A question about an 'extremal' surface

[Antarres]

I'm not sure this is a differential geometry question, but I think it is.

In general when we have a hypersurface(or in case of 3D space just a surface) it is defined with an equation $f(x^a)=0$ for some function $f$. Then the normal vector is the gradient of this function, if we want an unit vector we normalize it.

But what if this gradient is zero on the whole surface? How do we define the normal vector then? I called this surface 'extremal' because in black hole theory, horizons have this property in extremal solutions. But horizons are null hypersurfaces and we can avoid this for example using Killing vectors(because rigidity theorems tell us a Killing vector generates this surface as well).

But what do we do for an arbitrary surface?

Edit: By mistake I posted this in differential equations instead of differential geometry... I apologize for the mistake, the question can be moved to differential geometry maybe?

 

[andrewkirk]

A normal vector is defined relative to a surface, not relative to a function.

If hypersurface S is a submanifold of differentiable manifold U (the space in which S is embedded) then at point $p\in S$, the vector $v$ in the $T_p(U)$, tangent space of U at p, is a normal vector of S iff v is orthogonal to every vector in $T_p(S)$ (the tangent space of S at p). Note that this definition does not refer to any function f.

In some cases, we may be able to define the surface, locally if not globally, by another function, that enables us to find a normal vector.

Consider the following example. The bump function is infinitely differentiable at $x=1$ with all derivatives zero, but it is not constant in the neighbourhood of that point. We can piece together parts of that function to make a $C^\infty$ function g that is only zero at $x=1$ and has all zero derivatives there.

Now define function $f:\mathbb R^3\to \mathbb R$ by $f(x,y,z)=g(z)$. The hypersurface S defined by $f(x,y,z)=0$ is the plane $z=1$, and f has zero gradient everywhere on S. But we can also define the hypersurface by $h(x,y,z)=0$ where $h(x,y,z)=z-1$. The function h has nonzero gradient everywhere. Hence we can find vectors normal to S by using function h instead of function f.

In some cases there may be no global function without zero gradients to define the hypersurface but we may be able to make a patchwork of such functions that define it locally in each patch.

If we cannot do that at a point p then I suggest the answer is what simply follows from the definition of a normal vector. That is that every vector in $T_p(U)$ is normal to S, because every such vector has a zero inner product with every vector in $T_p(S)$.

For a black hole horizon, that feels intuitively correct to me. But I don't know much about black holes.

 

[Antarres]

I'm not sure about your example of a bump function, that is, I don't understand what you mean by piecing it together to put up the function $g$.

I understand that principally, you don't have to define the normal vector with respect to a function, but this function defines a surface, and it's standard that a gradient of this function is always normal to the surface. That doesn't stop it from being a zero vector, which is always normal to every vector, but in my eyes, every surface should have a normal vector defined, so this seemed vacuous to me. I was asking in a sense, when this surface is defined through this function, which is one common way to define it, how can I know all the tangent vectors and define normal by taking a vector orthogonal to them. Usually it's the other way around, I first find the normal, and then define a basis of vectors all orthogonal to this normal vector.
But while thinking about it, I came across this type of derivation. I'd like you to review it if you can.

Let's say, that a function $f(x,y,z) = 0$ defines a surface, and let's represent this as $f(\textbf{r})$, where $\textbf{r}$ is the position vector correspoding to the point $(x,y,z)$. Let's take a point $\textbf{r}_0$ on this surface, and take a random point $\textbf{r}$, also on the surface. Then $f(\textbf{r}) = f(\textbf{r}_0) = 0$ by definition of $f$. Then let's take that this point $\textbf{r}$ is infinitesimally close to $\textbf{r}_0$ and expand $f$ in Taylor series around $\textbf{r}_0$. Then since both $f(\textbf{r}_0)=0$ and it's gradient is also zero there(by hypothesis), we have that:
$$(\textbf{r}-\textbf{r}_0)^TH(\textbf{r}-\textbf{r}_0) + O((\textbf{r}-\textbf{r}_0)^3) = f(\textbf{r})=0$$
Where $H$ is the Hessian matrix.
But since this difference of vectors forms a tangent vector on this surface(they're infinitesimally close), we can say that $(\textbf{r}-\textbf{r}_0)^TH$ is the normal vector to this tangent vector. And since this $\textbf{r}$ is arbitrary, it's a vector normal to every tangent vector, hence it's a well defined normal.

Is this the correct reasoning? Now, obviously, this works in 3D Euclidean space, maybe it should be corrected in case of a more complicated manifold, I'm not sure, but it was an idea.

Edit: commenting on your further examination of the bump function, surely you can replace it by function $h$ as you said, but is it the case that you can always make such a substitution for an appropriate function that doesn't have a gradient zero?

 

[andrewkirk]

The Hessian approach won't achieve anything if the gradient is zero, as the Hessian will then also be zero. EDIT: That's wrong. Please ignore.

We'd need an example of an $(m-1)$-dimensional submanifold $S$ embedded in $\mathbb R^m$ such that:

1. $S$ is defined by $f(\mathbf x)=0$ for some $f:\mathbb R^m\to\mathbb R$
2. $f$ has zero gradient everywhere on $S$
3. $S$ cannot be alternatively defined either globally, or via a patchwork of local instances, as the null set(s) of alternative functions that do not have zero gradients on $S$
4. it seems reasonable to still expect that $S$ has a unique normal vector at every point

In the absence of such an example, there doesn't seem to be any problem to solve. I suspect that any hypersurface that satisfies 1-3 will be so pathological that we have no reason to expect it to have unique normals. It could be non-differentiable or even fractal.

 

[martinbn]

...
But what if this gradient is zero on the whole surface?
...

I don't think you mean this. In the context of relativity, which seems to be where you have seen it, the gradient is not zero. It has zero norm.

If the gradient is zero the function will be constant and will not define a surface.

 

[Antarres]

@martinbn In case of Kerr solution, the horizon is defined in the coordinates in the link by $\Delta=0$, and this is a quadratic function in the radial coordinate. In the extremal case, it has a double root. Taking the gradent(or just exterior derivative, to make it simpler), you'll find it's zero on the surface. Off the surface, of course it isn't zero, but that's not important if we want to look on the surface. And this holds for every quadratic function with a double root, but those functions can be redefined as @andrewkirk said, where we can take as a definition of the surface $\sqrt{\Delta} = 0$ and we're fine, we have non-zero gradient.

But I wasn't sure that every possible surface with this kind of property could be simply redefined, or if there is a way to work with it without redefinition in general. So I took as an example a function whose gradient is zero, but Hessian is non-zero(on the surface), which is the case in the case of a simple quadratic function with a double root. I wondered if such a method can generate a normal vector in general? Or in general I can always in every particular example redefine my defining function. @andrewkirk seems to say that this is always the case. I can accept that as an answer too, but I don't see a way to prove it.

 

[martinbn]

If a surface is given by a quadratic equation in one of the variables say $x^2-2ax+a^2=0$, and if it has a double root, the actual equation is $x-a=0$ and not $(x-a)^2=0$.

 

[Antarres]

Yeah, I'm aware of that, quadratic equation was just the simplest example I could think of. For example, if $f(x) = x^2 - ax + b$, and $f(x)=0$ is a surface defined through this equation, then for particular $a$ and $b$ this function will have a double root. In that case $\sqrt{f(x)} =0$ could be taken as the function that defines the surface. This is the replacement I mentioned, and that was mentioned by @andrewkirk earlier. I was just asking if we're considering examples that are not quadratic function, is this type of redefinition always possible? Or perhaps without resorting to it we could construct the normal like I tried with the Hessian(assumed to be non-zero on the surface, as is the case with quadratic example). I just didn't see a way to prove other examples couldn't exist, or that this type of replacement is always possible.