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!

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    $\begingroup$ $1$-volume is length, so I guess you're talking about an isometry? $\endgroup$ – Matthew Leingang Jun 29 '17 at 17:17
  • $\begingroup$ Well, I'm thinking of a general setting with $d$ fixed. In the particular case where $d=1$, I don't mean the same as an isometry. If $A$ is a 1-dimensional set in $\mathbb{R}^D$, I would like $1\mbox{-}\text{vol}(A)=1\mbox{-}\text{vol}(f(A))$, so for example a segment might be deformed in such a way that the points in it are not at the same distances anymore but the length of the segment remains constant. $\endgroup$ – Vokram8 Jun 29 '17 at 17:34
  • $\begingroup$ @Vokram8: If you map a segment to itself in a way that the points in it are not at the same distances anymore but the length of the segment remains constant, then the length of some subsegment is not preserved. $\endgroup$ – Andrew D. Hwang Jun 29 '17 at 19:28
  • $\begingroup$ Thanks for pointing that out Andrew, you are right! Then yes, for the $d=1$ case I am just talking about isometries. $\endgroup$ – Vokram8 Jun 29 '17 at 22:17

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.


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