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Let $\def\C{\mathbb{C}}T = (\C^*)^n$. A character of $T$ is defined to be a homomorphism from $T$ to $\C^*$. The characters of $T$ is of the form $f(t_1,\ldots,t_n)=t_1^{a_1}\cdots t_n^{a_n}$ for some $a_1,\ldots, a_n \in \mathbb{Z}$.

I think we can also consider the homomorphisms from a unipotent group $U$ to $\C^*$. Are there some references which study these homomorphisms? For example, describe the expressions of these homomorphisms like in the case of of torus $T$. Thank you very much.

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  • $\begingroup$ Such a morphism must be trivial on the commutator subgroup. I don't know the definition of a unipotent group by heart, but I imagine the quotient by its commutator subgroup must be rather simple, maybe just a sum of additive groups? $\endgroup$
    – Marc van Leeuwen
    Jun 10, 2014 at 7:56
  • $\begingroup$ Should not be just sufficient to note that if $U$ is a unipotent group and $\sigma : U \rightarrow \mathbb{C}^*$ a homomorphism then $\sigma(U)$ is a unipotent subgroup of $\mathbb{C}^*$, i.e. the trivial subgroup? $\endgroup$
    – wood
    Dec 13, 2022 at 14:12

1 Answer 1

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Suppose that $U$ is a unipotent linear algebraic group over $\mathbb{C}$, and that $f : U \to \mathbb{C}^*$ is a homomorphism of complex Lie groups. Since $\mathbb{C}^*$ is abelian, $f$ factors through the abelianisation $V = U/U'$ of $U$. Since the commutator subgroup $U'$ of $U$ is algebraic and the quotient map $p: U \to V$ is a homomorphism of algebraic groups, $V$ is also unipotent, by the preservation of Jordan decomposition under such homomorphisms. Therefore, $V$ is isomorphic to the additive group of a finite-dimensional vector space over $\mathbb{C}$. All holomorphic homomorphisms $\chi : V \to \mathbb{C}^*$ are of the form $\chi(z) = e^{\lambda(z)}$ for some linear function $\lambda \in V^*$.

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  • $\begingroup$ Right, that's about what I guessed when I wrote my comment. +1 $\endgroup$
    – Marc van Leeuwen
    Jun 10, 2014 at 10:19
  • $\begingroup$ @MarcvanLeeuwen: Sorry, I did not see your comment which pretty much makes my answer redundant. $\endgroup$
    – ivanpenev
    Jun 10, 2014 at 10:42

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