The following is taken from these lecture notes on page 8.16:
[...] $\mathfrak{so}(p+q)$ has as subalgebras $\mathfrak{so}(p)$ and $\mathfrak{so}(q)$ as well as their direct sum $\mathfrak{so}(p)\oplus\mathfrak{so}(q)$. How can this be seen from the Dynkin diagrams? Suppose $p,q >2$ are even, then the relevant Dynkin diagrams are all of the form:
By eliminating some nodes one can obtain $\mathfrak{so}(p)$ and $\mathfrak{so}(q)$, but not their direct sum. However, the direct sum is contained in the affine Dynkin diagram of $\hat{\mathfrak{d}}_{\frac{p+q}{2}}$:
This trick also works if one of $p,q$ is even and the other one is odd:
If both $p,q$ are odd, however, the subalgebra $\mathfrak{so}(p)\oplus \mathfrak{so}(q)$ has lower rank than $\mathfrak{so}(p+q)$ and it cannot be obtained in this way. Here, the relevant Dynkin diagram is the twisted affine $\mathfrak{so}(p+q)^{(2)}$:
Question
As I understand these statements, one should in principle be able to read off some subalgebras of a given simple Lie algebra directly from the Dynkin diagram of the affine algebra. Unfortunately, I don't really understand the here described technique. It seems that one has just to "eliminate some nodes" to get the result, but I don't really see if there are additional restrictions, or if I can just remove any node and get a valid subalgebra. Could someone maybe elaborate on how this is done?