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A polynomial map $f: \mathbb{C} \to \mathbb{C}^2$, $t \mapsto f(t):=(f_1(t),f_2(t))$ is called an embedding of $\mathbb{C}$ in $\mathbb{C}^2$ if $\mathbb{C}$ is isomorphic to its image under $f$, see A. van den Essen, page 2. (By a polynomial map we mean that $f_1(t),f_2(t) \in \mathbb{C}[t]$).

After Example 1 (also in page 2), A van den Essen presents a criterion from differential geometry: $f$ is an embedding if and only if $f'(t) \neq 0$ for all $t \in \mathbb{C}$ and the map $f: \mathbb{C} \to \mathbb{C}^2$ is injective.

Now, replace $\mathbb{C}$ by $\mathbb{R}$ and call a polynomial map as above an embedding of $\mathbb{R}$ in $\mathbb{R}^2$ (= $\mathbb{R}$ is isomorphic to its image under $f$).

Is the differential geometry criterion holds over $\mathbb{R}$?

Any hints are welcome!

Edit: (1) I do not see how this question or this question help in solving my question, although they are somewhat relevant (I am not sure if my question should be tagged also 'real analysis' or 'general topology'). (2) I guess that my question has a trivial positive answer? Namely, such a criterion is also valid over $\mathbb{R}$?

Another Edit: My question has a counterexample $\mathbb{R}[t^2,t+t^3] \subsetneq \mathbb{R}[t]$, but $f(t)= (t^2,t+t^3)$ satisfies the geometric conditions ($f'(t) \neq 0$ and $f$ is injective); please see this question. Therefore, I would like to ask:

Is it possible to find an additional condition (in addition to the two geometric conditions $f'(t) \neq 0$ and $f$ is injective) which will guarantee that $\mathbb{R}[f_1(t),f_2(t)]=\mathbb{R}[t]$? (Perhaps the additional condition will involve the second derivative $f''(t)$?).

Thank you very much!

user237522
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  • There is an injective immersion ($f'(t)\neq0$) that is not a topological embedding. See the figure in An injective immersion that is not a topological embedding. In my opinion, the author assumes $f$ is a polynomial. Then you can avoid this example. – ChoF Mar 19 '18 at 02:36
  • @ChoF, thank you very much for your comment! Unfortunately, I still do not understand what is the answer to my question. If I understand correctly, the example in the link is $f(t)=(\sin{2t},\sin{t})$, which is not a polynomial map (so perhaps my question has a positive answer after all?). – user237522 Mar 19 '18 at 03:13
  • If $f$ is polynomial, i.e., $f_1(t),f_2(t)\in\mathbb{R}[t]$, then the "if and only if" condition makes sense. Actually ($\Rightarrow$) holds always, but ($\Leftarrow$) may not. In case that $f$ is polynomial, ($\Leftarrow$) also holds, I think. – ChoF Mar 19 '18 at 03:29
  • The extra condition you need to get an embedding is properness. Can you prove that a polynomial map $\Bbb R\to\Bbb R^2$ is proper (preimage of compact is compact)? – Ted Shifrin Mar 19 '18 at 05:04
  • Thank you very much both of you! @TedShifrin, please: (1) Do you claim that ChoF's last comment is correct?, and (2) A polynomial map $\mathbb{R} \to \mathbb{R}^2$ is always proper? – user237522 Mar 19 '18 at 05:25
  • @ChoF: Sorry, I now read your first comment again and have noticed that I misread it before; when you wrote: ''In my opinion, the author assumes $f$ is a polynomial map" you referred to A. van den Essen, and I thought that you referred to J. Lee (since van den Essen explicitly wrote 'a polynomial map', so there is no doubt it is a polynomial map). – user237522 Mar 19 '18 at 05:36
  • @TedShifrin, I 'guess' that both my questions to you (1) and (2) have positive answers. Here is a link to properness: http://planetmath.org/polynomialfunctionisapropermap; in the link the polynomial map is $\mathbb{R} \to \mathbb{R}$, but I think it is immediate to adjust that proof to a map $\mathbb{R} \to \mathbb{R}^2$. – user237522 Mar 19 '18 at 05:44
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    @user237522 As Ted Shifrin pointed out, it is well-known that "Let $f\colon M\to N$ be a proper injective immersion of differentiable manifolds $M$ and $N$. Then $f$ is an imbedding and $f(M)$ is a regular submanifold and a closed subset of $N$, and conversely." You can find the proof here (p.8). – ChoF Mar 19 '18 at 06:14
  • @ChoF, thanks again! So my question is a special case of the well-known theorem you now quoted? – user237522 Mar 19 '18 at 06:18
  • This clarification for properness is very helpful. – user237522 Mar 19 '18 at 06:42
  • @user237522 My previous comment on $f$ being proper was not correct. We only know that $f^{-1}(C)\subset f_1^{-1}(p_1(C))\cap f_2^{-1}(p_2(C))\subset\mathbb{R}$. We already know that $f^{-1}(C)$ is closed since the compact set $C$ is closed. Furthermore, $f_1^{-1}(p_1(C))\cap f_2^{-1}(p_2(C))$ is bounded so that $f^{-1}(C)$ is also bounded. Thus $f^{-1}(C)$ is compact in $\mathbb{R}$. – ChoF Mar 19 '18 at 08:55
  • Thanks again. You and Ted Shifrin helped me a lot. – user237522 Mar 19 '18 at 15:08
  • @TedShifrin, please, can you explain to me in a few sentences why http://matwbn.icm.edu.pl/ksiazki/apm/apm58/apm5834.pdf is NOT relevant to my question. Is it because here $\mathbb{R} \to \mathbb{R}^2$ (finite?), while there $\mathbb{C}^n \to \mathbb{C}^n$ (What happens if in that paper we replace $\mathbb{C}$ by $\mathbb{R}$?). I really apologize if my comment is not reasonable; it is because I am not familiar with the algebraic geometry notions in that paper. – user237522 Mar 19 '18 at 21:56
  • @user237522: Polynomial mappings in more than one variable are generally not proper (definitely not when we work over $\Bbb C$) because the zero-sets are algebraic varieties and hence often unbounded (always over $\Bbb C$). – Ted Shifrin Mar 19 '18 at 23:02
  • @TedShifrin, thank you very much! – user237522 Mar 19 '18 at 23:23
  • @TedShifrin, I am afraid that we were talking about different definitions: My definition is $\mathbb{R}[f_1(t),f_2(t)]=\mathbb{R}[t]$. In my definition there is a counterexample to my question: $f_1(t)=t^2$, $f_2(t)=t+t^3$ which satisfies the geometric criterion ($f'(t) \neq 0$ and $f:\mathbb{R} \to \mathbb{R}^2$ is injective), but does not satisfy the algebraic condition ($\mathbb{R}[t^2,t+t^3] \subsetneq \mathbb{R}[t]$). See https://mathoverflow.net/questions/297555/abhyankar-moh-embedding-theorem-without-algebraic-closedness/297591?noredirect=1#comment740211_297591 – user237522 Apr 11 '18 at 18:37
  • @ChoF, can you please present your definition of an embedding? (You mentioned 'topological embedding'?). I guess that what you were claiming is that geometic embedding is quivalent to topological embedding. But I was interested in algebraic embedding.. Please see my above comment to Ted Shifrin (which is also a comment for you). – user237522 Apr 11 '18 at 18:45
  • Please, did you mean the following definition: https://ncatlab.org/nlab/show/embedding+of+topological+spaces – user237522 Apr 11 '18 at 18:51
  • @user237522 I mentioned that the iff criterion does not hold in the case of "topological embedding", and it may be possible to hold in the case of "polynomial (algebraic) embedding". But I cannot guarantee it. No proof. – ChoF Apr 11 '18 at 21:53
  • @ChoF, thank you for your explanation. – user237522 Apr 12 '18 at 06:47
  • @TedShifrin, Please, can you write the exact claim that you had in mind, concerning your definition of an embedding and 'my' algebraic definition. – user237522 Apr 12 '18 at 14:08
  • I think van den Essen's algebraic notion is rather unusual and is, apparently, unrelated to the usual differential topology notion. I don't see how you're going to detect the algebra with topology. My claim was only regarding the usual "proper 1-1 immersion = embedding" theorem in differential topology. – Ted Shifrin Apr 12 '18 at 17:36
  • @TedShifrin, thank you. Please, do you think that there is a reasonable additional condition to $f'(t)\neq 0$ and $f$ is injective, that will imply that $\mathbb{R}[f_1(t),f_2(t)]=\mathbb{R}[t]$? – user237522 Apr 12 '18 at 18:59

1 Answers1

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I'm going to write a brief answer here, rather than continuing in comments. The algebraic condition cannot be recovered by a differential topological criterion. Here's the proof.

I will write down a global diffeomorphism (which is definitely not algebraic) of $\Bbb R^2$ carrying the parametric curve $g(t)=(t^2,t)$ [for which the algebraic criterion holds, of course] to the parametric curve $f(t)=(t^2,t+t^3)$ [for which it fails]. To do this, let $\rho\colon\Bbb R\to [0,1]$ be a smooth function with $\rho(x) = 0$ for all $x\le -1/2$ and $\rho(1)=1$ for all $x\ge 0$. Define $F\colon\Bbb R^2\to\Bbb R^2$ by $$F(x,y) = \big(x,y(1+\rho(x)x)\big).$$ You can check that $F$ is a diffeomorphism (it's easy enough to write down the global inverse function and see it's smooth). And $F(g(t)) = F(t^2,t) = \big(t^2,t(1+\rho(t^2)t^2)\big) = (t^2,t+t^3) = f(t)$, as promised.

Ted Shifrin
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