You can find the answer in the following wikipedia article. It involves the metric on the surface: https://en.wikipedia.org/wiki/Laplace%E2%80%93Beltrami_operator

Dear Carlos,

Thanks very much for your answer. Unfortunately, I have never studied Riemannian manifolds and so I don't know how to apply the material in the wikipedia article to the case I'm interested in. It seems clear that in my case n=2, dx^1=dx, dx^2=dy and g = 1+(\nabla h)^2. In my problem, the vector field J is everywhere tangent to the surface. Can you tell me how I can compute J^1 and J^2 given the x, y and z components of J?

Best regards,

Mark

Dear R. Mark Bradley ,

h=h(x,y), assuming that h ∈ (C2 (D) ), D is the domain of definition.

the corresponding vector field on this surface is described

by it's gradient vector:

∇h= (hx, hy, -1)= (hx)i+ (hy)j - k

where (i, j, k) is the orthonormal basis of R³.

Now, we compute the divergence of ∇h,

Div(∇h)=((∂hx /∂x)+(∂hy/∂y)+(∂(-1)/∂z) = hxx + hyy

Best regards

Dear Issam,

Thanks for your answer. Actually, I am looking for the divergence of a vector field J defined on the surface, not of h itself.

---Mark

Dear Mark,

Any $J$ vector field being tangent to the surface $h=h(x,y)$ has the form $J = (J^x,J^y,J^x\partial_x h+J^y\partial_y h)$, otherwise it will have nonzero component along normal vector and can’t be tangent to the surface. If so, you can write it as $J = J^x e_1 + J^y e_2$, where $e_1 = \partial_x + \partial_x h\partial_z$ and $e_2 = \partial_y + \partial_y h\partial_z$. Hence, $\nabla J = e_1 J^x + e_2 J^y$. For example:

$$ h = x^2 + y^2, $$ $$J = y\partial_x + z\partial_y + 2(xy+yz)\partial_z$$

then

$$\nabla J = \partial_x y + 2x \partial_z y + \partial_y z + 2y \partial_z z = 2y$$

BR,

Sergei

Dear R. Mark Bradley ,

You didn't mention the relation between the vector field J and

the surface S defined by h=h(x,y)!

Best wishes

Issam Kaddoura, you have built just normal vector to the surface. But the question is about the vector fields which are the fibres of the tangent bundle. As I realised, the subject is how to compute the divergence for some section of the bundle determined by the surface.

BR

Here is Python script to compute at any point. To change the surface and vector field, just modify functions

def h(x,y):

return x**2+y**2

def hx(x,y):

return 2*x

def hy(x,y):

return 2*y

def J(x,y,z):

return [y,z,2*x*y+2*y*z]

def Jx(x,y,z):

return [0,0,2*y]

def Jy(x,y,z):

return [1,0,2*x+2*z]

def Jz(x,y,z):

return [0,1,2*y]

def divJ(x,y):

return Jx(x,y,h(x,y))[0]+hx(x,y)*Jz(x,y,h(x,y))[0]+Jy(x,y,h(x,y))[1]+hy(x,y)*Jz(x,y,h(x,y))[1]

x = float(input("x: "))

y = float(input("y: "))

print(divJ(x,y))

Sergei Zuev

The question doesn't show the required vector field!

The family of gradients ( Normal bundle) determines a vector field covered the surface. The gradient itself represents the normal to the tangent plane at each point. Apart from the abbreviations, which J(x,y,z) formula you used?

BR

Issam Kaddoura

J(x,y,z) is just an arbitrary vector with components J^1(x,y,z), J^2(x,y,z) and J^3(x,y,z) which are some differentiable functions. The only restriction is that the J(x,y,z) is tangent to the specified surface.

BR

Sergei Zuev

The tangent bundle should be embedded in R4 and not R3.

Also, how can you do the calculations without the explicit formula of the tangent vector field related to h = h(x,y)?

In other words, what conditions on J(x,y,z) to guarantee the tangency condition?

BR

Issam Kaddoura

Any vector field J=(J^x,J^y,J^x*dh/dx+J^y*dh/dy) will be tangent to the surface. You can check it by dot product with normal ...

Sergei Zuev

So it is equivalent to study the VF generated by the gradients at each point.

Notice that at each point we have a vector space with dimension one that generated by the gradient, on the other hand, the tangent VF has dimension

= dim (the manifold), which needs higher dimensions to embed.

In our case, where the manifold M is a surface, it is more practical to consider the VF generated by the gradient itself.

Anyway, it is based on the main goal of Bradley request.

BR

I have modified the question so that the vector field has a name, J. The question now also specifies that J is everywhere tangent to the surface.

Dear Sergei,

Thanks again for all of your help. I have a question for you.

You write $J = J^x e_1 + J^y e_2$, where $e_1 = \partial_x + \partial_x h\partial_z$ and $e_2 = \partial_y + \partial_y h\partial_z$. How can the basis vectors e_1 and e_2 be differentials? J itself is not infinitesimal.

Best,

Mark

R. Mark Bradley

Dear Mark,

As I know, any vector field which is lying in the tangent space, can be presented by linear combination of infinitesimal operators. Because the vector field itself is an operator that transforms a function to another function. As for J, it is, of course, infinitesimal and its relevant form is follows:

J = J^x\partial_x + J^y\partial_y + J^z\partial_z

BR,

Sergei

I agree with Sergei Zuev where the vector field is defined by derivations

with a basis is { ∂/∂x, ∂/∂y} with dimension two and not three.

In fact, there are more than one approach to describe the tangent VF at any point

of the surface z = h(x,y).

If we consider the tangent VF J(x,y,z)= (J1,J2,J3) that is defined by the plane perpendicular to the gradient ∇h, then J(x,y,z). ∇h= (J1,J2,J3)(hx, hy, -1) = 0

implies J = (J1,J2, J1 hx+J2 hy ) for arbitrary J1 and J2 .

Div(J) = ∂J1/∂x + ∂J2/∂y + (∂J1/∂z ) hx+ (∂J2/∂z ) hy

Best regards