Why can some gradient fields not be simply connected?

In summary, a vector field in a simply-connected region is conservative if it is curl-free. In a non-simply-connected region, the vector field may still be considered a gradient field if the integral in a closed curve that winds around any holes in the domain is zero. This can result in a multiple-valued function that is single-valued on each branch of the manifold.
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Homework Statement
Reference: https://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/3.-double-integrals-and-line-integrals-in-the-plane/part-c-greens-theorem/session-72-simply-connected-regions-and-conservative-fields/MIT18_02SC_MNotes_v5.pdf
(Last 2 pages)
Relevant Equations
As I understand it, being "simply connected" means that the closed curves in the domain region contain some area(s) that are not in the domain. In other words, the region has got some hole(s) in it.
For example,
$$\left\langle
\frac x {r^3},
\frac y {r^3}
\right\rangle
= \nabla \left(
-\frac 1 r
\right)$$
where ##r=\sqrt{x^2+y^2}##, is a gradient field even though it is undefined at the origion. I get that it is physically possible since it is similar to the equation of the electric field of a positive charge place at the origion, and electric field is the gradient of the gradient of the potential function. But what is the mathematical explanation here? Thanks.
 
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Leo Liu said:
Homework Statement:: Reference: https://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/3.-double-integrals-and-line-integrals-in-the-plane/part-c-greens-theorem/session-72-simply-connected-regions-and-conservative-fields/MIT18_02SC_MNotes_v5.pdf
(Last 2 pages)
Relevant Equations:: As I understand it, being "simply connected" means that the closed curves in the domain region contain some area(s) that are not in the domain. In other words, the region has got some hole(s) in it.

For example,
$$\left\langle
\frac x {r^3},
\frac y {r^3}
\right\rangle
= \nabla \left(
-\frac 1 r
\right)$$
where ##r=\sqrt{x^2+y^2}##, is a gradient field even though it is undefined at the origion. I get that it is physically possible since it is similar to the equation of the electric field of a positive charge place at the origion, and electric field is the gradient of the gradient of the potential function. But what is the mathematical explanation here? Thanks.
In the notation I am familiar with, it is regions ( open, connected sets) that may be simply/not simply -connected; not vector fields. I think a standard definition is that in a non-simply-connected region, curves may wind around points not in the set, as you said. But a Vector Field is a function ( that assigns n-ples of vectors) and not a region. It may be defined on a region, but it is not itself a region. I believe the result you are considering here is that every vector field in a simply-connected region is conservative/ independent of path. Edit: Vector Field on a simply-connected region is Conservative if it is curl-free
 
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WWGD said:
In the notation I am familiar with, it is regions ( open, connected sets) that may be simply/not simply -connected; not vector fields. I think a standard definition is that in a non-simply-connected region, curves may wind around points not in the set, as you said. But a Vector Field is a function ( that assigns n-ples of vectors) and not a region. It may be defined on a region, but it is not itself a region. I believe the result you are considering here is that every vector field in a simply-connected region is conservative/ independent of path. Edit: Vector Field on a simply-connected region is Conservative if it is curl-free
Thank you. But if the domain of the vector field given is not simply connected, and for any curve C (including those that enclose the undefined origin), $$\oint_C \vec F \cdot d \vec r$$ can we still conclude that it is a gradient field? Why?
 
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It is not the hole a the domain that necessarily causes a problem. You can take any gradient in a simply connected domain and arbitrarily say that it is not defined in a hole in the domain. It would still be a gradient on the rest of the domain.
The issue is whether an integral in a closed curve that winds around the hole is non-zero. If the vector field has a zero integral on every closed curve that does not wind around the hole, it can be considered to be a gradient on a Riemannian manifold. The integrals on curves that wind around the hole gives a multiple-valued function which is single-valued on each branch of the manifold.
 
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FactChecker said:
It is not the hole a the domain that necessarily causes a problem. You can take any gradient in a simply connected domain and arbitrarily say that it is not defined in a hole in the domain. It would still be a gradient on the rest of the domain.
The issue is whether an integral in a closed curve that winds around the hole is non-zero. If the vector field has a zero integral on every closed curve that does not wind around the hole, it can be considered to be a gradient on a Riemannian manifold. The integrals on curves that wind around the hole gives a multiple-valued function which is single-valued on each branch of the manifold.
What do you call the case in which the path integral of a closed curve C enclosing an undefined region is 0 (eg the gravitational field of a neutron star)?
 
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Leo Liu said:
Thank you. But if the domain of the vector field given is not simply connected, and for any curve C (including those that enclose the undefined origin), $$\oint_C \vec F \cdot d \vec r=0$$ can we still conclude that it is a gradient field? Why?

(Added an ##=0## at the end which I think you meant to put).
Yes, this is true. Pick any point ##p_0## in your domain. For any other point, ##p##, let ##C(p)## by any path from ##p_0## to ##p## and define ##f(p)=\int_{C(p)}\vec{F}\cdot d\vec{r}.## The answer you get is independent of which path you pick by the condition on ##\vec{F}## and you can check that ##\vec{F}## is the gradient of ##f.##
 
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Leo Liu said:
What do you call the case in which the path integral of a closed curve C enclosing an undefined region is 0 (eg the gravitational field of a neutron star)?
That makes the line integral between two points path independent. Therefore, it can be used to define a well-defined potential for which it is the gradient.
 
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1. Why is it important to understand simple connectedness in gradient fields?

Simple connectedness is important in gradient fields because it helps us understand the topology of the field and how it behaves. It also allows us to determine if a field is conservative or not, which is crucial in many applications.

2. What does it mean for a gradient field to be simply connected?

A simply connected gradient field is one where any closed path within the field can be continuously deformed into a single point without leaving the field. In simpler terms, it means that there are no "holes" or disconnected regions in the field.

3. What are the implications of a gradient field not being simply connected?

If a gradient field is not simply connected, it means that there are closed paths within the field that cannot be continuously deformed into a single point without leaving the field. This can affect the behavior of the field and make it more difficult to analyze or use in applications.

4. How can we determine if a gradient field is simply connected?

One way to determine if a gradient field is simply connected is by using the fundamental theorem of calculus. If the integral of the field along any closed path is equal to zero, then the field is simply connected. Another way is by using the concept of homotopy, where we can continuously deform a path into a point without leaving the field.

5. What are some examples of gradient fields that are not simply connected?

One example is the electric field surrounding a point charge, where there is a singularity at the point charge and the field is not defined. Another example is the magnetic field surrounding a wire, where the field lines are closed loops and cannot be continuously deformed into a single point without leaving the field.

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