TITLE: Is the space of volume-preserving maps path-connected?
QUESTION [14 upvotes]: This is a clarification of another post of mine.
Fix $n$ a positive integer.  Let $SL(n)$ have its usual matrix representation, so that it really is the codimension-one subset of $M(n) = \mathbb R^{n^2}$ cut out by the degree-$n$ condition that the determinant is $1$.  So we have $n^2$ coordinate functions $A^i_j$ on $SL(n)$, $i,j = 1,\dots,n$.
Let $U$ be a domain in $\mathbb R^n$, with coordinates $x_1,\dots,x_n$.  Consider the set $\mathcal S$ of smooth functions $f: U \to SL(n)$ satisfying the differential equation $\frac{\partial f^i_j}{\partial x^k} = \frac{\partial f^i_k}{\partial x^j}$ for each $i,j,k = 1,\dots,n$ (of course, $f^i_j = A^i_j \circ f$ is the $(i,j)$th coordinate of $f$).
(Why would you care about $\mathcal S$?  Because a smooth map $g: U \to \mathbb R^n$ is volume-preserving if and only if $\frac{\partial g^i}{\partial x^j} \in \mathcal S$, and every element of $\mathcal S$ arises this way; indeed, $\mathcal S$ is the space of volume-preserving maps up to translations.)
Let's agree that a smooth path in $\mathcal S$ is a smooth function $F: [0,1] \times U \to SL(n)$ such that for each $t\in [0,1]$, $F(t,-) \in \mathcal S$.
Question:  Is $\mathcal S$ smooth-path-connected?  I.e. given $f_0, f_1 \in \mathcal S$, does there exist a smooth path $F$ so that $F(0,-) = f_0$ and $F(1,-) = f_1$?
If the answer is "no" in general, is it "yes" for sufficiently nice domains $U$ (contractible, say, or with compact closure and require that each $f\in \mathcal S$ extend smoothly to a neighborhood of the closure, or...)?

REPLY [3 votes]: I'd like to point out a special case where there's a particularly appealing answer -- a variant of the Alexander trick. 
Consider the space of smooth volume-preserving immersions (or embeddings) $f : D^n \to \mathbb R^n$.  $D^n$ is the unit compact disc in $\mathbb R^n$. 
This space deformation-retracts to the subspace where $f(0)=0$. So we restrict to that subspace for brevity of notation. 
Given $t \in (0,1]$ define
$$F_t : D^n \to \mathbb R^n$$
by $F_t(x) = \frac{1}{t}f(tx)$ and define $F_0(x) = Df_0(x)$. 
Provided $f$ is $C^1$, this makes $F : [0,1] \times D^n \to \mathbb R^n$ continuous. 
And $F_t$ is volume preserving for all $t$ by the chain rule -- even better $D(F_t)_p = (Df)`_`{tp}$ 
for all $p \in D^n$. Moreover, if $f$ is an embedding, $F_t$ remains an embedding for all $t$. 
Since $F_1 = f$ and $F_0 \in SL_n\mathbb R$, and $SL_n\mathbb R$ is connected, we're done.