Timeline for is connected complex Lie group with a trivial center linear?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Mar 12, 2013 at 6:36 | vote | accept | Dima Sustretov | ||
| Dec 8, 2012 at 22:49 | comment | added | Venkataramana | Perhaps I should add that the image o an algebraic group under an algebraic homomorphism is also an algebraic group | |
| Dec 8, 2012 at 22:41 | comment | added | Venkataramana | of $e_1,e_2$. Hence the two algebraic characters $\chi _1, \chi _2$ of the image of $G$ (which is one dimensional) must be algebraically dependent i.e. there exist integers $m,m'$ such that $\chi ^m=\chi ' ^{m'}$. | |
| Dec 8, 2012 at 22:38 | comment | added | Venkataramana | Dear Dima, the question is "is a connected complex Lie group algebraic"? I understand it to mean that given a complex Lie group $G$, can you put the structure of a complex algebraic group $G'|$ such that as complex LIE groups, $G$ and $G'$ are isomorphic? Now suppose that the $G$ in the example is a complex ALGEBRAIC group (in the foregoing sense). $e_1,e_2$ may be thought of as elements of the Lie algebra, which is necessarily an algebraic representation of $G$, and are eigenvectors for $G$. But, $${\mathbb C}^2$ acts trivially, and hence $G$ acts via a one dimensional group on the span | |
| Dec 8, 2012 at 22:13 | comment | added | Dima Sustretov | Dear Aakumadula, could provide a bit more detail? I do not understand what you mean by having a structure of an algegbraic group. Does it mean being a closed subgroup of $GL_n(\mathbb C)$? How does that imply that the two characters should be algebraically independent? | |
| Dec 7, 2012 at 23:09 | history | edited | Venkataramana | CC BY-SA 3.0 |
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| Dec 7, 2012 at 23:03 | history | edited | Venkataramana | CC BY-SA 3.0 |
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| Dec 7, 2012 at 22:51 | history | answered | Venkataramana | CC BY-SA 3.0 |