Topology, lemiscate not an embedded submanifold

[BrainHurts]

Homework Statement

Show that the image of the curve Let β: (-π,π) → ℝ² be given by β(t) = (sin2t, sint)
is not an embedded submanifold of ℝ²

Homework Equations

The Attempt at a Solution

So I'm not too great with the topology. I do see that β'(t) = (2cos2t, cost) ≠ 0 for all t. So β is a regular curve and as a result β is an immersion.

So here's what I have to work with.

A subset S of M is an embedded submanifold of M if:
- S is a smooth Manifold ( the image of (-π,π) under β is smooth)
-the inclusion map S to M is a smooth embedding

So this means that the topology of S (in our case β(-π,π)) is the subspace topology coming from M (in our case ℝ²)

We declare a set U[itex]\subset[/itex]β(-π,π) is open if and only if there exists a set W[itex]\subset[/itex]ℝ² (namely the open discs) such that

U = W [itex]\cap[/itex] S

So to be more specific let's take U = β(-π/6, π/6), this set includes the origin.

Let W be the open unit disc B(0,1) which includes the origin. I've done a lot of reading where if we remove the origin we have trouble, not really seeing how W [itex]\cap[/itex] S will look like the letter X. Any thoughts?

 

[mighty2000]

Firstly, great job on recognizing that the image of β is a smooth curve and that β is an immersion. This is an important step in showing that the image is not an embedded submanifold.

To show that the image of β is not an embedded submanifold, we can use the following theorem:

Theorem: Let M and N be smooth manifolds and let F: M --> N be a smooth map. If F is an immersion, then the image of F is an immersed submanifold of N.

In our case, we know that β is an immersion, so the image of β is an immersed submanifold of ℝ2. However, to be an embedded submanifold, the inclusion map from the image of β to ℝ2 must also be a smooth embedding.

To show that this is not the case, we can use the fact that an embedded submanifold must have a well-defined tangent space at each point. In other words, the tangent space at any point on the submanifold must be equal to the tangent space of the ambient space at that point.

However, for the image of β, this is not the case. Consider the point (0,0) on the image of β. The tangent space at this point is given by the span of the vector (0,0) (since the derivative of β is zero at this point). However, the tangent space of ℝ2 at the point (0,0) is given by the entire ℝ2. Therefore, the tangent spaces do not match up, and the inclusion map is not a smooth embedding.

In conclusion, the image of β is not an embedded submanifold of ℝ2.