# Definition for shape operator

1. Feb 26, 2014

### Jhenrique

In wolframpage there is follows definition for shape operator in a given point by vector v:

I think that this equation means: $$S(\vec{v})=-\frac{d\hat{n}}{d \vec{v}}$$ correct, or not?

If yes, of according with the matrix calculus (https://en.wikipedia.org/wiki/Matrix_derivative#Numerator-layout_notation) the derivative of a vector in the space wrt another vector in the space results in a 3x3 matrix:

But, the shape operator is a 2x2 matrix. Conclusion, there is something wrong in my interpretation, what is?

2. Feb 27, 2014

### Staff: Mentor

I don't think so. $\hat{n}$ is a unit vector, while N is a normal vector, so isn't guaranteed to be a unit vector.
The matrix in the image is not a 3 X 3 matrix. It is an n X n matrix.

3. Mar 1, 2014

### Jhenrique

Moreover, the shape operator of a point is a scalar! But all my deduction results in a tensor of rank 2 or 3... I don't know what I'm doing of wrong.

4. Mar 1, 2014

### Staff: Mentor

You asked whether -DvN was the same as $-\frac{d\hat{n}}{d\vec{v}}$. I said I didn't think so, as $\hat{n}$ is a unit vector and N is not.

You also show what you called a 3 X 3 matrix, that is obviously meant to be m X n. (In my earlier reply I said that it was n X n, but on closer inspection, I see that it has m rows and n columns, so is m X n.)

The wiki link that you provided is a summary of matrix calculus, and doesn't have anything to do (that I can see) with shape operators.

If you want better answers, provide clearer descriptions of what you're asking.

5. Mar 1, 2014

### Jhenrique

This is an observation almost irrelevant when compared with the unique and principal question propossed in this topic for my, that is specifically about the nature (the rank) of a computation.

The image is general fomrula! In the 3D space it's become a 3x3 matrix, like I said, what is obvious! Another times you focused in the question of 2nd order. I don't think that this kind of observation is irrelevant, SINCE the principal question is answered, thing that you didn't did.

It's depends. If -DvN means the derivative of a vector wrt another vector (and I didn't affirm this, in my first post, I assumed that maybe can be this) so have to do with matrix calculus and with the link that I provided.

I can not be being clear with the my doubt, but you seem haven't will of to connect the points.

By definition, the shape operator is a 2x2 matrix (I hope that you know this) but the shape operator of a point is a scalar. I'd like of an explanation of how this process (from rank 2 to rank 1) happens. Wich is the definition of shape operator of a point? How compute this? What this formula -DvN means operationally? These are the principal questions in this topic.

6. Mar 1, 2014

### micromass

To define the shape operator of a surface $S$, we first need a unit normal field on the surface. This is a smooth function $N:S\rightarrow \mathbb{R}^3$ such that $\|N(p)\|=1$ for each $p\in S$ and such that $N_p\bot T_pM$ for each $p$. Such a unit normal field exists (essentially by definition) if $S$ is orientable.

Since $\|N(p)\|=1$, we see that $N(p)\in S^2$. Thus $N$ defines a smooth function between $S$ and the unit sphere $S^2$. The shape operator is by definition the differential of $N$.

So given a certain point $p\in S$, we have the shape operator which is $d_p N: T_p S\rightarrow T_pS^2$. We see also that $T_pS^2$ are all the vector orthogonal to $N(p)$, and this is exactly $T_pS$. So the shape operator actually is an operator
$$d_p N:T_pS\rightarrow T_pS$$
between $2$-dimensional spaces.

So in this case, the notation $d_pN$ (or as you write $D_v\mathbf{N}$) is the differential of a map between surfaces.

7. Mar 2, 2014

### Staff: Mentor

How is it irrelevant? You asked whether two things were the same, and my response was that they were not, and I gave a reason for why I thought so.

Yes, it obviously is a general formula, but if you describe something that is clearly an m X n matrix as a 3 X 3 matrix, one can only conclude that either you're being careless or you don't know what you're talking about. This is what I was talking about, about not being clear in what you're asking. We aren't mind readers here. If your question isn't clear, it's harder to figure out what you're asking.

Nowhere in your original post did you include a definition or description of a shape operator, or at least a link to the wolfram page that was discussing them. Instead, you showed two equations that are clearly different, and then a link to a wiki page on matrix calculus. Without knowing what a shape operator was, what you presented seemed incoherent to me.

It's not that I dont' have the will to "connect the dots", it's that I'm unable to read your mind.

It would have been better if you included this information in the first post of this thread.

8. Sep 8, 2014

### WWGD

It all comes down to knowing the (version of the) covariant derivative of a vector field along another vector field; the output here is a vector field, which is not necessarily a tangent vector field, i.e., it does not necessarily live in the tangent space to the point.
Under a choice of Riemannian metric, we use a vector space decomposition of the ambient space $\mathbb R^n$ into a sum of the tangent space and the normal space (the metric determines the normal space) then the covariant derivative here is the projection of the output vector field into the tangent space. In our case (see below for notation) of the Shape Operator we consider the covariant derivative of a normal vector field (which exists locally, since every manifold is locally-orientable ) in the direction of a tangent vector v, at a point p. This is $S_v(p)$

In coordinates, the covariant derivative of the vector field w in the direction of the vector field w at a point p : $∇_v w$, using the curve definition of tangent vector, we find a curve c(t) with c(0)=p and c'(0)=v . We then have :

$$∇_v w := w(c)'(0)$$

Another way of computing $∇_v w$ is by expressing w in terms of a basis , say ${E_i}; w= ∑w_iE_i$
and then we just have:

$$∇_v w := v[w_i] E_i$$ (remember that vector fields act as derivations/directional derivatives ; this v[w_i] is just the directional derivative operator acting on vector fields).

Since you ultimately get a linear map from $T_pM \rightarrow T_p M$, you can determine this map by evaluating at the basis vectors on $T_p M$ .

I will give you a few concrete examples as soon as I have more time.