This is from Hymphreys 'Introduction to Lie Algebras and Representation Theory', Lemma 9.1 in page 42.

Lemma. Let $\Phi$ be a finite set which spans $E$. Suppose all reflections $\sigma_\alpha$($\alpha \in \Phi$) leave $\Phi$ invariant. If $\sigma \in GL(E)$ leaves $\Phi$ invariant, fixes pointwise a hyperplane $P$ of $E$, and sends some nonzero $\alpha \in \Phi$ to its negative, then $\sigma=\sigma_\alpha$ (and $P=P_\alpha$).

$GL(E)$ is the set of all invertible endomorphism on $E$. In the proof of this Lemma, it says that

Let $\tau=\sigma\sigma_\alpha$. Then $\tau$ acts as the identity on $\mathbb R \alpha$ as well as on the quotient $E/\mathbb R\alpha$.

Next they continue the argument using eigenvalue and minimal polynomial. I understand why it acts as the identity. But I haver a question in the next step.

Since $P$ is a hyperplane and $\mathbb R \alpha$ is its complement, $E=P\oplus\mathbb R\alpha$. Since $\tau$ is the identity on both $\mathbb R \alpha$ and $E/\mathbb R \alpha\cong P$, isn't $\tau$ the identity on whole $E$? Why should we go further using the argument about minimal polynomial?

Any help would be appreciated. Thanks.


1 Answer 1


You seem to be thinking that if one has a vector space $V$ with a subspace $W$, and an endomorphism $f$ which is the identity on $W$ and induces the identity on $V/W$, then $f$ must be the identity on all of $V$.

That is not true: Take the vector space $V=\mathbb R^2$ with basis $e_1, e_2$ and consider the linear endomorphism given (in that basis) by $$\pmatrix{1 &1\\0&1}.$$ You will notice it is the identity on $\mathbb R e_1$ (in particular, leaves that subspace invariant), it also induces the identity on $V/\mathbb Re_1$, but it is not the identity on $V$.

In fancier terms, even though one can identify the vector space $V/W$ with any complement of $W$ in $V$, such an identification in general is not compatible with the action of $f$, and so we do not have $V\simeq W \oplus V/W$ as $K[f]$-modules.

In the case at hand, the result is true in hindsight, so of course one cannot come up with a counterexample, but one cannot just invoke a wrong general argument as above. In fact whatever argument works here would equivalently show that there is a hyperplane (complement of $\mathbb R\alpha$) which is invariant under the endomorphism in question.


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