TITLE: Why is the Riemannian Curvature tensor considered a $(3,1)$ type tensor?
QUESTION [0 upvotes]: 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??

REPLY [1 votes]: 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.
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