Question about volume preserving transformations I will denote by $d$-vol the $d$ dimensional volume in $\mathbb{R}^D$, where $D \geq d$. For example, if $A=\{(x,y,0)\in\mathbb{R}^3:\text{max}(|x|,|y|)\leq 1\}$, then $3\mbox{-}\text{vol}(A)=0$ but $2\mbox{-}\text{vol}(A)=4$. I know that if $f:\mathbb{R}^D \rightarrow \mathbb{R}^D$ is a differentiable and invertible function, the change of $D$-volume induced by $f$ is $|\text{det}(J)|$, where $J$ is the Jacobian of $f$. Thus, for $f$ to be a $D$-volume preserving transformation we need to ensure that $|\text{det}(J)|=1$. However, if $f$ is $D$-volume preserving it need not be $d$-volume preserving for $d<D$. I was wondering what the equivalent condition is for $f:\mathbb{R}^D \rightarrow \mathbb{R}^D$ to be $d$-volume preserving. Thank you very much!

By "zooming in" or "subdividing" (cf. Andrew's comment), this implies (and with a bit more effort is equivalent to) the Jacobian $J_x$ preserving $d$-volume at each point $x \in \mathbb R^D$. 
Fixing an $x$, we can use the singular value decomposition to choose oriented orthonormal bases $e_i$ for the domain and $v_i$ for the target so that the matrix representation of $J_x$ is diagonal: 
$$ M = [J_x]_{e \to v} = \left(\begin{matrix}\sigma_1 & & \\ &\ddots & \\ &&\sigma_D\end{matrix}\right).$$
Now, the action of $M$ on the unit $d$-cube aligned with the axes numbered $i_1,\ldots,i_d$ is simply to stretch each axis by the corresponding $\sigma$ factor (and maybe do some reflections); so it distorts the volume by the product $|\sigma_{i_1}\ldots\sigma_{i_d}|$. Thus your condition implies that the product of any $d$ of the $D$ numbers $|\sigma_i|$ must be equal to $1$. When $d=D$ this is just the single equation $$|\sigma_1\ldots\sigma_D| = |\det M| = |\det J_x| = 1,$$ as expected.
When $1 \le d < D$, however, there are many more equations; so many, in fact,
 that (Exercise!) we can conclude $|\sigma_i|=1$ for each $i$. So $M$ is just a composition of reflections, which means $J_x$ is just a composition of reflections and rotations; i.e. a linear isometry. Since this holds for each $x$, we conclude that $f$ is an isometry.
