I am aware of the following standard generating function for single-variable Legendre Polynomials: $$ \sum\limits_{n=0}^{\infty}P_n(x)z^n = \frac{1}{\sqrt{1-2xz+z^2}} $$ for $x \in \mathbb{R}, z \in \mathbb{C}$, with $|{z}| < 1$.
Is there any known generating function for the bivariate Legendre Polynomials?
For context, Sehili and Zerroug (2017) give the following Rodrigues formula: $$ P_{i,j}(x,y) = \frac{1}{2^{i+j}i!j!}\frac{\partial^i}{\partial x^i}(x^2 - 1)^i\frac{\partial^j}{\partial y^j}(y^2-1)^j $$ and a recurrence relation for the bivariate case I'm working with: $$ \begin{cases} P_{-1,j}(x,y) = 0\\ P_{i,-1}(x,y) = 0\\ P_{0,0}(x,y) = 1\\ P_{0,0}(x,y) = 1\\ P_{i+1,j}(x,y) = \frac{2i+1}{i+1}xP_{i,j}(x,y) - \frac{i}{i+1}P_{i-1,j}(x,y)\\ P_{i,j+1}(x,y) = \frac{2j+1}{j+1}yP_{i,j}(x,y) - \frac{j}{j+1}P_{i,j-1}(x,y)\\ \end{cases} $$ The paper can be found here.
The paper uses power $j$ instead of $2$ in $(y^2-1)$.
But with this definition, doesn't it just hold that $P_{i,j}(x,y) = P_i(x) P_j(y)$?
If I'm not missing something and it's true, you can just do $$ \sum\limits_{i=0}^\infty \sum\limits_{j=0}^\infty P_{i,j}(x,y) a^i b^j = \left(\sum\limits_{i=0}^\infty P_i(x) a^i\right)\left(\sum\limits_{i=0}^\infty P_j(y) b^j\right). $$ This, in turn, gives the generating function $$ =\frac{1}{\sqrt{1-2xa+a^2}}\frac{1}{\sqrt{1-2yb+b^2}}. $$