Coordinate dependence of the volume of parallelotope

Question

It is well known that for $n$ vectors $v_1, \ldots, v_n$ in $\mathbb R^n$, the determinant of the matrix $A = (v_1 \ldots v_n)$ [i.e. with the vectors as columns] is related to the volume of the parallelotope with $v_1, \ldots, v_n$ as corners, i.e. $$ \mbox{Volume Parallelotope} = |\det A|. $$ Suppose these vectors are lineary independent, then this value is non-zero. Now take these vectors as as basis for $\mathbb R^n$, then each of them has the coordinate representation $e_1, \ldots, e_n$, if I know make the same procedure I get $$ |\det E_n| = 1 $$ which in general is not the volume anymore. So this relationship depends heavily on the coordinates choosen. But this relationship is heavily used, for example in the substitution rule for multi-dimensional integrals, but I think such general formulas should not be coordinate dependet.

On the other side, for example it is well known that the determinant of an representation matrix of an endomorphism is independent of the coordinates I choose to represent the endomorphism, so this is a real invariant.

So my question, is this relationship just a strange coincide which works in a special coordinate system, or is there a coordinate-independent formulation of this relationship? And what makes the standard basis, interpreted as an orthonormal basis in $\mathbb R^n$, so special that it works here, but not in other bases?

Answer 1

What makes the standard basis special is that it's one in which the measure was defined.

It's not SO special, however: bases that different from it by an orthogonal change-of-basis matrix work fine, too. And to say things "should not be coordinate dependent" is a little odd -- is this a moral imperative, or just a principle based on "some other things behave this way, and I like it"? Sometimes coordinates matter. In geometry, the measure of curvature is coordinate independent, but one of the best ways to compute it involves the Christoffel symbols, which don't even transform nicely ("tensorially") wrt coordinates.

I think that if you ask this question not for $R^n$, but for $R$, where things are really pretty clear, you'll see that coordinate dependence of "size" (in this case, length) isn't so crazy -- it amounts to the chain rule.

Answer 2

As volume is measured in chosen units so the determinant is a scalar depending on the "unit", elementary k-vector, e.g. a bivector in $\mathbb{R}^2$, $u\wedge v=\det\bigl(\begin{smallmatrix} [u]_e&[v]_e\end{smallmatrix} \bigr)e_1\wedge e_2 $. You can make it 1 by changing the basis.