This is a clarification of [another post of mine](https://mathoverflow.net/questions/7817/normal-coordinates-for-a-manifold-with-volume-form).


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...)?



Is the space of volume-preserving maps path-connected?
The answer is yes, on a contractible domain $U$. Suppose that $f^i\_j$ satisfies $\partial\_k f^i\_j = \partial\_j f^i\_k = g^i\_{jk}$. Then $g$ satisfies $g^i\_{jk} = g^i\_{kj}$ and $g^i\_{ik} = 0$. Conversely, if $g^i\_{jk} = g^i\_{kj}$ and $g^i\_{ik} = 0$, then since $U$ is contractible, there exists $f^i\_j$ with $\partial\_kf^i\_j = g^i\_{jk}$. Indeed, pick a basepoint $u \in U$; then $f$ is determined by $g$ and $f(u)$.


However, the conditions on $g$ are linear. Let $f\_1,f\_2$ be two matrix-valued functions with $\partial\_k{f\_1}^i\_j = \partial\_j{f\_1}^i\_k$ and $\partial\_k{f\_2}^i\_j = \partial\_j{f\_2}^i\_k$. Then set ${g\_a}^i\_{jk} = \partial\_k{f\_a}^i\_j$ for $a=1,2$. Introduce a time variable $t$, and let $G^i\_{jk}(t) = (1-t){g\_1}^i\_{jk} + t{g\_2}^i\_{jk}$. Then for each $t$, $G(t)$ satisfies the integrability conditions that $g$ did in the previous paragraph. For the basepoint $u\in U$, let $F(u,t)^i\_j$ be any smooth path connecting the values $F(u,0) = f\_1(u)$, $F(u,1) = f\_2(u)$. Then we can find extend $F(u,-)$ to $F(-,-)$ satisfying $\partial\_k F(t)^i\_j = G(t)^i\_{jk}$. And this does the job.


