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I want to determine the Sullivan model of an odd sphere $S^{2n+1}$. Let $(A,d)$ be a cdga such that $H^*(S^{2n+1};\mathbb Q)\cong H^*((A,d_A))$ as graded algebras. Hence

$$H^*((A,d))\cong H^0(S^{2n+1})\oplus H^{2n+1}(S^{2n+1}) $$

where $H^0(S^{2n+1})$ is isomorphic to the field $\mathbb Q$ with basis $\{1\}$ and $H^{2n+1}(S^{2n+1})$ is a one dimensional $\mathbb Q$-vector space with basis denoted by $\{e\}$. On the other hand we compute $H^*((A,d_A))$ going back to the definition. Write $A=\bigoplus_{i\geq 0}A^i$ then we have

$$0\longrightarrow A^0 \xrightarrow{d_0} A^1 \xrightarrow{d_1}A^2 \xrightarrow{d_1}\cdots A^{2n} \xrightarrow{d_{2n}} A^{2n+1} \xrightarrow{d_{2n+1}} A^{2n+2} \cdots $$ Here $H^0((A,d))$ is always supposed to be isomorphic to the field $\mathbb Q$ with basis $\{1\}$. And $H^{2n+1}((A,d))=Ker(d_{2n+1})/Im(d_{2n})$ which we said is isomorphic to $H^{2n+1}(S^{2n+1})$. Hence $Ker(d_{2n+1})/Im(d_{2n})$ is a one dimensional $\mathbb Q$-vector space with basis the class $[v]$ of a cocycle $v\in Ker(d_{2n+1})\subset A^{2n+1}$. Let $V$ denote the vector space with basis the cocyle $v$. The free cdga on $V$ is an exterior algebra on $v$ denoted $\Lambda V$. This is a cdga with grading $ \Lambda V=\Lambda^0V\oplus\Lambda^1V$ where $\Lambda^0V$ is isomorphic to the field $\mathbb Q$ with basis $\{1\}$ and $\Lambda^1V$ is the one dimensional $\mathbb Q$-vector space $V$ with basis the cocycle $v$. We claim that the cdga $\Lambda V$ with zero differential is quasi-isomorphic to $(A,d)$. An extension result assures that the cdga inclusion $(V,0)\hookrightarrow (A,d)$ extends to a unique cdga morphism $$m:(\Lambda V,0)\hookrightarrow (A,d)$$ Now we show that the induced map in cohomology $H(m):H^*((\Lambda V,0))\longrightarrow H^*((A,d))$ is an isomorphism. where the cdga $(\Lambda V,0)$ is written $$0\longrightarrow \Lambda^0 V=\mathbb Q \xrightarrow{0} \Lambda^1 V=V \xrightarrow{0}0 \xrightarrow{0}\cdots 0 $$ Thus $H^*((\Lambda V,0))=H^0((\Lambda V,0))\oplus H^1((\Lambda V,0))$ while $H^*((A,d))=H^0((A,d))\oplus H^{2n+1}((A,d))$. The problem now is that $H(m)_1:H^1((\Lambda V,0))\longrightarrow H^1((A,d))=0$ is not an iso. Also $H(m)_{2n+1}:H^{2n+1}((\Lambda V,0))=0\longrightarrow H^{2n+1}((A,d))$ is not an iso.. thank you for correcting any detail in the above computations and to helping me prove that $m$ is a quasi-isomorphism.

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The grading on $\Lambda V$ is wrong -- you still want $\nu$ to be in degree $2n+1$. Fixing this solves both problems. All the other cohomology of $\Lambda V$ and $(A,d)$ is zero, so once you've established that $H(m)$ is an isomorphism in degrees $0$ and $2n+1$, it follows that $m$ is a quasi-isomorphism.

Here's a general algorithm for constructing Sullivan minimal models of simply connected dgas, half because I feel like you deserve a better answer and half because I just learned it. Inductively let $f_n:(M(n),d_{M(n)}) \to (A,d_A)$ be a morphism with $M(n)$ a minimal model and $f_n$ inducing an isomorphism on cohomology in degrees $<n$ and an injection on cohomology in degree $n$. Recall that the mapping cone of $f_n$ is defined as $C(n)^k = M(n)^k \oplus A^{k-1}$, with differential $d_{C(n)}(m,a) = (d_{M(n)}(m), f_n(m) - d_A(a))$. We define relative cohomology as $H^k(M(n), A) = H^k(C(n))$. If $V(n) = H^{n+2}(M(n), A)$, then $V(n)$ is a quotient of $Z^{n+2}(C(n))$; choosing a splitting of this map, and projecting to $M(n)^{n+2}$, gives a map $g:V(n) \to Z^{n+2}(M(n))$. We now define $M(n+1)$ to be the tensor product of $M(n)$ and the free cdga generated by $V(n)$ in degree $n+1$, with $d_{M(n+1)}(v) = g(v)$ for $v \in V(n)$. The colimit of these $M(n)$ will evidently be a Sullivan model for $A$.

It's an (only moderately difficult) exercise to show that each $M(n)$ is well-defined and that they have the described properties by induction. It's important that $A$ be simply connected, so that when adding the new generators in degree $n+1$ we don't get any undesired ones in degree $n+2$. If you use this method, then so long as you get the degrees right, you'll know you're on the right track since you'll always have the right cohomology up to a certain degree; if the cohomology of $A$ is bounded above, as in the case of $S^{2n+1}$, the process will actually stop. You can and should think of the $M(n)$ as a Postnikov tower for $A$.

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thank you Paul, i will read the details of your answer and try to understand it.. But first i want to know why the grading is wrong? a free cdga on a one dimensional vector space has only one grading $\Lambda V=\Lambda^0 V\oplus \Lambda^1V$ where $\Lambda^0 V=\mathbb Q$ and $\Lambda^1V\cong V$, do you suggest that $\Lambda V=\Lambda^0 V\oplus \Lambda^{2n+1}V$? is this possible, can we shoose any grading on $\Lambda V$? – palio Feb 13 '13 at 8:57
I believe you can since you are constructing the model. You are really picking a graded vector space and then applying $\Lambda$, if I recall correctly... – Sean Tilson Feb 13 '13 at 14:43
@palio: Don't think of it as a free cdga on a vector space. Think of it as a free cdga on a graded vector space. Since the generator $v$ has odd degree, this must be an exterior algebra. If $v$ had even degree, the free cdga on it would be a polynomial algebra, and you'd have to have extra classes to kill off its powers, which explains the difference between the odd and even spheres. – Paul VanKoughnett Feb 14 '13 at 5:32
The grading on $\Lambda V$ is defined by $\Lambda V=\oplus_{i\geq 0}\Lambda^iV$ where $\Lambda^iV$ consists of words of length $i$. Here $v$ is of length $1$ so it must be in $\Lambda^1V$, hence $\Lambda V=\Lambda^0V\oplus \Lambda^1V$ is the only possible grading in this case.. am i wrong???? – palio Oct 15 '13 at 10:07
You're right, but $V = \Lambda^1 V$ has degree $2n+1$, by definition. You can pick the grading on the vector space to begin with, and then apply the exterior algebra construction. – Paul VanKoughnett Oct 17 '13 at 2:39

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