# Checking whether the image of a smooth map is a manifold

I have a specific problem, but would also like to know how to tackle the general case. I will first state the genral question. Let $M$ be an embedded submanifold of $\mathbb{R}^n$ and let $F: \mathbb{R}^n \to \mathbb{R}^n$ be a smooth map. How do I go about checking whether $F(M)$ is a smooth embedded submanifold of $\mathbb{R}^n$ or not? The specific problem I have is the following :- Let $F:\mathbb{C}^2\to \mathbb{C}^2$ the map $(z_1,z_2) \mapsto (z_1+z_2,z_1z_2)$ and let $M$ be the unit sphere in $\mathbb{C}^2$, i.e., $\lbrace (z_1,z_2) : |z_1|^2 + |z|^2 = 1 \rbrace$. Is $F(M)$ an embedded submanifold of $\mathbb{C}^2$ (considered as $\mathbb{R}^4$)?

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The easiest way to play this is to find some 'regular' equations that characterise your image then see if they really are regular- i.e. their derivatives are surjective. Not sure about your specific example yet... Will think on and turn into an answer if I get round to it. – Tom Boardman Jun 15 '10 at 14:00
Maybe not so easy in this case... regular equations not forthcoming in degree 2. I'm sure they'll work in a high enough degree but this might be sledgehammer to crack a nut... – Tom Boardman Jun 15 '10 at 16:25

The specific $F(M)$ is not a smooth submanifold. Here is an argument.

To simplify formulas, I renormalize the sphere: let it be the set of $(z_1,z_2)\in\mathbb C^2$ such that $|z_1|^2+|z_2^2|=2$ rather than 1. Then, as Gregory Arone pointed out, $F(M)$ is the set of $(b,c)\in\mathbb C^2$ such that the roots $z_1,z_2$ of the equation $z^2-bz+c$ satisfy $|z_1|^2+|z_2^2|=2$. I claim that it is not a smooth manifold near the point $p:=(2,1)\in F(M)$.

Let us intersect $F(M)$ with two planes: $\{b=2\}$ and $\{c=1\}$. If it is is a smooth submanifold, at least one of the intersections should be locally (near $p$) a 1-dimensional smooth submanifold of the respective plane (otherwise both planes are tangent to $F(M)$ at $p$, but this is impossible since they span the whole space). But this is not the case:

If $b=2$, the equation is $z^2-2z+c=0$, hence $z_1+z_2=2$, then $|z_1|+|z_2|\ge 2$ and therefore $|z_1|^2+|z_2|^2\ge 2$. Equality is attained only for $z_1=z_2=1$, thus the intersection is a single point $c=1$, not a 1-dimensional submanifold.

If $c=1$, the equation is $z^2-bz+1$, hence $z_1z_2=1$, then $|z_1|\cdot|z_2|=1$ and therefore $|z_1|^2+|z_2|^2\ge 2$. The equality is attained if and only if $|z_1|=|z_2|=1$, so $z_1$ and $z_2$ are conjugate to each other. The set of $b$'s for which this happens is the real line segment $[-2,2]$ which is not a submanifold near 2.

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If you consider the real map $(x_1, x_2)$-->$(x_1+x_2, x_1x_2)$ then the image of the circle is not a submanifold. Infact your map is not injective: it is symmetric with respect to swapping the two coordinates, so the circle is folded once onto itself and its image is homeomorphic to a closed segment. In the complex case, my guess is that the image is homeomorphic to the quotient of the sphere by the involution that swaps the two complex variables... The fixed point set of this involution is the circle given by intersection with the line $(z,z)$. Near this circle the quotient should be like the quotient of R x C by the involution $(t, z) \mapsto (t, -z)$, which is still a manifold. So I would think F(M) is a manifold...

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Hey! Welcome to MO - from another ex-Warwick PhD. – Loop Space Jun 15 '10 at 20:55
Hi Andrew, nice to see you. I just started here, looks fun. – Diego Matessi Jun 16 '10 at 8:47

To elaborate on Diego Matessi's answer (and Boyarsky's comment which I only saw after writing this)... the map $F$ is symmetric, so it factors as a composite map $$C^2\longrightarrow (C^2)_{\Sigma_2}\longrightarrow C^2.$$ Here $\Sigma_2$ is the group with $2$ elements acting on $C^2$ by switching coordinates, the first map is the quotient map, and the second map is well-known to be a homeomorphism. It can be thought of as the map that sends an unordered pair $[z_1, z_2]$ to the unique monic complex polynomial whose roots are $-z_1$ and $-z_2$. So the map $S^3\to F(S^3)$ is not injective. Nevertheless, it is an easy exercise to show that $F(S^3)$ is homeomorphic to $S^3$. But then again, it {\em seems} to me that $F(S^3)$ should not be a smooth submanifold of ${\mathbb C}^2$. $F(S^3)$ can be identified with the space of monic complex quadratic polynomials $z^2+bz+c$ whose pair of roots gives a unit complex vector. So it is the space of pairs of complex numbers $(b, c)$ that solve the equation $$|\frac{-b+\mbox{first square root of } b^2-4c}{2}|^2+|\frac{-b+\mbox{second square root of } b^2-4c}{2}|^2=1.$$ It seems that the space of solutions of this equation should not be smooth, and that a singularity should occur where the discriminant $b^2-4c$ is zero - which corresponds to the diagonal in the original $C^2$, but I could be mistaken.

Edit Let me point out that the equation can be simplified. Let $\Delta=b^2-4c$. Then the equation is in fact equivalent to $$(*) \,\, |b|^2+|\Delta|=2.$$ Since the map $(b,c)\to (b, b^2-4c)$ is a diffeomorphism, the question of whether $F(M)$ is a smooth submanifold is equivalent to the question whether the space of solutions of the equation ($\ast$) is a smooth submanifold of $C^2$. I just wanted to point this out because equation ($\ast$) looks simpler than the one I originally wrote. In the meantime, Sergei Ivanov gave an argument showing that it is not smooth.

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