Timeline for Action of the endomorphism monoid on an irreducible GL-module
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 6, 2013 at 17:20 | comment | added | Vít Tuček | Right. I got confused by the notation; I thought for a moment that $m.v$ is the usual matrix multiplication. | |
| Oct 6, 2013 at 9:48 | comment | added | Jesko Hüttenhain | @VítTuček: Sure we are dealing with linear actions, but that does not mean that for a fixed $v$, the morphism $M\to V$ mapping $m\mapsto m.v$ is a linear map. It only means that for a fixed $m$, the morphism $V\to V$ defined by $v\mapsto m.v$ is a linear map. | |
| Oct 6, 2013 at 8:53 | history | edited | Rami | CC BY-SA 3.0 |
added 80 characters in body
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| Oct 6, 2013 at 8:47 | comment | added | Rami | Sorry, I made 2 mistakes. One of them is not crucial but the other seems to be so. As it written above the Lemma and its proof are wrong. :-( | |
| Oct 6, 2013 at 0:00 | comment | added | Vít Tuček | I'm confused. I thought we are dealing with linear representation, are we not? | |
| Oct 5, 2013 at 19:20 | comment | added | Dave Anderson | Unfortunately, the "lemma" is false. Take the map $(x,y) \to (x^2, xy)$. The image is the union of $(0,0)$ and the set where $x$ is nonzero, which is not even a variety. | |
| Oct 5, 2013 at 15:26 | comment | added | Jesko Hüttenhain | What is $\mathrm{Ker}(f)$? Do you mean the fiber over $0$? It does not have to be a sub-vectorspace. You might look at $\mathbb P(M\setminus f^{-1}(0))$, but that is probably no longer projective, so the argument will not work. | |
| Oct 5, 2013 at 14:31 | comment | added | Vít Tuček | Does it really matter? The "deprojectivization" is a cone anyway and so it does include zero. What I mean is that you can just replace $f$ by $\tilde{f} \colon \mathrm{End}(\mathbb{C}^n)/\mathrm{Ker}(f) \to V$. | |
| Oct 5, 2013 at 13:29 | comment | added | Jesko Hüttenhain | The problem is that $\bar f$ is usually not well-defined. There could be matrices $m\in M\setminus\{0\}$ such that $mv=0$. | |
| Oct 5, 2013 at 8:43 | history | answered | Rami | CC BY-SA 3.0 |