Hypercomplex Analysis Selected Topics

of initial polynomials for H 1 k (R3 ). Next we need to describe a decomposition
of the space I 1 k into irreducible Spin(2)-submodules. Put z = x1 + ix2,
z = x1 − ix2 and, for k ∈ N0, denote
p k j = zjz k−j
j!(k − j)! e3, j = 0, . . . , k and q k j = 1 k+1
e3∂(pj ), j = 0, . . . , k + 1.
2
(3.31)
Then the space I1 k decomposes into 1-dimensional irreducible Spin(2)-submodules
as follows:
I 1 k+1
k = 〈q k j 〉 ⊕
k
〈p k j 〉. (3.32)
j=0
To get the Gelfand-Tsetlin basis for H 1 k (R3 ) it is now sufficient to apply
the Cauchy-Kovalevskaya extension operator to the polynomials p k j and q k j .
Indeed, we have the following result (see [L9, 78]).
Proposition 3. Let k ∈ N0 and let the polynomials p k j and q k j be as in (3.31).
Then the polynomials
g k 2j+2 = e x3e3∂ (p k j ), j = 0, . . . , k and g k 2j+1 = e x3e3∂ (q k j ), j = 0, . . . , k + 1
form a Gelfand-Tsetlin basis of the irreducible Spin(3)-module H 1 k (R3 ).
It is not difficult to obtain explicit formulas for the basis elements in terms
of hypergeometric series. Recall that the hypergeometric series 2F1(a, b, c; y)
is given by
2F1(a, b, c; y) =
∞
s=0
j=0
(a)s(b)s
(c)s s! ys with (a)s = a(a + 1) · · · (a + s − 1).
Corollary 1. (See [78].) Let {g k j | j = 1, . . . , 2k + 3} be the Gelfand-Tsetlin
basis of the module H 1 k (R3 ) defined in Proposition 3. Let v± = (e1 ± ie2)/2.
Then we have that g k 1 = (z k /k!)v−, g k 2k+3 = (zk /k!)v+ and, in general,
g k 2j+2 =
g k 2j+1 =
Here |z| 2 = zz.
1
j!(k − j)! (2F1(−j, −k + j, 1
2 , − x23 |z| 2 ) zjz k−j e3 +
+2 2F1(−j + 1, −k + j, 3
2 , − x2 3
|z| 2 ) jx3z j−1 z k−j v+ +
+2 2F1(−j, −k + j + 1, 3
2 , − x23 |z| 2 ) (k − j)x3z j z k−j−1 v−);
1
j!(k + 1 − j)! (2F1(−j + 1, −k − 1 + j, 1
2 , − x23 |z| 2 ) jzj−1z k+1−j v+ +
−2 2F1(−j + 1, −k + j, 3
2 , − x2 3
|z| 2 ) j(k + 1 − j)x3z j−1 z k−j e3 +
+2F1(−j, −k + j, 1
2 , − x2 3
|z| 2 ) (k + 1 − j)zj z k−j v−).
34