Embeddings (how to prove them exactly)

Question

For which of the following sets is the statement: '$A$ can be embedded in $B$' true?

I can try to decide this intuitively but don't know if I'm right, and surely don't know how to formally prove it.

(i) $A = S^1 \times \mathbb{R}$, $B = \mathbb{R^2}$

$S^1$ with a point removed is homeomorphic to $\mathbb{R}$, so for a $x \in S^1$ we have $(S^1 - \{x\}) \times \mathbb{R}$ is homeomorphic to $\mathbb{R^2}$. Therefore it seems $A$ is to 'large' to be embedded in $R^2$, so I think this is false.

(ii) $A = S^1 \times S^1 \times S^1$, $B = \mathbb{R^4}$

Because $A$ is a compact topological manifold it can be embedded in $\mathbb{R}^N$ for some $N$. $A$ is the product of the torus and $S^1$, and because the torus can be embedded in $\mathbb{R}^3$ it feels like $A$ can be embedded in $\mathbb{R}^4$ so I think this is true.

(iii) $A = S^2$, $B = S^1$

I can remove two antipodal points from $A$ and it stays connected, but when I do that in $B$ it loses the topological property of connectedness. This then also needs to hold for the image of $A$ in $B$ so I think this is false.

(iv) $A = M$, $B = P^2$ (Where $M$ is the Möbius strip and $P^2$ is the real projective space of dimension $2$) The projective space of dimension two holds half the 'data' of the 2-sphere. When I think of the Möbius strip in $\mathbb{R}^3$ it seems to hold more data than the 2-sphere so I think this is false.

(v) $A = P^3$, $B = \mathbb{R}^6$

The space of all lines through the origin in $\mathbb{R}^4$ can be embedded in $\mathbb{R}^4$ itself, so then it must also be possible to embed it in $\mathbb{R}^6$ so I think this is true.

I'm sorry for the lack of rigor and hope someone can give some general advice to prove this formally.

Answer 1

If a manifold, $A$, does embed in $B$, the best way to show this is by constructing a map explicitly. In general, showing that a particular manifold does not embed in another is a difficult thing to do. However, in this problem, they can all be done directly.

(i) Your intuition on this one is incorrect. $S^1\times\mathbb{R}$ is a cylinder. This is topology, so we can stretch and deform our manifolds. Can you deform a cylinder so it lies flat? Try to write down the map $S^1\times\mathbb{R}\to\mathbb{R}^2$.

(ii) Try to construct a map. For starters, write explicitly the maps $S^1\to\mathbb{R}^2$ and $T^2\to\mathbb{R}^3$. Then see if you can generalize them. If you're having trouble, you can probably find this one online.

(iii) Your intuition is correct for this one. For a strategy, note that $S^2$ is a 2-manifold, while $S^1$ is a 1-manifold. Then use invariance of domain.

(iv) It turns out that $M$ does embed in $P^2$. Instead of thinking of $P^2$ as $S^2$ modulo the antipodal map, think of it as $D^2$ modulo the antipodal map on the boundary (i.e. the map $f\colon S^1\to S^1$ given by $x\mapsto -x$). Now can you "see" the mobius band? (It may help to draw $D^2$ as a square instead.)

(v) What you said here isn't exactly correct. The lines themselves may be embedded in $\mathbb{R}^4$, but that is different from the space of such lines. For example, $P^2$ is the space of lines through the origin in $\mathbb{R}^3$, but $P^2$ cannot be embedded in $\mathbb{R}^3$. Despite this, it is true that $P^3$ embeds in $\mathbb{R}^6$, however I don't know how to show this.

Answer 2

(i) true. While you are right that $S^1$ cannot be embedded in $\mathbb{R}$ (the images of $S^1$ and $S^1-\{x\}$, where $x$ is any point, have to be connected), $A$ can be embedded in $B$: consider the open set between two circles centered at the origin with radii $1$ and $2$ (first embed $\mathbb{R}$ into $(0,1)$, and then "flatten" $S^1\times(0,1)$).

(ii) true. Another way to see this: $S^1$ can be embedded in $\mathbb{R}^2$, so $A$ can be embedded in $S^1\times S^1\times \mathbb{R}^2$, and using (i) $S^1\times \mathbb{R}$ can be embedded into $\mathbb{R}^2$.

(iii) false. The image of $A$ as well as the image of $A$ with any finite number of points removed should be connected in $S^1$.

(iv) true. Think of the projective space as a square with opposite points being "equal", now cut it with two lines to get the Mobius strip. Also, consider this answer: Real projective plane and Möbius strip.

(v) true. But for a reason different from what you said. In general, any smooth real $n$-dimensional manifold can be embedded into $\mathbb{R}^{2n}$: http://en.wikipedia.org/wiki/Whitney_embedding_theorem. There you will find some expressions to construct such embedding explicitly.

Answer 3

[This began as a comment, but got a bit unwieldy.]

There are some good answers already, but perhaps it will help to have an answer that discusses the intuition behing these sorts of questions a bit more explicitly.

The first thing to emphasize is that you can improve your intuition! Here is an answer to a similar question with some remarks about how to do this. Just to repeat, and add to, some of the points made there:

One thing to do is learn some basic examples, such as $\mathbb R^2 \setminus \{0\} \cong S^1 \times \mathbb R_{> 0} \cong S^1 \times \mathbb R$. Note that the first isomorphism is just polar coordinates, which you surely are very familiar with, and the second is just log, identifying $\mathbb R_{> 0}$ with $\mathbb R$.

Question (iv) involves another basic example. E.g. if you look in old fashioned books, they will say that $\mathbb R P^2$ is obtained by gluing a cross-cap onto a disk. Gluing a cross-cap just means gluing a Mobius band along its boundary circle to a disk. (I'm not sure why the term "cross-cap" was used, but it seems to have been standard at one time.) If you know this then (iv) is immediate.

So basically, thinking about some particular examples will help a lot, because they come up again and again. Then you could think about some less familiar examples: what is the lowest dimensional Euclidean space in which you can embed $S^1 \times S^2$, or $S^2 \times S^2$.

Thinking in terms of dimension is very helpful in the contexts of manifolds; invariance of domain shows that you can't embed an open set from $\mathbb R^n$ into $\mathbb R^m$ if $m < n$, and so for $S^2 \times S^2$ this already puts a lower bound of $\mathbb R^4$. But if we could embed into $\mathbb R^4$, we would have a compact $4$-manifold ($S^2\times S^2$) in $\mathbb R^4$. Is this possible? If the answer isn't clear, think about the cases of $\mathbb R^1$, and $\mathbb R^2$. Can these contain a compact $1$-manifold, or a compact surface. Once you think you have an intuitive answer, can you prove it?

Connectedness was one tool used in your own ideas and the other answers; as I have just suggested, compactness is another.

One more idea is to think in $\mathbb R^4$ by treating the $4$th dimension as time. E.g. you have probably seen a picture of a Klein bottle immersed in $\mathbb R^3$; but this is not an embedding, as the "neck" of the bottle has to pass through itself. However, imagine now that each point has a time coordinate added to it. Then we separate the points where the intersection occurs by imagining they are at different times, and so get an immersion into $\mathbb R^4$.

One last comment: you should make sure you understand why your argument for (v) is quite wrong. Certainly $\mathbb R P^3$ can't be embedded into $\mathbb R^4$. Basically, thinking of $\mathbb R P^3$ as the space of lines through the original in $\mathbb R^4$ identifies it as a quotient of $\mathbb R^4$, and typically a quotient of some space won't embed back into it. (E.g $S^1$ is a quotient of $\mathbb R$, and you can't embed $S^1$ into $\mathbb R^1$.)