# A maximal subalgebra of $E_6$ !?

I'm puzzeled by the following sentence in one of Baez's posts:

The Lie algebra $E_6$ has a subalgebra of maximal rank isomorphic to $\mathfrak{so}(10)\oplus \mathfrak{u}(1)$.

However, I thought that the maximal subalgebras of a Lie algebra were obtained by deleting a dot in the $extended$ Dynkin diagram:

But I can only obtain $\mathfrak{so}(10)\oplus \mathfrak{u}(1)$ by deleting a dot in the (normal) Dynkin diagram of $E_6$...???

The subalgebra $\mathfrak{l} \simeq \mathfrak{so}(10) \oplus \mathfrak{u}(1)$ of $\mathfrak{g} = E_6$ obtained by removing the right-most (or the left-most) node from the Dynkin diagram is maximal among the regular subalgebras of $\mathfrak{g}$ which are reductive in $\mathfrak{g}$, but is not a maximal subalgebra of $\mathfrak{g}$ (even though it is of maximal rank). It is contained in the parabolic subalgebra $\mathfrak{p} = \mathfrak{l} + \mathfrak{b}$ of $\mathfrak{g}$ where $\mathfrak{b}$ is the Borel subalgebra of $\mathfrak{g}$ corresponding to the simple roots of the given Dynkin diagram. Also, by removing a node from the extended Dynkin diagram, one obtains maximal semi-simple subalgebras of $\mathfrak{g}$. The algebra $\mathfrak{l}$ is not semi-simple (it has a one-dimensional centre). The above is valid for any finite-dimensional simple Lie algebra, not only for $E_6$.