Timeline for What is the relationship between determinantal varieties?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| May 19 at 6:18 | vote | accept | zhjzwlys | ||
| May 10 at 12:17 | comment | added | Zach Teitler | Sorry, please disregard that comment. I think it's better to simply put $M_k=V(I_{k+1})$, then everything is fine. | |
| May 10 at 12:06 | comment | added | Zach Teitler | For the second statement you should put $M_2$ instead of $M_1$. In this notation $M_k$ are the matrices of rank $k-1$. $M_{k+1}$ is the $k$ secant variety of $M_2$. Other than that, aren't these statements true when $k=\min(m,n)$? | |
| May 9 at 11:18 | comment | added | Francesco Polizzi | @Sasha: fixed, thanks (it seems that [ACGH] forget to specify this, too...) | |
| May 9 at 11:17 | history | edited | Francesco Polizzi | CC BY-SA 4.0 |
added 257 characters in body
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| May 9 at 11:13 | comment | added | Francesco Polizzi | Then you must consider secant varieties. I will add it in the answer. | |
| May 9 at 10:11 | comment | added | zhjzwlys | Thanks for your answer. I know the result, but I don't understand how it can help me. I hope that smaller k can lead to larger k, but this seems to be the opposite. | |
| May 9 at 10:06 | comment | added | Sasha | Only for $k < \operatorname{min}(m,n)$. | |
| May 9 at 8:27 | history | answered | Francesco Polizzi | CC BY-SA 4.0 |