On Certain Tilting Modules for SL2

We give a complete picture of when the tensor product of an induced module and a Weyl module is a tilting module for the algebraic group $SL_2$ over an algebraically closed field of characteristic $p$. Whilst the result is recursive by nature, we give an explicit statement in terms of the $p$-adic expansions of the highest weight of each module.

Let k r be the one dimensional B module on which T acts via r ∈ Z, and let ∇(r) be the induced module Ind G B (k r ). Then ∇(r) is finite dimensional and is non-zero only when r is dominant, i.e. r ≥ 0. It is well known that ∇(r) = S r E, the r th symmetric power of the natural module E. Let ∆(r) be the Weyl module given by ∆(r) = ∇(r) * . By a tilting module we mean a module which has both a ∇-filtration (or good filtration) and a ∆-filtration (or Weyl filtration) as defined in [1]. We will denote by T (r) the unique indecomposable tilting module of highest weight r ∈ X + .
We will make use of the character Ch(V ) of a module V . This is given by inside the ring Z[x, x −1 ] of Laurent polynomials, where V r is the r weight space of V . We will write χ(r) for Ch(∇(r)) = Ch(∆(r)) and note that χ(1) = x + x −1 . From the action of the Weyl group on each weight space, we have in fact that Ch(V ) ∈ Z[χ(1)], which is a unique factorization domain.
The objects of interest in this article are the modules ∇(r) ⊗ ∆(s), for dominant weights r and s. The character of these modules is given by the well known Clebsch-Gordan formula (assuming r ≥ s) Ch(∇(r) ⊗ ∆(s)) = χ(r)χ(s) = s i=0 χ(r + s − 2i).

Tilting Modules
We will now give several useful results which will be used in the proceeding sections. In particular we will make extensive use of the following well known result, for which we have outlined a proof for the reader's convenience. Proof. That the sequence exists is clear by considering the ∇-filtration of ∇(r) ⊗ E = ∇(r) ⊗ ∆(1). If p does not divide r + 1, the result follows by considering the blocks (see [6,II.7.1]) for SL 2 (k). On the other hand, if p does divide r + 1, then the module E ⊗ ∇(r) is projective as a G 1 -module, while neither ∇(r − 1) nor ∇(r + 1) are, so the sequence cannot be split.
Proof. First we note that we can apply Theorem 1.1 equally well to E ⊗ ∆(r) since (E ⊗ ∇(r)) * = E ⊗ ∆(r). Now we consider ∇(np + 1) ⊗ E ⊗ ∆(np). Since the tensor product of tilting modules is also tilting (this follows from [7, Theorem 1]), this is a tilting module. Furthermore p does not divide np + 1, so using the above result we obtain Since the whole module is tilting, each summand on the right hand side is tilting, in particular the module ∇(np + 1) ⊗ ∆(np − 1). We can continue to propagate in this manner, tensoring E with both ∇(t) and ∆(t) for t ∈ {np, np + 1, . . . , (n + 1)p − 2}, until we reach np − 1 and (n + 1)p − 1, for which we can no longer apply the above result.
This result shows us that there are more tilting modules of the form ∇(r) ⊗ ∆(s) than those given in [8,Lemma 3.3] for every characteristic p. Before we delve further into this investigation, we prove a couple of useful lemmas concerning tilting modules. For the following lemma, G may be an arbitrary semisimple, simply connected algebraic group, over an algebraically closed field of prime characteristic. We will denote by ( , ) the usual positive definite symmetric bilinear form on the Euclidean space in which the root system of G lies. Lemma 1.2. Let T 1 and T 2 be tilting modules where T 1 is projective as a G 1 -module, then the tensor product T 1 ⊗ T F 2 is also a tilting module.
Proof. First notice that it's sufficient to prove that for any T 1 and T 2 satisfying the hypothesis, the tensor product T 1 ⊗T F 2 has a ∇-filtration. Then the dual module T * 1 ⊗(T F 2 ) * (a tensor product of tilting modules still satisfying the hypothesis) also has a ∇-filtration, or equivalently, T 1 ⊗ T F 2 has a ∆-filtration.
Now suppose that T 1 is a tilting module that is projective as a G 1 -module. Then each indecomposable summand of T 1 must be T (λ) for some λ as above (again using [1, Proposition 2.4]). Hence T 1 is a direct sum of modules with a ∇-filtration, and thus itself has a ∇-filtration.
We will use this lemma throughout the article, in conjunction with the facts that ∇(p − 1) = ∆(p − 1) is a projective G 1 -module [6, Proposition II.10.1], and that the tensor product of a projective G 1 -module with another G 1 -module is again projective.
Next we return to the case G = SL 2 (k).
Then W is a tilting module.
Proof. Since each tilting module has a unique decomposition (up to isomorphism) into indecomposable tilting modules T (m) with highest weight m, it suffices to prove this for V = T (m). We can split this into three separate cases, the first of which deals with 0 ≤ m ≤ p − 1. For such m we have T (m) = L(m) and so Considering the case t = p − 1 separately we get For the remaining cases we will use induction by writing m = p − 1 + t + pn for some n ∈ N and 0 ≤ t ≤ p − 1 so that we can write T (m) = T (p − 1 + t) ⊗ T (n) F . Taking the G 1 fixed points we get H 0 (G 1 , T (m)) = H 0 (G 1 , T (p − 1 + t)) ⊗ T (n) F which by the previous case gives us and is thus tilting.

Main Theorem
Before stating the main theorem of this paper, we will introduce some notation. Let r ∈ N and p a prime. We can write the base p expansion of r as where each r i ∈ {0, . . . , p − 1}, r n = 0 and for all j > n we have r j = 0. We will say that r has p-length n (or just length n if the prime is clear), and write len p (r) = n.
We define len p (0) = −1. Now given any pair (r, s) ∈ N 2 we can write where n = max (len p (r), len p (s)) so that at least one of r n and s n is non zero. Now let m be the largest integer such that r m = s m and let so that if r > s we have r m > s m andr >ŝ. Using this notation we may write Notice in particular that r −r = s −ŝ and denote this number by ε p (r, s) so that p m+1 divides ε p (r, s). We will call the pair (r,ŝ) the primitive of (r, s), and say that (r, s) is a primitive pair if (r, s) = (r,ŝ).
Lemma 2.1. For r and s as above with (r, s) = (r,ŝ), write r = pt + r 0 and s = pu + s 0 , then we have the following.
Proof. First we note that we have It's clear then that pt + r 0 =r, and pû + s 0 =ŝ. Furthermore we have that For the final statement, we first note that since r ≥ s we must have that r m > s m , and since s < s it follows that r m > s m . Hence we have that ε p (r, s) = ε p (r, s ). Now s =ŝ + ε p (r, s), so it follows that s =ŝ − 1 + ε p (r, s). On the other hand we have Hence the pair (r,ŝ ) is equal to (r,ŝ − 1).
This lemma will be helpful in proving the main theorem, which follows.
The following picture illustrates which of the ∇(r) ⊗ ∆(s) are tilting up to r, s ≤ 31 for p = 2 . Before beginning the proof of this theorem, we will say a quick word on the figure above. It's clear that the pairs (r, s) for which (r,ŝ) = (r, s) are given by the intervals [2 n , 2 n+1 − 1] × [0, 2 n − 1] for r ≥ s, and vice versa for s ≥ r. Of these, according to the theorem, only the modules ∇(2 n+1 − 1) ⊗ ∆(s) and ∇(r) ⊗ ∆(2 n − 1) are tilting (again, with r ≥ s). With the mirrored situation for s ≥ r, this accounts for the horizontal and vertical green lines appearing every 2 n − 1. To prove the theorem, we will first gather some elementary results on the modules ∇(r) ⊗ ∆(s).
Notice that in the case p = 2 we have that v = 0 = p − 2 − v, so this reduces to the sequence Tensoring the former with the latter gives the following short exact sequence Since both ∇(v) ⊗ ∆(p − 1) and ∇(p − 2 − v) ⊗ ∆(p − 1) are tilting and projective as It remains to determine which of the modules ∇(r) ⊗ ∆(s) are tilting when neither r nor s is congruent to p − 1 modulo p. It turns out that this only occurs in the cases given in Lemma 1.1. In order to show this we will make use of the character.
Lemma 2.4. Let G be a semisimple, simply connected algebraic group over k, and let T be a G-module that is projective as a G 1 -module. Then Proof. This follows immediately from [2, 1.2(2)], since T must also be a projective B 1 module.
We now revert to the case G = SL 2 (k) and obtain the following corollary.
so the roots of this equation are the (2r + 2) th roots of unity, except ±1. If χ(p − 1) divides χ(r) then, we must have the 2p th roots of unity are also (2r + 2) th roots of unity, which would imply that p divides r + 1, i.e. that r is congruent to p − 1 modulo p.
Hence we have shown that if both r and s are not congruent to p − 1 modulo p, the character χ(p − 1) does not divide Ch(∇(r) ⊗ ∆(s)) = χ(r)χ(s). Now suppose that ∇(r) ⊗ ∆(s) is tilting, and that |r − s| > p − 1. By considering its good filtration, we see that the decomposition of ∇(r) ⊗ ∆(s) into indecomposable tilting modules cannot contain any T (j) for j = 0, . . . , p − 1. By Corollary 2.1 its character is divisible by χ(p − 1) but the above calculation contradicts this. In summary: Lemma 2.5. For r and s both not congruent to p − 1 modulo p, and |r − s| > p − 1, the module ∇(r) ⊗ ∆(s) is not tilting.
There are now only a few more cases which we have not considered. These occur when |r − s| ≤ p − 1, but not both of r and s lie in the set given in Lemma 1.1 (for example, take r = np and s = np − 2). We can swiftly deal with these cases, by once again appealing to Theorem 1.1, but we must first make precise exactly which r and s we are considering. We will assume that r > s, but the argument works equally well for r < s.
We have now determined exactly which of the modules ∇(r) ⊗ ∆(s) are tilting for all primes p (recall that for r, s ∈ {0, 1, . . . , p − 1} the module ∇(r) ⊗ ∆(s) is tilting, so we can begin applying Lemma 2.3 to these modules). The figure below illustrates this for p = 3.
The pattern here is similar to that for the case p = 2, however we now have the extra complication that the values a and b from the theorem can be either 0 or 1 (whereas in the p = 2 case, we had that a = b = 0), and the coefficients in the base 3 expansion are in {0, 1, 2}. As such, we have that for a given n ∈ N, those blocks where (r, s) = (r,ŝ) can be broken up into the union

Proof of Theorem 2.1
We are now ready to prove Theorem 2.1, which we will do in two steps. The first is to show that for a primitive pair (r,ŝ), we have that ∇(r) ⊗ ∆(ŝ) is a tilting module if and only ifr andŝ are as described in the statement of the theorem. The second step is to show that for any pair (r, s) with primitive pair (r,ŝ), we have that ∇(r) ⊗ ∆(s) is tilting if and only if ∇(r) ⊗ ∆(ŝ) is tilting.
We first note that since the dual of a tilting module is also a tilting module, and we have the relation (∇(r) ⊗ ∆(s)) * = ∇(s) ⊗ ∆(r), it will be safe to assume that r ≥ s, and simply take the dual for the case r < s.
1.) We will begin the first step by assuming that, for a primitive pair (r, s), we have that ∇(r) ⊗ ∆(s) is tilting. We want to show that this forces r and s to be of the form in the statement of the theorem. We will assume that r ≥ s and proceed by induction on len p (r) = N . For N = 0 we have that r ≤ p − 1 and so r = ap N + p N − 1 for a = 0, . . . , p − 2, or in the case r = p − 1 we have r = p N +1 − 1. In each case we have that r is of the desired form, and s < p N +1 .
In the second case, since we are assuming the pair (r, s) to be primitive, we must have that one of r or s is equal to np − 1, and the other is not. In any case then, we may assume then r 0 = p − 1 so that r = pt + p − 1.
Now we have two cases to consider, the first is that s 0 = p − 1, and the second that s 0 = p − 1. Let's suppose that s 0 = p − 1, then by Lemma 2.2 we have that ∇(t) ⊗ ∆(u) is tilting. By induction we must have that t and u are of the form given in the statement of the theorem. Without loss of generality, we may assume t = ap N −1 + p N −1 − 1 for some a ∈ {0, . . . , p − 2}, and u ≤ p N − 1. Hence we have that and s ≤ p N − p + s 0 , which is strictly less than p N +1 since s 0 < p.
For the second case, we suppose that s 0 = p − 1 (the case r 0 = p − 1 and s 0 = p − 1 is obtained in the same manner), so that by Lemma 2.3 we have that ∇(t) ⊗ ∆(u) and ∇(t)⊗∆(u−1) are tilting By induction we have that the pairs (t, u) and (t, u−1) are both of the form in the theorem. Since we cannot have that both u and u − 1 are of the form ap N −1 +p N −1 −1, we must have that t is of this form, and we complete the proof as above.
Now we prove the converse statement, that is, if r = ap n + p n − 1 for some a ∈ {0, . . . , p − 2}, n ∈ N, and s < p n+1 , then ∇(r) ⊗ ∆(s) is tilting. Once again, we will use induction on n, with the case n = 0 being clear. For the inductive step, we have that if r = pt + p − 1 and s = pu + s 0 then t = ap n−1 + p n−1 − 1 and u < p n . Then by induction the modules ∇(t) ⊗ ∆(u) and ∇(t) ⊗ ∆(u − 1) are tilting, so by either Lemma 2.2 or Lemma 2.3 we have that ∇(r) ⊗ ∆(s) is tilting too.
2.) For the next step we prove the statement: ∇(r) ⊗ ∆(s) is tilting if and only if ∇(r) ⊗ ∆(ŝ) is tilting. First, let's assume that ∇(r) ⊗ ∆(s) is tilting. By Lemma 2.6 we have that either one of r and s is congruent to p − 1 modulo p or they lie in the set {np − 1, np, np + 1, . . . , (n + 1)p − 1} for some n ∈ N. Suppose that both r and s lie in the set {np − 1, np, np + 1, . . . , (n + 1)p − 1}. If neither are equal to np − 1 then it's clear thatr andŝ lie in the set {0, . . . , p − 1}, and so ∇(r) ⊗ ∆(ŝ) is tilting. Note that if we assume ∇(r) ⊗ ∆(ŝ) is tilting, we must also have that either r or s is congruent to p − 1 modulo p, as in step 1. We may then, only consider the case that at least one of r and s is congruent to p − 1 modulo p.
Let's suppose that r = pt + p − 1 and s = pu + s 0 , so that by Lemma 2.1 we havê r = pt + p − 1 andŝ = pû + s 0 . As in the previous step, there are two cases to consider: s 0 = p−1 and s 0 = p−1. In both cases we will proceed by induction on len p (r) with r ≥ s.
First, we assume s 0 = p − 1. For the base case len p (r) = 0 we have that r = s = p − 1 so (r,ŝ) = (0, 0), and the result is clear. Now by Lemma 2.2 we have ∇(r) ⊗ ∆(s) is tilting if and only if ∇(t) ⊗ ∆(u) is tilting. By induction then we have that this is tilting if and only if ∇(t) ⊗ ∆(û) is tilting, so applying Lemma 2.2 again (sincer = pt + p − 1 and s = pû + p − 1) we find that this is if and only if ∇(r) ⊗ ∆(ŝ) is tilting.