Riemann P-symbol

Remarks

The order of the columns, and the order of the two exponents within a column, do not matter. Two basic rules hold.

1. If $T$ is a Möbius transformation, then the P-symbol with points $a, b, c$ and variable $z$ equals the P-symbol with points $T(a), T(b), T(c)$, the same exponents, and variable $T(z)$.
2. If $a, b, c$ are finite, then for all $k, l \in \C$, $$\left(\frac{z - a}{z - b}\right)^{k} \left(\frac{z - c}{z - b}\right)^{l} P\begin{Bmatrix} a & b & c & \\ \alpha & \beta & \gamma & z \\ \alpha' & \beta' & \gamma' & \end{Bmatrix} = P\begin{Bmatrix} a & b & c & \\ \alpha + k & \beta - k - l & \gamma + l & z \\ \alpha' + k & \beta' - k - l & \gamma' + l & \end{Bmatrix}.$$

Example: the hypergeometric function satisfies $${}_2F_1(a, b; c; z) \in P\begin{Bmatrix} 0 & \infty & 1 & \\ 0 & a & 0 & z \\ 1 - c & b & c - a - b & \end{Bmatrix},$$ and the exponents add up to $0 + (1 - c) + a + b + 0 + (c - a - b) = 1$, as required.

Non-example: a table whose six exponents do not add up to $1$ is not a valid P-symbol. No second-order Fuchsian equation has exactly those three singular points with those exponents.

Rule 2 is consistent with the sum condition: the shifts $k$, $-k-l$, $l$ appear twice each and add up to $0$.