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I try to solve the following problem:For each $a\in \mathbb{R}$, let $M_a$ be the subset of $\mathbb{R}^2$ defined by $$ M_a=\{(x,y):y^2=x(x-1)(x-a)\}.$$ For which values of $a$ is $M_a$ an embedded submanifold of $\mathbb{R}^2$? For which values can $M_a$ be given a topology and smooth structure making it into an immersed submanifold?

My attempt: Let $F(x,y)=y^2-x(x-1)(x-a)$ so that $M_a=F^{-1}(0)$.\ Then $DF(x,y)=[-(3x^2-(2a+2)x+a)~~ 2y]$. Therefore, $0$ is a regular value of $F$ unless there is a point $(x,y)$ such that $$y=0,3x^2-(2a+2)x+a=0 ~\text{and}~ F(x,y)=-x(x-1)(x-a)=0.$$ In this case, $-x(x-1)(x-a)=0$ implies that $x=0$ or $x=1$ or $x=a$.\ When $x=0$, $3x^2-(2a+2)x+a=0$ implies that $a=0$. When $x=1$, $3x^2-(2a+2)x+a=0$ implies that $a=1$. The case $x=a$ gives the above values. Thus we have the following cases to consider: Case 1: $a=0, (x,y)=(0,0)$. When $a=0$, the point $(0,0)$ is local minimum of $F$, so $(0,0)$ is an isolated point of $M_0$, and hence $M_0$ can not be an embedded or immersed submanifold. my difficulty is in the following case: Case 2: $a=1, (x,y)=(1,0)$. when $a=1$, the point $(1,0)$ is saddle point of $F$, and thus the curve $M_1$ is self-intersecting at $(1,0)$. In such cases I saw examples saying it can not be an embedded submanifold but by giving an appropriate topology and smooth structure we can make $M_1$ an immersed submanifold of $\mathbb{R}^2$. My difficulty is to justify how it cannot be embedded submanifold, and how we define the topology and smooth structure to make it immersed submanifold. Clearly for $a\neq 0$ and $a\neq 1$, $M_a$ is an embedded submanifold.

JYM
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  • Are you insist that $M_a$ can be a embedding of $\mathbb{R}$ ? – HK Lee May 25 '18 at 06:21
  • For $a\neq 0, 1$ i think so but embedded submanifold of $\mathbb{R}^2$ not of $\mathbb{R}$. – JYM May 25 '18 at 06:34
  • In my understanding an immersed submanifold of a manifold $M$ is a pair $(N,f)$ where $N$ is a manifold and $f : N \to M$ is an Immersion. Alternatively one could understand it as the image of a manifold under an Immersion. This seems to be the interpretation in your question. So you have to find $N_a$ and an immersion $f : N_a \to \mathbb{R}^2$ such that $f(N_a) = M_a$. Thus, the question "For which values can $M_a$ be given a topology and smooth structure making it into an immersed submanifold?" probably means that $M_a$ can be retopologized (i.e. giving it another topology than that – Paul Frost May 25 '18 at 09:14
  • inherited from $\mathbb{R}^2$) producing a 1-dimensional manifold $N_a$ such that the injection of $N_a$ into $\mathbb{R}^2$ is an immersion. An example is the figure "8" which is the image of an immersion $f : \mathbb{R} \to \mathbb{R}^2$. – Paul Frost May 25 '18 at 09:17

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Define $f:\mathbb{R}\setminus \{-1\}\rightarrow \mathbb{R}^2$, by $f(t)=(t^2, t^3-t)$. Then $f$ is one to one injective immersion whose image is $M_1$. Thus $M_1$ is immersed submanifold with the topology and smooth structure inherited from $\mathbb{R}\setminus \{-1\}$.

JYM
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Concernig $M_1$ the problem is not the self-intersection in $(1,0)$.

It is clear that $M_1$ is the image of a map $m : \mathbb{R} \to \mathbb{R}^2$. On $[0,\infty)$ it is defined by $m(x) = \sqrt{x(x-1)^2}$ and produces the upper half $M_1^+ = \lbrace (x,y) \in M_1 \mid y \ge 0 \rbrace$ and on $(-\infty,0]$ the lower half $M_1^-$ via $m(x) = -\sqrt{\lvert x \rvert (\lvert x \rvert-1)^2}$. $M_1^\pm \backslash \lbrace (0,0) \rbrace$ are submanifolds of $\mathbb{R}^2$.

Let us consider the self-intersection. If you understand an immersed submanifold as an immersion $f : N \to \mathbb{R}^2$, then there is no problem at all since only the local behavior is relevant. If you want to find an injective immersion (which means to retopologize $M_1$), then the self-intersection can be resolved by cutting $M_1$ into two pieces which form the components of a new space $N_1$. The two pieces are $N_1' = m((-1,\infty))$ and $N_1'' = m((-\infty,-1))$ but in $N_1$ they are disjoint open sets.

Paul Frost
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  • @Poul Frost You are right he self-intersection in $M_0$ at (0,0). As you point out what I need is an injective immersion $F: N\rightarrow M$ such that $F: N\rightarrow F(N)=M_0$ is a diffeomorphism. Your explanation is good but I can't catch up your conclusion. – JYM May 25 '18 at 12:07
  • I just changed my answer since $M_1$ does not have a cusp at $(0,0)$ ($DF(0,0)$ is not $0$). So take $N = (-\infty,-1) \cup (-1,\infty)$ which is a non-connected submanifold of $\mathbb{R}$. Then $m \mid_N$ is an injective immersion with image $M_1$ (but not an embedding). $M_0$ is not an immersed submanifold as you already pointed out. – Paul Frost May 25 '18 at 12:20
  • $M_a$ has two components unless $a=1$. If $a \ne 1$, one component is a compact, the other unbounded. If $a = 0$ the compact component degenerates to a point. If $a \ne 0,1$, then the compact component is diffeomorphic to the circle $S^1$, the other to the line $\mathbb{R}$. – Paul Frost May 25 '18 at 12:28