# Identity on hyperplane and its complement, then identity on whole space? (Humphreys p.42 Lemma)

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.

• Jun 25, 2021 at 3:04
• @TorstenSchoeneberg Thanks a lot.
– user
Jun 25, 2021 at 3:50
• Also related now: math.stackexchange.com/q/4227594/96384. This lemma and its proof seem to confuse many people! Aug 18, 2021 at 21:03
• @TorstenSchoeneberg You're so kind. Thank you very much!
– user
Aug 26, 2021 at 8:32

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.