Conjugacy of CSAs: Proof in Humphries' Intro to Lie Algebras and Rep Theory

Question

I am reading Chapter 16 of the title mentioned. On the middle of page 82 the statement is made "it is clear that if $\phi: L \rightarrow L'$ is a surjective homomorphism (of Lie algebras) then $\phi(L_a(ad(y)))={L'}_a(ad(\phi(y)))$". Here $L_a(ad(y))$ means $ker(ad(y)-aI)^m$ for $m$ sufficiently large (so that the kernel is as large as possible). The forward inclusion seems clear to me but the reverse inclusion I do not see. Is there something obvious I am missing?

Answer 1

I think what we have here is this linear algebra set-up. There are finite-dimensional vector spaces $V$ and $V'$ ($L$ and $L'$ here) and a surjective linear map $\phi:V\to V'$. I'll let $K$ denote its kernel. One also has an endomorphism $A$ of $V$ (here, $\text{ad y}$) with $A(K)\subseteq K$ so that $A$ induces an endomorphism $A'$ of $V'$ (here, $\text{ad }\phi(y)$).

I'll assume we are working over $\Bbb C$ or a similar field. Then $V$ splits as a direct sum of generalised eigenspaces $V_a$ where $V_a$ consists of all vectors with $(A-aI)^mv=0$ for some $A$. Likewise $V'$ splits into a direct sum of $V_a'$ where these are now the generalised eigenspaces for $A'$.

As you note, $\phi(V_a)\subseteq V_a'$. So $\phi\left(\bigoplus_{b\ne a} V_b\right)\subseteq\bigoplus_{b\ne a}V_{b}'$. Call these direct sums $W_a$ and $W_a'$; they are complements to $V_a$ and $V'_a$ in $V$ and $V'$. If $u\in V_a'$, then $u=\phi(v+w)=\phi(v)+\phi(w)$ where $v\in V_a$ and $w\in W_a$. But then $\phi(w)=0$ since $\phi(w)\in W_a'$ which is a complement of $V_a'$ in $V'$. Then $u=\phi(v)\in \phi(V_a)$