Algorithm for Brauer lifting via Brauer tree?

I know this is a very old question, but for anyone else who finds this, here is a simple way to write an edge (=simple module) of the Brauer tree as an alternating sum of the vertices (=ordinary characters).

Pick your edge $e$. Removing this edge from the tree yields two connected components, so we choose the one that does not contain the exceptional node, for simplicity, say $X$. (Choose either if there is no exceptional node.)

We assign value +1 to the unique vertex $v$ in $X\cap e$, and value $(-1)^n$ to every vertex of $X$ of distance $n$ from $v$. The reduction modulo $p$ of this alternating sum yields the Brauer character labelling $e$.

To see this, note that the ordinary character of $v$ reduces to $e$ plus the other incident edges. These are subtracted off by removing, for each other edge incident to $v$ (hence in $X$) a copy of the other vertex incident to it (they have distance $1$), which gives us $e$ minus those edges in $X$ that are separated from $e$ by a single edge. The rest is an induction or standard inclusion-exclusion argument.

The only difference between this and fherzig's algorithm is I've used the fact that there are two different ways to write $e$ as an alternating sum of the vertices (one for each component of the tree once the edge is deleted) to avoid the exceptional node.

If I understand correctly, you have a tree with edges labelled by e1,...,en and vertices labelled by v0,...,vn,...v(n+m-1), where the last m >= 1 of these label the same (namely the exceptional) vertex. Relations among Brauer characters are given by: vi = sum of the ej, where ej runs through all the edges adjacent to vi. You would like to find an algorithm to solve this overdetermined system of equations for the ej. Here is one: pick your favourite label at the exceptional vertex and discard the rest (or pick your favourite Z-linear combination whose coefficients add up to 1). Then we are reduced to the case when m = 1. If an edge ej has a leaf vi as vertex, it's easy: we have ej = vi. Next pick an edge ej such that all the adjacent edges at one of its two vertices vi have already been 'lifted'. Then we can use the equation for vi to solve for ej. Inductively work your way further and further away from the leaves. This will give you what you wanted.

Incidentally, you get all other solutions by using the relations vn = ... = v(n+m-1) and sum over even vertices = sum over odd vertices (for this pick an arbitrary vertex v; then say that another vertex is even/odd if it has even/odd distance from v; the relation holds because each edge has precisely one even and one odd vertex).

Unfortunately I don't know about Green's lifting.

Update: I claim that the Brauer tree alone does not determine Green's lift. In particular there is no algorithm for Green's lift that uses only the Brauer tree. The point is that in some cases there is no lift at all that respects the symmetries of the tree.

Example: suppose that $G = S_3$, $p = 2$. The principal block has the cyclic 2-Sylow as defect group. It contains only the trivial representation. The tree thus has only one edge, and there is no exceptional vertex (e.g. as its multiplicity is $p-1 = 1$). More precisely the two characteristic zero representations of this block are the trivial character T and the sign character S (both reducing to the trivial). So all the lifts of the trivial mod p representation are of the form $nT-(n-1)S$ for some integer $n$, but none of these is symmetric w.r.t. transposition of the vertices T, S.