One can define the $latex j$-th Legendre polynomial $latex P_j$ in many ways, one of the easiest being to use the generating function
$latex \sum_{j\geq 0}{ P_j(x)T^j}=(1-2xT+T^2)^{-1/2}.$
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 $latex G=\mathrm{SU}_2(\mathbf{C})$ and its $latex (2j+1)$-dimensional irreducible representation on the space $latex V_{2j}$ of homogeneous polynomials in two variables of degree $latex 2j$, where the action of $latex G$ is by linear change of variable:
$latex (g\cdot P)(X,Y)=P((X,Y)g)=P(aX+cY,bX+dY)$
for
$latex g=\begin{pmatrix}a&b\\c&d\end{pmatrix}.$
Then, up to normalizing factors, $latex P_j$ “is” the matrix coefficient of $latex V_{2j}$ for the vector $latex e=X^jY^j\in V_{2j}$. Or, to be precise (since a matrix coefficient is a function on $latex G$, which is 3-dimensional, while $latex P_j$ is a function of a single variable), we have
$latex P_j(\cos\theta)=c_j \langle g_{\theta}\cdot e,e\rangle$
for the elements
$latex g_{\theta}=\begin{pmatrix}\cos(\theta/2)&i\sin(\theta/2)\\i\sin(\theta/2)&\cos(\theta/2)\end{pmatrix}$
(the inner product $latex \langle \cdot,\cdot\rangle$ used to compute the matrix coefficient is the $latex G$-invariant one on $latex V_{2j}$; since this is an irreducible representation, it is unique up to a non-zero scalar; the normalizing constant $latex c_j$ 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 $latex e$ is, up to a scalar, the only vector in $latex V_{2j}$ invariant under the action of the subgroup $latex T$ of diagonal matrices, one can also say that $latex P_j$ is “the” spherical function for $latex V_{2j}$ (with respect to the subgroup $latex T$).
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 $latex V_{2j}$, but now seen as a representation of the larger group $latex \mathbf{G}=\mathrm{SL}_2(\mathbf{C})$ (the action being given with the same linear change of variable): we have
$latex P_j(1+2t)=d_j \langle u_t\cdot e,u_1\cdot e\rangle$
where $latex d_j$ is some other normalizing constant, and now $latex u_t$ are unipotent elements given by
$latex u_t=\begin{pmatrix}1&t\\ 0 & 1\end{pmatrix}.$
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
$latex u_t\cdot e=X^j(tX+Y)^j,$
expand into binomial coefficients, use the formula
$latex \langle X^iY^{2j-i}, X^kY^{2j-k}\rangle=\binom{2j}{i}^{-1}\delta(i,k),$
for the invariant inner-product, and obtain a somewhat unwieldy polynomial which can be recognized as a multiple of the hypergeometric polynomial
$latex {}_2F_1(-j,1+j;1;-t),$
which is known to be equal to $latex P_j(1+2t).$ (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…)
