I was thinking about characters of unipotent groups, this is my question. Let's define a unipotent group as an algebraic linear group formed by unipotent elements, where an element is called unipotent if it acts unipotently on some regular representation. Let $U$ be a unipotent group and $g:U \rightarrow \mathbb{C}^\times$ a character of $U$. Since image of groups under homomorphisms are groups and the Jordan decomposition is preserved, we have that $g(U) \subset \mathbb{C}^\times$ is a unipotent subgroup. But, the only unipotent subgroup of $\mathbb{C}^\times$ is $\{1\}$. Therefore, we have only the trivial character. Is this argument right?
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I made it, thanks. Actually, in my question I was thinking about matrices. – wood Dec 13 '22 at 16:17
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Do you require a character to be algebraic? – Moishe Kohan Dec 13 '22 at 16:48
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Yes, I require it to be an homomorphism of algebraic groups – wood Dec 13 '22 at 17:12