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I know that not all algebraic variety are smooth manifold I would like to know an example of the inverse. What is a manifold that is not an algebraic variety?

LuckyS
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  • What is your definition of an algebraic variety? The most common definitions of algebraic varieties all include the property of being irreducible and hence also of being connected. Therefore any non-connected manifold would not be an algebraic variety. – Con Apr 04 '21 at 11:47
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    You've already been given an answer in a previous question. What about that was insufficient? – KReiser Apr 04 '21 at 11:56
  • There are several complex abelian variety which is not algebraic. To be more precise, there are complex abelian varieties which are not the complex points of an algebraic variety. You can find this in the discussion after the last corollary of Chapter 1, Abelian varieties by Mumford. – random123 Apr 04 '21 at 12:01
  • This is tricky to address as worded. An algebraic variety (projective? affine? over a particular field?) comes equipped with structure that "most" manifolds don't possess. In order to compare apples to apples, it might help to restrict to "real plane curves" or something. Then you have a nodal cubic $y^2=x^3-x^2$ (as in a comment to your other question) as a non-manifold algebraic curve, and a graph such as $y = e^x$ that's a manifold but not an algebraic curve. – Andrew D. Hwang Apr 04 '21 at 12:12

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