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I was interested to know whether set of singular points of a smooth map forms a manifold? for example if $f:M\rightarrow\mathbb{R}^2$ is a smooth map.

and I am trying to find the singular points and singular values of the map $\mathbb{R}^2\rightarrow\mathbb{R}^2$, $z\mapsto z^3-3\bar{z}$, could any one help me? thank you.

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Sard's theorem asserts that the set of singular values has Lebesgue measure zero. –  Yuchen Liu Feb 2 '13 at 12:23
    
jerry : any smooth submanifold of positive codimension will have Lebesgues measure 0 –  Glougloubarbaki Feb 2 '13 at 14:19

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up vote 1 down vote accepted

The following example is "typical" for maps ${\bf f}:\ \Bbb R^2\to\Bbb R^2$:

Consider the map $${\bf f}:\quad (x,y)\mapsto\left\{\eqalign{u(x,y)&:={1\over2}(x^3-3xy) \cr v(x,y)&:=y\cr}\right.\quad.$$ One computes the Jacobian $$J_{\bf f}(x,y):=\det d{\bf f}(x,y)={3\over2}(x^2-y)\ ,$$ whence ${\bf f}$ is singular along the parabola $y=x^2$. The image $\gamma$ of this parabola can be parametrized by $x$ in the form $$\gamma: \quad x\mapsto {\bf f}(x,x^2)=(-x^3,x^2)\qquad(-\infty<x<\infty)\ .$$ Therefore $\gamma$ is a Neil parabola and has a cusp at the origin.

I suggest you draw pictures showing what's going on in the $(x,y)$-plane and in the $(u,v)$-plane. If you can read German look here, example 2 on p. 106:

http://www.math.ethz.ch/~blatter/Analysis_12.pdf

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In your example your map is much more than smooth, it is polynomial. I'm pretty sure algebraic geometry can tell you a lot in this context. However it won't be a smooth manifold in the sense of differential geometry (the zeroes of a function define a submanifold when the function is a submersion near its zeroes, and there are counterexamples when it's not).

If you are just looking at smooth functions, however, almost nothing can be said. Indeed let $F$ be a closed arbitrary subset of $\mathbb{R}$. It is a classical fact that there is a function $f : \mathbb{R} \rightarrow \mathbb{R}$ smooth such that the set of its zeroes is exactly $F$. Now if you take a primitive $g$ of $f$, you have a smooth function whose singular points are exactly $F$.

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to whoever downvoted : please discuss, this is the point of this site after all... –  Glougloubarbaki Feb 2 '13 at 14:12

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