I am confused about jacobians, wedge products, tensors, and more. How do I use multilinear algebra to derive the jacobian from basics?

Question

Question

I am confused about differentials and deriving Jacobians. Can someone try to understand this rambling mess and help identify what parts I am missing?

I understand that this may take some patience to read. I have done my best to keep this simple and clear. I have spoken to my professors and I usually just get ignored or left on read regarding this confusion. So I leave it here.

I want to state upfront that I am aware that differentials are more formally treated as one-forms, and that we can take their wedge products and this should all be easy-peasy. Maybe it's just that I didn't take a math degree and took a physics degree instead, but the cross product of two scalars, or differentials, isn't defined in typical vector algebra. Since the wedge product is supposed to be an extension of the cross product to $\mathbb{R}^n$ it still doesn't jive with me to actually state something like $dx \wedge dy$ unless $dx$ is being used in lieu of $\hat{e}_1 dx$.

I am going to make this very simple. When I was introduced to Jacobians, a lot of hand waving occurred and we were told to just ignore some of the questions we had for now. Having taken vector algebra and several other courses I realize I still never have gotten a good answer.

So let's begin with polar coordinates in 2D.

The Jacobian is derived from the coordinate equations $x=x(r,\theta)=r\cos\theta$ and $y=y(r,\theta)=r\sin\theta$ by considering the differential of these functions $$ \left\{\begin{array} .dx =\frac{\partial x}{\partial r}dr+\frac{\partial x}{\partial \theta}d\theta\\ dy =\frac{\partial y}{\partial r}dr+\frac{\partial y}{\partial \theta}d\theta \end{array}\right. \Rightarrow \left\{\begin{array} .dx =\cos\theta dr-r\sin\theta d\theta\\ dy =\sin\theta dr+r\cos\theta d\theta \end{array}\right. $$ If we just simply multiply these together we get something close to the right answer. $$ dxdy = \cos\theta\sin\theta dr^2 + r\cos^2\theta drd\theta - r\sin^2\theta d\theta dr - r^2\sin\theta\cos\theta d\theta^2 $$ Applying the rule that 'something small squared is approximately zero' we simplify. $$ dxdy = r\cos^2\theta drd\theta - r\sin^2\theta d\theta dr $$ And then we must add a new arbitrary rule $d\theta dr = -dr d\theta$ so that this works out correctly. $$ dxdy = r( \cos^2\theta + \sin^2\theta )drd\theta = r drd\theta $$ But now we have a new rule which is not present in standard calculus courses where exchanging the order of integration introduces a negative sign. This is directly contradicted by the fact that when we integrate over a square domain the integral is independent of integration order. example: integrating over the area of a circle is a separable integral in polar coordinates $$ \iint_{\text{circle radius R}} dxdy = \int_{-R}^R \int_{-\sqrt{R^2-y^2}}^{\sqrt{R^2-y^2}} dx dy = \int_{-R}^R \int_{-\sqrt{R^2-x^2}}^{\sqrt{R^2-x^2}} dy dx = \int_{0}^{2\pi}\int_0^R rdrd\theta $$ Integrating the cartesian integral requires a lot of patience, but integrating the polar integral can be done either first on $r$ or $\theta$ or by separation. $$ \int_{0}^{2\pi}\int_0^R rdrd\theta = \int_0^R 2\pi rdr = \int_0^{2\pi} \frac{r^2}{2} d\theta = \int_0^{2\pi} d\theta \int_0^R rdr = \pi R^2 $$

At some point I saw someone factor the equations of differentials into matrix equations which neatly presented the Jacobian matrix which I had never observed as a 'derived' matrix before. $$ \begin{bmatrix}dx\\ dy\end{bmatrix} = \begin{bmatrix}\cos\theta&-r\sin\theta\\\sin\theta&r\cos\theta\end{bmatrix} \begin{bmatrix}dr\\ d\theta\end{bmatrix} $$ In this presentation the Jacobian is merely the absolute value of the determinant of the partial derivatives matrix. This, like many things in calculus, is given without explanation or derivation.

This is just the matrix factored version of the above equations. But now there is no natural way to simply multiply things together. On the left hand side, we have one object with two components and there is no standard matrix operation to multiply the first component to the second without first reducing the matrix equation to a system of separated equations. This is fine in principle, but I feel there should be such a method.

To this end, I noticed that if we right multiply the transpose of this equation we obtain the following. $$ \begin{align} \begin{bmatrix}dx\\dy\end{bmatrix} \begin{bmatrix}dx\\dy\end{bmatrix}^T &= \begin{bmatrix}dx\\dy\end{bmatrix} \begin{bmatrix}dx&dy\end{bmatrix}\\ &= \begin{bmatrix}dx^2&dxdy\\dydx&dy^2\end{bmatrix}\tag{1} \end{align} $$ This is so close and yet so far away from the identity Jacobian for the Cartesian coordinates. Proceeding, we obtain for the righthand side. $$ \begin{bmatrix} \cos\theta&-r\sin\theta\\ \sin\theta& r\cos\theta \end{bmatrix} \begin{bmatrix}dr\\d\theta\end{bmatrix} \left( \begin{bmatrix} \cos\theta&-r\sin\theta\\ \sin\theta& r\cos\theta \end{bmatrix} \begin{bmatrix}dr\\d\theta\end{bmatrix} \right)^T = \begin{bmatrix} \cos\theta&-r\sin\theta\\ \sin\theta& r\cos\theta \end{bmatrix} \begin{bmatrix}dr\\ d\theta\end{bmatrix} \begin{bmatrix}dr & d\theta\end{bmatrix} \begin{bmatrix} \cos\theta& \sin\theta\\ -r\sin\theta& r\cos\theta \end{bmatrix} \\ = \begin{bmatrix} \cos\theta&-r\sin\theta\\ \sin\theta& r\cos\theta \end{bmatrix} \begin{bmatrix}dr^2& dr d\theta\\ d\theta dr & d\theta^2\end{bmatrix} \begin{bmatrix} \cos\theta& \sin\theta\\ -r\sin\theta& r\cos\theta \end{bmatrix} $$ But simplifying this last matrix multiplication is pretty obviously excessively long and convoluted out type out and in the process I have made several mistakes and while a CAS can do it in microseconds, I cannot easily copy and paste the result here. In addition this is a pretty obviously wrong computation.

My professors waved away questions about the value of $dx^2$ and $dy^2$ as being zero because $dx$ is already small so $dx^2$ is basically zero. Or rather they did this hand-wave with the real approximations ($\Delta x$) to the infinitesimals, $\Delta x \approx dx$. There is also the version of all of this that comes up in some other books in terms of the virtual displacements $\delta x$ and so forth. Moreover, Google has returned results on the value of square differentials in integrations, albeit perhaps with niche applications.

Additionally there is another problem, it should be the case that $dxdy=dydx$, but if we take the determinant of the (1) we get the zero as the determinant. If we take the square of the differentials to be zero, we get the wrong sign for the matrix. It must be the case that $dxdy=-dydx$ for the computation to work out. We also must throw in a square root so we aren't left with the squares going to zero again. This just reiterates the issues from above.

In other words I think it's best to express this all with a bit more formalism that I have read and seen elsewhere. Let $\vec{r}=\vec{r}(x,y)=\vec{r}(r,\theta)$ be out position/displacement vector. $$ \vec{r} =\hat{e}_1 x +\hat{e}_2 y =\hat{e}_1 r\cos\theta +\hat{e}_2 r\sin\theta $$ The differential of position is $$ d\vec{r} = \frac{\partial \vec{r}}{\partial x}dx+\frac{\partial \vec{r}}{\partial y}dy = \frac{\partial \vec{r}}{\partial r}dr+\frac{\partial \vec{r}}{\partial \theta}d\theta $$ Using the same matrix factoring method $$ \begin{align} d\vec{r} &= \hat{e}_1 dx + \hat{e}_2 dy\\ &= \begin{bmatrix}\hat{e}_1&\hat{e}_2\end{bmatrix} \begin{bmatrix}dx\\dy\end{bmatrix} = \begin{bmatrix}dx&dy\end{bmatrix} \begin{bmatrix}\hat{e}_1\\\hat{e}_2\end{bmatrix}\\ &= (\cos\theta\hat{e}_1+\sin\theta\hat{e}_2) dr + (-r\sin\theta\hat{e}_1+r\cos\theta\hat{e}_2) d\theta\\ &= \begin{bmatrix}\hat{e}_1&\hat{e}_2\end{bmatrix} \begin{bmatrix} \cos\theta&-r\sin\theta\\ \sin\theta& r\cos\theta \end{bmatrix} \begin{bmatrix}dr\\d\theta\end{bmatrix}\\ \end{align} $$ So far nothing is truly new. The only objection I can think of to this presentation is that 'Matricies can only hold numbers' which is true I suppose but then the question becomes what exactly a number is? Since all fields are vector spaces with the ability to multiply the elements, this point is moot. Further I will also point to the determinant method for taking the vector cross product in $\mathbb{R}^3$.

Regardless, I know I am not unique in using this method to describe linear combinations since EigenChris uses this same representation in his series on Tensors for Beginners.

In fact, these thoughts come back nearly full circle to the following. Consider taking the previous process but instead of simple matrix multiplication, apply the wedge product in between. In fact, with the basis vectors written inside matricies, there is no need to use the transpose operation to make anything work out neatly. We can just use the commutativity of the basis matrix and the component matrix to make any operation more convenient to write out. $$ \begin{align} \begin{bmatrix}\hat{e}_1&\hat{e}_2\end{bmatrix} \begin{bmatrix}dx\\dy\end{bmatrix} \wedge \begin{bmatrix}\hat{e}_1&\hat{e}_2\end{bmatrix} \begin{bmatrix}dx\\dy\end{bmatrix} =& \begin{bmatrix}dx&dy\end{bmatrix} \begin{bmatrix}\hat{e}_1\\\hat{e}_2\end{bmatrix} \wedge \begin{bmatrix}\hat{e}_1&\hat{e}_2\end{bmatrix} \begin{bmatrix}dx\\dy\end{bmatrix}\\ =& \begin{bmatrix}dx&dy\end{bmatrix} \begin{bmatrix} \hat{e}_1 \wedge \hat{e}_1 & \hat{e}_1 \wedge \hat{e}_2\\ \hat{e}_2 \wedge \hat{e}_1 & \hat{e}_2 \wedge \hat{e}_2 \end{bmatrix} \begin{bmatrix}dx\\dy\end{bmatrix}\\ =& \begin{bmatrix}dx&dy\end{bmatrix} \begin{bmatrix} 0 & \hat{e}_1 \wedge \hat{e}_2\\ - \hat{e}_1 \wedge \hat{e}_2 & 0 \end{bmatrix} \begin{bmatrix}dx\\dy\end{bmatrix}\\ =& \begin{bmatrix}dx&dy\end{bmatrix} \begin{bmatrix} 0 + \hat{e}_1 \wedge \hat{e}_2 dy\\ - \hat{e}_1 \wedge \hat{e}_2 dx + 0 \end{bmatrix}\\ =& dx \hat{e}_1 \wedge \hat{e}_2 dy - dy \hat{e}_1 \wedge \hat{e}_2 dx \\ =& (dxdy - dydx) \hat{e}_1 \wedge \hat{e}_2 \end{align} $$ So now we have solved one issue, where we were stuck with a matrix and had to arbitrarily take the determinant and replaced it with another where either $dydx=dxdy$ and we have a zero differential area element, or $dydx=-dxdy$ and then we have to introduce an arbitrary half.

What am I missing? I feel as though there should be a neat resolution here where this all just simplifies quite nicely but I have never seen it.

Answer 1

The relation $$\mathrm{d}x \wedge \mathrm{d}y = -\mathrm{d}y \wedge \mathrm{d}x$$ isn't "a new arbitrary rule", it captures the (physicist) familiar (right-hand rule) fact that $$ \vec{a} \times \vec{b} = - \vec{b} \times \vec{a} \text{.} $$ To rephrase: cross products and wedge products anticommute.

Alternatively, if you are using the right-hand rule to orient infinitesimal areas (for a surface for applying Gauss's Law, for instance), the normals point inward if you order the infinitesimals one way and outward if you swap the order. So of course the integral is negated under exchange of variables. (In undergraduate land, we aren't careful about this. We manually point the normals outward because symmetry means we can intuit which way the field is passing through the surface patches. In graduate land, one may have surfaces with patches that are not so easy to manually orient, so define your infinitesimals carefully and let the wedge product track that for you.)

It may help to realize that cross products don't actually give you vectors, they give axial vectors.

When you say "the Jacobian is merely the absolute value of the determinant of the partial derivatives matrix". You're repeating the undergraduate land version of "you always know how to properly orient things". Generically,

• The Jacobian matrix is the linear coefficient in the Taylor expansion of a multivariable function, also called the gradient. (The Hessian matrix is the second coefficient.)
• The Jacobian determinant is the determinant of the Jacobian matrix.
• When you swap two rows or two columns of a matrix (corresponding to reversing the handedness of one of the input or the output coordinate systems), the determinant picks up a factor of $-1$, again tracking orientations for you. (See second bullet at Determinant:Properties:Immediate Consequences. In jargon, the determinant is alternating. This might be a good time to slowly read alternating form.)

When you take $\mathrm{d}x\, \mathrm{d}y = \mathrm{d}y \, \mathrm{d}x$, you throw away all information about orientation. The only way this can be true for oriented quantities is if $\mathrm{d}x\, \mathrm{d}y = 0 = \mathrm{d}y \, \mathrm{d}x$. Then some of what you write about happens. But if we track orientations, the correct version of this is $$|\mathrm{d}x\, \mathrm{d}y| = |\mathrm{d}y \, \mathrm{d}x|$$ or, equivalently, $$|\vec{a} \times \vec{b}| = |\vec{b} \times \vec{a}| \text{,} $$ where we explicitly delete the orientation information.