Therefore there exists a semisimple element x in mathfrak{g}. Consider S={lambda x|lambda in mathbb{C}}. By section 1.7 this is Abelian. Now either S is maximal, or we can find elements that commute with x and are themselves semisimple. So we can continually add these elements to form a basis of our Cartan Subalgebra. This process must terminate due to the finite dimensional nature of mathfrak{g}, so we can have at most Dim(mathfrak{g}) elements to form the basis of our Cartan subaglebra. end{proof} begin{framed} begin{theorem} Let G be a Lie group with semisimple Lie...

Therefore there exists a semisimple element x in mathfrak{g}. Consider S={lambda x|lambda in mathbb{C}}. By section 1.7 this is Abelian. Now either S is maximal, or we can find elements that commute with x and are themselves semisimple. So we can continually add these elements to form a basis of our Cartan Subalgebra. This process must terminate due to the finite dimensional nature of mathfrak{g}, so we can have at most Dim(mathfrak{g}) elements to form the basis of our Cartan subaglebra.

end{proof}

begin{framed} begin{theorem} Let G be a Lie group with semisimple Lie algebra mathfrak{g}. If mathfrak{h} is a Cartan subalgebra, then gmathfrak{h}g^{-1}={ghg^{-1}|g in G, h in mathfrak{h}} is also a Cartan subalgebra for any element g in G end{theorem}end{framed} We won’t go into too much detail here, but we can show some examples of this in the next section. The most important thing to note is that whilst Cartan Subalgebras are not unique, we can establish all the properties we require from examining one of them.

section{Examples of Cartan Subalgebras}

For the families of Lie algebras outlined at the end of chapter 1, we can give details of their Cartan Subalgebras.

subsection{Cartan Subalgebra of mathfrak{sl}_{n+1}}

For the algebra mathfrak{sl}_{n+1} we can consider the traceless diagonal matrices which we encountered in chapter 2. These have basis h_i=E_{i,i}-E_{i+1,i+1} for i=1,cdots, n. This has dimension n. These are clearly diagonalisable (and thus semisimple), and as diagonal matrices commute this subalgebra is Abelian.

To show maximality, we calculate the commutator of [H,E_{i,j}] for a traceless diagonal matrix H. The commutator [H,E_{i,j}]=(e_i(H)-e_j(H))E_{i,j} shows us that for some matrix that is not in the Cartan subalgebra, we in fact recover mathfrak{sl}_{n+1}! Thus our subalgebra is maximal, and hence a Cartan Subalgebra.