# Rational homotopy theory of a punctured manifold

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 $M$?

[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.]

• A similar question is to recover rational cohomology of N from rational cohomology of M. Mayer-Vietoris helps in this task. Is there Mayer-Vietoris for DGA's? – Igor Belegradek Apr 9 '10 at 21:54
• Igor -- yes, one can do Mayer-Vietoris for cdga's, the problem is that models one gets in this way are usually non-Sullivan, let alone minimal. – algori Apr 10 '10 at 15:08

This is an old question but I hope the following is still of interest.

If $M^n$ is a closed simply connected manifold then the inclusion $M^n\backslash \{ pt\}\hookrightarrow M^n$ is $(n-1)$-connected which means that the induced map of minimal models is an isomorphism through dimension $(n-2)$. Next note that $M^n\backslash \{ pt\}$ 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 $M^n\backslash \{ pt\}$ which can be more explicitly described as follows.

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 $M^n\backslash \{ pt\}$. In practice it's easier to directly compute the minimal model of $A$ by the general procedure outlined above.

Here are a couple of examples.

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 $\mathbb{CP}^2\backslash\{pt\}$ is a Hopf disk bundle over $\mathbb{CP}^1$.

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 $\mathbb S^3\times\mathbb S^5\backslash \{ pt\}$ 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 $\mathbb S^3\times\mathbb S^5\backslash \{ pt\}$ is rationally hyperbolic.

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 $M^n\backslash \{ pt\}$ is obtained by simply removing a single generator from $V$ corresponding to the fundamental class of $M$.

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".

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 $M$.

I hope this helps.

• Somnath -- yes, going the other way is standard, but the model one gets in this way is not minimal so it is not clear how to go back to $N$. – algori Apr 10 '10 at 15:00

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 relative 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).

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).

Note that $(\Lambda V\otimes\Lambda Z,D)$ is not minimal, as $D$ has a linear part in general, but I think it's a Sullivan algebra?

• Mark -- thanks. It seems that what is "left as an exercise" here is basically the original question itself. Re the proposal for the Sullivan fiber: I think there may be problems with it: the Sullivan fiber models the Serre fiber; if one takes $M=S^3$, the recipe you mention gives $H^2$ of the fiber of rank 0, so when one considers the Serre spectral sequence of the open embedding $\mathbf{R}^3\subset S^3$, there will be no one to kill $H^3$ of the sphere. This would give a nontrivial cohomology class on $\mathbf{R}^3$. – algori Sep 9 '10 at 21:21
• @algori -- oops, I forgot the disjoint basepoint! I meant the fibre to have the same cohomology as the Thom space of the trivial $\mathbb{R}^n$ bundle over $\Omega M$, shifted by one. I'll edit it. With regards the exercise, let $\omega\in \Lambda V$ be a fundamental cocycle. You need a $w\in Z$ to kill $\omega$. Now for any cocycle $1\neq x\in \Lambda V$ we have $x\omega = d(b)=D(xw)$, for some $b\in \Lambda V$, so you need a $\beta\in Z$ with $D(\beta)=b-xw$. Done systematically, this should give the answer (although I haven't worked out all the details)! – Mark Grant Sep 10 '10 at 8:48
• Mark -- thanks, that's interesting. Do you think that the Serre fiber itself will be $\Omega M+\wedge S^{n-1}$? – algori Sep 10 '10 at 15:23
• No. Your example with $M=S^n$, $N=\mathbb{R}^n$ has homotopy fibre $\Omega S^n$. What I should have said was that the Serre fibre $F$ and $\Omega M_+\wedge S^{n-1}$ have isomorphic reduced cohomology {\em groups} (but not rings). There is a fibration $\Omega M\to F\to N$ (which is the restriction of the based path fibration $\Omega M\to PM\to M$ to $N\subseteq M$). This fibration is trivial when $M$ is a sphere, but doesn't even have a section if the LS-category of $M$ is greater than one. So I don't know if we can identify the homotopy type of $F$ in general. Worthy of a separate MO qn? – Mark Grant Sep 14 '10 at 7:47
• Mark -- you are right, I got carried away. And yes, explicitly identifying the Serre fiber of an open embedding may be the topic of a separate question. Maybe I'll have a go at it some day. – algori Sep 16 '10 at 1:04