# Reading: Conformal Parameterization (Due 11/28)

Your next reading assignment is Chapter 7 of the course notes, which will be critical for doing your conformal parameterization assignment. (Hand-in instructions are identical to previous reading assignments.)

## 14 thoughts on “Reading: Conformal Parameterization (Due 11/28)”

1. plav says:

I want to make a point clear in ch.7 before continue my reading…

In the pp.99-100, it says
“More precisely, if $\alpha$ is a 1-form on a surface $M$ then we can define the 1-form Hodge star $\star$ via
$\star \alpha(X) := \alpha(\jmath X).$”

Until this point, it is pretty clear, as the rotation is defined between the derivatives of the maps $f$ or $z$ (immersed surface and parameterization). Now, we are dealing with a differential 1-form on a plane and we are defining a hodge star based on the $\jmath$. I might be doing the computation wrong, but I think that in $\mathbb{R}^2$ a 1-form measures another vector by something similar to the dot product, right?

Let’s see an example:
$X = 2 \frac{\partial}{\partial x} + 3 \frac{\partial}{\partial y}$
$\jmath X = -3 \frac{\partial}{\partial x} + 2 \frac{\partial}{\partial y}$
$\alpha = 2dx + dy$
$\star \alpha = -dx + 2dy$

$\alpha(\jmath X) = ( 2dx + dy)(-3 \frac{\partial}{\partial x} + 2 \frac{\partial}{\partial y}) = -6 + 2 = -4$
$\star \alpha(X) = ( -dx + 2dy)(2 \frac{\partial}{\partial x} + 3 \frac{\partial}{\partial y}) = -2 + 6 = 4$

I feel there is something missing, as there is no map anymore to associate with the $\jmath$. Is this the right way to think about it?

1. yangxio says:

The fourth line regarding $\star\alpha$ is inconsistent. The Hodge star operator $\star$ is DEFINED via $\star\alpha(X) := \alpha(\mathcal{J}X), \forall X$, not by counterclockwise rotation. Note that one has to make a choice or convention regarding the orientation when defining the Hodge star. Adopting counterclockwise rotation of the one-form is inconsistent with the definition of $\mathcal{J}$ being counterclockwise rotation of the vector field. Based on your calculation, $\star\alpha = dx-2dy$. In algebra, $\star$ is the adjoint of $\mathcal{J}$. What is the adjoint of counterclockwise rotation (say represented as a $2\times 2$ matrix)? Well, it is the clockwise rotation, represented by the (conjugate) transpose of the former matrix.

The geometric interpretation of the definition $\star\alpha(X) := \alpha(\mathcal{J}X), \forall X$ is as follows: $\star\alpha$ uses $\alpha$ to measure the counterclockwise rotation of the input vector field. What is the equivalent one-form that measures the original vector field? Well, one should instead rotate the one-form clockwise and then measure the original vector field.

2. plav says:

I am really confused with the notion of the exterior differential. I will post all the questions in my mind now, from the basics …

1)Are all the following interpretations correct?
– $dx$ is a differential
– $dx$ is a base of a the differential form
– $dx$ is the exterior derivative applied to $x$.
– $df$ is a map from vectors to vectors on the map f

2)Are all the following interpretations correct?
– $ddx$ is the exterior derivative of a differential
– $ddx$ is the exterior derivative of a base of a the differential form
– $ddx$ is the exterior derivative applied twice to $x$.
– $ddf$ is a map from vectors to vectors tangent to the map f

3) What is the reason we always treated $ddx = 0$. What is the reason? Is it because it represents the change of x along dx? Is it this still valid for complex-valued k-forms?

4) I understand that u(X) represents the differential complex-valued 1-form measuring the base X. Therefore it results in a complex-valued 0-form (a complex number). In this case, the exterior derivative du(X) is equivalent to the directional derivative, and results in a complex-valued 1-form, right? What does it represent?

1. plav says:

5) in exercise 33, $\mathcal{A}(z) = -\frac{i}{2}\int_M d \overline z \wedge dz$.

I assume that $dz$ is the map that takes a vector from $M$ to a vector in $\mathbb{C}$. In this case, what is $d \overline z$? Should it be a map that takes a vector from $M$ to a vector in $\mathbb{C}$ that is reflected compared to the vectors generated by $dz$? For instance, if $dz(3 \frac{\partial}{\partial x} + 1\frac{\partial}{\partial y} ) = 5 + 3i$, should $d\overline z(3 \frac{\partial}{\partial x} + 1\frac{\partial}{\partial y} ) = 5 – 3i$?

2. Josh says:

1) Those looks ok.
2) I’m not sure what you mean for $ddf$, but the others look ok.
3) Pedantically, the reason $ddx = 0$ is that it is part of the definition of the exterior derivative. 😉 I think in lecture some reasons were give, for example that the curl of the gradient is always 0. This still applies for complex-valued k-forms.
4) Is $u$ a 0-form or a 1-form? In the first case, $u(X)$ does not make sense, but $du(X)$ is indeed the directional derivative (but with complex numbers).
5) $d\overline{z}(X) = \overline{dz(X)}$, so indeed your example is correct.

2. plav says:

I want to clarify some points regarding differential forms…

-In exercise 27 and 28, we assume that the “1-form” is actually a 1-form and not a vector-valued 1-form, right? In other words, can we support the proofs on basic notion of wedge product with real number multiplication, or should we extrapolate and consider the case of cross-products?

-What is the difference between a vector-valued k-form and the map f and df we have being using to find the surface and tangents? I am confused because p. 100 says that “we viewed map $f \rightarrow \mathbb{R}^3$ as an $\mathbb{R}^3-valued \space 0-form$ and its differential $df: TM\rightarrow \mathbb{R}^3$ as an $\mathbb{R}^3-valued \space 1-form$.” However, the initial notion of the differential form was associated with measurement, and I do not see how these maps measure geometry… they actually are rules to generate geometry.

-What is TM? It is used in the image of p.99 and cited in p.100, but I haven’t found further explanations…

1. Josh says:

1) Yes, 27 and 28 are for real valued 1-forms.

2) Good question. $f$ is a 0-form, it just gives the positions of the points of the manifold in $\mathbb R^3$ (no notion of measurement here). Since $df$ is a 1-form, there is a notion of measurement, if $X$ is a tangent vector in $M$, then $df(X)$ gives a vector-valued measurement which is the physical direction $X$ is pointing in three space.

3) TM is what as known as the tangent bundle. Intuitively, it is the vector space of all vectors tangent to a given point on the manifold M.

3. plav says:

I am also confused with the notion of 2-forms in $\mathbb{C}$.

1) For example, if I have $\alpha = 3 + i$ and $\beta = 1 + 3i$
If I treat it as a real 2-form in the coordinates of the complex space:
(a) $\alpha \wedge \beta = 8 dx \wedge dy$
But if I treat as a simple multiplication of complex numbers,
(b) $\alpha \wedge \beta =10i$

2) Besides, what is $i(\alpha \wedge \beta)$?
(a) $= (i \alpha \wedge \beta) = -6dx \wedge dy$
(b) $= (\alpha \wedge i \beta) = 6dx \wedge dy$
(c) $= (i \alpha \wedge \beta) = (\alpha \wedge i \beta) = i(10i) = -10$ (the order does not matter)
(d) None of the options above.

1. Josh says:

In this case, $\alpha$ and $\beta$ are both complex valued 0-forms, so $\alpha \wedge \beta = (3+i)(1+3i) = 3 + i + 9i – 3 = 10i$. In the other case you mention, $i\alpha \wedge \beta = i\alpha\beta = -10$. You should not be getting any $dx \wedge dy$ terms unless $\alpha$ and $\beta$ have $dx, dy$ terms. Does this makes sense?

1. plav says:

Yes in terms of computation, but now 0-form, 1-form, 2-form all will be represented by the same a + bi…. so I get completely lost with the basis.

1. Josh says:

Ok, I think I see what your problem is. $a + bi$ can mean different things depending on what $a$ and $b$ are. For example, if $a = dx$ and $b = dy$, then $a + bi$ is a 1-form. In your example above, your $a$s and $b$s are scalar values, so you instead have 0-forms. You have to carefully look at the contest to determine what kind of form a complex expression is.

4. intrepidowl says:

I don’t understand the use of notation on page 102 in the section on Conformal Parameterization.
When we say $$\star dz=idz,$$ why is this is not $$d\star z=idz?$$ The description throughout chapter 7 centers around the fact that “rotating then mapping is the same as mapping then rotating,” but the statement seems inconsistent with that fact.

1. Josh says:

If $z : M \to \mathbb C$, then $dz$ maps tangent vectors to $M$ to elements of $\mathbb C$, which can also be thought of as a vector. There are two notions of rotating by $\pi/2$, rotating the tangent vector before mapping or rotating the mapped complex number. The first is done by the hodge star and is written as $\star d z$ and the latter is done by multiplying by $i$ and is written as $i dz$. A conformal map is when these two perspectives are consistent: $\star dz = idz.$
The confusing part is that $\star$ and $i$ are written in identical parts of the equation, even though they are doing very different things from a type-theory perspective. If we add in the tangent vector $X$, then the equation becomes
$(\star dz)(X) = dz({\mathcal J} X) = i(dz(X))$,
where $\mathcal J X$ is $X$ rotated by $\pi/2.$ This last equation better reflects “rotate then map” and “map then rotate.”