Timeline for Is the conjugation action linearizable?
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| Apr 25, 2017 at 8:16 | comment | added | a_g | You are completely right. I think that, actually, it can be done directly. Embed $G \subset GL(n, k) \subset k^2$. Now, seen as matrices, given any $g \in G$, the action of $g$, $g \cdot: G \to G$ extends to a linear map $g \cdot: k^{n^2} \to k^{n^2}$ giving us an action of $G$ on $k^{n^2}$ that restricts to the conjugation action on $G \subset k^{n^2}$. | |
| Apr 4, 2017 at 1:20 | history | answered | Marc Hoyois | CC BY-SA 3.0 |