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Are the multiplicative one-parameter subgroups of the general linear group (i.e., morphisms $\lambda:\Bbbk^\times\to\mathrm{GL}_n\Bbbk$ of algebraic groups) completely classified? The obvious candidates are the $$\lambda(t) := \mathrm{diag}(t^{\alpha_1},\ldots,t^{\alpha_n})$$ for certain $\alpha=(\alpha_1,\ldots,\alpha_n)\in\mathbb Z^n$. Then, we can also consider conjugates of these, i.e. $$(g.\lambda)(t) = g\lambda(t)g^{-1}.$$ Are these all?

Thoughts: One-parameter subgroups are commutative. Hence, if they consist of diagonalizable elements, they can be diagonalized simultaneously. Thus, if the image of $\lambda$ consists of diagonalizable elements, it is of the above form. However, are there one-parameter subgroups that are not diagonalizable?

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  • $\begingroup$ In the characteristic $0$ case the answer is evidently that you found them all; $t^A$ will contain logs of $t$ if $A$ is non-diagonalizable. But there should be a nice answer independent of char $\Bbbk$. $\endgroup$ – user8268 Sep 11 '13 at 8:20
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Assume that $k$ is algebraically closed and let $\lambda : k^\times \rightarrow \mathrm{GL}_n(k)$ be a morphism of algebraic groups. Let $R = \{x \in k : x^m = 1 \text{ for some $m \in \mathbb{N}$}\}$.

If $x \in R$ satisfies $x^m = 1$ then $\lambda(x)^m = I_n$ and so $\lambda(x)$ is diagonalizable. An infinite family of commuting linear maps on a finite-dimensional vector space may be simultaneously diagonalized, so there is a basis for $k^n$ in which all the maps $\lambda(x)$ for $x \in R$ are diagonal. Since $R$ is Zariski dense in $k^\times$, it follows that all $\lambda(x)$ are diagonal in this basis.

Finally if $\lambda(x) = \mathrm{diag}(\lambda_1(x),\ldots,\lambda_n(x))$ then each $\lambda_i$ is a morphism of algebraic groups $k^\times \rightarrow k^\times$ and so is of the form $x \mapsto x^r$ for some $r \in \mathbb{Z}$. This shows that every one-parameter subgroup is of the form you expected.

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  • $\begingroup$ Nice! The density argument is really elegant, also thanks for answering such an old question. $\endgroup$ – Jesko Hüttenhain Jun 9 '14 at 17:16
  • $\begingroup$ Thank you! You might be interested in these notes of mine: ma.rhul.ac.uk/~uvah099/Maths/PolyRepsRevised.pdf. They have an extension of the argument (due to Darij Grinberg) to the case when $k$ is not algebraically closed. $\endgroup$ – Mark Wildon Jun 9 '14 at 19:28

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