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Reimundo Heluani
Reimundo Heluani
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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.

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

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

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.

Source Link
Reimundo Heluani
Reimundo Heluani
  • 3.2k
  • 22
  • 33

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

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