Timeline for Ring of invariants of some special type of subgroups of $GL_3(\mathbb C)$
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| Dec 1, 2018 at 1:25 | history | bounty ended | user521337 | ||
| Dec 1, 2018 at 1:25 | vote | accept | user521337 | ||
| Nov 28, 2018 at 22:30 | comment | added | Robert Bryant | @user521337: To your first question: $L:\mathbb{C}^3\to\mathbb{C^3}$ is a linear isomorphism, and I regard $x_1,x_2,x_3$ as linear functions (coordinates) on $\mathbb{C}^3$, i.e., $x_i:\mathbb{C}^3\to\mathbb{C}$ is linear. Then $L^*x_i$ means the 'pullback' linear function $x_i\circ L$, etc. Of course, $L^*$ extends uniquely to a ring isomorphism $L^*:\mathbb{C}[x_1,x_2,x_3]\to \mathbb{C}[x_1,x_2,x_3]$, so $(L^*f_1)^3 = L^*(f_1)^3$, which answers your seccond question. | |
| Nov 28, 2018 at 19:57 | comment | added | user521337 | btw, what does $L^*$ stand for ? | |
| Nov 27, 2018 at 12:36 | comment | added | Robert Bryant | I just noticed that where I wrote $\xi_1$, $\xi_2$, and $\xi_3$ in the above comment, I should have written $\xi_0$, $\xi_1$, and $\xi_2$ for consistency with the answer. I can't change it now. (Or, if you prefer, you can just assume that $\xi_3=\xi_0$.) | |
| Nov 27, 2018 at 12:33 | history | edited | Robert Bryant | CC BY-SA 4.0 |
Simplified or expanded some arguments and removed irrelevant remarks.
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| Nov 27, 2018 at 1:56 | history | edited | Robert Bryant | CC BY-SA 4.0 |
Fixed a notation clash
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| Nov 27, 2018 at 1:51 | comment | added | Robert Bryant | @user521337: Clearly, $G$ permutes the $\xi_i$ and multiplies them by unit complex numbers, so $G$ preserves the Hermitian form $F=|\xi_1|^2{+}|\xi_2|^2{+}|\xi_3|^2$. Now, if you substitute what the $\xi_i$ are in terms of the $x_i$, you get $F = 3(|x_1|^2{+}|x_2|^2{+}|x_3|^2)$. QED. $H$ is the subgroup of $G$ you get when $\pi$ is the identity permutation, so these are $L^*(\xi_i) = \lambda_i\xi_i$; obviously, this is an abelian group. $G$ is not abelian: The elements $L^*(\xi_1,\xi_2,\xi_3)=(s\xi_1,\xi_2,\xi_3)$ and $L^*(\xi_1,\xi_2,\xi_3)=(\xi_2,\xi_3,\xi_1)$ don't commute. | |
| Nov 27, 2018 at 0:36 | comment | added | user521337 | I can't quite see why matrices in $G$ preserves $\sum |x_i|^2$ ... are you certain $H$ is abelian ? Could you please elaborate a proof of that ? And is $G$ abelian ? | |
| Nov 27, 2018 at 0:21 | comment | added | Robert Bryant | @user521337: Yes, I showed that $G$ is unitary; it preserves the Hermitian form $|x|^2+|y|^2+|z|^2$. Also, $H$ is abelian, it's just the product of three copies of $\mathbb{Z}_3$. | |
| Nov 26, 2018 at 23:33 | comment | added | user521337 | thanks, I still have to read and grasp your whole answer thoroughly ... I don't see at a glance if you proved whether every matrix in $G$ is unitary or not ... also, let me ask, $H$ is non-abelian right ? | |
| Nov 26, 2018 at 23:15 | history | edited | Robert Bryant | CC BY-SA 4.0 |
Completely replaced the wrong answer with a corrected answer.
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| Nov 26, 2018 at 23:05 | history | edited | Robert Bryant | CC BY-SA 4.0 |
Completely replaced the wrong answer with a corrected answer.
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| Nov 26, 2018 at 12:20 | history | edited | Robert Bryant | CC BY-SA 4.0 |
fixed another sign error
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| Nov 25, 2018 at 19:24 | history | edited | Robert Bryant | CC BY-SA 4.0 |
fixed a sign error made throughout and a few other misprints
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| Nov 25, 2018 at 18:40 | history | answered | Robert Bryant | CC BY-SA 4.0 |