Conjugate of Lie subalgebra What does it mean that "all Cartan subalgebras of a semisimple Lie algebra are conjugates"? I know this refers to adjoint action but I don't know exactly what it means.
The most obvious definition to me is that if $\mathfrak{h}_1$ and $\mathfrak{h}_2$ are two Cartan subalgebras, then there exists some $x \in \mathfrak{g}$ such that $[x,\mathfrak{h}_1] = \mathfrak{h}_2$. However I don't see why $[x,\mathfrak{h}_1]$ is even a subalgebra at all, so maybe this doesn't make sense.
 A: Two subalgebras $\mathfrak{h}_1$ and $\mathfrak{h}_2$ in a semisimple Lie algebra $\mathfrak{g}$ of a Lie group $G$ are conjugate if there is some element $g\in G$, s.t. $Ad(g)(\mathfrak{h}_1)=\mathfrak{h}_2$. In other words, if $\mathfrak{h}_1$ can be mapped to $\mathfrak{h}_2$ by a certain automorphism $Ad(g)$, called conjugation.
Here $Ad(g)$ is the following. The group $g$ acts on $G$ itself by conjugation: $c_g\colon G\to G, \;h\mapsto ghg^{-1}$. If we take derivative of this map at $e$, we get $dc_g\colon \mathfrak{g}\to \mathfrak{g}$. We define $Ad(g)$ to be this $dc_g$. Since your algebra is semisimple, $Ad(g)$ will actually be an automorphism.
A: In this context, $\mathfrak{h}_1$ and $\mathfrak{h}_2$ being conjugate means that there is an (inner) automorphism $\varphi$ of $L$ such that $\varphi(\mathfrak{h}_1) = \mathfrak{h}_2$.
I put the "inner" in parantheses, as I am not certain everyone will require such an automorphism to be inner in order to call the Cartans conjugate (as it happens, they are conjugate by an inner automorphism, so in this case it is not so important). 
