Given a function $f:\Delta^n \to Y$ from a simplex into a riemmannian manifold. Furthermore given a point $x \in \Delta^n$ we can send an orthonormal basis at $x$ using $D_x f$ to a set of vectors defining a parallelepiped. I know that the volume of a parallelepiped can be calculated using the absolute value of the determinant of its defining vectors.

So it would be convenient to write $\left|\det(D_x f)\right|$. But the Jacobian $D_x f$ is not necessarily a square matrix.

Is there a notational way to fix this?

EDIT: What I am looking for is a simple notation to write this down.

  • $\begingroup$ what is $Y$'s dimension? $\endgroup$ – janmarqz Jul 8 '15 at 18:19
  • $\begingroup$ @janmarqz it is only necessary for Y to have a dimension $\geq n$ $\endgroup$ – Loreno Heer Jul 8 '15 at 19:36

It is possible to calculate the volume by $n$ vectors $\{\vec{v_i}\}$ in a $m$-dimensional coordinate system with $m > n$. Actually they must lie in some $n$ dimensional subspace. Denote the matrix with $\{\vec{v_i}\}$ as rows $V$.

Apply an orthogonal transformation $O$ to get them into the first $n$ coordinates, we get $\{O V\}$, and multiply with a $E_{m,n}$ which is an $m*n$ matrix with $1$ on $(1, 1), (2,2),\cdots$ and $0$ elsewhere. Then the desired volume is $\det(E_{m,n}OV)$.

Now lets get rid of $E_{m,n}$ and $O$.

Remember that $\det(A^T)=\det(A)$, and $\det(A^T A)=\det(A)^2$. So in this case we have $$\mathrm{Volume}=\sqrt{\det((E_{m,n}OV)^T(E_{m,n}OV))} = \\ \sqrt{\det(V^TO^TE_{m,n}^TE_{m,n}OV)}$$

It is nice that $O^TO=E$ and $E_{m,n}^TE_{m,n}=E$, so $$\mathrm{Volume}=\sqrt{\det(V^TV)}$$, which is only dependent of $V$, or $\{\vec{v_i}\}_n$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.