Xrfgtjtk

There is this property about Cartan subalgebras that is not clear to me.

Suppose $mathfrak{g}$ is a semisimple Lie algebra. Then I know we can decompose it uniquely as $mathfrak{g}=bigoplusmathfrak{g}_i$, where the $mathfrak{g}_i$ are simple ideals. If $mathfrak{h}$ is a Cartan subalgebra (maximal toral subalgebra), then $mathfrak{h}=bigoplus(mathfrak{h}capmathfrak{g}_i)$. The $supseteq$ direction is clear.

Why does the other follow? I thought of something like this, take $xinmathfrak{h}$, and write $x=sum x_i$ for $x_iinmathfrak{g}_i$. I want to show $x_iinmathfrak{h}$ for each $i$. I did this by induction on the number of nonzero summands. If there is only one summand so $x=x_i$, the claim is clear. If there's more than one summand, I'm not sure how to reduce it to imply the induction, if this is the right idea.

Suppose that $K$ is the base field. Fix $hinmathfrak{h}$. Let $h_iinmathfrak{g}_i$ be such that $h=sumlimits_i,h_i$ (the $h_i$'s are unique). We claim that $h_iinmathfrak{h}$ for all $i$. Let $tinmathfrak{h}$. Write $t=sumlimits_{i},t_i$ with $t_iinmathfrak{g}_i$. For $ineq j$, $$left[t_i,h_jright]inmathfrak{g}_icapmathfrak{g}_j={0},,$$ so $left[t_i,h_jright]=0$. Ergo, $$sumlimits_{i},left[t_i,h_iright]=[t,h]=0,.$$ Since $left[t_i,h_iright]inmathfrak{g}_i$, we conclude that $left[t_i,h_iright]=0$ for every $i$. Thus, $left[h_i,tright]=left[h_i,t_iright]=0$ for every $tinmathfrak{h}$.

Because $h$ is a semisimple element of $mathfrak{g}$, $h_i$ is also a semisimple element of $mathfrak{g}$ (to be proven below). If $h_inotinmathfrak{h}$, then $mathfrak{h}oplus Kmathfrak{h}_i$ is a strictly larger toral subalgebra containing $h$, which is a contradiction. Thus, $h_iin mathfrak{h}$ for all $i$, so $h_iinmathfrak{h}capmathfrak{g}_i$. Therefore, $$h=sumlimits_i,h_iinbigopluslimits_i,left(mathfrak{h}capmathfrak{g}_iright),,$$ whence $mathfrak{h}subseteq bigopluslimits_i,left(mathfrak{h}capmathfrak{g}_iright)$. Since it is trivial that $mathfrak{h}supseteq bigopluslimits_i,left(mathfrak{h}capmathfrak{g}_iright)$, $mathfrak{h}= bigopluslimits_i,left(mathfrak{h}capmathfrak{g}_iright)$, as required.

To show that $h_i$ is a semisimple element of $mathfrak{g}$, we let $V^lambda$ be the eigenspace of $text{ad}_mathfrak{g}(h)$ with the eigenvalue $lambdain K$. For each $i$, take $V_i^lambda$ to be the intersection $V^lambdacapmathfrak{g}_i$. Trivially, $bigopluslimits_{lambdain K},V^lambda_i=mathfrak{g}_i$. However, as $V^lambda_i$ is the eigenspace with eigenvalue $lambda$ of $text{ad}_{mathfrak{g}_i}left(h_iright)$, we see that $h_i$ is a semisimple element of $mathfrak{g}_i$. Now, define $tilde{V}^lambda_i:=V^lambda_i$ if $lambdaneq 0$, and $tilde{V}^0_i:=V^0_ioplusleft(bigopluslimits_{jneq i},mathfrak{g}_jright)$. Then, $tilde{V}^lambda_i$ is the eigenspace of $text{ad}_mathfrak{g}left(h_iright)$ with eigenvalue $lambda$. Since $bigopluslimits_{lambda in K},tilde{V}^lambda_i=mathfrak{g}$, we conclude that $h_i$ is a semisimple element of $mathfrak{g}$.