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Oct 22, 2021 at 16:19 comment added David E Speyer The previous comment works for $n \equiv 3 \bmod 4$; I don't see a comparable issue when $n \equiv 1 \bmod 4.
Oct 22, 2021 at 15:57 vote accept Jef
Oct 22, 2021 at 15:56 comment added David E Speyer To be concrete, in the type $A$ example, $\Lambda/2\Lambda$ is $\{(x_1, x_2, \dots, x_{n+1}) \in \mathbb{F}_2^{n+1} : \sum x_i=0 \}$. For $n$ odd, the fixed $F$ space of $w$ is spanned by $(1,1,\ldots,1)$. But $(1,0,1,0,\ldots)$ is fixed in $\Lambda/(2 \Lambda+F)$.
Oct 22, 2021 at 11:39 history edited Andrei Smolensky CC BY-SA 4.0
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Oct 22, 2021 at 11:28 history edited Andrei Smolensky CC BY-SA 4.0
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Oct 22, 2021 at 11:27 history undeleted Andrei Smolensky
Oct 22, 2021 at 11:26 history deleted Andrei Smolensky via Vote
Oct 22, 2021 at 11:20 history answered Andrei Smolensky CC BY-SA 4.0