Differentiating the scalar curvature $R_g$ w.r.t. a family $\{g_t\}_t$ of Riemannian metrics Let $(M,g)$ be a Riemannian manifold, $R$ the scalar curvature of $M$, $C^\infty(T^*M\odot T^*M)$ the space of smooth symmetric $2$-tensor fields on $M$, and $C^\infty(M)$ the space of smooth functions on $M$. I'd like to know why the codomain of the following operator $DR|_g$ is $C^\infty(M)$.



The above material comes from the book Geometric Relativity by Dan A. Lee. As to $g_t$, my guess would be that it is smooth as a map from an open interval containing $0$ to $C^\infty(T^*M\odot T^*M)$.
That $R\in C^\infty(M)$ makes sense to me, but how could its derived quantity, $\left.\frac{d}{dt}\right|_{t=0}R_{g_t}$, be still in $C^\infty(M)$? Take a curve $\gamma:(a,b)\to N$ in a smooth manifold $N$ as an example. The map $\gamma$ may describe the position of an object at every instant,but once we differentiate $\gamma$, we are left with the velocity of that object, not with the position. With that in mind, how could one see that $\left.\frac{d}{dt}\right|_{t=0}R_{g_t}$ is a function on $M$? This part just doesn't make sense to me. Does anyone have an idea? Thank you.
 A: First of all, let me say that your sentence "differentiating $R_g$ with respect to a family $\{g_t\}$" is misleading.
I would rather express the idea as "differentiating the familly of smooth functions $\{R_{g_t}\}$ with respect to $t$".
The scalar curvature can be thought of as an operator
$$
R \colon \mathcal{C}^{\infty}(\Sigma^2_{++}M) \to \mathcal{C}^{\infty}(M),
$$
where $\Sigma^2_{++}M$ is the open subset of the vector bundle of symmetric 2-tensors $\Sigma^2M= T^*M\odot T^*M$, with the additional restriction that they are definite positive.
This operator is smooth: this can be shown using the complicated formulae in coordinates for $R_g$, which depends smoothly on $g$ and its derivatives.
Coarsely, as a smooth function, $R$ has a differential
$$
DR\colon T(\mathcal{C}^{\infty}(\Sigma^2_{++}M))\longrightarrow T(\mathcal{C}^{\infty}(M)) 
$$
such that for $g\in \mathcal{C}^{\infty}(\Sigma^2_{++}M)$, it yields a linear map
$$
DR|_g \colon T_g(\mathcal{C}^{\infty}(\Sigma^2_{++}M)) \longrightarrow T_{R_g}(\mathcal{C}^{\infty}(M)).
$$
Now, since $\mathcal{C}^{\infty}(\Sigma^2_{++}M)$ is an open subset of the vector space $\mathcal{C}^{\infty}(\Sigma^2M)$, its tangent space at $g$ is canonically identified with $\mathcal{C}^{\infty}(\Sigma^2M)$.
Similarly, since $\mathcal{C}^{\infty}(M)$ is a vector space, its tangent space at $R_g$ is canonically identified with itself.
Finally, we have defined a map
$$
DR|_g \colon \mathcal{C}^{\infty}(\Sigma^2 M) \longrightarrow \mathcal{C}^{\infty}(M).
$$
To compute the value $DR|_gh$, you just have to find a path $g(t)$ in $\mathcal{C}^{\infty}(\Sigma^2_{++}M)$ with $g(0)=g$ and $g'(0) = h$: for instance, $g(t) = g + th$.
Then the Taylor formula says that
$$
R_{g(t)} \underset{t\to 0}{=} R_g + t(DR|_g)(h) + O(t^2).
$$
(Note that fixing $t_0$, this latter expression confirms that $R_{g(t_0)}$, $R_g$ and $DR|_g(h)$ are of the same nature, namely, they are functions on $M$.)
Setting $h=\dot g$, it follows that $DR|_g(\dot g) \in \mathcal{C}^{\infty}(M)$, that is, it is a smooth function on $M$.
