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Let $G$ be a linear algebraic group of dimension one.

The proof that I am looking at, in t.a springer's book (thm 3.4.9) proceeds by showing that $G$ must be either equal to its semisimple part $G_s=\{g \in G | g \text{ is diagonalizable in some( and hence every ) embedding into GL} \}:=\{g \in G| g \text{ is semisimple } \}$

or its unipotent part =$\{g \in G | g \text{ is unipotent} \}$.

In the case where $G=G_s$ she mentions that $G_s$ can be embedded into the $D_n$ the group of $n-$ dimensional diagonal matrices.

I don't understand this because $G_s$ is not by hypothesis $\textbf{simultaneously}$ diagonalizable. One can't choose one $P$ that will diagonalize all of the elements in $G$. I would be grateful if someone could point out why this not a problem.

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  • $\begingroup$ In my edition (the second one), it is proved first in the same proposition (3.1.3) that $G$ must be commutative. $\endgroup$ – Captain Lama Apr 20 '16 at 8:40
  • $\begingroup$ yes. i had no problem with this part of the proof. $\endgroup$ – user062295 Apr 20 '16 at 13:08
  • $\begingroup$ I don't have the text, but I think it's worth mentioning that (unless this is implicit in Springer's assumptions) that $G$ should be connected and smooth, else it's easy to cook up counterexamples (in removing either of the hypotheses) to the claim. $\endgroup$ – Alex Youcis Apr 20 '16 at 14:05
  • $\begingroup$ Take $O(2)$ for example $\endgroup$ – user062295 Apr 20 '16 at 16:02
  • $\begingroup$ @user062295 Those do satisfy both of the conditions that I said (at leas in characteristic not $2$), but is not one of the above examples if $k\ne\overline{k}$. (or, more operatively, doesn't contain a root of $-1$). $\endgroup$ – Alex Youcis Apr 20 '16 at 21:39
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It seems that the missing link is that commuting diagonalizable elements are simultaneously diagonalizable, since then we may conclude by the fact that $G$ is commutative.

We may invoke the general Lie theorem about solvable Lie algebras, but the proof is quite elementary for commuting elements. So let $(x_i)_{i\in I}$ be a family of pairwise commuting elements of $End_K(V)$ for some finite-dimensional $K$-vector space $V$, such that each $x_i$ is diagonalizable. We want to show that the $x_i$ are simultaneously diagonalizable.

Then up to extracting a basis of $Vect(x_i)$, we may assume that $I$ is finite and prove the result by induction on $|I|$. If $|I|=1$, the result is trivial. If $|I|>1$, write $I=\{0\}\cup J$. then the $(x_j)_{j\in J}$ are simultaneously diagonalizable : let $V = \bigoplus_{k\in X} V_k$ be a decomposition such that all $x_j$ with $j\in J$ act as scalars on each $V_k$. Then since $x_0$ commutes with the $x_j$ it stabilizes the $V_k$ and its restriction to each $V_k$ is still diagonalizable, and we can write $V_k = \bigoplus_{l\in Y_k} V_{k,l}$ such that $x_0$ acts on each $V_{k,l}$ as a scalar. We conclude by the fact that $V = \bigoplus_{k\in X}\bigoplus_{l\in Y_k} V_{k,l}$ and all $x_i$ with $i\in I$ act as scalar on each $V_{k,l}$.

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