A (complex) del Pezzo surface is a smooth projective complex surface with ample anticanonical line bundle. Such surface has a degree defined as the self intersection of the canonical divisor. It is well known that the possible degrees run between $d=1$ and $d=9$ and that topologically del Pezzo surfaces are determined by their degree except for $d=8$. If $d$ is different from $8$, then a del Pezzo surface of degree $d$ has to be a generic blow up of the projective plane $\mathbb{P}^2$ in $9-d$ points. But if $d=8$, there are two choices: $\mathbb{P}^2$ blown up in one point and $\mathbb{P}^1 \times \mathbb{P}^1$.

On the other hand, the homotopy group $\pi_4(S^3)$ (which is also $\pi_4(S^2))$ is cyclic of order 2.

My question is: is there a natural relation between the fact that there are two del Pezzo surfaces of degree $8$ and the fact that $\pi_4(S^3)$ is of order $2$ ?

On the face of it, this question might sound vague: it seems that I am looking for an explanation of the existence of a bijection between two randomly choosen sets with two elements. But it is not the case: in fact, there exists a natural relation between these two sets with two elements but it is in the realm of theoretical physics (consider M theory compactified on a Calabi-Yau 3-fold containing a shrinking del Pezzo surface, decouple gravity, the effective low energy five dimensional theory is, for $d=8$, a $SU(2)$ gauge theory, and the various possible choices are parametrized from this point of view by a discrete theta angle, i.e. by the topological choice of a principal $SU(2)$-bundle on $S^5$, i.e. by $\pi_4(SU(2))$) and I would like to know if there exists a more geometric/topological way to understand this relation.

  • 2
    $\begingroup$ Both del Pezzo surfaces of degree 8 are $\mathbb{P}^1$-bundles over $\mathbb{P}^1$, i.e. $S^2$-bundles over $S^2$ (these are special Hirzebruch surfaces). This seems relevant, but I don't know enough topology to know if it actually is. $\endgroup$ Sep 4 '15 at 12:20
  • 5
    $\begingroup$ The relevant homotopy calculation for $S^2$-bundles over $M$ is $Map(M,BDiff(S^2)) = Map(M,BO(3))$, so for $M$ also $S^2$ it's $Map(S^2,BO(3)) = \pi_2(BO(3)) = \pi_1(O(3)) = \pi_1(SO(3)) = \pi_1(SU(2)/Z_2)$. I don't know how to get from there to $\pi_4(SU(2))$. $\endgroup$ Sep 5 '15 at 11:13
  • 5
    $\begingroup$ $\pi_4(S^3)\cong \lim\pi_{k+1}(S^k)$ is the stable homotopy group of spheres. By Pontrjagin construction $\lim\pi_{k+1}(S^k)$ is the same as the group of framed $1$-cobordisms $\Omega_1^{fr}$. a framing on a circle is just an element of $\pi_1(SO)=\pi_1(SO(3))=Z_2$. This gives a natural iso of $\pi_4(S^3)$ and $\pi_1(SO(3))$ which as Allen explained classifies $S^2$ bundles over $S^2$. Can’t speak to the rest of the question. $\endgroup$ Sep 5 '15 at 18:05
  • $\begingroup$ And they are indeed topologically distinct $S^2$-bundles over $S^n$. The intersection form of $\mathbb P^1 \times \mathbb P^1$ is even, while the intersection form of $\mathbb P^2$ blown up at a point is odd. $\endgroup$
    – Will Sawin
    Sep 8 '15 at 22:23

Borrowing from 4 different comments to make a complete answer (and making it community wiki):

We can construct a bijection between the set of del Pezzo surfaces and $\pi_4(S_3)$ using topology. However, the proof that this is a bijection depends on the classification of del Pezzo surfaces and does not give an explanation for why there are exactly two.

Both del Pezzo surfaces of degree $8$ are $S^2$-bundles over $S^2$, by the projection map $\mathbb P^1 \times \mathbb P^1 \to \mathbb P^1$, or by projection from the blown-up point in $\mathbb P^2$ onto a line.

$S^2$-bundles over a manifold $M$ are classified by $$Map(M, BDiff(S^2))= Map(M, BO(3))$$ Thus $S^2$ bundles over $S^2$ are classified by $$Map(S^2, BO(3))= \pi_2 (BO(3)) = \pi_1(O(3)) = \pi_1(SO(3)) = \mathbb Z/2$$

Furthermore the map from del Pezzo surfaces to $S^2$-bundles on $S^2$ is bijective, because the two surfaces are topologically distinct. We can check this by computing the intersection form. By the Kunneth formula $H^2(\mathbb P^1 \times \mathbb P^1)$ has generators $H$ and $V$ with $H \cdot H = 0$, $H \cdot V = 1$, $V \cdot V=0$. On the other-hand by the blow-up formula $H^2(Bl_x( \mathbb P^2))$ has generators $H$ and $E$ with $H \cdot H = 1$, $H \cdot E = 0$, $E \cdot E=-1$. The first intersection form is even, while the second is odd, so the two spaces are not homoeomorphic.

Finally $\pi_4(S^3)$ is the first stable homotopy group of spheres, which by the Pontrjagin construction is the same as the group of framed $1$-cobordisms $\Omega_1^{fr}$. $\pi_1( SO(3))$. A framing on a circle is just an element of $\pi_1(SO) = \pi_1(SO(3))= \mathbb Z/2$. In fact this gives an isomorphism between the stable homotopy group and $\pi_1(SO(3))$.

Composing these bijections, we get a bijection between degree 8 del Pezzo surfaces and $\pi_1(SO(3))$.

  • $\begingroup$ Maybe one thing to add is that Hirzebruch surfaces, that is $\mathbf{P}^1$-bundles over $\mathbf{P}^1$ all have the form $\mathbf{F_n} = \mathbf{P}(\mathcal{O}+\mathcal{O}(n))$, $n \in \mathbf{Z}$, and their diffeomorphism type depends only on $n \mod{2}$: map.mpim-bonn.mpg.de/Hirzebruch_surfaces Finally, $\mathbf{F}_0 = \mathbf{P}^1 \times \mathbf{P}^1$ and $\mathbf{F}_1 = Bl_{x}(\mathbf{P^2})$. $\endgroup$ Mar 20 '20 at 23:20

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