# Exploring Non-Orientable Submanifolds in $\mathbb{R}^{m+1}$

• Kreizhn
In summary: For the given conversation, the summary is:In summary, the question is whether there exists a non-orientable compact boundary-less submanifold of \mathbb{R}^{m+1}. However, it is uncertain if this is a trick question or not, and there are conflicting ideas on how to approach the problem. One possible clue is the example of the Klein bottle, which is non-orientable but not a submanifold of \mathbb{R}^{m+1}. It is suggested that the answer may be no for all m, but a complete solution has not been found.
Kreizhn

## Homework Statement

Is there a non-orientable compact m-dimensional boundry-less submanifold of $\mathbb{R}^{m+1}$?

## The Attempt at a Solution

It should be noted that in the context of the situation, we've assumed that the manifolds we're dealing with are Hausdorff.

But I'm wondering if this isn't a trick question since all compact subspaces of a Hausdorff space are necessarily closed, and as such have a boundary.

And if it isn't a trick question, does anybody have any clues as how I can go about showing that there is/isn't one. I suspect that there isn't since m-dimensional submanifolds are often hypersurfaces that can be expressed as the preimage of a regular point of a homogeneous polynomial, and are therefore automatically orientable.

I've also thought about defining an orientation preserving map between the submanifold, say M, and $\mathbb{R}^{m+1}$ via $\{v_1, ... , v_m\} \rightarrow \{v_1, ..., v_m, N(p) \}$ where N(p) is a normal vector field at a point p mapping M to the tangent bundle. This would show that non-orientability in M would imply non-orientability of $\mathbb{R}^{m+1}$, a contradiction. The only problem here is that we would require N(p) to be everywhere non-vanishing.

Any ideas?

After looking around a bit, I've seen that you can have boundaryless compact manifolds, and so that relieves the issue of it being a trick question - though I'm still somewhat uncertain as to why the whole compact -> closed -> boundaryless doesn't work. But I'm still not sure as how to proceed with the actual question.

I'm unfamiliar with "Hausdorff", but a sphere is compact, closed, and boundaryless. It is orientable, however.

An example of a nonorientable compact, unbounded 2-manifold is the Klein bottle; but it is a submanifold of R4, not R3.

I suppose that might be a clue, but I'm not familiar enough with topology to tell for certain.

After Wiki-ing "Hausdorff", it appears that a manifold being Hausdorff ought to imply that it has no self-intersections. I think that completes the clue (Klein bottle has self-intersections in R3, but not R4).

It's been awhile since I've done any topology, so I must be screwing something up in my association between closed and having a boundary.

I'm not entirely sure as to what clue you're suggesting, since haven't you just shown that the Klein bottle under our demand that manifolds (and consequently submanifolds) be Hausdorff isn't a submanifold of $\mathbb{R}^3$? It can obviously be embedded in $\mathbb{R}^4$ via the Whitney embedding theorem, but that doesn't help since the codimension would be 2, and we want a hypersurface of codim = 1.

There is a piece missing from the question: namely, is this for all m, or for any m?

My clue would suggest that "no", there is no m for which one can satisfy the question.

I'm pretty sure the question pertains to any m, but your clue isn't really a clue in that case. I too suspect that the answer is no, but providing an example that doesn't work isn't really a clue.

It's like saying

"Is there a countable non-standard model of arithmetic?"

And the clue being "Well, here's a non-standard model, but it's uncountable, so it's a clue that the answer is no."

Edit: I apologize for the example, I couldn't think of anything more relevant at the time.

I meant, the Klein bottle might provide some clue as to how to prove the statement in higher dimensions by induction. I can't say for sure, though.

Well, I think we may have deliberately been given $\mathbb{R}^{m+1}$ since it is very easy to consider an ordered basis orientation over the tangent space. Thus, if I could define a non-vanishing normal vector, I could show that all such hypersurfaces are orientable - this is clearly not possible by the Hairy Ball Theorem. Thus, I need to find some other way to find a natural orientation on any submanifold.

## 1. What is a non-orientable submanifold?

A non-orientable submanifold is a subset of a higher-dimensional space (such as $\mathbb{R}^{m+1}$) that behaves like a lower-dimensional space (such as a plane or a curve) but cannot be assigned a consistent orientation. This means that there is no consistent way to define a "right-hand rule" for the tangent vectors at every point on the submanifold.

## 2. How do you explore non-orientable submanifolds?

Exploring non-orientable submanifolds involves studying their properties and characteristics, such as dimensionality, curvature, and topology. This can be done through mathematical analysis and visualization, as well as by applying techniques from differential geometry and topology.

## 3. What are some real-world applications of studying non-orientable submanifolds?

Non-orientable submanifolds have many applications in physics, engineering, and computer graphics. For example, they can be used to model the behavior of physical systems with non-trivial topologies, such as Möbius strips or Klein bottles, and to create more realistic and visually interesting 3D objects in computer graphics.

## 4. Can non-orientable submanifolds exist in lower-dimensional spaces?

No, non-orientable submanifolds can only exist in higher-dimensional spaces. This is because lower-dimensional spaces, such as the 2D plane, can always be consistently oriented using a "right-hand rule", whereas higher-dimensional spaces require more complex rules to define orientation.

## 5. How does the concept of non-orientability relate to the concept of orientation-reversing transformations?

Non-orientable submanifolds can be thought of as spaces that cannot be consistently oriented under any transformation, including orientation-reversing transformations. This is because orientation-reversing transformations change the direction of tangent vectors, making it impossible to define a consistent "right-hand rule" for the submanifold.

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