# Rational homotopy groups of a projective hypersurface

Let $X$ be a smooth projective hypersurface in $\mathbb{P}^n$. Has anyone computed the rational homotopy groups $\pi_i(X)\otimes \mathbb{Q}$ of $X$? I tried Google, but did not find anything.

One could use the formality and Sullivan's minimal model to compute these groups up to some level by hand. And in principle this can be done for all the groups. But is there any nice way of organising these things? For example, give the ranks using some generating functions, or express the generators in a nice way?

• any hypersurface becomes a hyperplane section after a suitable Veronese embedding of the ambient projective space. So by Lefshetz hyperplane theorem for Homotopy groups, for $i<n-1, \pi_{i}(X)$ is isomorphic to $\pi_{i}(\mathbb{CP}^{n})$. These Homotopy groups can be calculated by applying the long exact sequence to the $S^{1}$-bundle $S^{2n+1} \rightarrow \mathbb{CP}^{n}$. – Nick L May 16 '18 at 9:33
• Does this also work for étale homotopy groups over arbitrary fields? – TKe May 16 '18 at 13:02
• I don't know, sorry. – Nick L May 16 '18 at 13:11
• For the étale fundamental group: hypersurfaces are almost always simply connected. – TKe May 16 '18 at 13:13
• I found something for étale homotopy groups I can't read: repository.kulib.kyoto-u.ac.jp/dspace/bitstream/2433/61752/1/… – TKe May 16 '18 at 13:54

Let $$X$$ be a degree $$d$$ hypersurface in $$\mathbb{C}\mathbb{P}^{n+1}$$, and denote its topological Euler characteristic by $$\chi$$ (note that $$\chi$$ depends only on $$d$$ and $$n$$). If $$\chi = n+1$$, then the only rationally non-zero homotopy groups of $$X$$ are $$\pi_2 (X) \otimes \mathbb{Q} = \pi_{2n+1} (X) \otimes \mathbb{Q} \cong \mathbb{Q}$$. If $$\chi \neq n+1$$, then for any $$j\geq 1$$ we have $$\dim\pi_{j+1}(X)\otimes \mathbb{Q} = \frac{(-1)^j}{j} \sum_{d|j, \ \ d\geq 1} (-1)^d \mu\bigl(\frac{j}{d}\bigr)\sum_{\alpha=1}^{2n-2} \xi_\alpha^{-d},$$
where $$\xi_\alpha$$ are the roots of $$1-(-1)^n(\chi-n-1)(1+z)z^{n-1} + z^{2n-1}$$ that are not the root $$-1$$, and $$\mu$$ is the Möbius function.
For example, in the case of a $$K3$$ surface, $$n=3$$ and $$\chi = 24$$, and we get $$\pi_3(K3)\otimes \mathbb{Q} \cong \mathbb{Q}^{252}$$, $$\pi_4(K3)\otimes \mathbb{Q} \cong \mathbb{Q}^{3520}$$, $$\pi_5(K3)\otimes \mathbb{Q} \cong \mathbb{Q}^{57960}$$, $$\pi_6(K3)\otimes \mathbb{Q} \cong \mathbb{Q}^{1020096}$$, $$\pi_7(K3)\otimes \mathbb{Q} \cong \mathbb{Q}^{18664800}$$, $$\pi_8(K3)\otimes \mathbb{Q} \cong \mathbb{Q}^{351204480}$$, $$\ldots$$. This is consistent with Corollary 4.10 in Samik Basu and Somnath Basu's Homotopy groups and periodic geodesics of closed 4-manifolds and the answers to this question What are the higher homotopy groups of a K3 suface?
For convenience, here is a formula for the topological Euler characteristic of a degree $$d$$ hypersurface in $$\mathbb{C}\mathbb{P}^{n+1}$$ (simplified thanks to abx's comment); $$\chi = \frac{1}{d}((1-d)^{n+2} + d(n+2) -1).$$
• Note that the formula for $\chi$ can be expressed as $\frac{1}{d}\left[ (1-d)^{n+2}+d(n+2)-1\right]$, which seems easier to compute... – abx Sep 25 '18 at 15:43