Why do we need wedge product for $dxdy$ to polar coordinates when $dx$ and $dy$ are perpendicular to each other?

Question

Why can't we convert $dx$ and $dy$ to polar and straight out multiply the answer to get the area element? By multiply I mean, taking x=rcos(theta) and y=… and then taking differentials dx and dy and multiplying them. Answers here point out that wedge is supposed to be used but this is just a rectangle so do we still need it and why? Thank you.

Answer 1

As was pointed out in the comments to this question the multiplication of numbers and of differential forms are different. E.g. number multiplication obeys the commutativity rule $AB=BA$, while multiplication of differentials does not in general. That is why we use the $\wedge$-sign to indicate this different nature of multiplication.

Differentials have an anticommutativity rule: $dx\wedge dy=-dy\wedge dx$. Here is a conceptual way to understand this. Differentials like $dx$ with one component are used in measuring length, differentials like $dx\wedge dy$ with two components are relevant to measuring area. If you swap $dy\wedge dx$ then you change the "orientation" of the area in question and the area is measured with negative weight.

This comes in handy if you think about very elementary situations: If you have an L-shape that it is obtained by removing a rectangle from a larger one at one of the corners you want to compute the area of the L-shape as the difference of the areas of the two rectangles, i.e. the sum of the two areas but with the area of the smaller rectangle having negative weight. The anticommutativity of the $\wedge$-product helps in such situations and is useful there. Because of this, multiplication of differentials is different in nature than multiplication of numbers.

If you want to see how $\wedge$-products work out computationally in an explicit example you can go to the link provided by peek-a-boo or wikipedia.

You can also compare with multiplication of matrices or of quaternions: for these commutativity doesn't hold in general either. However the motivation and details why and how commutativity is violated in these cases is different from differentials.

Answer 2

This becomes less mysterious if you just consider the situation in pure linear algebra.

Suppose you are in the plane $\mathbb{R}^2$ and you want to measure areas; suppose you want to measure the area spanned by the orthogonal vectors $\hat{i}=(1,0)$ and $\hat{j}=(0,1)$. One way way to do this is to just multiply their sides to get $1$. Now suppose you express $\hat{i}$ and $\hat{j}$ in a different basis. Let this other basis constitute the vectors $v=(1,1),w=(0,2)$. Write the same old vectors in terms of this basis: $$\hat{i}=v-\dfrac{w}{2}$$ $$\hat{j}=\dfrac{w}{2}$$

Your goal now is to write the Area spanned by $\hat{i},\hat{j}$ denoted by $A(\hat{i},\hat{j})$ in terms of area spanned by $w,v$ denoted by $A(v,w)$ (notice how this is analogous to writing the area differential $dxdy$ in terms of $drd\theta$). Next you put the components of $\hat{i},\hat{j}$ (w.r.t. $v,w$ basis) in a matrix and take the determinant to compute $A(\hat{i},\hat{j})$ in terms of $A(v,w)$; this should hopefully be an intuitive statement. Draw a picture for this to become obvious:

$$A(\hat{i},\hat{j})=\text{det} \bigg(\begin{bmatrix} 1 & 0 \\ 1/2 & 1/2 \\ \end{bmatrix}\bigg)A(v,w)=\dfrac {A(v,w)}{2}=1$$

This could alternatively be written in terms of the wedge product $\wedge$ since it exactly mimics the behavior of the determinant: $$A(\hat{i},\hat{j})=\hat{i} \wedge \hat{j} = (v-\dfrac{w}{2}) \wedge (\dfrac{w}{2}) = \dfrac{1}{2} v \wedge w - \dfrac{1}{4} w \wedge w= \dfrac{1}{2} Area(v,w)$$

In exact analogy you do this for the polar/cartesian differentials. The chart map induces a basis and dual basis at each point on your manifold, hence you can do the above at each tangent space. so we literally just repeat the steps:

$$dx=cos(\theta)dr-rsin(\theta)d\theta$$ $$dy=sin(\theta) dr +rcos(\theta) d\theta$$

$$dx \wedge dy = A(dx,dy)=\text{det} \bigg(\begin{bmatrix} cos(\theta) & -rsin(\theta) \\ sin(\theta) & rcos(\theta) \\ \end{bmatrix}\bigg)A(dr,d\theta)=r dr \wedge d\theta$$

The key idea is this: If you want a way to compute areas under arbitrary change of bases you need to use the determinant rather than the native rule of "multiplying side lengths" since this only works in the canonical basis. But in differential geometry we work in many different and crazy charts which induce a change of basis at every point and hence requires a notion of area that works in arbitrary basis. But as you can see, this is pure linear algebra and it is not unique to differential geometry.

Also note: technically $dx$,$dy$ are covectors and not vectors; this has its own reasons but doesn't affect the main point of this post.

Edit:

You're on the right track to understanding this but there's one thing you're missing.

All what you say is true, $x+dx=rcos\theta +drcos\theta -rsin\theta d\theta$ and the same for $y$; now comes the key idea: Notice that now you are in the $r-\theta$ plane. Hence the elements $dx$ and $dy$ (written in terms of increments of $dr$ and $d\theta$) are oblique in the $r-\theta$ space. So to clarify: $dx$ does not point in the right direction, and $dy$ does not point in the up direction since you are in the $r-\theta$ space. This can be seen since you have written them in terms of $r$ and $\theta$. In the $r-\theta$ plane $dr$ is to the right and $d\theta$ points up. This is what i've been trying to tell you all along: You are now seeing things from a different basis. What was once right and up is no longer up and right and hence you can't just naively multiply their sides, you must use the determinant. Basically the problem is that whenever you express $dx$, $dy$ in terms of $dr$, $d\theta$ then you are trying to compute the area spanned by $dxdy$ in the $r \theta$ plane and in terms of $dr$ and $d\theta$ (which are along right and up directions in $r-\theta$ plane). This is equivalent to the problem I adressed in my post.

So I suggest you do the following: draw explicitly $dx$ and $dy$ in the $r-\theta$ plane and you should immediately realize where your problem lies.