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Let $\phi$ be an algebraic group homomorphism from $G_{m}$ to $GL_{n}(\mathbb C)$,where $G_{m}= \mathbb C^{*} $. Then image of $\phi$ lies in the $D(n,\mathbb C) \cap GL_{n}(\mathbb C)$. Moreover each diagonal entries will be of the form $t^{m_{i}}$ where $1\leq i\leq n$ and $t\in G_{m}$.

Now if the image lies in the diagonal matrices then it is clear that each diagonal entries will be of the required form as each algebraic group homomorphism from $G_{m}$ to itself will be of the form $t^{m}$ where $t\in G_{m}$. To show the image lies in the diagonal matrices I have only information that it is an abelian closed subgroup of $GL_{n}(\mathbb C)$.

  • The claim is not quite true as stated. You can compose $\phi$ with an inner automorphism of $GL_n$, and as the diagonal matrices don't form a normal subgroup, it is possible that the image will contain non-diagonal matrices. However, this is the only problem, and the image will be diagonal "up to a choice of basis". – Jyrki Lahtonen Apr 21 '13 at 18:29
  • Yes,agree.So it should be that there exist $P \in GL_{n}(\mathbb C)$ such that $P\phi(G_{m})P^{-1}$ lies in the diagonal matrices.Certainly there exist $P$ such that $P\phi(G_{m})P^{-1}$ lies in set of all upper triangular matrices. –  Apr 21 '13 at 18:47
  • If $z$ is a root of unity, then $\phi(z)$ is of finite order. With your choice of $P$ that implies that $P\phi(z)P^{-1}$ is diagonal for all such $z$ (If there are non-diagonal Jordan blocks, then the matrix has infinite order). The roots of unity are dense on the unit circle, so by continuity the unit circle is mapped to diagonal matrices. Don't know about the rest? – Jyrki Lahtonen Apr 21 '13 at 18:56
  • well,I need to think about it. –  Apr 21 '13 at 18:58
  • I guess it revolves around the word algebraic homomorphism. If that means that the mapping is algebraic (as opposed to any continuous homomorphism), then the preimage of diagonal matrices must be Zariski-closed. But the Zariski closure of the unit circle in $G_m$ is all of $G_m$. – Jyrki Lahtonen Apr 21 '13 at 19:01

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As Jyrki Lahtonen mentioned, problem is up to a choice of basis. Use that $G_m$ (any torus) is consisting of semisimple elements and semisimple elements goes to semisimple under $\phi$. Hence the image $\phi(G_m)$ is consisting of commuting semisimple elements. So we are done.

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