The written portion of Assignment 2 can be found below. It takes a look at the curvature of smooth and discrete surfaces, which we’ll cover in the next few lectures. **Note that you are required to complete only two problems from each section!**

## 26 thoughts on “Assignment 2 (Written): Investigating Curvature”

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Would you mind posting the tex source file, and the original questions in the homework pdf file so that it is self-contained instead of something like “do exercise n in chapter m of the course note” (unless you really want us to read the course notes)? It will be more convenient for those texing the solutions.

Thank you.

Hi Yangxio,

I really *do* want you to read the course notes for those problems—there is important stuff in there! 🙂

Best,

Keenan

Regarding question 2: I could not find a definition of the Gauss map in the notes or in the slides posted online so far – is it simply the map from $f \left( \mathbb{R}^2 \right)$ to $\mathbb{R}^3$ which assigns to every point on the surface the corresponding unit normal vector?

Yep, that’s correct. You can find a brief discussion of the Gauss map in Section 2.1 of the notes (and there’s not much more to it than what you just said!).

Oh, how embarrassing – it’s indeed right there in section 2.1: “we can actually think of $N$ as a continuous map $N$ : M \to S^2$ (called the Gauss map) which associates each point with its unit normal”.

Thank you!

The chapter defines the metric induced by is a dot product of 2 vectors $df(A)$ and $df(B)$.

I am confused in the question 2(c). It asks us to compute the metric $g$ induced by $f$, but this requires not only the $f$, but also 2 vectors. As we do not have these vectors, are we:

– Supposed to provide a general formula with the coordinates of random vectors?

$A = \begin{bmatrix} u_1 \\ v_1 \end{bmatrix}$

$B = \begin{bmatrix}u_2 \\ v_2 \end{bmatrix}$

$g(A,B) = $ some formula using $u_1, v_1, u_2$, and $v_2$.

– Supposed to compute for any vectors we want?

In Euclidean Space, I think any surface inherits induced metric which assigns every p in surface an ordinary scalar product of X and Y, from the tangent space of the surface, in R^3 (i.e.〈,〉:TuM×TvM→IR). I wasn’t sure how we could go about describing X, Y however.

Yes, exactly: you should give an expression that can be applied to any two vectors; not just evaluate it on a specific pair of vectors.

When you see a definition like

\[

g(X,Y) := df(X) \cdot df(Y),

\]

and no particular \(X\) or \(Y\) has been specified, this implicitly means that “for

any\(X\) and \(Y\), this is what \(g\) does.”1) The reading states that a normal is a vector that $df(x).u = 0$. I think that any vector in the plane orthogonal to the tangent is going to satisfy this relation. However, the function in the slides $\left( \tfrac{\tfrac{\partial}{\partial}T}{\left |\tfrac{\partial}{\partial}T \right|} \right)$ is the one that provides a unique map to the normal.

2) With that said, the assignment says “compute the Gauss map $N$ induced by $f$”. Is there any difference between a Gauss map of $f$ and a Gauss map induced by $f$? The term induced seems to be relevant only in the sense that the normal is based on the tangent (that was pushed forward by $f$).

1) It’s a subtle difference in wording: any vector \(u\) is

normal to the surfaceif \(df(X) \cdot u = 0\) for all \(X\). Thesurface normalorGauss mapis in particular aunitvector normal to the surface. (That is perhaps why we use such a fancy word, “Gauss map,” specifically meaning a canonical choice ofunitnormal that is consistently oriented.)2) No, there is no difference between the Gauss map

of\(f\) and the Gauss mapinduced by\(f\). The term “induced by” is just a strong reminder that \(N\) comes from \(f\).I’m getting slightly confused by the reading when I try to relate it to what we have learnt in exterior calculus (for example, is there a way to think about the differential of an immersion as an exterior derivative?). The reading says ‘there will be a more algebraic treatment of this when we talk about exterior calculus’, but I couldn’t find it in the long version of the notes. When can I find this stuff in the language of exterior calculus?

We talked about this connection in our lectures on curves and vector-valued differential forms: for instance, rather than writing tangents of different length as \(a\gamma^\prime\) (for some \(a \in \mathbb{R}\)), we can write them as \(d\gamma(a\tfrac{d}{ds})\),

i.e., compute the exterior derivative of the map \(\gamma\) and apply it to a tangent vector on the domain \([0,L]\).Nothing changes for surfaces: if we take the exterior derivative \(df\) of an immersion \(f\), then for any tangent vector \(X\) on the domain, \(df(X)\) is tangent to the surface. The quantity \(df\) is a vector-valued 1-form (specifically taking values in \(\mathbb{R}^3\)).

For the Exercise 19, I am not sure the gradient of the total surface area w.r.t. the position of one of the vertices should be $\frac{1}{2}\sum_{ij \in E}(cot \alpha_{ij}+cot \beta_{ij})(f_j – f_i)$ or $\frac{1}{4}\sum_{ij \in E}(cot \alpha_{ij}+cot \beta_{ij})(f_i- f_j)$ ?

I think the direction of $(f_{i} – f_{i})$ matters.

Maybe the number $\frac{1}{2}$ or $\frac{1}{4}$ is not important?

Thanks!

The sign isn’t a big deal (for instance, when you write code, you run it once; if it’s going the wrong direction, you flip the sign!). But sure, you should try to get it right. An easy way to check whether you have the right sign is to consider what direction your gradient points for a vertex contained in only a single triangle (like at the “corner” of a mesh). Does this vector look like it

increasesarea? If so, it’s pointing in the right direction.As for the factor 1/2, also not a big deal, but sure, may as well figure it out. If I remember correctly (it’s been a while since I’ve done this calculation directly…) you start with a factor 1/2 from the area gradient, and then another factor 1/2 appears when you re-express things in terms of cotangents. But I’d double-check if I were you.

P9 of the written asks us to prove the gradient of a triangle with respect to a point P is 1/2 the perpendicular to the side S opposite to P. In class, we argued why the gradient has to be perpendicular to S. But I’m not sure what its magnitude means or why it’s half the magnitude of P. Could you give me a hint or fill me in on something from class that I missed? Thanks.

The area of a triangle is $A = \frac{1}{2}bh$.

@maliang’s hint is a good one. 🙂

Basically you want to ask: if I move the vertex a unit distance along the gradient, how much does the area change? What, therefore, is the rate of change in area per unit length?

For Q1, do we need to get the final numeric result or just the equation is OK?

You should just give a general expression; not a numerical result. (Because from there, anybody could plug in a value of $s$ to see what these quantities are at a particular point.)

In section 3.4 of the problem set, the variable $H_{ij}$ is used in the formula of $\mathcal{V}(r)$, but it isn’t defined anywhere. Only $H_i$ is defined for each vertex $i$.

Sorry about that; should be just

\[ H_{ij} := \ell_{ij} \theta_{ij}. \]

Regarding arguing geometrically for the gradient of a quantity, and in particular finding its magnitude once the direction has been found: the example in the assignment / in class (i.e. the gradient of the length of a vector) seems to be a special case since the length varies linearly (when moving an endpoint tangentially).

When the quantity whose gradient we wish to obtain geometrically does not vary linearly, are we allowed to compute the difference quotient and then take a limit, or is there another more geometric / less computational approach?

The whole point of the gradient is to find the “best linear approximation” of how a quantity is changing. For instance, even in the case of a segment, the quantity you’re taking the gradient is a nonlinear function of the location of the endpoints:

\[ \ell_{ij} := \sqrt{ (b-a) \cdot (b-a) }. \]

Though as you point out, following the gradient will take you along a straight trajectory in this case. Consider for instance the case of the angle made by two vectors. If you follow the gradient for more than just an instant, it will trace out a circular arc rather than a straight line. But you can still ask the same question: how does the angle change relative to the length of this circular arc? Your trajectory does not have to be straight in order for this kind of argument to work.

So, in the case of the length of a segment, the change in the length of the segment $\Delta l$ is related to the change $\Delta a$ in the norm of the endpoint $a$ by $\Delta l = \Delta a$ (when me move the endpoint $a$ tangentially to the segment).

In the case of the angle $\phi$ between two vectors $u$ and $v$, the change in the angle $\Delta \phi$ is related to the change $\Delta u$ in the norm of the vector $u$ by $\Delta \phi = f (\Delta u)$ for some function $f$ which is now nonlinear (I guess because the trajectory we get by following the gradient is not a straight line).

So at this point we have to compute the limit of that ratio $$\frac{\Delta\phi}{f(\Delta u)}$$ as $\Delta u$ goes to zero to obtain the magnitude of the gradient (i.e. compute the linear approximation to $f$ about $0$)? I’m essentially not sure about when it’s ok to end the geometric argument and just defer to calculus to evaluate the limit.

Basically the whole point of this assignment is to avoid reverting to calculus. 🙂

If you find yourself tending toward using limits, derivatives, etc., then you are working too hard!

You should just ask at a more basic level: if I move my point along the trajectory that most quickly increases this quantity, then

1. How far did the point move (along a line, arc, etc.)?

2. How much did the quantity change?

The magnitude of the gradient is how much the quantity changed, divided by how far the point moved (along the line, arc, etc., i.e.,

notthe point-to-point distance between start and end).“The magnitude of the gradient is how much the quantity changed, divided by how far the point moved (along the line, arc, etc., i.e., not the point-to-point distance between start and end).”

Ah ok ok, that’s a good way to do it – thanks!

And just to make sure I understand how to reconcile this with calculus’ definition of a gradient: we can find the gradient this way because, in the limit, moving the point along the gradient will be like moving along a straight line?