I have trouble understanding this proof from Humpreys: Linear Algebraic Groups page 104.


Let $G$ be a connected and diagonalizable linear algebraic group (i.e. there exists a closed embedding into a torus $D_n(\mathbb C)$). Then $G$ is a torus.


The inclusion $G \to D_n$ induces a surjection $\mathbb C[D_n] \to \mathbb C[G]$ (restriction map) which induces a surjection $\chi(D_n) \to \chi(G)$ (restriction map) between the group characters. Since $D_n$ is a torus $\chi(D_n) = \mathbb Z^n$, and since $G$ is diagonalizable and connected $\chi(G) = \mathbb Z^r$. Thus $$ \mathbb Z^n = \chi(D_n) \to \chi(G) = \mathbb Z^r $$ gives us a decomposition $$ \chi(D_n) = Ker(\to) \oplus\mathbb Z^r. $$ We can then pick a basis for the summands, $\chi_1, \cdots, \chi_{n-r}$ and $\chi_{n-r+1}, \cdots, \chi_n$.

Then I can't understand the conclusion.


  • $\begingroup$ As I understand, $\chi(G)=\mathbb{Z}^r$ already implies that $G$ is a torus, and the rest few lines are there to show how you can choose a "natural" basis for $G$. $\endgroup$ – Yanko Jan 29 '18 at 14:20
  • $\begingroup$ @yanko is it true that $\chi(G) = \chi(G')$ implies $G = G'$? $\endgroup$ – Maffred Jan 29 '18 at 14:30
  • $\begingroup$ It follows from the pontryagin duality en.wikipedia.org/wiki/Pontryagin_duality (Though, when $G'=\mathbb{Z}^r$ it might be possible to prove this without using this result) $\endgroup$ – Yanko Jan 29 '18 at 14:31
  • $\begingroup$ @yanko maybe in general if $G$ and $G'$ are d-groups (groups for which $\chi(G)$ as a set linearly generates $\mathbb C[G]$), then $\chi(G)= \chi(G') \implies \mathbb C[G] = \mathbb C[G'] \implies G=G'$. $\endgroup$ – Maffred Jan 29 '18 at 15:03
  • $\begingroup$ we are talking about isomorphisms, so technicaly $\mathbb{C}[H]=\mathbb{C}[G]$ whenever $|G|=|H|$ (right?) $\endgroup$ – Yanko Jan 29 '18 at 16:09

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