Let $E\to M$ be a real vector bundle of finite rank over a closed differentiable manifold $M$. Let $C^{\infty}(E)$ denote the space of smooth sections of $E$ and let $e\in C^{\infty}(E)$ be a section. I often see statements of the type:

"The tangent space to $C^{\infty}(E)$ at any given section $e\in C^{\infty}(E)$ is isomorphic to $C^{\infty}(E)$ itself, namely $T_{e}C^{\infty}(E)\simeq C^{\infty}(E)$."

I wonder what is the precise formulation of the statement above.

I know that if we only take smooth sections, $C^{\infty}(E)$ admits the structure of an infinite dimensional Frechet manifold with respect to the appropriate topology. The statement should be then that the isomorphism $T_{e}C^{\infty}(E)\simeq C^{\infty}(E)$ holds understanding $C^{\infty}(E)$ as a Frechet manifold and taking $T_{e}C^{\infty}(E)$ to be the tangent space of $C^{\infty}(E)$ as a Frechet manifold? Or rather, should it be understood by assuming that we have implicitly Sobolev-completed $C^{\infty}(E)$ into $H_{s}(E)$ using some appropriate Sobolev norm as to make $H_{s}(E)$ into a smooth Hilbert manifold and then the isomorphism that actually holds is $T_{e}H_{s}(E)\simeq H_{s}(E)$?

My second question is: can one make sense of an isomorphism of the type $T_{e}C^{\infty}(E)\simeq C^{\infty}(E)$ if the base is non-compact?

And lastly, let $Q\to M$ be a smooth fiber bundle over a closed manifold $M$ with typical fiber $F$ given by a smooth manifold. How should be understood the tangent space at a point of the space of sections of $Q$?

References are welcome.

Thanks.

Let $E\to M$ be a real vector bundle of finite rank over a closed differentiable manifold $M$. Let $C^{\infty}(E)$ denote the space of smooth sections of $E$ and let $e\in C^{\infty}(E)$ be a section. I often see statements of the type:

"The tangent space to $C^{\infty}(E)$ at any given section $e\in C^{\infty}(E)$ is isomorphic to $C^{\infty}(E)$ itself, namely $T_{e}C^{\infty}(E)\simeq C^{\infty}(E)$."

I wonder what is the precise formulation of the statement above.

I know that if we only take smooth sections, $C^{\infty}(E)$ admits the structure of an infinite dimensional Frechet manifold with respect to the appropriate topology. The statement should be then that the isomorphism $T_{e}C^{\infty}(E)\simeq C^{\infty}(E)$ holds understanding $C^{\infty}(E)$ as a Frechet manifold and taking $T_{e}C^{\infty}(E)$ to be the tangent space of $C^{\infty}(E)$ as a Frechet manifold? Or rather, should it be understood by assuming that we have implicitly Sobolev-completed $C^{\infty}(E)$ into $H_{s}(E)$ using some appropriate Sobolev norm as to make $H_{s}(E)$ into a smooth Hilbert manifold and then the isomorphism that actually holds is $T_{e}H_{s}(E)\simeq H_{s}(E)$?

My second question is: can one make sense of an isomorphism of the type $T_{e}C^{\infty}(E)\simeq C^{\infty}(E)$ if the base is non-compact?

And lastly, let $Q\to M$ be a smooth fiber bundle over a closed manifold $M$ with typical fiber $F$ given by a smooth manifold. How should be understood the tangent space at a point of the space of sections of $Q$?

References are welcome.