Lifting to non-Integral Idempotents Geoffrey R. Robinson School of Mathematics and Statistics University of Birmingham Edgbaston Birmingham B15 2TT England

AMS SUBJECT CLASSIFICATION: 20C20

INTRODUCTION: Let G be a finite group, p be a prime divisor of the order of G. Let R be a complete discrete valuation ring ( with valuation group ℤ ) whose field of fractions 𝕂 is a splitting field ( of characteristic 0) for G and all its subgroups. Let (π) be the unique maximal ideal of R. Set F = R/(π). We assume that F has characteristic p. Set A = Z(RG), and let J = {a ∈ A : a^{n} ∈ πA for some n ∈ ℕ}. Then J is an ideal of A containing πA, and is the full pre-image of the Jacobson radical of A/πA.

When we discuss convergence in A, we mean with respect to the metric naturally induced on A by the metric on R obtained via the valuation ν_{π}.

The procedure for lifting idempotents of Z(F G) to idempotents of Z(RG) is well-known.In the first section, we illustrate how to produce all central idempotents of 𝕂G, from knowledge of certain distinguished elements of A/π^{2t}J, where t is a positive integer less than or equal to ν_{π}(|G|).

It will be evident from the description below that the method of the first section would work in much more general situations. However, our present motivation lies in the fact that knowledge of the primitive central idempotents of 𝕂G is equivalent to knowledge of the irreducible characters of 𝕂G, and these correspond naturally to the complex irreducible characters of G. Consequently, in the first section, we content ourselves with explicitly describing only this particular case and remarking that these constructions work “block-by-block” with trivial modifications.

In the second section, we outline another approach to lifting elements closely related to idempotents in more general contexts, which may be of wider interest.

Among other things, we regard the first section of this paper as a contribution to the attempt to understand heights and defects of complex irreducible characters. The results there illustrate that the greater the defect of an irreducible character, the larger the factor ring of A it is necessary to consider in order to determine the associated idempotent. On the other hand, we note that R may be chosen so that A/π^{s}J is finite for every non-negative integer s.

Let us consider an idempotent e ∈ Z(𝕂G). We call the smallest non-negative integer t such that π^{t}e ∈ A the deficiency of e. By consideration of irreducible characters, |G|e ∈ A, so that t ≤ ν_{π}(|G|). We set X = π^{t}e. Then X^{2} = π^{t}X.

Conversely, if we have an element Y ∈ A\πA such that Y^{2} = π^{t}Y for some non-negative integer t, then e = π^{−t}Y is an idempotent of deficiency t of Z(𝕂G). REMARK: We point out that, more precisely, if e is primitive in Z(𝕂G), and χ is the associated irreducible character, then the deficiency of e is

ν_{π}(|G|) − ν_{π}(χ(1)) − min{ν_{π}(χ(g)) : g ∈ G}.

This may also be written as d_{χ}ν_{π}(p)− min{ν_{π}(χ(g)) : g ∈ G}, where d_{χ} is the defect ( with respect to p) of χ in the now usual sense. Notice, however, that this deficiency can be strictly less than d_{χ}ν_{π}(p) - for example, it always will be when G is a p-group and χ is non-linear. More generally, strict inequality holds precisely when χ(g) ∈ π for all g ∈ G ( in fact, it suffices to consider p-regular g). This happens if and only if all decomposition numbers associated to χ are divisible by p. In particular if χ is an irreducible character of maximal defect in its p-block ( equivalently, has height 0), then the associated idempotent does have deficiency d_{χ}ν_{π}(p). This is the maximal possible deficiency for any primitive idempotent ( of Z(𝕂G)) in that block, and only primitive idempotents associated to irreducible characters of height zero in that block can have that deficiency.

This raises the question of what happens if we work with the case A = Z(B) for some block B instead of A = Z(RG). Since blocks of very different groups can have isomorphic centres, we may wish to consider the commutative algebra A as an algebra in its own right, without reference to a particular group. In those circumstances, what information about the block can be read from invariants of A ? Clearly, the number of irreducible characters in any block B with centre isomorphic to A is the R-rank of A, as was already known to Brauer.

The discussion above shows that all blocks B which have centre ( isomorphic to ) A have defect groups of the same order, and that the number of irreducible characters of height zero in any such block B is determined by A, as has already been observed ( with a somewhat different proof) by Cliff,Plesken, and Weiss in [1].

As E.C. Dade brought to our attention, it does not seem immediately obvious that the defects (or heights) of other irreducible characters in the block can be determined from the knowledge of the isomorphism type of A alone.

Section 1 : The Main Result

We say that the elements X_{0} + π^{2t}J and Y_{0} + π^{2t}J of A/π^{2t}J are orthogonal if X_{0}Y_{0} ∈ π^{2t}J. It will be useful in the proof below to define ( for each natural number n ∈ ℕ) the ideal J_{n} of A

as the set of elements a ∈ A such that aλ ∈ π^{n} for each linear character λ of Z(𝕂G). We note that J_{1} = J, and that J^{n} ⊆ J_{n}, but equality need not hold in general. However, setting h = ν_{π}(|G|), we note that J_{h+r} ⊆ π^{r}A for each non-negative integer r. Hence a sequence (a_{n}) of elements of A is Cauchy with respect to the given metric if and only if for each positive integer n, there is an integer N(n) such that a_{r} − a_{s} ∈ J_{n} whenever r, s > N(n).

Our main result is : THEOREM:Let t be a non-negative integer and let X_{0} + π^{2t}J and Y_{0} + π^{2t}J be elements of A/π^{2t}J with neither X_{0} nor Y_{0} in πA. Suppose that (X_{0} + π^{2t}J)^{2} = π^{t}X_{0} + π^{2t}J, and that (Y_{0} + π^{2t}J)^{2} = π^{t}Y_{0} + π^{2t}J. Then there is a unique element, X, of A\πA such that X ≡ X_{0} ( modJ_{t+1}) and X^{2} = π^{t}X. Similarly, there is a unique element Y of A\πA such that Y ≡ Y_{0} ( modJ_{t+1}) and Y^{2} = π^{t}Y. Furthermore, π^{−t}X and π^{−t}Y are orthogonal idempotents if X_{0} + π^{2t}J and Y_{0} + π^{2t}J are orthogonal. PROOF: Suppose first that we have elements X and Y which lie in A\πA such that

X^{2} = π^{t}X, Y^{2} = π^{t}Y, X − Y ∈ J_{t+1}.

Set E_{1} = π^{−t}X, E_{2} = π^{−t}Y, both of which are idempotents of deficiency t in Z(𝕂G).

Let λ be any linear character of Z(𝕂G). Now E_{i}λ ∈ {0, 1} for i = 1, 2, while E_{1}λ ≡ E_{2}λ ( mod π) by hypothesis, so that E_{1}λ = E_{2}λ. Since λ was arbitrary, E_{1} = E_{2}. Hence X = Y.

Now we have an element X_{0} ∈ A\πA such that

X_{0}^{2} ≡ π^{t}X_{0} (mod π^{2t}J). Suppose that for some non-negative integer n we have found an element X_{n} ∈ A\πA such that

X_{n}^{2} ≡ π^{t}X_{n} (mod J_{2t+2^{n}},) X_{n}^{2} ≡ π^{t}X_{n} (mod π^{2t}J) and X_{n} ≡ X_{0} ( mod J_{t+1}).

Notice that

(2X_{n} − π^{t}1_{A})^{2} = π^{2t}1_{A} + 4(X_{n}^{2} − π^{t}X_{n}),

an element of J_{2t}\J_{2t+1}. This last element is also

π^{2t}[1_{A} + 4π^{−2t}(X_{n}^{2} − π^{t}X_{n})].

By hypothesis, 1_{A} + 4π^{−2t}(X_{n}^{2} − π^{t}X_{n}) is a unit of A, which we denote by U_{n} for convenience. Furthermore, we note for later convenience that 2X_{n} − π^{t}1_{A} ∈ J_{t}.

Now (X_{n}^{2} − π^{t}X_{n})(2X_{n} − π^{t}1_{A}) is certainly in π^{2t}J_{t+1}. We set

X_{n+1} = X_{n} − U_{n}^{−1}π^{−2t}(X_{n}^{2} − π^{t}X_{n})(2X_{n} − π^{t}1_{A}),

an element of A. Furthermore, X_{n+1} ≡ X_{n} (mod J_{t+2^{n}}). Notice that when t = 0, this inductively implies that X_{n+1} ∉ J. In particular, X_{n+1} ∉ πA. When p = 2 and t > 0, we note that X_{n+1} ≡ X_{n} ( mod πA).

Now consider X_{n+1}^{2} − π^{t}X_{n+1}. Notice that

X_{n+1} = π^{−2t}U_{n}^{−1}[π^{2t}X_{n}U_{n} − (X_{n}^{2} − π^{t}X_{n})(2X_{n} − π^{t}1_{A})]
= π^{−2t}U_{n}^{−1}([2X_{n} − π^{t}1_{A}][X_{n}(2X_{n} − π^{t}1_{A}) − (X_{n}^{2} − π^{t}X_{n})]),

making use of the fact that (2X_{n} − π^{t}1_{A})^{2} = π^{2t}U_{n}. Hence X_{n+1} = π^{−2t}U_{n}^{−1}[(2X_{n} − π^{t}1_{A})X_{n}^{2}]. Consequently,

(X_{n+1}^{2} − π^{t}X_{n+1}) = π^{−4t}U_{n}^{−2}[X_{n}^{4}(2X_{n} − π^{t}1_{A})^{2}] − π^{−t}U_{n}^{−1}[(2X_{n} − π^{t}1_{A})X_{n}^{2}].

Again, we recall that (2X_{n} − π^{t}1_{A})^{2} = π^{2t}U_{n}, so the above expression simplifies to

U_{n}^{−1}π^{−2t}(X_{n}^{2} − π^{t}X_{n})^{2}.

By the hypotheses on X_{n}, this last expression lies in J_{2t+2^{n+1}}, and lies in π^{2t}J. Hence the sequence (X_{n}) so constructed is Cauchy by the remarks above, so has a limit X ∈ A. Furthermore, X ≡ X_{0} (mod J_{t+1}), and X satisfies X^{2} = π^{t}X.

We claim that X_{n+1} lies outside πA. When p = 2 or t = 0, we have seen this already, so we suppose that p is odd, that t > 0, and that X_{n+1} ∈ πA, ( we recall that X_{n} ∉ πA). We note that, in this case, 1_{A} + U_{n} is a unit of A, and hence so is 1_{A} + U_{n}^{−1}. Notice that since

X_{n+1} = π^{−2t}U_{n}^{−1}[(2X_{n} − π^{t}1_{A})X_{n}^{2}],

we may conclude that

X_{n} + U_{n}^{−1}π^{−t}X_{n}(2X_{n} − π^{t}1_{A}) ∈ πA.

Since X_{n}^{2} − π^{t}X_{n} ∈ π^{2t}J, this last expression is congruent to (1_{A} + U_{n}^{−1})X_{n} ( mod πA), a contradiction, as X_{n} ∉ πA and (1_{A} + U_{n}^{−1}) is a unit of A. Hence X ∉ πA in any case.

Returning to the proof, the argument at the beginning shows that if X^{(1)}+π^{2t}J and X^{(2)}+π^{2t}J satisfy (X^{(i)})^{2} − π^{t}X^{(i)} ∈ π^{2t}J and X^{(i)} ∉ πA, but X^{(1)} − X^{(2)} ∈ J_{t+1}, then X^{(1)} and X^{(2)} must each “lift” to the same solution X of X^{2} = π^{t}X, (X ∉ πA). Conversely, elements which are incongruent (modJ_{t+1}) must certainly “lift” to distinct elements.

Finally, suppose that we have elements X_{0} and Y_{0} which lie in A\πA such that

X_{0}^{2} − π^{t}X_{0} ∈ π^{2t}J, Y_{0}^{2} − π^{t}Y_{0} ∈ π^{2t}J, X_{0}Y_{0} ∈ π^{2t}J.

We inductively construct the convergent sequences (X_{n}) and (Y_{n}) as before, with respective limits X and Y. We prove by induction that X_{n}Y_{n} ∈ J_{2t+2^{n}}, this being the case by assumption when n = 0. Suppose then that n ≥ 0 and that the the assertion is known to hold for n. We have seen that

X_{n+1} = π^{−2t}U_{n}^{−1}[(2X_{n} − π^{t}1_{A})X_{n}^{2}]

and similarly,

Y_{n+1} = π^{−2t}V_{n}^{−1}[(2Y_{n} − π^{t}1_{A})Y_{n}^{2}]

for an appropriate unit, V_{n} of A. Thus

X_{n+1}Y_{n+1} = π^{−4t}U_{n}^{−1}V_{n}^{−1}(2X_{n} − π^{t}1_{A})(2Y_{n} − π^{t}1_{A})X_{n}^{2}Y_{n}^{2}.

We have seen that

(2X_{n} − π^{t}1_{A})(2Y_{n} − π^{t}1_{A}) ∈ J_{2t}.

Hence, by the inductive hypothesis, X_{n+1}Y_{n+1} lies in J_{2t+2^{n+1}}. Hence the claim is established by induction, so that, taking limits, XY = 0 and the proof is complete. REMARKS: It is easy to recognise which “lifted” solutions X with X^{2} = π^{t}X, (X ∉ πA) are such that π^{−t}X is a primitive idempotent of Z(𝕂G), since this is the case precisely when XA has R-rank 1.

We remark that we have shown that for each non-negative integer t, the number of primitive idempotents of Z(𝕂G) which have deficiency t ( or less) is completely determined by the structure of the ring A/π^{2t}J. Hence to find the primitive idempotents associated to irreducible characters of a given defect s of 𝕂G, it is certainly sufficient to consider the ring A/p^{2s+1}A.

Section 2 : Polynomials and Idempotents

The method of the first section is a modification of a particular case of Newton’s method. The results of this section evolved in attempting to accelerate convergence of an earlier version of the method of the first section.

Now let R denote the ring ℤ[x, y], where x and y are commuting, ( algebraically independent) indeterminates.

We inductively define polynomials p_{n}(x, y) as follows: p_{1}(x, y) = 1, and for n > 0,

xp_{n+1}(x, y) = (2x − y)^{2}p_{n}(x, y) − p_{n}(0, 1)(y − x)^{n+1} … (1).

It follows easily by induction that p_{n}(xy, y) = y^{n−1}p_{n}(x, 1) for all n. In particular, we see that p_{n}(0, y) = y^{n−1}p_{n}(0, 1) for each n ( notice that this fact for p_{n} is necessary for the definition of p_{n+1}, since it is this which ensures that the right hand side of the equation defining p_{n+1} vanishes when evaluated at (0, y)).

We set q_{n}(x, y) = (−1)^{n+1}p_{n}(y − x, y) for each n. Then replacing x by y − x in the above shows that

(x − y)q_{n+1}(x, y) = (2x − y)^{2}q_{n}(x, y) − p_{n}(0, 1)(−x)^{n+1} … (2),

which may also be written as

(x − y)q_{n+1}(x, y) = (2x − y)^{2}q_{n}(x, y) − q_{n}(1, 1)x^{n+1} … (3).

We have x^{2}p_{1}(x, y) = (2x − y)x − (x^{2} − yx), while for n > 0 we have

x^{n+2}p_{n+1}(x, y) = (2x − y)^{2}x^{n+1}p_{n}(x, y) − p_{n}(0, 1)(xy − x^{2})^{n+1} … (4).

Similarly,

(x − y)^{n+2}q_{n+1}(x, y) = (2x − y)^{2}(x − y)^{n+1}q_{n}(x, y) − p_{n}(0, 1)(xy − x^{2})^{n+1} … (5).

It follows by induction that we have

x^{n+2}p_{n+1}(x, y) = y(2x − y)^{2n+1} + (x − y)^{n+2}q_{n+1}(x, y) … (6)

for all n ≥ 0. Hence we see that

x^{n+2}p_{n+1}(x, y) − (y − x)^{n+2}p_{n+1}(y − x, y) = y(2x − y)^{2n+1} … (7)

for all such n.

Now suppose that we are given a ring S with 1. Suppose further that S has an ideal J such that ∩_{n=1}^{∞}J^{n} = {0} and such that S is complete with respect to the J-adic topology.

Let a be an element of S which is not a zero divisor in the subring C_{S}(a). We say that a non-zero element s of C_{S}(a) is a-potent if s^{2} = as ( notice that if S embeds in some larger ring in which a becomes invertible, then a^{−1}s becomes an idempotent, which, as will be apparent from the first section, is our motivation for considering such elements). We are interested in locating non-trivial a-potent elements of C_{S}(a) and we first show that it suffices consider images (mod a^{4}J) to do this.

Suppose that we have an element b ∈ C_{S}(a) with b^{2} ≡ ab ( mod a^{4}J ). We will use b to produce an a-potent element of C_{S}(a). It suffices to work within the closure of the subring generated by a and b, which is commutative. Hence we assume from now on that S is commutative, and that a is not a zero divisor in S.

Notice that the hypotheses tell us that (2b − a)^{2} = a^{2}(1 + j) for some j ∈ J, and that 1 + j is a unit of S. Hence 2b−a is not a zero divisor in S. Notice furthermore that (2b−a)^{−2}(b^{2}−ab) is a welldefined element of S, and lies in the ideal a^{2}J. Hence for all positive integers n, (2b−a)^{−2n}(b^{2}−ab)^{n} is an element of a^{2n}J^{n}. Multiplying this element by (2b − a), we see that (2b − a)^{1−2n}(b^{2} − ab)^{n} is a well-defined element of J for all such n.

We have b^{2} = (2b − a)b − (b^{2} − ab), while for n > 0,

b^{n+2}p_{n+1}(b, a) = (2b − a)^{2}b^{n+1}p_{n}(b, a) − p_{n}(0, 1)(ab − b^{2})^{n+1},

so it follows by induction that

(2b − a)^{−1−2n}b^{n+2}p_{n+1}(b, a)

is a well-defined element of S whenever n is a non-negative integer. We label this element b_{n+1}. We remark that

b_{n+1} = b − (2b − a)^{−1}(b^{2} − ab) − ∑_{k=1}^{n} p_{k}(0, 1)(ab − b^{2})^{k+1}(2b − a)^{−1−2k}

for all n > 0.

Since

x^{n+2}p_{n+1}(x, y) = y(2x − y)^{2n+1} + (x − y)^{n+2}q_{n+1}(x, y)

for all n ≥ 0, we conclude that

b_{n+1} = b^{n+2}p_{n+1}(b, a)(2b − a)^{−1−2n} = a + (b − a)^{n+2}q_{n+1}(b, a)(2b − a)^{−1−2n}

for all n ≥ 0.

Notice then that

b_{n+1}^{2} − ab_{n+1} = (b^{2} − ab)^{n+2}p_{n+1}(b, a)q_{n+1}(b, a)(2b − a)^{−2−4n}.

Now (2b − a)^{2+4n} has the form a^{4n+2}u for some unit u of S. Consequently, b_{n+1}^{2} − ab_{n+1} lies in a^{6}J^{n+2}. The sequence (b_{n}) is Cauchy with respect to the metric discussed earlier, so has a limit in S, say b_{0}, which satisfies b_{0}^{2} = ab_{0}, and is thus a-potent.

We have

b_{0} = b − (2b − a)^{−1}(b^{2} − ab) − ∑_{k=1}^{∞} p_{k}(0, 1)(ab − b^{2})^{k+1}(2b − a)^{−1−2k},

so it also follows that b_{0} ≡ b ( mod a^{2}J). Notice that we have given an explicit infinite series expression for b_{0} in terms of a and b.

We note further that if c is an element of C_{S}(⟨a, b⟩) with c^{2} ≡ ac ( mod a^{4}J), and we have bc ∈ a^{4}J, then we have b_{n+1}c_{n+1} ∈ a^{6}J^{n+2} for all n. Hence we see that the respective limits b_{0} and c_{0} satisfy b_{0}c_{0} = 0.

Returning to the question of accelerated convergence, we point out that the above methods allow us to produce iterative procedures which converge at any desired rate.

Recall that

b_{n+1} = (2b−a)^{−1−2n}b^{n+2}p_{n+1}(b, a) = b−(2b−a)^{−1}(b^{2} −ab)−∑_{k=1}^{n} p_{k}(0, 1)(ab−b^{2})^{k+1}(2b−a)^{−1−2k}

and that

b_{n+1}^{2} − ab_{n+1} = (b^{2} − ab)^{n+2}p_{n+1}(b, a)q_{n+1}(b, a)(2b − a)^{−2−4n}.

Let us fix a choice of n, and define a Cauchy sequence (b^{(m)}) as follows: b^{(1)} = b, and b^{(m+1)} = (b^{(m)})_{n+1}, with notation as above. Then an easy induction argument shows that

b^{(m+1)} ≡ b^{(m)}(moda^{2}J^{(n+2)^{m−1}})

and that

(b^{(m)})^{2} − ab^{(m)} ∈ a^{4}J^{(n+2)^{m−1}}

for all m.

REFERENCES

[1] Cliff,G., Plesken,W., & Weiss,A. ; in P. Fong (ed.),The Arcata Conference on Representations of Finite Groups,Proc. Symp. Pure Math. 47 (1), American Math. Soc.,Providence, (1987), 413-420 .