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Is it true that every smooth variety (over $\mathbb{R}$ or $\mathbb{C}$ ) is a (real or complex) manifold? I have tried to show this using the implicit function theorem but I am not getting anywhere.

I think I understand it if I have a complete intersection variety,

If $X=V(f_{1},...,f_{m})\subset\mathbb{A}^{n+m}$ and $dimX=n$ then we have a function, $f:\mathbb{R}^{n+m}\to\mathbb{R}^{m}$, and the implicit function theorem gives a local function $g:\mathbb{R}^{n}\to\mathbb{R}^{m}$ such that $f(x,g(x))=0$. Thus, $X$ is the graph of $y=g(x)$.

The problem is that this uses the fact that $n+m$ is the dimension of the ambient space.

Bates
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    The zero locus of $m$ polynomials is always of codimension $n$ over an algebraically closed field, so over $\mathbb{C}$ your argument will go through. I think that over $\mathbb{R}$ you get a potentially lower-dimensional variety that's nonetheless a manifold, but don't quote me on that. – Kevin Carlson Jun 07 '14 at 18:15
  • @KevinCarlson should your $n=m$ ? Further are you saying that every variety over $\mathbb{C}$ is the complete intersection ? – Rene Schipperus Jun 07 '14 at 18:17
  • @KevinCarlson, are you sure this is true? What about the twisted cubic? – Bates Jun 07 '14 at 18:49
  • See http://mathoverflow.net/questions/7439/algebraic-varieties-which-are-also-manifolds and http://math.stackexchange.com/questions/9010/what-is-the-difference-between-a-variety-and-a-manifold. – Dietrich Burde Jun 07 '14 at 18:53
  • Yes you are right, but doesnt OP's proof depend on having complete intersection ? But as I write this it occurs to me you dont need complete intersection, you just need that the variety is a component of a complete intersection. – Rene Schipperus Jun 07 '14 at 18:53
  • I think that is always true a variety is a component of a complete intersection. – Rene Schipperus Jun 07 '14 at 19:01

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It suffices to work with affine varieties over a field $K$ (where $K={\mathbb R}$ or $K={\mathbb C}$). I will work with $K={\mathbb R}$ for concreteness. Let $V\subset {\mathbb R}^N$ be an affine subvariety. Smoothness of $V$ means that generators $f_1,...,f_n$ of its ideal $I_V$ satisfy the property that the rank $r$ of the derivative $DF$, $F=(f_1,...,f_n)$, is constant on $V$ (i.e. the Zariski tangent space of $V$ has constant dimension). Now, let us apply the Constant Rank Theorem, see say, here or here for a proof. The constant rank theorem states that for each point $x\in V$ there exists a smooth change of coordinates near $x$ and $F(x)$, after which the map $F$ locally has the form $$ F(x_1,...,x_N)= (x_1,...,x_r, 0,...,0). $$ Then, clearly, $F^{-1}(0)$ is a smooth submanifold in ${\mathbb R}^N$ of dimension $N-r$. This shows that $V$ is a smooth (real) manifold. In the case $K={\mathbb C}$ you use the holomorphic version of the constant rank theorem (proven in the same fashion as the real one; all what you need is the complex inverse function theorem) to conclude that $V$ is a complex submanifold.

Needless to say, complete intersections are irrelevant here.

Moishe Kohan
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  • I don't understand, what if my variety is defined by a number of polynomials? How would I use the constant rank theorem? – Bates Jun 08 '14 at 18:02
  • @Bates: Your (affine) variety is defined by a finite set of polynomials $f_1,...,f_k$, which determines a function $F=(f_1,...,f_k)$. If every point is smooth then $F$ has constant rank. Now, apply the CRT to $F$. – Moishe Kohan Jun 09 '14 at 15:37