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When an algebraic variety, affine or projective, is a manifold ( on $\mathbb{R}$)? And when the inverse is true? I think that each algebraic variety is a manifold but for the projective? I would think the same but I would like to have an explanation about the relathionship of these structures

LuckyS
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    It is not true that all real affine varieties are manifolds. For example, the variety defined by $((x+1)^2 + y^2 - 1)((x-1)^2 + y^2 - 1) = 0$ in $\mathbb{R}^2$ is a wedge of two circles. – diracdeltafunk Mar 28 '21 at 00:28
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    Perhaps your definition of "variety" requires irreducibility, in which case the above does not work. If so, consider $y^2 = x^3 + x$ instead. This defines an irreducible variety in $\mathbb{R}^2$ which is not a manifold: here's a picture of it. – diracdeltafunk Mar 28 '21 at 00:32

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There is actually a nontrivial theorem regarding this issue:

Definition. 1. A smooth manifold $M$ is tame if $M$ is diffeomorphic to the interior of a smooth compact manifold $N$ with (possibly empty) boundary.

  1. A smooth manifold $M$ is algebraic if it is diffeomorphic to a nonsingular real-algebraic subset of ${\mathbb R}^n$ for some $n$.

Theorem. A smooth manifold is tame if and only if it is algebraic.

Both directions are nontrivial. See Corollary 4.3 in

Akbulut, Selman; King, Henry C., The topology of real algebraic sets with isolated singularities, Ann. Math. (2) 113, 425-446 (1981). ZBL0494.57004.

Moishe Kohan
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