Several years ago, I mentioned offhandedly to a colleague that I had noticed that, if you extend the $\mathsf E\_n$ series downwards in the natural way, by removing nodes from the long arm of $\mathsf E\_8$, then the fundamental group of $\mathsf E\_n$ (i.e., the quotient of the weight lattice by the root lattice) is cyclic of order $9 - n$ for $3 \le n \le 8$. (Specifically, $\mathsf E\_5 = \mathsf D\_5$, $\mathsf E\_4 = \mathsf A\_4$, and $\mathsf E\_3 = \mathsf A\_2 + \mathsf A\_1$.) I thought that this was just a funny coincidence, but my colleague said that there was actually more to it, and gave me a beautiful explanation of the deeper reason.


Unfortunately, as is the way of such things, I did not write down the beautiful explanation and so forgot it; and, upon reaching out to my colleague several years later, I find that they can no longer re-construct their beautiful explanation. Can you provide any explanation other than "that's how it works out"?



Why is the fundamental group of $\mathsf E_n$ cyclic of order $9 - n$?
If we define the $E\_n$ lattice in the algebraic geometer's way as the orthogonal complement of $(1,\dots, 1, 3)$ in the Lorentzian unimodular lattice $I\_{n,1}$, i.e. in $\mathbb Z^{n+1}$ with intersection form $$\langle(x\_1,\dots, x\_{n+1}), (y\_1,\dots, y\_{n+1})\rangle =x\_1y\_1 + x\_2 y\_2 + \dots + x\_n y\_n - x\_{n+1} y\_{n+1}, $$


then there is an immediate proof: because $(1,\dots, 1, 3)$ is a primitive vector of norm $9-n$ and the complement of a primitive vector of norm $k$ in a unimodular lattice always has fundamental group of norm $k$ (with the isomorphism of the dual lattice modulo the lattice to $\mathbb Z/k$ on the dual lattice given by the pairing with that vector).


One can check that the lattice given by this construction has the $E\_n$ Dynkin diagram for all $n\geq 3$ using the roots given by all vectors with a series of $0$s, then a $1$, then a $-1$, then a series of at least one $0$, and the vector $(0,\dots, 0,1,1,1,1)$.




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Another perspective:


For any series of Dynkin diagrams where we add one vertex after a number in a line, a simple recurrence shows the determinant of the Cartan matrix (= order of the fundamental group) of the $n$th member of the series must be a linear function of $n$. Finding the right linear function for $E\_n$ then requires computing two examples.


Given that the fundamental group has order $9-n$ for $n\geq 3$, the fact that it is cyclic is nontrivial only for $n=5$, so not much of a coincidence.




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A third perspective: A $k$-torsion element of the dual lattice modulo the lattice is the same thing as an element of the lattice tensor $\mathbb Z/k$ with zero dot product with every vector of the lattice, i.e. a $\mathbb Z/k$-linear combination of the simple roots with zero dot product with any root. These can be calculated directly from the Dynkin diagram.


On an $E\_n$-type lattice, if the value at the vertices near the end are


$$ \begin{array}{ccc} a & b & c \\ d & & \end{array}$$


then $a=2d$ and $a+c=2b$ and $b=2c$ so $a=3c$ so because $2d=3c$ we have $c=2(d-c)$, $b= 4(d-c)$, $a=6(d-c)$, and $c=3(d-c)$, and then, continuing left, every vertex is a multiple of $d-c$. This shows every element of the fundamental group is a multiple of


$$ \begin{array}{ccc} \dots & 2 & 3 & 4 & 5 & 6 & 4 & 2 \\&&&&& 3 & & \end{array}$$


showing the fundamental group is cyclic, and this element is $9-n$-torsion since it is only satisfies the orthogonality condition in $\mathbb Z/k$ if the label $9-n$ of the unique vertex of $E\_{n+1}$ that's missing in $E\_n$ is divisible by $k$, showing the fundamental group is cyclic of order $9-k$.


This proof, despite appearances, works for $n=3$, where we just have the additional relation $a=0$ to work with.




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All three arguments work for $n>9$ as well as $n\leq 8$, with the first and third arguments still determining the fundamental group and the second argument determining its order.


