Timeline for Why is the fundamental group of $\mathsf E_n$ cyclic of order $9 - n$?
Current License: CC BY-SA 4.0
10 events
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| Sep 30, 2022 at 0:31 | comment | added | Will Sawin | @LSpice If you define the fundamental group as the quotient of the duall lattice by the lattice, now allowing indefinite lattices, it means the fundamental group has order $|9-n|$ for all $n \neq 9$. | |
| Sep 30, 2022 at 0:15 | comment | added | LSpice | By the way, I only realised on much later reflection: what does it mean that the argument for the fundamental group having order $9 - n$ works for $n > 9$? | |
| Sep 19, 2022 at 13:24 | vote | accept | LSpice | ||
| Sep 19, 2022 at 13:14 | comment | added | Will Sawin | @LSpice You have interpreted the second argument in exactly the spirit it was meant. | |
| Sep 19, 2022 at 13:13 | comment | added | Will Sawin | @LSpice Why? The original vector was orthogonal to every root but the last one, truncating only changes the dot product with roots of $E_n$ adjacent to roots you deleted, there's only one root you deleted adjacent to roots of $E_n$, and so the dot product is only changed by the coefficient of that root, which after dividing is $1$, so $0$ after modding out by integers. The argument can be generalized to show that for every graph made by gluing three paths at a point, the "fundamental group" is cyclic if the lengths of two of the paths are relatively prime. | |
| Sep 19, 2022 at 13:10 | comment | added | Will Sawin | @LSpice Let me say it like this. Take the highest root of $E_8$. Restrict the linear combination to $E_n$ by throwing away the coefficients of roots that are not in $E_n$. You now have a vector of $E_n$. You now divide by the coefficient of the unique root in $E_{n+1}$ that's not in $E_n$. You now have a vector in $\mathbb Q E_n$, which as a vector in $\mathbb Q E_n/ E_n$ is orthogonal to all the roots of $E_n$. | |
| Sep 19, 2022 at 13:07 | comment | added | LSpice | (Perhaps, as you say, that $\pi_1(\mathsf E_n) = \operatorname C_{9 - n}$ is mildly more surprising than that $\pi_1(\mathsf D_n)$ always has order $4$, since $\pi_1(\mathsf D_n)$ is only sometimes cyclic.) | |
| Sep 19, 2022 at 13:04 | comment | added | LSpice | Other than that, of these the third argument strikes me as somewhat closer to showing that it's true, but not really as much why it's true—but of course this will always be a matter of taste. $\newcommand\s{\mathsf}\newcommand\C{\operatorname C}$The second argument is more in the 'why' spirit, and I guess suggests I shouldn't see it as more surprising that $\pi_1(\s E_n) = \C_{9 - n}$ than that $\pi_1(\s A_n) = \C_{n + 1}$ or perhaps even that $\pi_1(\s B_n) = \pi_1(\s C_n) = \C_2$ and that all the $\pi_1(\s D_n)$'s have order $4$. And the first argument is definitely a winner. Thank you! | |
| Sep 19, 2022 at 12:59 | comment | added | LSpice | I am a bit confused by the third argument, doubtlessly entirely my fault. Your generator of the fundamental group is the highest root of $\mathsf E_8$, which means it's not orthogonal to the $\mathsf E_8$ weights (but maybe the dots indicate that I should extend it in some way so it is?), and, more importantly, it seems to live in the wrong space—it is written as a linear combination of roots involving the "$8$th simple root" of $\mathsf E_8$ non-trivially, so can't possibly lie in $\mathbb Q\mathsf E_n$ for $n < 8$. What am I misunderstanding? | |
| Sep 19, 2022 at 2:23 | history | answered | Will Sawin | CC BY-SA 4.0 |