Why is the Riemannian Curvature tensor considered a $(3,1)$ type tensor?

Question

Why is the Riemannian Curvature tensor considered a $(3,1)$ type tensor?

Because to me, it looks like you input three vectors and get a real number... Wouldn't a $(3,1)$ tensor mean you input 3 vectors and a covector??

Answer 1

By definition $$ R(X,Y)Z = \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z - \nabla_{[X,Y]} Z $$ which is a vector. Thus, $R$ takes in $3$ vectors and outputs a vector.

Answer 2

On an holonomic basis you have $$R(\partial_i,\partial_j)\partial_k=R^s{}_{ijk}\partial_s.$$ So pairing as \begin{eqnarray*} \langle dx^t,R(\partial_i,\partial_j)\partial_k\rangle&=&\langle dx^t,R^s{}_{ijk}\partial_s\rangle\\ &=&R^s{}_{ijk}\delta^t{}_s\\ &=&R^t{}_{ijk} \end{eqnarray*} you get your result.