Let $A$, $B$ be two smooth vector bundles of finite rank over a smooth manifold $M$. Let $Diff(A,B)$ be the space of differential operators from $A$ to $B$. Can I talk about "the space of smooth maps from $[0,1]$ to $Diff(A,B)$?
2 Answers
You could use the bijective correspondence between differential operators $L: \Gamma(A) \to \Gamma(B)$ and vector bundle homomorphisms $\tilde{L}: J^k A \to B$ to topologize your space $Diff(A,B)$. Then $C^1$curves in $Diff(A,B)$ are defined as usual, that is, the differential quotient has to exist and be continuous. Iterating yields the notion of smooth curves.

$\begingroup$ If the order of operators in $Diff(A,B)$ is not a priori bounded (and in particular not bounded by $k$), one should use the infinite jet bundle $J^\infty A$ instead of $J^k A$. $\endgroup$ Nov 24, 2013 at 22:21

$\begingroup$ I was hoping to get an answer which would in the end match with differential operators $Diff(p^*A,p^*B)$ that are linear in the $[0,1]$ direction where $p: M\times [0,1]\rightarrow M$ is the projection map. Is this intuition correct, with your topology? $\endgroup$– tujunwuNov 25, 2013 at 5:07
One way to introduce the space you are interested in is: $$ \widetilde{\textrm{Diff}}(A,B):=C^\infty([0,1])\otimes_{\mathbb{R}}\textrm{Diff}(A,B). $$ I'm sure this approach is equivalent to the topological one, suggested by @Tobias, at least for operators of bounded order.
You can also characterize it as a distinguished subspace $$ \widetilde{\textrm{Diff}}(A,B)\subseteq \textrm{Diff}(p^*A,p^*B).\quad\quad\quad (*) $$ Indeed, if $\{\alpha^i\}$ (resp., $\{b_j\}$) is a basis of $\Gamma(A)^*$ (resp., $\Gamma(B)$), then $$ L=L_{i\, I\, l}^j\cdot b_j\otimes \frac{\partial^{I+l}}{\partial x^I\partial t^l}\circ \alpha^i $$ is the local expression of a generic element $L$ belonging to the righthand side of $(*)$, which sits in the lefthand side if and only if $L_{i\, I\, l}^j=0$ for $l>0$, i.e., as you said, $L$ is of order 0 along the fibers of $p$.
P.S. I used coordinates $x^i$ on $M$ and $t$ on $[0,1]$; $I$ is a multiindex of length $I\leq k$, and $L_{i\, I\, l}^j\in C^\infty(M\times [0,1])$; I tacitly used the same bases for $A$ and $B$ and their pullbacks (it is only the algebra of scalars which grows bigger).

$\begingroup$ Just noticed a minor bug in my answer: $\widetilde{\mathrm{Diff}}(A,B)$ should be defined as $C^\infty(M\times[0,1])\otimes_{C^\infty(M)}\mathrm{Diff}(A,B)$. The rest is unaltered. $\endgroup$ Nov 25, 2013 at 17:30

$\begingroup$ Is $Diff(A,B)$ a Frechet space? Because then your first tensor product would make sense. $\endgroup$– tujunwuNov 25, 2013 at 18:33