Background: Given a finite group $G$ and a prime $p$ dividing its order, Brauer theory compares the ordinary characters of $G$ with the Brauer characters arising from $p$-modular representations. On the character level there is a well-defined "reduction mod $p$", which Brauer showed to be surjective in the sense that each irreducible Brauer character lifts to a virtual character ($\mathbb{Z}$-linear combination of ordinary irreducible characters). Part III of Serre's textbook works this out in the framework of Grothendieck groups. The lifting is not unique, though J.A. Green's 1955 work here provides a simple character formula for one lifting. (This is developed in a more sophisticated way by Lusztig, Ann. of Math. Studies No. 81, 1974.)

An important subtheme, initiated about 70 years ago, concerns blocks (indecomposable 2-sided ideals of the modular group algebra) which have cyclic defect groups: here a defect group is a certain $p$-subgroup of $G$ determined up to conjugacy. To such a block is associated a Brauer tree, a graph with edges labelled by irreducible Brauer characters and vertices labelled by one (or exceptionally several) irreducible ordinary characters. Chapter VII of Feit's 1982 book has an extensive treatment, while his 1984 paper studies the possible Brauer trees using the classification of finite simple groups.

For finite groups of Lie type with defining characteristic $p$ (simple or close to simple), the only family giving rise to Brauer trees consists of the groups $\mathrm{SL}(2,p)$ (say for $p>3$). Leaving aside the Steinberg character (with trivial defect group), there are two blocks with defect group of order $p$. Consider just the block containing the trivial ordinary (and modular) character, or the corresponding block of $\mathrm{PSL}(2,p)$. Here the tree is an open polygon with $1_G$ at one end and two exceptional characters at the other end. The edges correspond to even highest weights $0, p-3. 2, p-5, \dots$. It's easy to specify Brauer liftings here by following vertices to the left or right with alternating signs.

For an arbitrary $G$ having a block with a nontrivial Brauer tree, is there a similar algorithm for Brauer lifting (in particular, an algorithm for Green's special lifting)?

I don't know all the literature well enough to sort this out, but the idea would be somewhat in the spirit of Green's "walk around the Brauer tree" (J. Austral. Math. Soc. 17, 1974) for producing a projective resolution (necessarily infinite, but periodic). For this purpose the Brauer tree simultaneously encodes the projective covers needed, via Brauer reciprocity.

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    $\begingroup$ You might try contacting Raphael Rouquier and/or David Craven. I think they have some new perspectives on Brauer Trees. $\endgroup$ – Geoff Robinson Sep 28 '12 at 19:19

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.

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  • $\begingroup$ Thanks for the proposed algorithm, which I'll have to sort out. I've relied mainly on the general method of Green (and Lusztig) and will meanwhile add a link to Green's 1955 paper which gives his approach to Brauer lifting. $\endgroup$ – Jim Humphreys Sep 30 '12 at 14:56
  • $\begingroup$ You are welcome. i was able to find green's result in his 55 paper (in fact i had come across it as grad student but forgot). he constructs a character starting from a symmetric polynomial -- if you take the degree 1 symmetric polynomial you indeed get an ordinary character that extends the given Brauer character on p'-elements. it's not clear to me that this character can be read off purely from the tree, i.e. without using the structure of the group. if i had the time, i would take some (small) examples, e.g. open polygons that arise for general linear but also for alternating groups, ... $\endgroup$ – fherzig Sep 30 '12 at 18:08
  • $\begingroup$ ...and see what happens there. $\endgroup$ – fherzig Sep 30 '12 at 18:08
  • $\begingroup$ Your comments make me more cautious about what can be expected in the way of a uniform algorithm. But leaving aside Green's earlier viewpoint for a moment, I'm still impressed by his ability to find all projective resolutions using the Brauer tree. Meanwhile I don't yet understand your Update and Example, since I'm not expecting symmetry but rather asymmetry in the choice of a lifting. In your example, I could lift the trivial Brauer character in two ways to an ordinary (not just virtual) character of degree 1, but prefer the lifting to the 1-character. $\endgroup$ – Jim Humphreys Oct 1 '12 at 16:42

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.

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  • $\begingroup$ I meant to accept the answer by Herzig many years ago, but then forgot to return to this question. Your version is simpler but comes late, so the credit for an answer should still go to him. Thanks for the clarification, which is helpful. $\endgroup$ – Jim Humphreys May 3 '18 at 0:38

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