Are there explicit relations between hypergeometric functions over finite fields of different size?
In particular, if we consider the function 2F1(A,B,C|x) over Fp (i.e., with characters A,B and C defined over Fp), and the same hypergeometric function but over Fp^2 (i.e., with characters defined over Fp^2) or any other field extension of Fp, can we relate them in any explicit way?


Thank you very much!



Finite field hypergeometric functions.
Assuming by the hypergeometric function, you mean the trace of Frobenius the hypergeometric sheaves, there is a nice relation. 


First I will mention the right way to move a character to a higher degree finite field. If $A$ is a character of $\mathbb F\_q^\times$, then the corresponding character of $\mathbb F\_{q^n}^\times$ is $A$ composed with the norm from $\mathbb F\_{q^n}$ to $\mathbb F\_q$. This is correct because both characters correspond to the same sheaf.


Thus a hypergeometric function $2F1(A,B,C|x)$ over $\mathbb F\_{q^2}$ corresponds to a hypergeometric function $2F1(A',B',C'|x)$ over $\mathbb F\_{q^2}$ because they come from the same sheaf, if $A$ corresponds to $A'$, $B$ corresponds to $B'$, and $C$ corresponds to $C'$.


But what does this say about the hypergeometric function? Well the function is just the trace of Frobenius. Since $Frob\_{q^n}=\left(Frob\_q\right) ^n$, we can use an identity about the traces of powers of matrices. Since the sheaf has rank $2$, the matrices are two-dimensional, and we have the identities:


\[ \operatorname{tr}(M^2) = \operatorname{tr}(M)^2 - 2 \det(M)\]


\[ \operatorname{tr}(M^3) = \operatorname{tr}(M)^3 - 3 \det(M) \operatorname{tr}(M)\]


and so on, using the Chebyshev polynomials. Here $ \operatorname{tr}(M)$ corresponds to the hypergeometric function of a point over $\mathbb F\_q$, $ \operatorname{tr}(M^2)$ to the corresponding function of the same point over $\mathbb F\_{q^2}$, and $ \operatorname{tr}(M^3)$ to the corresponding function of the same point over $\mathbb F\_{q^3}$. It remains to give a nice formula for the determinant. 


I'm pretty sure there is one, although I don't know exactly what it is. It should be something like $p^3 A\left(\frac{x}{x-1}\right)B\left(\frac{x}{x-1}\right)C(x-1)$. I'm not sure precisely what notation you are using. Without an exact formula, you still get a relation among $\mathbb F\_p$, $\mathbb F\_{p^2}$, and $\mathbb F\_{p^3}$, or between any three finite fields, using the symmetric functions.


