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Mar 14, 2020 at 16:32 comment added L. Xie @Enrico Thanks, if I take $x_{12}=x_{34}$, $x_{13}=x_{45}$, $x_{23}=x_{15}$, then by a calculation I get two nonintersecting lines: one joining $( x_{14}=1,x_{ij}=0)$ and $( x_{12}=1,x_{ij}=0)$, the other joining $( x_{24}=1,x_{ij}=0)$ and $( x_{25}=1,x_{ij}=0)$. This should solve the question.
Mar 14, 2020 at 14:31 comment added Enrico @L.X by the way, this might help mathoverflow.net/questions/323095/…
Mar 14, 2020 at 14:06 comment added Enrico With these choices (if you pick the coefficients general enough) the correspondent variety will be a smooth, generic example of $V_5$
Mar 14, 2020 at 13:48 comment added L. Xie @Enrico You mean picking the hyperplanes in this way define the same variety V5?
Mar 14, 2020 at 13:43 comment added Enrico @L.X in this case you should be safe by picking three of $x_{i,j}+x_{k,l}$ (with $i \neq j \neq k \neq l$, all five indexes involved and appropriate different coefficients).
Mar 14, 2020 at 13:21 comment added L. Xie @Enrico $V(5)$ can be obtained as the intersection of $Gr(2,5)$ with $3$ hyperplanes in general position. Which hyperplanes should they be?
Mar 12, 2020 at 21:36 answer added L. Xie timeline score: 4
Mar 9, 2020 at 16:35 review Close votes
Mar 14, 2020 at 3:04
Mar 9, 2020 at 14:12 vote accept CommunityBot
Mar 8, 2020 at 22:13 comment added Enrico Just an extra comment: the five quadrics you are mentioning are way non general by obvious reasons. They are given by the 4-Pfaffians of the 5x5 skew matrix with the linear form $x_{i,j}$ at the $(i,j)$ entry. This is not directly connected to the question (which Sasha already solved), but might be relevant if you keep working with this particular variety.
Mar 8, 2020 at 20:56 answer added Sasha timeline score: 4
Mar 8, 2020 at 20:41 history asked anon CC BY-SA 4.0