Hypercomplex Analysis Selected Topics

are mutually inequivalent, see [44]. The key step for constructing the Gelfand-
Tsetlin bases is to understand the branching of the module H s k (Rm ).
Theorem 10. Let m ≥ 3, s = 0, . . . , m and k ∈ N0. Denote by N s,m
k the
set of pairs (t, j) ∈ {0, . . . , m − 1} × {0, . . . , k} such that t ∈ {s − 1, s} and,
if t ∈ {0, m − 1} then j = 0. Then, under the H-action of P in(m − 1), we
have the following multiplicity free irreducible decomposition
H s k(R m ) =
Here the embedding factors X s,t,m
k,j
(t,j)∈N s,m
k
X s,t,m
k,j Ht j(R m−1 ). (3.24)
are defined as the polynomials
X s,t,m
k,j (x) = X (k−j)
m,j (x)e s−t
m + α X (k−1−j)
m,j+1 (x)(β t−s (x ∧) + β t−s+1 (x •)) e s−t+1
m
where x = (x, xm) ∈ R m , X (k−j)
m,j
β = −(j + m − 1 − t)/(j + t) and
α =
are given in (3.11), X (−1)
m,k+1 = 0,
−(j + 1)/(m + 2j − 1) unless t = 0, m − 1;
0 if t = 0, m − 1.
In particular, this decomposition is orthogonal.
Proof. In [L9], a proof is given by the Cauchy-Kovalevskaya method. Let Is k
be the set of initial polynomials for the space Hs k (Rm ), that is, Is k is a subset
of Pk(Rm−1 , Cℓm) such that CK(I s k ) = Hs k (Rm ). Then, in [L9], it is shown
that
I s k = Ker s k ∂ + ⊕ (Ker s−1
k ∂ − )em
where Ker s k ∂ ± = {u ∈ Pk(R m−1 , Cℓ s m) | ∂ ± u = 0}. The branching (3.24) for
Hs k (Rm ) is obtained by applying the operator CK = exmem∂ to the irreducible
decomposition of the module Is k under the H-action of P in(m − 1). See [L9]
for details.
Now we give an alternative proof. It is a well-known fact from representation
theory that, under the H-action of P in(m − 1), the module Hs k (Rm )
possesses the decomposition
H s k(R m ) =
(t,j)∈N s,m
k
into irreducible submodules ˜ H t j equivalent to H t j(R m−1 ). To get (3.24) we
need to describe explicitly the pieces ˜ H t j in the decomposition. To do this,
we recall that, under the H-action of P in(m), the space Mk(R m , Cℓm) decomposes
into inequivalent irreducible pieces as
Mk(R m
m
, Cℓm) = H s k(R m
) ⊕
s=0
m−1
s=1
30
˜H t j
((x ∧) + β s,m
k−1 (x •)) Hs k−1(R m )
(3.25)