Let $M$ be a simply connected topological 4manifold with intersection form given by the E8 lattice. Does anyone know of examples of continuous selfmaps of $M$ of degree 2 or 3? Or of degree any other prime for that matter?
3 Answers
There are plenty of integer matrices $A$ with $A^T E_8 A = d E_8$, which give plenty of maps, as in Oscar's answer.
First, for $d=1$, these are the automorphisms of the $E_8$ lattice. There are 696729600 of these.
Second, for any odd prime $p$, the E_8 quadratic form splits mod $p$, hence we can find a rank $4$ isotropic subspace. Consider the sublattice of elements that are in that subspace mod $p$. It is an index $p^4$ lattice, hence has discriminant $p^8$, and the quadratic form on it is divisible by $p$. Dividing by $p$, we obtain a unimodular lattice. It remains even and positive definite, so it is the $E_8$ lattice again. Fixing an isomorphism with $E_8$ describes a degree $p$ endomorphism of $E_8$.
In fact the number of such subspaces should be $2(p^6+p^5+p^4+2p^3+p^2+p+1)(p^4+p^3+2p^2+p+1)(p^2+2p+1)$, so this gives $696729600\cdot 2(p^6+p^5+p^4+2p^3+p^2+p+1)(p^4+p^3+2p^2+p+1)(p^2+2p+1)$ selfmaps of degree $p$.
I think something similar can be done as well to produce degree $2$ endomorphisms, just with a little more care.



$\begingroup$ The following paper gives a complete answer: [arxiv.org/abs/math/0207177 ] (On the Existence of Similar Sublattices, J.H. Conway, E.M. Rains, N.J.A. Sloane). $\endgroup$– degreeNov 8, 2017 at 0:36
Such a map $f : M \to M$ of degree $d >0$ satisfies, with respect to the cupproduct pairing $\langle ,  \rangle$, $$\langle f^*(x), f^*(y) \rangle = d \langle x, y \rangle.$$
Conversely, I claim that any integer matrix $A$ satisfying $$A^T E_8 A = d E_8$$ arises as $A = f^*$ for a map $f : M \to M$, necessarily of degree $d$. To see this, build $M$ as a CWcomplex from $\bigvee_8 S^2$ by attaching a 4cell along the map $g : S^3 \to\bigvee_8 S^2$ dictated by the $E_8$ form. The composition $$S^3 \overset{g}\to \bigvee_8 S^2 \overset{A}\to \bigvee_8 S^2$$ is the map dictated by the form $A^T E_8 A = d E_8$, so is $d \cdot g$. In particular it becomes nullhomotopic when composed with $\bigvee_8 S^2 \subset M$, giving a map $$f: M = (\bigvee_8 S^2) \cup_g D^4 \to M$$ which induces $A$ on second cohomology.
Taking e.g. $A=\lambda \cdot \mathrm{Id}$ yields a selfmap of degree $d=\lambda^2$. I am not sure whether it is possible to produce nonsquare degrees.
This is more like a long comment than a real answer.
This seems like it's an algebra problem that is probably hard to solve. If you had such a map, of degree $d$, then you would get the following. Choose a basis for the homology, so that the induced map on $H_2(M)$ is written as a matrix $A$. Let J be the matrix for the intersection form with respect to this same basis. Then I think you are asking for $A^TJA = dJ$ (1). I don't have much of a feel for this but somehow it seems unlikely that there is any such matrix.
Conversely, if you had such a matrix, then perhaps you could define a map of degree $d$ from M to itself. I would use the approach of the WhiteheadMilnor theorem; namely up to homotopy, you can write $M$ as a wedge of spheres union a $4$cell, where the attaching map for the $4$cell is determined by the matrix $J$. The matrix $A$ tells you how to map the $2$skeleton to the $2$skeleton, and I would guess that you get an extension over the $4$cell with degree $d$ if (1) holds.