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Set is given by $M = (x,y) \in \mathbb{R}^2 \vert \underbrace{e^{x^2+2\,y^2+2}}_{f} = c$.

Instantly I noticed $M$ can be counted as Manifold if $c \in(e^2,\infty)$ because $f \geq e^2$. Now apparently this is not complete, so I wonder which additional c I have to take into consideration. Is it the boundary $[e,\infty)$ ? Or even other sets.

Leon
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1 Answers1

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We have $M=\{(x,y)\in \mathbb{R}^{2} \colon x^{2}+2y^{2}=d\}$, where $d=\operatorname{log}(c)-2$. Let $g=x^{2}+2y^{2}$. We have that $\operatorname{grad}(g)=(2x,4y)$ vanishes at $(0,0)$. Now let's look at some cases:

If $d<0$ (which is the same as $c<e^{2}$), $M=\emptyset$ isn't a manifold.

If $d=0$ (i.e. $c=e^{2}$) then $M=\{(0,0)\}$ is a point, which is a manifold as long as you admit zero-dimensional manifolds.

If $d>0$ (which is the same as $c>e^{2}$), then $\operatorname{grad}(g)(x,y)\neq 0$ for every $(x,y)\in g^{-1}(c)$, so by the regular level set theorem, $g^{-1}(c)$ is a ($1$-dimensional) embedded submanifold of $\mathbb{R}^{2}$.

Hope this helps!