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All K3 surfaces have the same homotopy type. What are their higher homotopy groups?

I know that $\pi_1$ is trivial, and $\pi_2$ is $\mathbb{Z}^{22}$.

Even if the answer isn't known in all degrees, I'll accept an answer if someone can give me $\pi_3$.

According to this question, you can equivalently tell me the higher homotopy groups of an Enriques surface.

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    $\begingroup$ "Even if the answer isn't known in all degrees..." I already can't tell you all the higher homotopy groups of $\mathbb{CP}^1$! $\endgroup$ – Qiaochu Yuan Nov 3 '14 at 19:46
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    $\begingroup$ But the higher homotopy gorups of an elliptic curve is easy to find, and this is definitely a better analogy for a K3 surface. $\endgroup$ – Sasha Nov 3 '14 at 19:56
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    $\begingroup$ @Sasha: as Reimundo's answer makes clear, homotopy-theoretically a K3 surface behaves very differently from an elliptic curve. In particular, it's rationally hyperbolic, so the dimensions of its rational homotopy groups grow exponentially quickly. $\endgroup$ – Qiaochu Yuan Nov 3 '14 at 20:26
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    $\begingroup$ A new entry in the encyclopedia of integer sequences, perhaps? $\endgroup$ – Marty Nov 3 '14 at 23:57
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In the paper http://arxiv.org/abs/1303.3328 by Samik Basu and Somnath Basu, it is claimed (Theorem A) that the homotopy groups of a simply connected closed 4-manifold $M$ are determined by the second Betti number $k$. In particular, if $k \geq 1$ and $j \geq 3$, $\pi_j (M) = \pi_j (\#^{k-1} S^2 \times S^3)$.

For example, using what we know about homotopy groups of spheres, they prove (Corollary 4.10) that if the second Betti number of $M$ is $k+1$ (sorry for the shift from $k$ to $k+1$, I keep the notations of the paper) then

$\pi_3(M) = \mathbb{Z}^{k(k+3)/2}$

$\pi_4(M)=\mathbb{Z}^{(k-1)(k+1)(k+3)/3} \oplus (\mathbb{Z}_2)^{2k}$

For a K3 surface $M$, $k+1=22$ so

$\pi_3(M) = \mathbb{Z}^{252}$

$\pi_4(M) =\mathbb{Z}^{3520} \oplus (\mathbb{Z}_2)^{42}$

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Take a look at Example 6 in [1] which computes $rk \,\pi_2 = 22$, $rk\: \pi_3= 252$ and $rk \: \pi_4 = 3520$ simply using the fact that a quartic in $\mathbb{P}^3$ is a K3 surface and that for such quartics is easy to compute the second betti number $b_2$. The main theorem in Terzic's article is about computing the ranks of $\pi_{2,3,4}$ in terms of $b_2$.

[1]Terzić, Svjetlana. On rational homotopy of four-manifolds. Contemporary geometry and related topics, 375–388, World Sci. Publ., River Edge, NJ, 2004.

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    $\begingroup$ The arxiv link to Terzic paper: arxiv.org/abs/math/0309076 $\endgroup$ – Reimundo Heluani Nov 3 '14 at 20:15
  • $\begingroup$ True, but it would take a while to summarize the methods used in the article. It was easier to point to the example. The full Theorem uses properties of the minimal model and the main Theorem computes the ranks of lower homotopy groups of 4-manifods as a polynomial in the second betti number $\endgroup$ – Reimundo Heluani Nov 3 '14 at 20:23
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Here are some details on how to compute the rank of $\pi_3$. A K3 surface $X$ is, in addition to being simply connected, a compact Kähler manifold, and such spaces are known to be formal in the sense of rational homotopy theory; this means that their rational homotopy can be computed by finding a Sullivan minimal model of their rational cohomology rings, and in particular only depends on the rational cohomology ring. (Terzić's paper linked to in Reimundo's answer uses instead that a compact oriented simply connected $4$-manifold is formal.)

Here is, briefly, how this computation works out, at least if I'm not misreading something. The goal is to build a graded rational vector space $V^{\bullet} = \bigoplus_{k \ge 2} V^k$ and a differential $d$ on the exterior algebra $\Lambda^{\bullet}(V)$ such that

  • the cohomology of $(\Lambda^{\bullet}(V), d)$ agrees with $H^{\bullet}(X, \mathbb{Q})$,
  • $dV$ is contained in $\Lambda^{\ge 2}(V)$.

The machinery of rational homotopy theory, together with the fact that $X$ is formal and has homology of finite type, then guarantees that we have a natural identification

$$\pi_{\bullet}(X) \otimes \mathbb{Q} \cong \text{Hom}_{\mathbb{Q}}(V^{\bullet}, \mathbb{Q}).$$

In particular, $\dim \pi_{\bullet}(X) \otimes \mathbb{Q} = \dim V^{\bullet}$. So to compute $\dim \pi_3(X) \otimes \mathbb{Q}$ it suffices to figure out how many elements we need in $V^3$.

We already know that we need $\dim V^2 = b_2$, where $b_2 = \dim H^2(X, \mathbb{Q}) = 22$. The cup product $H^2(X, \mathbb{Q}) \times H^2(X, \mathbb{Q}) \to H^4(X, \mathbb{Q})$ takes the form

$$\alpha \cup \beta = Q(\alpha, \beta) \gamma$$

where $Q(\alpha, \beta)$ is the intersection form and $\gamma$ is a generator of $H^4(X, \mathbb{Q})$. The only way to impose these relations on the cohomology of $(\Lambda^{\bullet}(V), d)$ is to introduce elements in $V^3$ whose differentials will impose those relations. Explicitly, let $e_1, e_2, \dots e_{22}$ be an orthogonal basis for $H^2(X, \mathbb{Q})$ with respect to the intersection form, so that $Q(e_i, e_j)$ is some nonzero multiple of $\delta_{ij}$. For $i \neq j$ we need to introduce $\frac{b_2(b_2 - 1)}{2} = 231$ new elements of $V^3$, call them $f_{ij}, i \neq j$, so that we can impose the relations

$$d f_{ij} = e_i \cup e_j.$$

For $i = j$ we need to introduce introduce $b_2 - 1 = 21$ new elements of $V^3$, call them $f_i, 1 \le i \le 21$, so that we can impose the relations

$$d f_i = \frac{e_i \cup e_i}{Q(e_i, e_i)} - \frac{e_{i+1} \cup e_{i+1}}{Q(e_{i+1}, e_{i+1})}.$$

(We cannot introduce the generator of $H^4(X, \mathbb{Q})$ into $V^4$ because we cannot impose a relation that is linear in this generator, so instead we impose the relation that all of the $e_i$ square, up to a normalization, to the same thing.)

Altogether, we get

$$\dim V^3 = {b_2 \choose 2} + (b_2 - 1) = {b_2 + 1 \choose 2} - 1 = 252$$

as expected from the other answers.

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    $\begingroup$ I guess you could say that $V_3$ should be the kernel of the multiplication map $S^2H^2(X,Q) \to H^4(X,Q)$, so its dimension is $b_2(b_2+1)/2 - 1 = 252$. Of course this is equivalent to your computation, but does not require choosing a basis and a bit simpler. $\endgroup$ – Sasha Nov 4 '14 at 12:25

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