One can define the -th Legendre polynomial in many ways, one of the easiest being to use the generating function

Like many “classical” special functions (which one might call “the functions in Whittaker & Watson” — I find it charming, incidentally, that the PDF of this edition has exactly 628 pages), these can also be defined using representation theory. This is done by considering the group and its -dimensional irreducible representation on the space of homogeneous polynomials in two variables of degree , where the action of is by linear change of variable:

for

Then, up to normalizing factors, “is” the matrix coefficient of for the vector . Or, to be precise (since a matrix coefficient is a function on , which is 3-dimensional, while is a function of a single variable), we have

for the elements

(the inner product used to compute the matrix coefficient is the -invariant one on ; since this is an irreducible representation, it is unique up to a non-zero scalar; the normalizing constant involves this as well as the normalization of Legendre polynomials.) For full details, a good reference is the book of Vilenkin on special functions and representation theory, specifically, Chapter 3.)

Note also that, since is, up to a scalar, the only vector in invariant under the action of the subgroup of diagonal matrices, one can also say that is “the” spherical function for (with respect to the subgroup ).

This seems to be the most natural way of recovering the Legendre polynomials from representation theory. Just a few days ago, while continuing work on the lecture notes for my class on the topic (the class itself is finished, but I got behind in the notes, and I am now trying to catch up…), I stumbled on a different formula which doesn’t seem to be mentioned by Vilenkin. It is still related to , but now seen as a representation of the larger group (the action being given with the same linear change of variable): we have

where is some other normalizing constant, and now are *unipotent* elements given by

It’s not quite clear to me where this really comes from, though I suspect there is a good explanation. Searching around the web and Mathscinet did not lead, in any obvious way, to earlier sightings of this formula, but it is easy enough to get thoroughly unenlightening proof: just use the fact that

expand into binomial coefficients, use the formula

for the invariant inner-product, and obtain a somewhat unwieldy polynomial which can be recognized as a multiple of the hypergeometric polynomial

which is known to be equal to (Obviously, chances of a computational misake are non-zero; I certainly made some while trying to figure this out, and stopped computing only when I got this nice interesting result…)