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This question is in the continuation of this question. As it is cleared from the comments of the respective question that an infinitely long cylinder which is also a right circular, is a smooth $3$D mathematical object which can be proved mathematically with the help of Differential Geometry as shown at the page no. 8 in this pdf file.

I was trying to find a $3$D mathematically smooth object which can have edges, but no corners. I have defined an "edge" as a "curved edge" here and I am looking at the surface of this smooth 3-dimensional object from the outside.

Since, an infinitely long cylinder in $\mathbb{R}^3$ i.e. $x^2 + y^2 = 1$ for all $z \in \mathbb{R}$ is a smooth mathematical object, Now I want to know the answers for the following two questions:

$1.$ Does "curved edge" exist for this object ?
$2.$ Theoretically, can we see the "curved edge" of this object which is infinitely long in $z$ direction ?

According to me, "curved edge" exist for this infinitely long cylinder because if we see the equation $x^2 + y^2 = 1$ then it means we "always" get a unit-circle (circle with radius $1$) in $XY$ plane and so, we always get a "curved edge" which mathematically be at infinity.

And, theoretically, we can see this infinitely long right circular cylinder from outside because light coming from a light source, can be reflected through this object whose curved edges might be at infinity and go to our eye but I think practically it might not be possible for a human.

Though Describing "infinity" is difficult but still I want the answers for my above two questions from both mathematics and physics point of view (if possible).

Any help would be appreciated. (I am confused whether this question should be asked in MSE or here. Pardon me if this question fit in MSE and so, you can close it, if you think so.)

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    $\begingroup$ An infinitely long cylinder … is a smooth 3D mathematical object. It’s intrinsically a 2D surface. The fact that it is embedded in 3D Euclidean space is less interesting. $\endgroup$
    – Ghoster
    Mar 10 at 18:15
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    $\begingroup$ You may wish to think about this through the lens of Riemannian geometry. Note that your cylinder $C$ is diffeomorphic to the "finite length" cylinder $$C^\prime := \{(x,y,z) \in \mathbb{R}^3 : x^2+y^2=1, z \in (-\pi/2,\pi/2)\}$$ via the diffeomorphism $f : (x,y,z) \mapsto (x,y,\arctan(z))$. On the one hand, if you equip your cylinder $C$ with the restriction of the flat Riemannian metric on $\mathbb{R}^3$, you get a complete Riemannian manifold, which means that every geodesic on $C$ is defined for all time $t \in \mathbb{R}$. (1/2) $\endgroup$
    – Branimir Ćaćić
    Mar 10 at 21:09
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    $\begingroup$ On the other hand, if you equip $C$ with the pullback by $f$ of the analogous Riemannian metric on $C^\prime$, you'll find that $C$ is not geodesically complete: geodesics parallel to the $z$-axis will only be defined on finite time intervals, i.e., they "reach $\pm \infty$ in finite time." Indeed, this is obvious if you consider the corresponding geodesics on $C^\prime$, which are straight lines of the form $t \mapsto (x,y,t)$. Thus, a failure of geodesic completeness may be a hint that your Riemannian manifold is isometric to the interior of an honest manifold with boundary. (2/2) $\endgroup$
    – Branimir Ćaćić
    Mar 10 at 21:19
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    $\begingroup$ "I was trying to find a 3D mathematically smooth object which can have edges, but no corners." You may be looking for manifolds with boundaries, in this case 2 dimensional. You can read about them in Lee's "Introduction to Smooth Manifolds" $\endgroup$
    – Esgeriath
    Mar 10 at 21:59
  • $\begingroup$ Thank you so much to all of you for your effort and time :) $\endgroup$
    – ankit
    Mar 11 at 3:06

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