Why is the Riemann tensor $C^{\infty}$-multilinear? According to my textbook on differential geometry, the Riemann tensor  $R(\cdot, \cdot)$ is $C^{\infty}$-multilinear.
I suppose this means that if $M$ is a manifold, $p \in M$ and $x_1,x_2, y, z \in T_pM$, then for any $C^{\infty}$-function $f: M \longrightarrow \mathbb{R}$ it holds that
$$R( fx_1 + x_2, y )z = fR(x_1, y)z + R(x_2,y)z$$
and analogously for the second argument. Now I was wondering, why this property is restricted to $C^{\infty}$-functions and not to $C^2$-functions, since the Riemann tensor  contains only covariant derivatives of the second order. Of course each $C^{\infty}$-function is indeed $C^2$, however the converse statement is not true.
 A: No, this is not what it means. There are usually two (equivalent) ways of describing tensor fields.

*

*Let $M$ be a smooth manifold, and $T^r_s(TM):= \bigcup\limits_{p\in M}T^r_s(T_pM)$ be the $(r,s)$ tensor bundle (for any vector space $V$ over $\Bbb{R}$, $T^r_s(V):= V^{\otimes r}\otimes (V^*)^{\otimes s}$ is the space of $(r,s)$ tensors over $V$, which is canonically isomorphic to the space of multilinear maps $\underbrace{V^*\times \cdots \times V^*}_{\text{$r$ times}}\times \underbrace{V\times \cdots \times V}_{\text{$s$ times}}\to\Bbb{R}$). Then, a smooth tensor field on $M$ of type $(r,s)$ is a smooth section $\xi:M\to T^{r}_s(TM)$ (i.e a smooth map such that for every $p\in M$, $\xi(p)\in T^r_s(T_pM)$).


*Another way of describing it is that a smooth $(r,s)$ tensor field is a mapping $\xi: \Gamma(T^*M)^r\times \Gamma(TM)^s\to C^{\infty}(M)$ which is $C^{\infty}(M)$-multilinear. Meaning, $\xi$ has to eat $r$ smooth covector fields and $s$ smooth vector fields and output a smooth function.
I personally prefer the first definition because conceptually it formalizes the statement "a tensor field is a smooth assignment of a tensor to every point". The second definition is sometimes more convenient to quickly give definitions of certain tensor fields. I would suggest you read John Lee's books (intro to smooth manifolds) where this equivalence is proven in the chapter on tensors. Also, check out Abraham Marsden Ratiu's book  on tensor analysis and manifolds.
Going from definition $1$ to definition 2 is easy; in fact if in definition of $1$ we only consider $C^k$ maps, then we end up with $C^k$-multilinearity. To show the converse of definition 2 implying definition 1 however, we have to assume $C^{\infty}$. Roughly speaking this restriction has to do with the fact  that when you want to "localize things" (i.e going from the level of fields in definition 2 to the pointwise level in definition 1), one has to use bump functions and their derivatives. If we only assume a finite amount of differentiability, then the derivatives have one lower degree of differentiability, and this messes things up slightly, so one has to be very careful in making statements. This loss of differentiability is why in "introductory" books we always assume $C^{\infty}$, simply to avoid the burden of always counting the degree of smoothness. The book by Abraham, Marsden, Ratiu is more detailed with such technical details about smoothness, and what happens if we weaken hypotheses etc (see remarks after theorem 4.2.16).
For example, many books state Stokes theorem in the $C^{\infty}$ case simply because it's easier. Of course the theorem itself is valid under MUCH more general situations, but the hypotheses are much more involved.

For a manifold $M$ with a (affine) connection $\nabla$, one can define for each smooth vector field $X,Y$, an object $R(X,Y)$ which eats a smooth vector field $Z$ and outputs a smooth vector field $[R(X,Y)](Z)$ defined (perhaps up to some obscure sign convention) as
\begin{align}
[R(X,Y)](Z)&=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z
\end{align}
By dualizing appropriately, one can think of this as a $(1,3)$ tensor field according to definition 2. So for example, for any smooth function $f:M\to \Bbb{R}$ and smooth vector fields $X_1,X_2,Y,Z$, we have
\begin{align}
[R(fX_1+X_2,Y)](Z)&= f\cdot \bigg([R(X_1,Y)](Z)\bigg) + [R(X_2,Y)](Z).
\end{align}

If you want to think in terms of the first definition, one has to reword the definition of $R$ slightly almost into a theorem: there exists a well-defined $(1,3)$ tensor field $R$ on $M$ such that for each $p\in M$, $x,y,z\in T_pM$, if we take any smooth vector fields $X,Y,Z$ on $M$ such that $X(p)=x,Y(p)=y,Z(p)=z$, then
\begin{align}
[R_p(x,y)](z)&= \bigg(\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z\bigg)(p)
\end{align}
In this case, one has for any $p\in M, x_1,x_2,y,z\in T_pM$ and $c\in\Bbb{R}$, that
\begin{align}
[R_p(cx_1+x_2,y)](p)&=c\, [R(x_1,y)](z)+[R(x_2,y)](z).
\end{align}
i.e if you're working with definition 1 (which is at a pointwise level) then we only talk about linearity over the field $\Bbb{R}$.
