Complex coordinates $ds^2 = dzdz̄$ in 2d

Question

I have a very elementary question about complex coordinates in two dimensions. When we have a 2D Euclidean space,

$$ds^2 = dx^2 +dy^2$$

and we go to complex coordinates:

$$z = x + iy$$ $$z̄ = x - iy$$

We have $$ds^2 = dzdz̄$$

The differentials are related like

$$dzdz̄ = |det J| dxdy$$

which gives $dzdz̄ = 2dxdy$ ($J$ is the Jacobian). Now, what is the relation between these two equations for $dzdz̄$? We cannot equate them, right? Since that would mean $dx^2 +dy^2 = 2dxdy$, which would simply state $dx=dy$ which of course can't be true. It probably has something to do with the fact that $z$ and $z̄$ are related while $x$ and $y$ are not, but for some reason I'm still a bit fuzzy about this.

Answer 1

It helps to notationally distinguish between skew-symmetric exterior products and ordinary tensor products:
$$ dz \wedge d\bar z= (dx+idy)\wedge (dx-idy) = -2i dx\wedge dy $$ for the volume element, and $$ ds^2 =dx\otimes dx +dy \otimes dy\\ = \frac 12 d\bar z \otimes dz+ \frac 12 dz \otimes d\bar z\\ $$ for the metric, so $g_{zz}=g_{\bar z\bar z}=0$ and $g_{\bar z z}= g_{z\bar z}= \frac 12$.

Answer 2

Consider $z=x+i y$ and $\bar z=x-iy$ ($x$ and $y$ are real). It is easy to show that: $$ dz\land d\bar z=(dx+i\,dy)\land(dx-i\,dy) =dx\land dx+i \, dy\land dx -i \, dx\land dy-dy\land dy. $$ However, thanks to the wedge product properties, we are left with $dz\land d\bar z=-2i\,dx\land dy$: this is proportional to the area element of the plane (the $-2i$ is sometimes dropped, see e.g. this and this). You can find more on the Wiki page for Complex differential form and on several MathSE posts, e.g. How many differential forms on the complex plane?.

Edit: As stressed in Mike Stone's answer, $ds^2$ is just a symbol to indicate the metric tensor $g$: it is not a 2-form (like the area element), it is a symmetric rank-2 tensor. There is no precise mathematical meaning cooked within the symbol $ds^2$ ($ds^2=2s\,ds$ or maybe $ds^2=ds\land ds=0$?), it can be conveniently seen as an alternative name for $g$. However, in relativity, people may sometimes attribute to $g$ and $ds^2$ different interpretations. See, e.g., this PSE answer and What is the difference between the metric and the invariant interval?.

Answer 3

The other answers are perfectly fine, but I think they might be a bit too advanced for what you're looking for (if I could read your misunderstanding correctly).

The difficulty you're having is with the fact that those two different expressions use different meanings of the products of differentials. One of them is what we call a tensor product (denoted $\otimes$) and the other is what we call a wedge product ($\wedge$). The wedge product is the antisymmetrized tensor product and it is only defined for some specific kinds of tensors, known as forms. For example, $$\mathrm{d}x \wedge \mathrm{d}y = \mathrm{d}x \otimes \mathrm{d}y - \mathrm{d}y \otimes \mathrm{d}x.$$ (There is a chance I'm missing a proportionality factor, but I think this is the correct expression).

When writing $\mathrm{d}s^2$, what you mean is that you are writing a symmetric $(0,2)$ tensor. This means you're using the $(0,1)$ tensors $\mathrm{d}x$ and $\mathrm{d}y$ (for example) to build a larger tensor. The correct expression is $$\mathrm{d}s^2 = \mathrm{d}x \otimes \mathrm{d}x + \mathrm{d}y \otimes \mathrm{d}y.$$

Now, when we are performing integrals, we are dealing with other types of tensors. Namely, we always integrate forms, which are completely antisymmetric $(0,q)$-type tensors (i.e., all indices are covariant). The natural product of forms is the wedge product. Hence, your last expression actually reads $$\mathrm{d}z \wedge \mathrm{d}\bar{z} = |\det J| \mathrm{d}x \wedge \mathrm{d}y.$$

Hence, the main point in your confusion is that there are two different products of differentials written in the same way. One of them is used to build tensors, the other is used to build forms.