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)”

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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?

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.

Thanks for the explanation, Yangxio.

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?

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 $d\overline z(3 \frac{\partial}{\partial x} + 1\frac{\partial}{\partial y} ) = 5 – 3i$?

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.

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) 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.

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.

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?

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.

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.

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.

I think this was answered in class, but here’s another answer.

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

where \(\mathcal J X\) is \(X\) rotated by \(\pi/2.\) This last equation better reflects “rotate then map” and “map then rotate.”