Timeline for What is the subgroup of $\mathrm{SL}(n,\mathbb{C})$ which preserves the discriminant?
Current License: CC BY-SA 4.0
18 events
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| Nov 15, 2020 at 19:06 | vote | accept | Malkoun | ||
| Nov 15, 2020 at 15:43 | answer | added | abx | timeline score: 9 | |
| Nov 15, 2020 at 15:09 | comment | added | Malkoun | @abx, very nice argument. It answers my question, so could you please post it as an answer? | |
| Nov 15, 2020 at 8:07 | comment | added | abx | View $\mathcal{P}_{n}$ as the space of homogeneous polynomials of degree $n$ in 2 variables. Let $\Delta _p\subset \mathcal{P}_p$ be the locus of polynomials with one linear factor of multiplicity $\geq p$. It should be well-known that the singular locus of $\Delta _p$ is $\Delta _{p+1}$. This implies that your group preserves $\Delta _n$; up to homotheties, this is the group of automorphisms of $\mathbb{P}^{n}$ preserving $\mathbb{P}^1$ embedded by the $n$-th Veronese embedding. This is easily seen to be $\operatorname{PGL}(2,\mathbb{C})$. | |
| Nov 15, 2020 at 7:04 | history | edited | Malkoun | CC BY-SA 4.0 |
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| Nov 15, 2020 at 6:53 | history | edited | Malkoun | CC BY-SA 4.0 |
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| Nov 15, 2020 at 5:35 | comment | added | YCor | You need to convert into a system of linear equations, to solve it any mathematical software would do the job. And computing these equations, essentially only involve computing partial derivatives of the discriminant. | |
| Nov 15, 2020 at 5:09 | comment | added | Malkoun | @YCor, this is interesting. Which software could I use for that please? | |
| Nov 15, 2020 at 5:08 | history | edited | Malkoun | CC BY-SA 4.0 |
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| Nov 15, 2020 at 5:07 | comment | added | YCor | With a computer you could easily compute the Lie algebra of the group you're looking for, for small values of $n$, and in particular its dimension. | |
| Nov 15, 2020 at 5:05 | comment | added | Malkoun | @NoamD.Elkies, do you know what happens for higher degrees please? (By the way, I restricted my Lie group to be $SL(n,\mathbb{C})$, so $O(3,\mathbb{C}) \cap SL(3,\mathbb{C})$ is $SO(3,\mathbb{C})$). | |
| Nov 15, 2020 at 5:00 | history | edited | YCor | CC BY-SA 4.0 |
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| Nov 15, 2020 at 4:59 | comment | added | LSpice | @NoamD.Elkies, isn't it $\operatorname{SO}(2, 1)$? (EDIT: Oh, sorry, complex.) | |
| Nov 15, 2020 at 4:57 | history | edited | Malkoun | CC BY-SA 4.0 |
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| Nov 15, 2020 at 4:54 | comment | added | Noam D. Elkies | That's almost right, but there are more scalars, as you can already see for $n=3$: the image of ${\rm SL}_2$ is the special orthogonal group ${\rm SO}_3$ of the quadratic form $b^2-4ac$, but the full automorphism group is the full orthogonal group ${\rm O}_3$. | |
| Nov 15, 2020 at 4:52 | comment | added | Malkoun | Yes. And it is unique in this case up to conjugation by an element in $SL(n,\mathbb{C})$. | |
| Nov 15, 2020 at 4:51 | comment | added | LSpice | What is the principal homomorphism? One corresponding to a regular nilpotent element? | |
| Nov 15, 2020 at 4:39 | history | asked | Malkoun | CC BY-SA 4.0 |