A nice/simple relationship between the Chevalley generators of $\mathfrak{sp}_n$ and the Chevally generators of $\mathfrak{sl}_n$? The Lie algebra $\mathfrak{sl}_n$ is defined to be the trace free matrices in $M_n(\mathbb{C})$. The Lie algebra $\mathfrak{so}_n$ is defined to be the matrices $A$ in $M_n(\mathbb{R})$ satisfying $A + A^{\mathrm{T}} = 0$. This means that the elements of  $\mathfrak{so}_n$ have zero diagonal entries, i.e they are trace-free. Thus there is an inclusion
$$
\mathfrak{so}_n \to \mathfrak{sl}_n.
$$
With respect to this inclusion is there a nice/simple relationship between the Chevalley generators of $\mathfrak{so}_n$ and the Chevalley generators of $\mathfrak{sl}_n$?
For a related question: the Lie algebra of $\mathfrak{sp}_n$ also embeds into $\mathfrak{sl}_n$. With respect to this inclusion is there a nice/simple relationship between the Chevalley generators of $\mathfrak{sp}_n$ and the Chevalley generators of $\mathfrak{sl}_n$?
 A: Note that by writing everything in terms of matrices you are imposing a specific inclusion but there are many. Any choice of symmetric bilinear form gives a different but isomorphic copy of $\mathfrak{so}_n$ (the set of endomorphisms which are skew symmetric for the given form) all of which sit inside $\mathfrak{sl}_n$ (which more generally we think of as tracefree endomorphisms).
A Chevalley basis is built out of root vectors in such a way as to make all the structure constants integers. In particular it depends on a choice of a Cartan subalgebra and on a choice of simple roots. Writing the $\mathfrak{sl}_n$ as a set of matrices implies a special choice of these with the CSA the set of diagonal matrices and the simple root spaces given by the "just" off diagonal elementary matrices. However the given copy of $\mathfrak{so}_n$ doesn't contain any diagonal matrices so we don't have a special choice of CSA here and certainly not one that interacts neatly with the choices on $\mathfrak{sl}_n$.
It is of course possible to choose a different copy of $\mathfrak{so}_n$ where the diagonal matrices do form a CSA and then we might get a nice relationship between the two bases. In this case though we are almost deliberately choosing the bilinear form to produce this.
The same is true for the symplectic story where we replace symmetric bilinear forms with symplectic ones.
