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Let $X,Y$ be smooth manifolds and consider the infinite-dimensional manifold $$ C^\infty(X,Y) $$ of smooths maps $f: X \to Y$.

Note that there is an infinite-dimensional vector bundle $E$ over this space whose fibers are given by $f^\ast(TY)$-valued 1-forms. The derivative $$ f \mapsto df $$ gives a section of this bundle.

I want to describe explicitly the derivative of $df$--i.e., the linearization of the "differential" map at a point $f$. This means that at any point $f$, we will have a linear map $$ D(d): T_f C^\infty(X,Y) \to T_{(f,df)} E $$ between infinite-dimensional vector spaces.

I tried to compute $D(d)$, and chose a connection on $Y$ to do it. But it seems my answer depends on the choice of connection. Can someone point out my error? Below is the work that I did.

The Euclidean case.

If $X$ and $Y$ are both Euclidean space, this seems deceptively simple. Choose a map $f: X \to Y$. The tangent space at $f$ is given by the space of vector fields $$ \Gamma(f^* TY). $$ Since $X$ and $Y$ are Euclidean, we can cheat and identify $f^*TY \cong X \times Y$. There is some trickery going on here, but we move on.

Then a tangent vector to $f$ is simply a map $\xi: X \to Y$. The linearization of $d$ acts on $\xi$ as follows---one can take the test curve $$ f_t: \mathbb{R}_t \to C^\infty(X,Y) \qquad f_t = f + t\xi $$ using the linear structure of Euclidean space $Y$, and compute simply that the derivative at $t=0$ is given by $$ d\xi. $$ That is, the derivative of $d$ acts by taking the derivative of the vector field $\xi$. In other words, $$ D(d): \Gamma(f^*TM) \to \Gamma(T^*M \otimes f^*TM), \qquad \xi \mapsto d\xi $$ where we are treating $\xi$ as a map $\xi: X \to Y$.

Non-Euclidean case.

There was some trickery above. Namely, the trivialization of $f^* TY$ is not canonical. Moreover, the derivative should really take values in the tangent space of the total space of the infinite-dimensional vector bundle, and not in its vertical tangent space. (Note that the vertical tangent space of the vector bundle is indeed given by $f^*TY$-valued 1-forms on $X$.)

The crux of the trickery is when we wrote the test curve $f_t$, where we used a trivialization to write down the `addition' $f + t \xi$.

Normally one would need to actually choose a connection to do such a thing; giving a section of $f^*TY$ doesn't give you a 1-parameter family of maps $f_t$ unless you tell someone how to keep deforming the curve past time $0$. By putting a connection on $f^*TY$, we can write down an integral curve $f_t$ whose derivative at a time $t$ is given by covariant differentiation.

But then, tautologically, it would seem that the linearization of $d$ is given by the formula $$ \xi \mapsto \nabla_\bullet \xi. $$ Here, $\nabla_\bullet \xi \in \Gamma(T^*X \otimes f^*TY)$ is indeed a $f^*TY$-valued 1-form.

But this "linearization" of d doesn't depend on a choice of connection to define it. I'm seeming to get a result that says that the linearization does depend on a choice of connection, so my answer must be wrong. Is there a dumb mistake I'm making, and can someone point me to a more correct "derivation"?

I suspect that the error occurs because I am implicitly splitting the tangent space of $E$, and this splitting depends on my choice of connection. In other words, it seems I'm trying to study some map $$ T_f C^\infty(X,Y) \to T_{(f,df)} E \cong T_f C^\infty(X,Y) \oplus \Omega^1(f^*TY) \to \Omega^1(f^*TY) $$ but the isomorphism in the middle is not canonical unless I am along the zero section of $E$, so I suspect this is where the "dependence" on the connection appears. So is there a more choice-free way of writing down this linearization?

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This question is really rather long, and I doubt you'll get anyone reading it. Moreover, it doesn't seem to ask a definite question. This site is designed for people to post definite questions with clear cut answers. I would suggest that you redraft this question and cut out almost all of the theory. Anyone able to answer your questions isn't going to need a seven paragraph textbook introduction. Make sure you're clear in your own mind about what you want to know, and then formulate the question. –  Fly by Night Apr 11 '13 at 20:28
Thanks! I put the question closer to the top; I left in my work so someone would have an idea of what I did. –  user54535 Apr 12 '13 at 14:27
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