The Unapologetic Mathematician

Mathematics for the interested outsider

Immersions and Embeddings

As we said before, the notion of a “submanifold” gets a little more complicated than a naïve, purely categorical approach might suggest. Instead, we work from the concepts of immersions and embeddings.

A map f:M^m\to N^n of manifolds is called an “immersion” if the derivative f_{*p}:\mathcal{T}_pM\to\mathcal{T}_{f(p)}N is injective at every point p\in M. Immediately we can tell that this can only happen if m\leq n.

Notice now that this does not guarantee that f itself is injective. For instance, if M=\mathbb{R}^1 and N=\mathbb{R}^2, then we can form the mapping f(t)=(t-t^3,1-t^2). Using the coordinates t on M and {x,y} on N, we can calculate the derivative in coordinates:

\displaystyle f_{*t}:\frac{\partial}{\partial t}(t)\mapsto(1-3t^2)\frac{\partial}{\partial x}(f(t))-2t\frac{\partial}{\partial y}(f(t))

The second component of this vector is only zero if t itself is, but in this case the first component is 1, thus f_{*t} is never the zero map between the tangent spaces. But f(1)=f(-1)=(0,0), so f is not injective in terms of the underlying point sets of M and N.

Courtesy of Wolfram Alpha, we can plot this map to see what’s going on:

Self-Intersection

The image of the curve crosses itself at the origin, but if we restrict ourselves to, say, the intervals (-2,0) and (0,2), there is no self-intersection in each interval.

There is another, more subtle pathology to be careful about. Let M be the open interval (0,2\pi), and left f(t)=(\sin(t),\sin(2t)). We plot this curve, stopping just slightly shy of each endpoint:

Immersion

We see that there’s never quite a self-intersection like before, but the ends of the curve come right up to almost touch the curve in the middle. Going all the way to the limit, the image of f is a figure eight, which includes the crossing point in the middle and is thus not a manifold, even though the parameter space is.

To keep away from these pathologies, we define an “embedding” to be an immersion where the image f(M)\subseteq N — endowed with the subspace topology — is homeomorphic to M itself by f. This is closer to the geometrically intuitive notion of a submanifold, but we will still find the notion of an immersion to be useful.

As a particular example, notice (and check!) that the inclusion map of an open submanifold, as defined earlier, is an embedding.

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April 18, 2011 - Posted by | Differential Topology, Topology

6 Comments »

  1. [...] both of our pathological examples last time, the problems were very isolated. They depended on two separated parts of the domain manifold [...]

    Pingback by Immersions are Locally Embeddings « The Unapologetic Mathematician | April 19, 2011 | Reply

  2. [...] subspace topology from — then we say that is a submanifold of if the inclusion map is an embedding. If the inclusion is only an immersion, we say that is an “immersed submanifold” of [...]

    Pingback by Submanifolds « The Unapologetic Mathematician | April 20, 2011 | Reply

  3. [...] Armstrong: The Implicit Function Theorem, Immersions and Embeddings, Immersions are locally [...]

    Pingback by Fourth Linkfest | April 23, 2011 | Reply

  4. [...] map, and every point is a regular value. Its preimage is a submanifold diffeomorphic to . The embedding realizing this diffeomorphism is . The tangent space at a point on the submanifold is mapped by [...]

    Pingback by The Tangent Space of a Product « The Unapologetic Mathematician | April 27, 2011 | Reply

  5. [...] it is surjective, and if every point is a regular point of . Despite the similarity of the terms “immersion” and “submersion”, these are very different concepts, so be careful to keep them [...]

    Pingback by Submersions « The Unapologetic Mathematician | May 2, 2011 | Reply

  6. [...] But this time it’s not going to be quite so general; we will only extend our notion of an embedding, and particularly of an embedding in codimension [...]

    Pingback by Orientation-Preserving Mappings « The Unapologetic Mathematician | September 1, 2011 | Reply


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