Is a perturbation of a tensor field a tensor field? Let say I take some $2$-tensor field $T_{\mu\nu}$ on some pseudo-Riemannian manifold. Now, often, we are interested in its linearization, which means that we take a family of tensor fields $T_{\mu\nu}(t)$ such that $T_{\mu\nu}(0)=T_{\mu\nu}$. Then, we expand in a Taylor series which yields
$$T_{\mu\nu}(t)=T_{\mu\nu}+t T_{\mu\nu}^{\prime}+\mathcal{O}(t^{2})$$
where the perturbation or linearization is defined by $T_{\mu\nu}^{\prime}:=\partial_{t}T_{\mu\nu}(t)\vert_{t=0}$. How do I show that $T_{\mu\nu}^{\prime}$ is a tensor field? I know that it has to be a tensor field, since this is for example used in linearized gravity, where one takes $T$ to be the metric $g$ and derives the linearized Einstein equations for the perturbation $g^{\prime}$.
My attempt:
If we ignore all terms of order $\mathcal{O}(t^{2})$, we get
$$T^{\prime}_{\mu\nu}\propto T_{\mu\nu}(t)-T_{\mu\nu},$$
which is the difference between two tensor fields, however, I don't think that I am in general allowed to ignore all additional terms of higher orders....
 A: Yes, in fact perturbations of all orders, $\frac{d^k}{dt^k}(T(t))$ are still tensor fields. Let me first address this in more generality, then provide special cases, and other ways of thinking about this.
Let $(E,\pi,M)$ be a smooth vector bundle. Recall that a section of this vector bundle is by definition a mapping $\psi:M\to E$ such that $\pi\circ \psi=\text{id}_M$ (so far I only gave the definition of a section; you can also define smooth sections by requiring $\psi$ to be smooth). More intuitively, a vector bundle means you have a base manifold $M$ (think spacetime, or a configuration space of some mechanical system), and at each point $x\in M$, we have a vector space $E_x$ “attached at the point $x$”, and that the family of vector spaces $\{E_x\}_{x\in M}$ “vary smoothly” as you vary $x$. A section just means for each point $x\in M$, you have a vector $\psi_x\in E_x$ “attached at the point $x$”. The most important special case is when $E_x=T_xM$ is the tangent space to the manifold at the point $x$, in which case $E=TM$ is the tangent bundle.
Now, let us say we have a smooth one-parameter family of sections of $(E,\pi,M)$, $t\mapsto \Psi(t)$. So, for each $t\in\Bbb{R}$ and each point $x\in M$, we have a vector $\Psi_x(t)\in E_x$ (we can formulate smoothness in terms of $t$ alone, or jointly in $(t,x)$). Now, fix the point $x$, and let us vary $t$; so we have the mapping $\Psi_x:\Bbb{R}\to E_x$, whose value at a parameter $t$ is the vector $\Psi_x(t)\in E_x$. This is a very simple type of mapping because the domain is simply $\Bbb{R}$, and the target space is $E_x$, which is a real vector space. So, we can differentiate as usual (up to any order we like) to get  the mapping $\frac{d^k\Psi_x}{dt^k}:\Bbb{R}\to E_x$, which we can further evaluate at $t=0$ you get the vector $\frac{d^k\Psi_x}{dt^k}\big\rvert_{t=0}\in E_x$. Since we have a vector at each point $x$, we have a section $\frac{d^k\Psi}{dt^k}\big\rvert_{t=0}$ of $E$, as desired. Thus, we have shown that derivatives with respect to the parameter still give us sections (and if you initially assume smoothness with respect to $(t,x)$, then that smoothness is still preserved).
In your special case, we’re considering $E=T^0_2(TM)$, the $(0,2)$ tensor bundle on the manifold. Sections of this bundle are by definition the $(0,2)$ tensor fields, so everything above can be applied here. The bottom line is that once you fix the point $x$, you’re working within a single vector space so it is basic calculus as usual. In particular with $k=1$, you see that the linearization of any tensor field is still a tensor field

If you want to think in terms of components and transformation laws, you can do that as well. Given a smoothly varying 1-parameter family of (say) $(0,2)$ tensor fields $T(t)$, in terms of a coordinate chart, we can write it as $T(t)=T_{\mu\nu}(t,x)\,dx^{\mu}\otimes dx^{\nu}$. We can then differentiate the components $T_{\mu\nu}(t,x)$ with respect to $t$ as many times as we like, and evaluate at $0$ to get $\frac{\partial^kT_{\mu\nu}}{\partial t^k}(0,x)$. Now, you can change coordinates and verify that the transformation laws still hold. This is because the $\mu\nu$ transformation law of the tensor are “not bothered” by the $t$-dependence. Explicitly, we have the following equation because each $T(t)$ is a $(0,2)$ tensor field
\begin{align}
T_{\mu\nu}(t,x)&=\tilde{T}_{\alpha\beta}(t,y)\frac{\partial y^{\alpha}}{\partial x^{\mu}}\frac{\partial y^{\beta}}{\partial x^{\nu}}.
\end{align}
Now, since the coordinate changes do not depend on the parameter $t$, we can differentiate both sides with respect to $t$ as many times as we wish, then evaluate at $0$ and the same relation holds:
\begin{align}
\frac{\partial^kT_{\mu\nu}}{\partial t^k}(0,x)&=\frac{\partial^k\tilde{T}_{\alpha\beta}}{\partial t^k}(0,y)\frac{\partial y^{\alpha}}{\partial x^{\mu}}\frac{\partial y^{\beta}}{\partial x^{\nu}}.
\end{align}
This is the component proof that perturbations of all orders of tensor fields are again tensor fields (the smoothness with respect to $x$ being the same as before).
