric $g$) up to <em>isomorphism</em> ?</p> http://mathoverflow.net/questions/104458/vector-fields-on-path-spaces Vector fields on path spaces Manuel Rivera 2012-08-11T02:28:01Z 2012-08-12T14:35:33Z <p>I've been reading Chen's original works on iterated integrals and in order to consider differential forms on the path space $PM$ of a smooth manifold $M$ he gives $PM$ the following "differentiable space" structure:</p> <p>Let $N$ be a smooth manifold. A continuous map $\alpha: N \to PM$ (where $PM$ has the compact open topology) is said to be <em>smooth</em> if the adjoint map $\tilde{\alpha}: N \times I \to M$ defined by $(n,t)\mapsto \alpha(n)(t)$ is smooth in the usual sense. The smooth map $\alpha$ is a <em>plot</em> if $N$ is an open convex subset of $\mathbb{R}^n$ for some $n$. Thus, we are modeling $PM$ locally with plots of varying dimension.</p> <p>Chen defines a differential n-form $\omega$ on $PM$ as a rule which assigns to every plot $\alpha: U \to PM$ a differential form $\omega_{\alpha} \in \Omega^n(U)$. We define <code>$(d\omega)_{\alpha}=d\omega_{\alpha}$</code>. It turns out that for Chen's purposes one does not need to develop more calculus tools on $PM$. He shows a De Rham type result: the cohomology of the complex $\Omega^*(LM)$ (where $LM$ is the free loop space) is isomorphic to the real singular cohomology of $LM$.</p> <p>However, for other purposes it is useful to consider forms as alternating tensors and to do this in this context we need a notion of vector fields on $PM$. I've always thought of a tangent vector at $\gamma \in PM$ as a vector field $T_\gamma$ along $\gamma$ on $M$, so a vector field on $PM$ assings each point $\gamma \in PM$ a vector field along $\gamma$. However, following Chen, the natural way to define vector fields to make it compatible with his notion of differential forms is as follows: a vector field $T$ on $PM$ is a rule which assings to each plot $\alpha: U \to PM$ a vector field $T_{\alpha}$ on $U$. </p> <p>How do we reconcile these two notions of vector fields on $PM$? Are they equivalent? </p> http://mathoverflow.net/questions/88114/free-loop-space-and-invariant-forms free loop space and invariant forms Manuel Rivera 2012-02-10T16:22:35Z 2012-02-15T11:10:47Z <p>Cartan proved that for a connected compact Lie group $G$ the left invariant differential forms yield the correct cohomology of $G$. The same argument works for a connected compact $G$-manifold: the idea is to "average" left invariant forms on $G$ using a Haar measure.</p> <p>Can we extend this result for non compact infinite dimensional manifolds? In particular, consider the free loop space $LM$ of a manifold $M$; this is an infinite dimensional $S^1$-manifold. Is there a way to compute the cohomology of $LM$ using a model of "invariant forms" and the idea of averaging?</p> <p>By a result of Chen, we know that iterated integrals of differential forms in $M$ yield the correct cohomology of $LM$. Is this model related to Cartan's story of invariant forms?</p> <p>These questions are a bit vague, but I guess how to make them precise is part of my question.</p> http://mathoverflow.net/questions/79041/cartan-weil-model-for-equivariant-cohomology Cartan-Weil model for Equivariant Cohomology Manuel Rivera 2011-10-25T00:23:57Z 2011-10-25T01:02:44Z <p>Let $\mathfrak{g}$ be the Lie algebra of a Lie group $G$ which acts on a manifold $M$. It is quite standard that the basic forms in <code>$\Omega^*(M) \otimes W(\mathfrak{g}^*)$</code> form a model for the singular equivariant cohomology of $M$. However, I have never seen a proof and it is not straightforward to me. Could someone give a sketch or a reference of the proof of this fact? It is probably in one of Cartan's papers but I haven't been able to find it.</p> <hr> <p>Here goes some background:</p> <p>We define its Weil algebra by <code>$W^*(\mathfrak{g}^*)=S^*(\mathfrak{g}^*) \otimes \wedge^*(\mathfrak{g}^*)$</code> there is also a natural differential operator $d_W$ which makes $W*(\mathfrak{g}^*)$ into a complex. We define $d_W$ as follows:</p> <p>Choose a basis $e_1,...,e_n$ for $\mathfrak{g}$ and let $e^*_1,...e^*_n$ its dual basis in $\mathfrak{g}^*$. Let $\theta_1,...,\theta_n$ be the image of $e^*_1,...e^*_n$ in $\wedge(\mathfrak{g}^*)$ and let $\Omega_1,...,\Omega_n$ be the image of $e^*_1,...e^*_n$ in $S(\mathfrak{g}^*)$. Let $c_{jk}^i$ be the structure constants of $\mathfrak{g}$, that is $[e_j,e_k]=\sum_{i=1}^nc_{jk}^ie_i$. Define $d_W$ by \begin{eqnarray} d_W\theta_i=\Omega_i- \frac{1}{2}\sum_{j,k} c_{jk}^i \theta_j \wedge \theta_k \end{eqnarray} and \begin{eqnarray} d_W\Omega_i=\sum_{j,k}c_{jk}^i\theta_j \Omega_k \end{eqnarray} and extending $d_W$ to $W(\mathfrak{g})$ as a derivation.</p> <p>We can also define interior multiplication $i_X$ on $W(\mathfrak{g}^*)$ for any $X \in \mathfrak{g}$ by \begin{eqnarray} i_{e_r}(\theta_s)=\delta^r_s, i_{e_r}(\Omega_s)=0 \end{eqnarray} for all $r,s=1,...,n$ and extending by linearity and as a derivation. </p> <p>Now consider <code>$\Omega^*(M) \otimes W(\mathfrak{g}^*)$</code> as a complex. Using this definition of interior multiplication, together with the usual definition of interior multiplication on forms, we define the basic complex of $\Omega^<em>(M) \otimes W(\g^</em>)$:</p> <p>We call <code>$\alpha \in \Omega^*(M) \otimes W(\mathfrak{g}^*)$</code> a basic element if $i_X(\alpha)=0$ and $i_X(d \alpha)=0$. Basic elements in <code>$\Omega^*(M) \otimes W(\mathfrak{g}^*)$</code> form a subcomplex which we denote by <code>$\Omega^*_G (M)$</code>.</p> <p>The claim is that <code>$H^*(\Omega^*_G (M))=H^*(M \times_G EG)$</code> where the right hand side denotes the singular equivariant cohomology of $M$.</p> http://mathoverflow.net/questions/66508/infinite-configuration-of-lines infinite configuration of lines Manuel Rivera 2011-05-31T00:53:09Z 2011-06-01T09:33:50Z <p>I was looking at some random problems and questions I liked when I was in high school and I found this one which I still cannot prove.</p> <p>Does there exist a configuration of a countable number of straight lines in the plane such that:</p> <p>1) no two are parallel</p> <p>2) no three are concurrent</p> <p>3) any bounded subset of the plane is intersected by a finite number of lines</p> <p>4) the area of every minimal polygon is equal, where a minimal polygon is a polygon formed by a finite subset of the set of lines such that no lines pass through the inside of the polygon.</p> <p>The answer is certainly no, but it is not that easy to prove. Any ideas?</p> http://mathoverflow.net/questions/61814/ask-for-recommendations-for-textbook-on-mathematical-logic/61887#61887 Answer by Manuel Rivera for Ask for recommendations for textbook on mathematical logic Manuel Rivera 2011-04-16T02:11:58Z 2011-04-16T02:11:58Z <p>I was going to recommend the English translation of the two volume sequence by Cori and Lascar. But after reading again your message it is highly possible that this is the text you used. I really like these two introductory books. </p> http://mathoverflow.net/questions/61878/geometric-interpretation-of-cartans-structure-equations Geometric interpretation of Cartan's structure equations Manuel Rivera 2011-04-15T23:00:42Z 2011-04-16T01:14:55Z <p>Given a linear connection on a Riemmanian manifold $M$ and $\phi^1,...,\phi^n$ a local frame for $T^*M$ we can define the connection 1-forms $\omega^j_i$. We define the curvature 2-forms by $\Omega_i^j=\frac{1}{2}R_{klij}\phi^k \wedge \phi^l$.</p> <p>We have the following identities also known as Cartan's first and second structure equations:</p> <p>i) $d\phi^j=\phi^i \wedge \omega_i^j + \tau^j$ where $\tau^1,...,\tau^n$ are the torsion 2 forms.</p> <p>ii) $\Omega_i^j=d\omega_i^j-\omega_i^k \wedge \omega_k^j$</p> <p>I have two questions: </p> <p>1)Is there a geometric meaning attached to these equations?</p> <p>2) Why are these equations important and what are they useful for?</p> http://mathoverflow.net/questions/60877/geometric-meaning-of-torsion-in-homotopy-groups Geometric meaning of torsion in homotopy groups Manuel Rivera 2011-04-07T02:49:12Z 2011-04-07T13:07:36Z <p>It is not too hard to understand the geometric meaning of torsion in homology groups of CW complexes. However, I thought it would be interesting to hear how people describe/think of the geometric meaning of torsion in the homotopy groups of a CW-complex.</p> http://mathoverflow.net/questions/24144/is-there-an-uncountable-non-discrete-hausdorff-toronto-space Is there an uncountable, non-discrete, Hausdorff Toronto space? Manuel Rivera 2010-05-10T19:54:06Z 2011-02-02T22:45:30Z <p>We call a topological space $X$ a <em>Toronto space</em> if for any subspace $Y \subseteq X$ such that $Y$ and $X$ have the same cardinality it follows that $Y$ is homeomorphic to $X$. </p> <p>Does anybody know what is known about the following question?:</p> <p>Is there an uncountable, non-discrete, Hausdorff Toronto space?</p> <p>It is not hard to show that if $X$ is countable, Hausdorff and Toronto then $X$ has the discrete topology. I have been thinking about the uncountable case for a while and it turns out it is a much harder question.</p> http://mathoverflow.net/questions/43623/classification-problems Classification Problems Manuel Rivera 2010-10-26T03:06:10Z 2010-11-09T05:22:14Z <p>I was thinking about the famous question in philosophy of mathematics: "When are two proofs the same?" and I was wondering if we could somehow "classify" proofs by establishing some sort of functorial relationship between proofs and other mathematical objects which we can classify (like for example, surfaces; my initial idea was to somehow capture the logical structure of a proof in a graph and then classify graphs by their topological structure). I searched MO and found <a href="http://mathoverflow.net/questions/3776/when-are-two-proofs-of-the-same-theorem-really-different-proofs" rel="nofollow">this</a> interesting post which contained some similar ideas.</p> <p>However, I was wondering if we can come up with a list of examples of classification problems in mathematics which have been answered using category theoretic tools by functorially "translating" the original problem into a different category in which we can classify the corresponding objects... and everything works in a nice way. The natural place to start is obviously algebraic topology.</p> http://mathoverflow.net/questions/43889/proof-synopsis-collection/43911#43911 Answer by Manuel Rivera for Proof synopsis collection Manuel Rivera 2010-10-28T00:08:06Z 2010-10-28T00:08:06Z <p>Brouwer's Fixed Point Theorem: Every map $f: D^n \rightarrow D^n$ has a fixed point.</p> <p>$S^{n-1}$ is not a retract of $D^n$, otherwise we could then factor the identity map $H_{n-1}(S^{n-1}) \rightarrow H_{n-1}(S^{n-1})$ through the trivial group $H_{n-1}(D^n)$. If $f$ had no fixed point we could then define a retraction $r: D^n \rightarrow S^{n-1}$ by letting $r(x)$ to be the point in the intersection of $S^{n-1}$ and the ray in $\mathbb{R}^n$ starting at $f(x)$ and passing through $x$.</p> http://mathoverflow.net/questions/41514/non-natural-iso-between-homotopy-and-homology "non natural" iso between homotopy and homology Manuel Rivera 2010-10-08T14:39:45Z 2010-10-09T07:05:22Z <p>Can we classify all finite CW complexes $X$ such that for each $i$ there is <em>some</em> isomorphism $\pi_i(X) \rightarrow H_i(X)$? Note that it is not hard to classify all complexes for which each isomorphism is given by the Hurewicz map.</p> http://mathoverflow.net/questions/34878/x-not-simply-connected-and-x-x-contractible X not simply connected and X-x contractible Manuel Rivera 2010-08-08T02:10:40Z 2010-08-08T15:40:20Z <p>Hello,</p> <p>I was wondering if there is a nice counterexample to the following question.</p> <p>Suppose $X$ is a CW-complex which is not simply connected and there is a point $x\in X$ such that $X-x$ is contractible. Is $X$ homotopy equivalent to a wedge of circles? Maybe we do not even need the CW-complex condition.</p> http://mathoverflow.net/questions/103620/geometric-information-on-transferred-structure Comment by Manuel Rivera Manuel Rivera 2012-12-05T02:54:22Z 2012-12-05T02:54:22Z Note that my question is metric dependent. http://mathoverflow.net/questions/103620/geometric-information-on-transferred-structure Comment by Manuel Rivera Manuel Rivera 2012-12-05T01:32:49Z 2012-12-05T01:32:49Z Let me see if I can make sense of my question. I agree that it is not clear. The transfer of structure depends on many choices. Suppose you have two metrics $g$ and $g'$ on $M$. Then these would lead to different $C_{\infty}$ structure maps $m^g_i$ and $m^{g'}_i$ on $H^*(M)$. However, when you take homology of these, i.e. when you form the bar construction and take homology of that you should get rational homotopy. My question is, do the transferred maps $m_i^g$ and $m_i^{g'}$ reflect any geometric information? http://mathoverflow.net/questions/38219/intuition-on-finite-homotopy-groups Comment by Manuel Rivera Manuel Rivera 2012-08-23T14:38:01Z 2012-08-23T14:38:01Z Check out this post: <a href="http://mathoverflow.net/questions/60877/geometric-meaning-of-torsion-in-homotopy-groups" rel="nofollow" title="geometric meaning of torsion in homotopy groups">mathoverflow.net/questions/60877/&hellip;</a> http://mathoverflow.net/questions/104458/vector-fields-on-path-spaces/104564#104564 Comment by Manuel Rivera Manuel Rivera 2012-08-13T00:27:17Z 2012-08-13T00:27:17Z Thanks Konrad. This was really helpful. http://mathoverflow.net/questions/104458/vector-fields-on-path-spaces Comment by Manuel Rivera Manuel Rivera 2012-08-11T20:13:54Z 2012-08-11T20:13:54Z This makes sense. For these purposes we should then define an n-form on $PM$ to be a function which assigns to each $\gamma \in PM$ an alternating n-tensor on $T_{\gamma}PM$, as usual. Then we can define interior multiplication in the usual sense. If we want to obtain forms in Chen's sense we can do by pulling back along plots. http://mathoverflow.net/questions/104458/vector-fields-on-path-spaces Comment by Manuel Rivera Manuel Rivera 2012-08-11T17:40:14Z 2012-08-11T17:40:14Z Right, but shouldn't we be able to describe vector fields locally via plots? For example how would you define operations like interior multiplication, i.e. if $T$ is a vector field on $PM$, $\omega$ an n-form, and $\alpha: U \to PM$ a plot how would you define $(\iota_T \omega)_{\alpha}$? http://mathoverflow.net/questions/88114/free-loop-space-and-invariant-forms/88510#88510 Comment by Manuel Rivera Manuel Rivera 2012-02-15T12:13:07Z 2012-02-15T12:13:07Z Note that the second integral is $\int_0^1 i(T)e_s^*(\omega)ds$, where $i$ denotes interior multiplication, which is an integral in the space of forms: you are averaging the pullback of $\omega$ with respect to the circle action just as we do for compact $G$-manifolds! Moreover, higher iterated integrals can be thought about as averaging over a simplex... http://mathoverflow.net/questions/88114/free-loop-space-and-invariant-forms/88510#88510 Comment by Manuel Rivera Manuel Rivera 2012-02-15T12:06:32Z 2012-02-15T12:06:32Z Yes, I actually wrote this question when I started to read Getzler's paper and now that I got to the section you cite it makes more sense. Also, I realized that we can also think about iterated integrals as averaging forms with respect to the circle action. For example, the simplest case is the following: Given a 1-form $\omega$ we have a function on $LM$ given by $\gamma \mapsto \int_{\gamma} \omega$ this function is the same as $\gamma \mapsto \int_0^1 \omega(e_{s*}T_{\gamma})ds$ where $T$ is the vector field on $LM$ generated by the circle action and $e_s: LM \to M$ is evaluation at $s$. http://mathoverflow.net/questions/88114/free-loop-space-and-invariant-forms Comment by Manuel Rivera Manuel Rivera 2012-02-15T03:45:30Z 2012-02-15T03:45:30Z Oops! Sorry, I wanted to consider the free loop space. I will edit it now. http://mathoverflow.net/questions/84521/a-question-on-generalized-gauss-bonnet-theorem Comment by Manuel Rivera Manuel Rivera 2012-01-06T12:41:06Z 2012-01-06T12:41:06Z There is a very nice exposition of this topic in Morita's Geometry of Differential Forms. http://mathoverflow.net/questions/66508/infinite-configuration-of-lines/66618#66618 Comment by Manuel Rivera Manuel Rivera 2011-06-02T14:17:11Z 2011-06-02T14:17:11Z Thanks Yaakov, this is nice. http://mathoverflow.net/questions/66508/infinite-configuration-of-lines Comment by Manuel Rivera Manuel Rivera 2011-06-01T00:05:33Z 2011-06-01T00:05:33Z @Roland Bacher By minimal polygon I mean a polygon which does not have any segments inside of it. Which is the same as to say minimal with respect to area, i.e. a polygon D is minimal if there is no polygon of smaller area contained in D. http://mathoverflow.net/questions/41514/non-natural-iso-between-homotopy-and-homology Comment by Manuel Rivera Manuel Rivera 2010-10-08T17:11:12Z 2010-10-08T17:11:12Z @ Somnath - Yes, this is what &quot;motivated&quot; the question. I am one of the new guys at Sullivan's class at CUNY and after class last wednesday I was wondering if we can get any interesting information about the geometry of the spaces - not the geometric meaning of the isomorphisms- for which some of these isomorphisms are not the given by the natural Hurewicz map... http://mathoverflow.net/questions/41514/non-natural-iso-between-homotopy-and-homology Comment by Manuel Rivera Manuel Rivera 2010-10-08T15:29:00Z 2010-10-08T15:29:00Z Yes, certainly the question sounds like it lacks of motivation, since crazy maps between homotopy and cohomology and not particularly useful. I was just trying to think what is happening geometrically when you have these &quot;non natural&quot; isomorphisms. http://mathoverflow.net/questions/34878/x-not-simply-connected-and-x-x-contractible/34880#34880 Comment by Manuel Rivera Manuel Rivera 2010-08-08T13:19:06Z 2010-08-08T13:19:06Z What if we strengthen the condition such that $X$ is not simply connected and $X-x$ is contractible for all $x \in X$. Can we now classify $X$ up to homotopy equivalence?