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Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Dec 18, 2015 at 1:08 vote accept Keivan Karai
Dec 16, 2015 at 6:15 answer added Ramin timeline score: 0
Dec 16, 2015 at 5:26 comment added nfdc23 For non-degenerate quadratic spaces $(V, q)$ and $(V', q')$ of dimension $n \ge 3$ over a field $k$, any $k$-group isomorphism ${\rm{SO}}(q) \simeq {\rm{SO}}(q')$ arises from a conformal isometry $q \simeq q'$ that is unique up to $k^{\times}$-scaling on $V$. (Indeed, Hilbert 90 reduces this to the case $k=k_s$, so then we may assume $q' = q$ and use Dieudonne's theorem (or better: the Isomorphism Theorem for split reductive groups) that the natural map ${\rm{PGO}}(q) \rightarrow {\rm{Aut}}_{{\rm{SO}}(q)/k}$ is an isomorphism.) Thus, ${\rm{SO}}(q)$ "remembers" $q$ up to conformal isometry.
Dec 16, 2015 at 0:34 history asked Keivan Karai CC BY-SA 3.0