Rational homotopy theory of a punctured manifold - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T16:48:05Z http://mathoverflow.net/feeds/question/20868 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/20868/rational-homotopy-theory-of-a-punctured-manifold Rational homotopy theory of a punctured manifold algori 2010-04-09T21:29:40Z 2011-12-27T04:06:37Z <p>Let $M$ be a smooth simply connected manifold and let $N$ be $M$ minus a point. Is it possible to construct an explicit Sullivan model for $N$ (i.e. a commutative differential graded algebra (cdga) which is connected to the algebra of $\mathbf{Q}$-polynomial forms on $N$ by a chain of cdga quasi-isomorphisms) starting from the minimal Sullivan model for $M$?</p> <p>[upd: in principle the above question is a very particular case of the one discussed in the paper Algebraic models of Poincar\'e embeddings by P. Lambrechts and D. Stanley, AGT 5, 2005. That paper discusses general polyhedra that satisfy some connectivity/codimension assumptions, which are certainly true when the polyhedron is a point. But the general construction involves some non-canonical choices and I was wondering if there is a cleaner and more canonical'' construction when the polyhedron to be thrown away is simply a point.]</p> http://mathoverflow.net/questions/20868/rational-homotopy-theory-of-a-punctured-manifold/20873#20873 Answer by Somnath Basu for Rational homotopy theory of a punctured manifold Somnath Basu 2010-04-09T23:35:38Z 2010-04-09T23:35:38Z <p>One can certainly go the other way if you're wiling to let $N$ be $M$ minus a disk. I'll just recall what is already done in Felix-Halerpin-Thomas's "Rational Homotopy Theory". </p> <p>Take a minimal model $(\Lambda V,d)$ for $N$. Since $M$ can be thought of as attaching a disk on a certain boundary $S^{n-1}$ of $N$, let $f:(S^n,p_0)\to(N,x_0)$ be the attaching map. Then $M=N\cup_{f} D^{n}$. Define a commutative cochain algebra $(\Lambda V\oplus \mathbb{Q} u,D)$ such that (i) $U$ has degree $n$, (ii) $\Lambda V$ is a subalgebra with $u^2=0=u\cdot \Lambda^{+}V$ and (iii) $Du=0$ and $Dv=dv+\left\langle v,f\right\rangle u$ if $v\in V$. Here $v$ is identified with an element of $A_{PL}(X)$, the space of polynomial differential forms, and the pairing is the usual pairing of cohomology and homology. Then $(\Lambda V\oplus \mathbb{Q} u,D)$ is a commutative model for $M$. </p> <p>I hope this helps.</p> http://mathoverflow.net/questions/20868/rational-homotopy-theory-of-a-punctured-manifold/38186#38186 Answer by Mark Grant for Rational homotopy theory of a punctured manifold Mark Grant 2010-09-09T15:32:05Z 2010-09-10T08:51:03Z <p>Perhaps the best you can do is write down the minimal model of the inclusion map $i\colon N\hookrightarrow M$. This will take the form of a minimal <em>relative</em> Sullivan algebra $$(\Lambda V,d)\hookrightarrow(\Lambda V\otimes\Lambda Z,D)\stackrel{\sim}{\to}(\Lambda W,d)$$ where the first map is an inclusion, the second a quasi-isomorphism, and $(\Lambda V,d)$ and $(\Lambda W,d)$ are the minimal models of $M$ and $N$ respectively. (See Chapter 14 of the big yellow book by FHT).</p> <p>Writing down $Z$ and the differential $D$ is left as an exercise! But I believe when you're done the Sullivan fibre $(\Lambda Z,\overline{D})=\mathbb{Q}\otimes_{\Lambda V}(\Lambda V\otimes\Lambda Z,D)$ should have the same cohomology as $S^{n-1}\wedge\Omega M_+$ (where $_+$ denotes a disjoint base point). </p> <p>Note that $(\Lambda V\otimes\Lambda Z,D)$ is not minimal, as $D$ has a linear part in general, but I <em>think</em> it's a Sullivan algebra?</p> http://mathoverflow.net/questions/20868/rational-homotopy-theory-of-a-punctured-manifold/84348#84348 Answer by Vitali Kapovitch for Rational homotopy theory of a punctured manifold Vitali Kapovitch 2011-12-27T01:17:24Z 2011-12-27T04:06:37Z <p>This is an old question but I hope the following is still of interest.</p> <p>If $M^n$ is a closed simply connected manifold then the inclusion <code>$M^n\backslash \{ pt\}\hookrightarrow M^n$</code> is $(n-1)$-connected which means that the induced map of minimal models is an isomorphism through dimension $(n-2)$. Next note that <code>$M^n\backslash \{ pt\}$</code> has zero homology in degrees above $n-2$. It's a general fact that given a minimal model up to dimension $k$ of a space whose cohomology vanishes in degrees above $k$ the rest of the minimal model is determined uniquely (and constructively) from the model up to degree $k$. This provides an easy recipe for computing the minimal model of <code>$M^n\backslash \{ pt\}$</code> which can be more explicitly described as follows.</p> <p>If $(\Lambda V, d)$ is a minimal model of $M^n$ then consider the following dga $(A,d)=(\Lambda V\oplus\Lambda \langle z\rangle/(z^2), d)$ with $\deg z=n-1, V\cdot z=0$ and $dz=[M]$ - the fundamental class of $M$. This is a model (non-minimal and even a non-free one!) of <code>$M^n\backslash \{ pt\}$</code>. In practice it's easier to directly compute the minimal model of $A$ by the general procedure outlined above.</p> <p>Here are a couple of examples.</p> <p>Let $M^4=\mathbb {CP}^2$. Its minimal model is $(\Lambda \langle x,y\rangle,d)$ with deg x=2, deg y=5, dx=0, dy=x^3. Up to degree $n-2=2$ this is simply given by $\Lambda\langle x\rangle$ with dx=0. Next we need another generator to make $H^4=0$ (which is currently generated by $[M]=x^2$) so we add $z$ of deg $3$ such that $dz=x^2$. Now the model $(\Lambda \langle x,z\rangle,d)$ with deg x=2, deg z=3, dx=0, dy=x^2 already has $H^i=0$ for $i\ge 4$ so we don't need to add anything else. The resulting model is easily recognized as the model of $\mathbb S^2$ which is of course not surprising since <code>$\mathbb{CP}^2\backslash\{pt\}$</code> is a Hopf disk bundle over $\mathbb{CP}^1$.</p> <p>A more interesting example: Let $M=\mathbb S^3\times\mathbb S^5$. Its minimal model is generated by $x,y$ with $\deg x=3, \deg y=5$ and $dx=0,dy=0$. Applying our recipe the model of <code>$\mathbb S^3\times\mathbb S^5\backslash \{ pt\}$</code> will be the same through dimension 6. Next, we need to kill off cohomology in degree 8 which is currently generated by $[M]=xy$. So we need another generator $z$ of degree 7 with $dz=xy$. However, adding such generator introduces more cohomology in degrees 10 and 12 generated by $xz$ and $yz$. So we need two more generators $a$ and $b$ with $da=xz, db=yz$. However, adding those introduces yet more cohomology and we need to keep adding more generators. This will continue forever because <code>$\mathbb S^3\times\mathbb S^5\backslash \{ pt\}$</code> is rationally hyperbolic.</p> <p>Lastly, let me mention that operations such as cell attachments (or in this case cell deletions) are usually easier handled by Quillen Lie algebra models which are better suited to work with cofibrations (while Sullivan models are better suited for fibrations). In this particular case it's especially easy. If $(\mathbb L_V,d)$ is a minimal Quillen Lie model of $M^n$ then the model of <code>$M^n\backslash \{ pt\}$</code> is obtained by simply removing a single generator from $V$ corresponding to the fundamental class of $M$.</p>