Beiträge zur Algebra und Geometrie Contributions to Algebra and Geometry Volume 35 (1994), No. 2, 193-204.

Remarks to the Dimension of an Atypical Irreducible Representation of the

Special Lie Superalgebra sl(1,n)

Hartmut Schlosser

Fachbereich Mathematik/Informatik, Universität Greifswald Friedrich-Ludwig-Jahnstraße 15a, D – 17489 Greifswald schloss@math-inf.uni-greifswald.d400.de

1. Introduction.

In [6] and [7] Palev gave explicit expressions for the action of the generators of the Lie superalgebra sl(1, n) at an (irreducible) representation. His description of the representation space W of an atypical irreducible representation in terms of highest weights of gl(n) representations give a possibility to compute the dimension of W numerically.

In this paper we will proof an explicit formula for the dimension of an atypical representation of sl(1, n) using numbers H_{l} which connect with the Bernoulli numbers B_{l}.

The results of this paper are announced in [8].

2. Facts from the representation theory of sl(1,n)

We follow the terminology of Palev ([6],[7]). Here we briefly recall the points of his papers which are important for the following. We will use both the language of representations and the language of modules.

The even part of sl(1, n) is isomorphic to gl(n). Let be

[m]_{n} = [m_{1},…, m_{n}]

the highest weight of an irreducible representation of gl(n). The m_{i} are the coordinates of a highest weight considered as functional with respect to the basis of the dual H^{∗} which is dual to the standard basis of the Cartan algebra H. So we will refer to [m]_{n} as highest weight and the corresponding representation we will denote with [m]_{n} also (analogous for sl(1, n)).

The m_{i} are in general complex numbers such that m_{i} − m_{j} ∈ Z_{+} for all i, j with i < j, i, j = 1,…, n. Furthermore, [m]_{n} corresponds to a gl(n) fidirmod (=finitedimensional irreducible module) V_{0}([m]_{n}). To V_{0}([m]_{n}) we can find an induced sl(1, n)

0138-4821/93 $ 2.50 \begin{matrix}c \\ ◯ \end{matrix}1994 Heldermann Verlag, Berlin

194 H. Schlosser: Remarks to the Dimension of an Atypical Irreducible Representation

module W̄ ([m]_{n+1}) where [m]_{n+1} = [m]_{n}; the index at the right bracket we use to differ between the representations of gl(n) and sl(1, n) resp.

We denote as Θ-tuple a n-tuple Θ = (Θ_{1},…, Θ_{n}) with Θ_{j} ∈ {0, 1} for j = 1,…, n. If we consider W̄ ([m]_{n+1}) as a gl(n) module then W̄ ([m]_{n+1}) decomposes into a direct sum of gl(n) fidirmods W ([m(Θ)]_{n}) with highest weights

[m(Θ)]_{n} = [m_{1}(Θ),…, m_{n}(Θ)]_{n}, m_{i}(Θ) = m_{i} + Θ_{i} − ∑_{j=1}^{n} Θ_{j}.  (1)

The sum is taken over all Θ-tuple such that m_{i}(Θ) − m_{j}(Θ) ∈ Z_{+} for all i, j with i < j (admissible Θ-tuple). The description of the [m(Θ)]_{n} by the Θ-tuple (Θ_{1},…, Θ_{n}) is oneby-one.

In the socalled typical case the sl(1, n) module W̄ ([m]_{n+1}) is irreducible. The atypical case is characterized by the condition m_{k} = k−1 for a certain k ∈ {1,…, n}. If m_{k} = k−1, m_{j} = j − 1 and k < j then m_{k} − m_{j} = k − j < 0 which contradicts to m_{k} − m_{j} ∈ Z_{+}. So, in a highest weight it exists almost one k with the condition of atypicity.

If m_{k} = k − 1 then W̄ ([m]_{n+1}) is an indecomposable sl(1, n) module which is not irreducible. W̄ ([m]_{n+1}) contains a maximal invariant subspace Ī which we can describe as a direct sum of gl(n) fidirmods:

Ī = ⊕W ([m_{1} + Θ_{1} − ∑_{j=1}^{n} Θ_{j},…, m_{n} + Θ_{n} − ∑_{j=1}^{n} Θ_{j}])

with

Θ_{k} = 1; Θ_{j} ∈ {0, 1}, j ≠ k, j = 1,…, n.  (2)

If the Θ-tuple is not admissible we drop the corresponding summand.

The factor module W̄ ([m]_{n+1})/Ī is isomorphic to an analogous direct sum where in (2) the condition Θ_{k} = 1 is replaced by Θ_{k} = 0. The factor module is irreducible. This factor module and the corresponding representation is said to be atypical (or nontypical) and is characterized by the highest weight [m]_{n+1}.

We denote the irreducible (atypical) representation given by W̄ ([m]_{n+1})/Īwith [m]_{n+1}^{irr} in difference to [m]_{n+1} denoting the indecomposable representation given by W̄ ([m]_{n+1}).

Furthermore, we set for all possible κ = 0, 1,…

\begin{matrix}[m − (κ + 1)]_{n+1}^{k−κ,...,k} = & [m_{1} − (κ + 1),…, m_{k−κ−1} − (κ + 1), m_{k−κ} − κ,…, \\ m_{k} − κ, m_{k+1} − (κ + 1),…, m_{n} − (κ + 1)]_{n+1}. \end{matrix}

Similar, we set for all possible λ = 0, 1,…

\begin{matrix}[m + λ + 1]_{n+1}^{k,...,k+λ} = & [m_{1} + λ + 1,…, m_{k−1} + λ + 1, m_{k} + λ,…, \\ m_{k+λ} + λ, m_{k+λ+1} + λ + 1,…, m_{n} + λ + 1]_{n+1}. \end{matrix}

Ī carries an atypical representation, too. This representation is irreducible (cf. [3], [4]). The highest weight of this representation is given in general by [m−1]_{n+1}^{k}. This corresponds to the Θ-tuple with Θ_{k} = 1, Θ_{i} = 0 elsewhere.

But it is possible that this Θ-tuple is not admissible . Then exist a κ > 0, κ < k such that m_{k−κ} = … = m_{k} = k − 1 and m_{k−κ−1} > m_{k−κ} if k − κ > 1. Here we have the highest weight [m − (κ + 1)]_{n+1}^{k−κ,...,k}.

3. The polynomial solution of the functional equation g(x_{1},…, x_{n}) + g(x_{1} − 1,…, x_{n} − 1) = 2^{n+1}x_{1} ⋯ x_{n}

The computation of the dimension of an atypical representation will be reduced in the following to the solution of an equation of the form

g(x_{1},…, x_{n}) + g(x_{1} − 1,…, x_{n} − 1) = 2^{n+1}x_{1} ⋯ x_{n}.  (3)

For the solution of this difference equation we use some ideas from the theory of difference equations in one variables and some own ideas. First we develop a suitable formalism.

We define for a function f(x_{1},…, x_{n}) in n variables with a suitable domain for all i, j ∈ Z_{+}:

f^{(i)} = f(x_{1} − i,…, x_{n} − i); (f^{(i)})^{(j)} = f^{(i+j)}  (4)

and

Δ^{0}f(x_{1},…, x_{n}) = f(x_{1},…, x_{n}), Δ^{i}f(x_{1},…, x_{n}) = Δ^{i−1}f^{(0)} − Δ^{i−1}f^{(1)}.  (5)

It is easy to see by induction that

Δ^{i}f(x_{1},…, x_{n}) = ∑_{j=0}^{i}(−1)^{j}\binom{i}{j}f^{(j)}.  (6)

For n = 1 and f(x) = x^{i} in [2] is computed at x = 0 the value

Δ^{l}f(0) = Δ^{l}0^{i} = ∑_{j=0}^{i}(−1)^{i−j}\binom{i}{j}j^{l}.  (7)

It is valid:

Δ^{l}0^{i} = 0, i > l.  (8)

Lemma 1: For n > 0 let be f(x_{1},…, x_{n}) = 2^{n}x_{1} ⋯ x_{n}. Then Δ^{i}f(x_{1},…, x_{n}) = 0 for all i > n.

Proof: Let σ_{l}(x_{1},…, x_{n}) denote the l-th elementar symmetric function in n variables:

σ_{l}(x_{1},…, x_{n}) = x_{1} ⋯ x_{l} + x_{1} ⋯ x_{l−1}x_{l+1} + ⋯ + x_{n−l+1} ⋯ x_{n}.

It is easy to see that

f^{(j)} = 2^{n}(x_{1} − j) ⋯ (x_{n} − j) = 2^{n} ∑_{l=0}^{n}(−1)^{l}j^{l}σ_{n−l}(x_{1},…, x_{n}).  (9)

196 H. Schlosser: Remarks to the Dimension of an Atypical Irreducible Representation Now we have by (6) and (9)

Δ^{i}f(x_{1},…, x_{n}) = ∑_{j=0}^{i}(−1)^{j}\binom{i}{j}2^{n} ∑_{l=0}^{n}(−1)^{l}j^{l}σ_{n−l}(x_{1},…, x_{n})
= 2^{n} ∑_{l=0}^{n}(∑_{j=0}^{i}(−1)^{j}\binom{i}{j}j^{l})(−1)^{l}σ_{n−l}(x_{1},…, x_{n}).  (10)

From (8) follows that the expression (−1)^{i}Δ^{l}0^{i} in the brackets is equal to zero for all i > l. Because l ≤ n the proof is finished. Remark: It is possible to prove the Lemma directly by induction. Lemma 2: Let be again f(x_{1},…, x_{n}) = 2^{n}x_{1} ⋯ x_{n}. Then

g(x_{1},…, x_{n}) = ∑_{i=0}^{n} \frac{1}{2^{i}} Δ^{i}f(x_{1},…, x_{n})

is a solution of (3). For the proof we use Lemma 1 and the following arguments:

a) Because \binom{i}{j} = 0 for j > i we can replace ∑_{j=0}^{i} \binom{i}{j}… by ∑_{j=0}^{n} \binom{i}{j}…, if n > i.

b) For n > 0, 0 < j ≤ n + 1 is valid: ∑_{i=0}^{n} \frac{1}{2^{i}} (\binom{i}{j} − \binom{i}{j−1}) = −\frac{1}{2^{n}} \binom{n+1}{j} (proof by induction). Now we prove Lemma 2:

∑_{i=0}^{n} \frac{1}{2^{i}} Δ^{i}f(x_{1},…, x_{n}) + ∑_{i=0}^{n} \frac{1}{2^{i}} Δ^{i}f^{(1)}(x_{1},…, x_{n})

= ∑_{i=0}^{n} \frac{1}{2^{i}} (∑_{j=0}^{i}(−1)^{j}\binom{i}{j}f^{(j)}) + ∑_{i=0}^{n} \frac{1}{2^{i}} (∑_{j=0}^{i}(−1)^{j}\binom{i}{j}f^{(j+1)})

= ∑_{i=0}^{n} \frac{1}{2^{i}} f^{(0)} + ∑_{j=1}^{n} ∑_{i=0}^{n} \frac{1}{2^{i}} (−1)^{j}\binom{i}{j}f^{(j)} + ∑_{j=1}^{n+1} ∑_{i=0}^{n} \frac{1}{2^{i}} (−1)^{j−1}\binom{i}{j − 1}f^{(j)}

= ∑_{i=0}^{n} \frac{1}{2^{i}} f^{(0)} + ∑_{j=1}^{n+1}(−\frac{1}{2^{n}})(−1)^{j}\binom{n + 1}{j}f^{(j)}

= ∑_{i=0}^{n} \frac{1}{2^{i}} f^{(0)} − \frac{1}{2^{n}} (Δ^{n+1}f(x_{1},…, x_{n}) − f^{(0)})

= (∑_{i=0}^{n} \frac{1}{2^{i}} + \frac{1}{2^{n}})f^{(0)} = 2f^{(0)} = 2^{n+1}x_{1} ⋯ x_{n}.

Lemma 2 and (10) give the way to solve the difference equation. But for an explicit formula we have to look to sums

H(n, l) = ∑_{i=0}^{n} \frac{1}{2^{i}} (−1)^{l} ∑_{j=0}^{i}(−1)^{j}\binom{i}{j}j^{l} = ∑_{i=0}^{n} \frac{1}{2^{i}} (−1)^{i+l}Δ^{l}0^{i}  (11)

for l = 0,…, n which are the coefficients by 2^{n}σ_{n−l}. (8) shows that H(n, l) = H(l, l) if l < n. So we must compute the coefficients H_{l} = H(l, l) only for l = 0, 1,…

Computations with a computer give the following values:

\begin{matrix}l & H_{l} & l & H_{l} \\ 0 & 1 & 6 & 0 \\ 1 & 1/2 & 7 & −17/8 \\ 2 & 0 & 8 & 0 \\ 3 & −1/4 & 9 & 31/2 \\ 4 & 0 & 10 & 0 \\ 5 & 1/2 & 11 & −691/4 \end{matrix}

It seems that H_{l} = 0 if l is even; the H_{l} couldt be similar to the Bernoulli numbers B_{l}. Furthermore, for the given numbers in the table we have

H_{l} = \frac{2(2^{l+1} − 1)}{l + 1}B_{l+1}  (12)

So, we come to the idea that a generating function belongs to the H_{l}.

Lemma 3:

2\frac{e^{t}}{e^{t} + 1} = ∑_{l=0}^{∞} H_{l} \frac{t^{l}}{l!}

Proof: We adopt the following equation from [2]:

(e^{t} − 1)^{j} = ∑_{l=1}^{∞} Δ^{l}0^{j} \frac{t^{l}}{l!}

for j = 1, 2,… and small t. Then we get for such small t:

\frac{e^{t} − 1}{e^{t} + 1} = \frac{e^{t} − 1}{2} \frac{1}{1 − \frac{1−e^{t}}{2}} = ∑_{j=0}^{∞}(−1)^{j}(\frac{e^{t} − 1}{2})^{j+1}

= ∑_{j=1}^{∞}(−1)^{j−1} \frac{1}{2^{j}} ∑_{l=1}^{∞} Δ^{l}0^{j} \frac{t^{l}}{l!}

= ∑_{l=1}^{∞}(∑_{j=1}^{∞}(−1)^{j−l} \frac{1}{2^{j}} Δ^{l}0^{j})(−1)^{l+1} \frac{t^{l}}{l!} = ∑_{l=1}^{∞}(−1)^{l+1}H_{l} \frac{t^{l}}{l!} .

Because

\frac{e^{t} − 1}{e^{t} + 1} = −\frac{e^{−t} − 1}{e^{−t} + 1}

(odd function) all H_{l} with even index vanishes; further, if l is odd we can drop the factor (−1)^{l+1}; moreover, with H_{0} = 1 we have

\frac{e^{t} − 1}{e^{t} + 1} + 1 = 2\frac{e^{t}}{e^{t} + 1} = ∑_{l=0}^{∞} H_{l} \frac{t^{l}}{l!} .

Lemma 3 is proved.

Now we can show the relation (12) between H_{l} and the Bernoulli numbers B_{l}. Because it is valid: B_{1} = −1/2 and for small t

\frac{t}{e^{t} − 1} = ∑_{l=0}^{∞} B_{l} \frac{t^{l}}{l!}  (13)

we get

2\frac{e^{t}}{e^{t} + 1} = 2\frac{e^{t}(e^{t} − 1)}{(e^{t} + 1)(e^{t} − 1)} = \frac{2}{t}(t − \frac{t}{e^{t} − 1} + \frac{2t}{e^{2t} − 1})

= 2 − \frac{2}{t} ∑_{l=0}^{∞} B_{l} \frac{t^{l}}{l!} + \frac{2}{t} ∑_{l=0}^{∞} B_{l} \frac{(2t)^{l}}{l!}

= 2 − \frac{2}{t}(∑_{l=2}^{∞}(B_{l} − 2^{l}B_{l})\frac{t^{l}}{l!} ) − 2B_{1} + 4B_{1}

= 1 + ∑_{l=1}^{∞} 2(2^{l+1} − 1)B_{l+1} \frac{t^{l}}{(l + 1)!} = ∑_{l=0}^{∞} H_{l} \frac{t^{l}}{l!}  (14)

and

H_{0} = 1, H_{l} = \frac{2(2^{l+1} − 1)}{l + 1}B_{l+1}  (15)

for l = 1, 2,… It is possible to get the H_{l} by computation directly. If we differentiate the function

h(x) = 2\frac{e^{x}}{e^{x} + 1}

then H_{l} = h^{(l)}(0). Unfortunately, the differentiation does not give a general expression for the function h^{(l)}(x). But it is easy to see that

h^{(l)}(x) = 2p_{l}(e^{x})\frac{e^{x}}{(e^{x} + 1)^{l+1}}

where p_{l} ∈ Z[y]. Hence the values 2^{l}H_{l} = 2^{l}h^{(l)}(0) are integers for all l = 0, 1,. . ..

Similar to the Bernoulli numbers the H_{l} satisfy some recursive relations which can be used to calculate the numbers H_{l}. For instances, we obtain:

a)From

2e^{t} = (e^{t} + 1) ∑_{l=0}^{∞} H_{l} \frac{t^{l}}{l!}

and development in power series we get by comparison of coefficients 2H_{0} = 2, 2H_{1} +H_{0} = 2, 2H_{2} + 2H_{1} + H_{0} = 2 etc. and in general

∑_{l=0}^{k} \binom{k}{l}H_{l} = 2 − H_{k}.

b)From

2 = (e^{−t} + 1) ∑_{l=0}^{∞} H_{l} \frac{t^{l}}{l!}

and development in power series we get

H_{0} = 1, ∑_{l=0}^{k}(−1)^{k−l+1}\binom{k}{l}H_{l} = H_{k}.

Furthermore, it is easy to express the H_{l} by aid of the important values B_{l}(1/2) and B_{l}(1) of the Bernoulli polynoms B_{l}(x).

Proposition:

g(x_{1},…, x_{n}) = 2^{n} ∑_{l=0}^{n} H_{l}σ_{n−l}(x_{1},…, x_{n})

is the unique polynomial solution of (3).

Proof: By the foregoing we have to prove the uniqueness only. For this we look to the equation

g(x_{1},…, x_{n}) + g(x_{1} − 1,…, x_{n} − 1) = 0  (16)

and show that the only polynomial solution is the nullpolynom. From this fact it is easy to get the uniqueness of the solution of (16) with an arbitrary polynomial on the right side.

Let be cx_{1}^{t_{1}} ⋯ x_{n}^{t_{n}} a summand in g with maximal degree t_{1} + ⋯ + t_{n} and

g(x_{1},…, x_{n}) = cx_{1}^{t_{1}} ⋯ x_{n}^{t_{n}} + d(x_{1},…, x_{n}).

On the left side of (16) we get

\begin{matrix}cx_{1}^{t_{1}} ⋯ x_{n}^{t_{n}} + c(x_{1} − 1)^{t_{1}} ⋯ (x_{n} − 1)^{t_{n}}+ & d(x_{1},…, x_{n}) + d(x_{1} − 1,…, x_{n} − 1) \\ = 2cx_{1}^{t_{1}} ⋯ x_{n}^{t_{n}}+ & h(x_{1},…, x_{n}) \end{matrix}

where h contains only summands of the degree less or equal to t_{1} + ⋯ + t_{n} and the summands with this maximal degree are linear independent from the first summand. Then c must be zero and g has no summand of maximal degree.

Remark: If we assume that the solution g(x_{1},…, x_{n}) of the equation (3) has the form from the proposition then by a (not short) calculation we get the recursive relation b) for the H_{l} directly. For this it is necessary to use the equation (proof by induction)

σ_{l}(x_{1} − 1,…, x_{n} − 1) = ∑_{j=0}^{l}(−1)^{l−j}\binom{n − j}{l − j}σ_{j}(x_{1},…, x_{n}).

For later applications we note:

Corollary: To every k ∈ {1,…, n} and for every natural number κ > 0, κ < k the function g(x_{1},…, x_{n}) satisfy the following equation:

g(x_{1},…, x_{k−κ−1}, x_{k−κ} = κ,…, x_{k−2} = 2, x_{k−1} = 1, x_{k},…, x_{n})+ (−1)^{κ}g(x_{1} − (κ + 1),…, x_{k−κ} = −1,…, x_{k−2} = −κ + 1, x_{k−1} = −κ, x_{k} − (κ + 1),…, x_{n} − (κ + 1))

= 2^{n+1}κ! ∏_{i=1,...,k−κ−1;k,...,n} x_{i}

Proof: From the proposition we know that

g(x_{1},…, x_{n}) = 2^{n} ∑_{l=0}^{n−1} H_{l}σ_{n−l}(x_{1},…, x_{n})

is the unique polynomial solution of

g(x_{1},…, x_{n}) + g(x_{1} − 1,…, x_{n} − 1) = 2^{n+1}x_{1} ⋯ x_{n}.

If we replace x_{i} by x_{i} − 1 and x_{k−1} by 1 we get

g(x_{1} − 1,…, x_{k−1} = 1,…, x_{n} − 1) + g(x_{1} − 2,…, x_{k−1} − 2 = −1,…, x_{n} − 2) =
2^{n+1}(x_{1} − 1) ⋯ (x_{n} − 1) = 0.

and

g(x_{1},…, x_{k−1} = 1,…, x_{n}) − g(x_{1} − 2,…, x_{k−1} − 2 = −1,…, x_{n} − 2) = 2^{n+1}x_{1} ⋯ x_{n}.

Analogous, by the assumption x_{k−κ} = κ,…, x_{k−2} = 2, x_{k−1} = 1 we can produce a chain of equations

g(x_{1} − s,…, x_{n} − s) + g(x_{1} − (s + 1),…, x_{n} − (s + 1)) = 0,

s = 1,…, κ, such that for x_{k−κ} = κ,…, x_{k−1} = 1 we have

g(x_{1},…, x_{n}) + (−1)^{κ}g(x_{1} − (κ + 1),…, x_{n} − (κ + 1)) = 2^{n+1}κ! ∏_{i=1,...,k−κ−1;k,...,n} x_{i}.

Remark: In general, g(x_{1},…, x_{n}) is not determined in an unique way by the equation of the corollary.

4. The dimension of an irreducible representation of sl(1,n)

We use the notations of 2.

The dimension D of W̄ ([m]_{n+1}) is determined by the gl(n) fidirmod V_{0} with the same highest weight [m]_{n}. The dimension of V_{0} is given by the expression (cf. [5])

dimV_{0} = ∏_{i<j, i,j=1,...,n} \frac{m_{i} − m_{j} + j − i}{j − i} = ∏_{i<j, i,j=1,...,n} \frac{r_{ij}}{j − i}  (17)

where r_{ij} = m_{i} − m_{j} + j − i (for the following we extend this definition of r_{ij} on the case i > j also) and

D = 2^{n}dimV_{0}.

So we can consider the dimension of a typical sl(1, n) fidirmod as known.

On the other side by the sum decomposition given by (1) it is possible to describe D in the following way: Let for a given Θ-tuple (Θ_{1},…, Θ_{n}) be r_{st}(Θ) = m_{s}−m_{t}+t−s+Θ_{s}−Θ_{t}. Then

D = ∑_{Θ} ∏_{s<t, s,t=1,...,n} \frac{r_{st}(Θ)}{t − s}  (18)

where the summation goes over all admissible Θ−tuple. If W̄ ([m]_{n+1}) is indecomposable and m_{k} = k − 1 then the dimensions D_{1} = dimĪ and D_{2} = dimW̄ ([m]_{n+1})/Ī are given by the same formula (18) with the difference that the summation goes over all admissible Θ-tuple with Θ_{k} = 1 and Θ_{k} = 0 resp. (cf. (2)). This gives a possibility for the numerical calculation of the dimension of an atypical sl(1, n) fidirmod. In the following we will prove an explicit formula for such dimensions.

We have to differ between two cases. In the first one we assume that k = 1 or k > 1 and m_{k−1} > m_{k} = k − 1. We will refer to this case as the general one. In the second case (the exceptional case) we assume that exist a κ > 0 such that m_{k−κ} = … = m_{k} = k − 1 and if k − κ > 1 we have m_{k−κ−1} > m_{k−κ}.

We look to the maximal invariant subspace Ī of W̄ ([m]_{n+1}). Ī itself is an atypical sl(1, n) fidirmod with the highest weight

[m′]_{n+1} = [m_{1} − 1,…, m_{k},…, m_{n} − 1] = [m − 1]_{n+1}^{k;irr}

and the same k as in W̄ ([m]_{n+1})/Ī in the general case. In the exceptional case the highest weight is given by

[m′]_{n+1}^{irr} = [m_{1} − κ − 1,…, m_{k−κ−1} − κ − 1, m_{k−κ} − κ,…,
m_{k} − κ, m_{k+1} − κ − 1,…, m_{n} − κ − 1] = [m − (κ + 1)]_{n+1}^{k−κ,...,k;irr}.  (19)

Firstly, we consider the general case. We look to the exceptional case after the theorem.

D_{1} and D_{2} we can compute by the same formula replacing m_{i} by m′_{i} and r_{st} by r′_{st}, where r′_{sk} = r_{sk} − 1 for s < k, r′_{kt} = r_{kt} + 1 for t > k and r′_{st} = r_{st} in all other cases. On the other side from the analogous formula to (18) we see that D_{2} is a polynomial f(r_{12},…, r_{n−1,n}). Because D_{2} +D_{1} = D we have to look to a polynomial f which satisfied the functional equation

f(r_{12},…, r_{n−1,n}) + f(r′_{12},…, r′_{n−1,n}) = 2^{n} ∏_{s<t, s,t=1,...,n} \frac{r_{st}}{t − s}.  (20)

Because r_{st} = r_{sk} + r_{kt} for arbitrary s, t, k we look to the polynomial f as a polynomial in n − 1 variables only. We choose x_{1} = r_{1k},…, x_{k−1} = r_{k−1,k}, x_{k+1} = r_{k,k+1},…, x_{n} = r_{kn} as such variables which are independent. On the right side of (20) we separate the factor

2^{n} ∏_{s=1,...,k−1} r_{sk} ∏_{t=k+1,...,n} r_{kt} = 2^{n} ∏_{i=1, i≠k}^{n} x_{i}

from the rest R_{kn} and consider the equation

f_{k}(x_{1},…, x_{k−1}, x_{k+1},…, x_{n})+f_{k}(x_{1} − 1,…, x_{k−1} − 1, x_{k+1} + 1,…, x_{n} + 1)
= 2^{n} ∏_{i=1, i≠k}^{n} x_{i}  (21)

and for k = n

f_{n}(x_{1},…, x_{n−1}) + f_{n}(x_{1} − 1,…, x_{n−1} − 1) = 2^{n} ∏_{i=1,...,n−1} x_{i}.  (22)

It is easy to see that R_{kn}f_{k}(r_{1k},…, r_{k−1,k}, r_{k,k+1},…, r_{kn}) is a solution of (20) where f_{k} solves (21). Furthermore, the solutions of (20) and (21) resp. are unique by the same arguments as in the proof of the proposition of 3.

Now we compare (21) for arbitrary k with the case k = n. If we replace r_{kt} by −r_{kt} = r_{tk} and r_{kt}(Θ) by −r_{kt}(Θ) = r_{tk}(Θ) we see by (18) that must be valid

\begin{matrix}R_{kn}f_{k}(r_{1k},…, r_{k−1,k}, r_{k,k+1},…, r_{kn}) = \\ (−1)^{n−k}R_{kn}f_{n}(r_{1k},…, r_{k−1,k}, −r_{k,k+1},…, −r_{kn}). \end{matrix}

So it is enough to solve (22) obtaining the solution of (20). But (22) is solved by the proposition of 3.

Theorem: Let Y_{k}([m]_{n+1}) = W̄ ([m]_{n+1})/Ī be an atypical finitedimensional irreducible sl(1,n) module with highest weight [m_{1},…, m_{k} = k − 1,…, m_{n}]_{n+1}^{irr}. For all possible s,t,k we set

r_{st} = m_{s} − m_{t} + t − s

and

R_{kn} = ∏_{s<t, s,t=1,...,n} \frac{1}{t − s} ∏_{i<j,i≠k,j≠k,i,j=1,...,n} r_{ij}.

The dimension of Y_{k}([m]_{n+1}) we get by the following formula:

dim(Y_{k}([m]_{n+1}) = 2^{n−1}R_{kn}∗

∑_{l=0}^{n−1}(−1)^{n−k}H_{l}σ_{(n−1)−l}(r_{1k},…, r_{k−1,k}, −r_{k,k+1},…, −r_{kn})

where the σ_{(n−1)−l} are the elementary symmetric functions of the first kind and the H_{l} are the coefficients determined in 3. More exact, because m_{k} = k − 1 we have

r_{sk} = m_{s} − s + 1, −r_{kt} = m_{t} − t + 1

such that

dim(Y_{k}([m]_{n+1})) = 2^{n−1}R_{kn} ∑_{l=0}^{n−1}(−1)^{n−k}H_{l}∗

σ_{(n−1)−l}(m_{1}, m_{2} − 1,…, m_{k−1} − (k − 2), m_{k+1} − k,…, m_{n} − (n − 1)).

Proof: The general case is proved above so we have only to consider the exceptional case. In this one it is possible to verify that

dim(Y_{k}([m]_{n+1})) + dim(Y_{k−κ}([m′]_{n+1})) = 2^{n} ∏_{s<t, s,t=1,...,n} \frac{r_{st}}{t − s}

([m′] is given by (19)) by aid of the corollary of 3.

The remark to this corollary said that the solutions are not unique in this case. We get the uniqueness in the following way:

I. If m_{k+1} < k − 1 or k = n we can look to the admissible representation [m̄]_{n+1}^{irr} = [m + 1]_{n+1}^{k;irr} which is atypical but not exceptional; [m]_{n+1}^{irr} and [m̄]_{n+1}^{irr} are paired in [m̄]_{n+1} and because the dimension formula is correct for [m̄]_{n+1} this must be also for [m]_{n+1}^{irr}.

II.If for some κ > 0 we have m_{k+1} = … = m_{k+κ} = k − 1 (κ maximal chosen) we define [m̄]_{n+1}^{irr} = [m+κ+1]_{n+1}^{k,...,k+κ}. [m̄]_{n+1}^{irr} is atypical with m̄_{k+κ} = k +κ−1 but in general exceptional. We repeat this process; by finitely many steps we come to I. and must apply the arguments on the constructed chain of highest weights in a successive way.

As examples we list the explicit expressions for the dimensions of the atypical irreducible sl(1, 3) modules:

dimY_{3}([m_{1}, m_{2}, 2]) = 4 ∗ 1/2(m_{1} − m_{2} + 1)(H_{0}σ_{2}(m_{1}, m_{2} − 1) + H_{1}σ_{1}(m_{1}, m_{2} − 1) + H_{2}) = (m_{1} − m_{2} + 1)(2m_{1}m_{2} − m_{1} + m_{2} + 1), dimY_{2}([m_{1}, 1, m_{3}]) = 4 ∗ 1/2 ∗ (−1)(m_{1} − m_{3} + 2)(H_{0}σ_{2}(m_{1}, m_{3} − 2) + H_{1}σ_{1}(m_{1}, m_{3} − 2) + H_{2}) = (m_{1} − m_{3} + 2)(−2m_{1}m_{3} + 3m_{1} − m_{3} + 2),

dimY_{1}([0, m_{2}, m_{3}]) = 4 ∗ 1/2(m_{2} − m_{3} + 1)(H_{0}σ_{2}(m_{2} − 1, m_{3} − 2) + H_{1}σ_{1}(m_{2} − 1, m_{3} − 2) + H_{2}) = (m_{2} − m_{3} + 1)(2m_{2}m_{3} − 3m_{2} − m_{3} + 1).

References:

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[2] Gelfond, A.O.: Differenzenrechnung, Dt. Verlag der Wissenschaften, Berlin 1958.

[3] Hughes, J.W.B.; King, R.C.; Van der Jeugt, J.: On the composition factors of Kacmodules for the Lie superalgebras sl(m/n), J. Math. Phys. 33 (1992), 470–491.

[4] Van der Jeugt, J.; Hughes, J.W.B.; King, R.C.; Thierry-Mieg, J.: A character formula for singly atypical modules of the Lie superalgebra sl(m/n), Communications in Algebra 18 (1990), 3453–3480.

[5] Kac, V.G.: Representations of classical Lie superalgebras, Lecture Notes Math. 626 (1978), 597–626.

[6] Palev, T.D.: Finite-dimensional representations of the special linear Lie superalgebra sl(1,n). I. Typical representations, J. Math. Phys. 28,10 (1987), 2280–2303.

[7] Palev, T.D.: Finite-dimensional representations of the special linear Lie superalgebra sl(1,n). II. Nontypical representations, J. Math. Phys. 29,12 (1988), 2589–2598.

[8] Schlosser,H.: Atypical representations of the Lie superalgebra sl(1,n) for n > 1, Seminar Sophus Lie, 3, no. 1 (1993), 15-24.

Received: January 26, 1994, revised version June 3, 1994.