# Variety vs. Manifold

In the ambit of differential geometry the aim is to study smooth manifolds. Why the objects studied in algebraic geometry are called algebraic varieties and not for example algebraic manifolds? I am a fan of the uniqueness of mathematics, so I think that these differences of terminology or notation may mislead the student.

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Historical accident. Things are different in France, where smooth manifolds are variétés différentielles. –  Qiaochu Yuan May 21 '12 at 18:34
IMO, the objects and techniques feel so different that it would be misleading to use the same term for both sorts! –  Hurkyl May 21 '12 at 18:37
Maybe relevant: math.stackexchange.com/a/40154/3217 –  Georges Elencwajg May 21 '12 at 19:13
Varieties tend to use the Zariski topology, which is wildly different from the topology of anything typically called a manifold. –  MartianInvader May 21 '12 at 21:40
@QiaochuYuan, nitpicking, differentiable manifolds are called as you say (or "variétés différentiables"), "smooth manifolds" is usually rather "variétés lisses". But you made a very good point that there is a historical component. Also the term "algebraic manifold" is highly used (and useful) to refer to algebraic varieties which are also manifolds. –  plm May 22 '12 at 11:33

A variety does not qualify as a manifold for more reasons other than smoothness. For example the $xy$-plane union the $z$-axis is a variety. But, there isn't even a well-defined dimension there. You would need a sufficiently broad definition of manifold to include varieties that are not smooth and don't have a dimension. At that point, the word "manifold" would not be very useful.

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Many algebraic varieties are not manifolds. For example, the coordinate axes in $\mathbb{R}^2$ are an algebraic variety, but not a manifold because it isn't locally homeomorphic to $\mathbb{R}$ at the origin.
The issue here is topology, not smoothness: the origin does not have a neighbourhood homeomorphic to $\mathbb R$. But there are other cases where smoothness is the issue, e.g. $y^2 - x^3 = 0$. –  Robert Israel May 21 '12 at 18:04