Resolving a Confusion About Implicit Function Theorem and Manifolds
Friday 27 February 2026 at 16:03
Visible to anyone in the world
"How do I show that a unit sphere is a smooth submanifold of ?"
That is the question I was pondering about when trying to solve exercise 3.1.1 of the book "Vector Calculus, Linear Algebra and Differential Forms - A Unified Approach" by John and Barbara Burke Hubbard. The problem asks for a proof of the above question, but trying to prove that the unit sphere is a submanifold of from the definition of a manifold is very cumbersome. For reference, this is how a manifold embedded in is defined as in the said book:
A subset is a smooth -dimensional manifold if locally it is the graph of a mapping expressing variables as functions of the other variables.
Here, "locally" means that every point has a neighborhood such that (the part of in ) is the graph of a mapping expressing of the coordinates of each point in in terms of the other .
So, in order to prove that the unit sphere is a manifold in , I would have to specify
a "chunk" of the (some variable, some variable) plane, such as
a "part" of the (remaining variable)-axis, such as
a function giving the graph of in the open neighborhood "(chunk) direct product (part of axis)". In this case, is the open neighborhood we're looking for, and is the graph of the function .
Quite tedious, but this is not over. Since this method only specifies an open neighborhood of when (in this case) , we have to also do the same thing using all of and ., and the functions:
Wow.
The implicit function theorem, on the other hand, gives a neat way to check whether
is a smooth manifold. We just have to check that the derivative of is surjective at every point on . In this case,
where , so since
by the implicit function theorem we get that is a smooth manifold in .
Looking back, the thing I was first misunderstanding was that, I mistakengly believed "I have to find a local representation of at every point first, then I can use the implicit function theorem". But doing so would defeat the whole purpose of the implicit function theorem, because by that point I would have already proved that is a manifold!
Also, I was having some difficulties grasping why we have to check the case for discussed earlier, why just being globally describable as two graphs is not enough to show that is a manifold. First, the case does not define an open neighborhood; it's just going to be a two dimensional plane. But at a deeper level, it taught me that manifold structure is a local condition at every point. This is exactly why the implicit function theorem is so powerful; we just have to check one condition and it automatically guarantees that a local representation as a graph exists at every point!
In the end, what felt like a technical shortcut turned out to be something deeper: the implicit function theorem is not a computational trick, but a precise machine that guarantees local manifold structure from a single global condition.
Resolving a Confusion About Implicit Function Theorem and Manifolds
"How do I show that a unit sphere is a smooth submanifold of ?"
That is the question I was pondering about when trying to solve exercise 3.1.1 of the book "Vector Calculus, Linear Algebra and Differential Forms - A Unified Approach" by John and Barbara Burke Hubbard. The problem asks for a proof of the above question, but trying to prove that the unit sphere is a submanifold of from the definition of a manifold is very cumbersome. For reference, this is how a manifold embedded in is defined as in the said book:
A subset is a smooth -dimensional manifold if locally it is the graph of a mapping expressing variables as functions of the other variables.
Here, "locally" means that every point has a neighborhood such that (the part of in ) is the graph of a mapping expressing of the coordinates of each point in in terms of the other .
So, in order to prove that the unit sphere is a manifold in , I would have to specify
Quite tedious, but this is not over. Since this method only specifies an open neighborhood of when (in this case) , we have to also do the same thing using all of and ., and the functions:
Wow.
The implicit function theorem, on the other hand, gives a neat way to check whether
is a smooth manifold. We just have to check that the derivative of is surjective at every point on . In this case,
where , so since
by the implicit function theorem we get that is a smooth manifold in .
Looking back, the thing I was first misunderstanding was that, I mistakengly believed "I have to find a local representation of at every point first, then I can use the implicit function theorem". But doing so would defeat the whole purpose of the implicit function theorem, because by that point I would have already proved that is a manifold!
Also, I was having some difficulties grasping why we have to check the case for discussed earlier, why just being globally describable as two graphs is not enough to show that is a manifold. First, the case does not define an open neighborhood; it's just going to be a two dimensional plane. But at a deeper level, it taught me that manifold structure is a local condition at every point. This is exactly why the implicit function theorem is so powerful; we just have to check one condition and it automatically guarantees that a local representation as a graph exists at every point!
In the end, what felt like a technical shortcut turned out to be something deeper: the implicit function theorem is not a computational trick, but a precise machine that guarantees local manifold structure from a single global condition.