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The $\mathbb{Z}_2$ topological degree of a (non-constant) polynomial in one variable, clearly, coincides with its degree as a polynomial, $\mod 2$.

Consider further a polynomial self-mapping $F$ on $\mathbb{R}^2$, and assume it is a proper map (in case, even more generally a map in higher dimension)

Is there a short way to decide what's the parity of the topological degree of $F$?

E.g. it's odd if $F$ is an odd map, or more generally, if $F$ can be transformed into an odd map by a proper homotopy. Actually: is there a short way to understand if a polynomial map is a proper map? What about the case of a gradient map (I mean, the gradient of a polynomial $f\in\mathbb{R}[x,y]$)?

I'm somehow confident that there may be a simple criterion known, at least in $\mathbb{R}^2.$ After all, what is required is just a one-bit information (well this argument doesn't convince me either).

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(whether $\gamma^\pi$ is rational or not is also a one-bit answer) –  Pietro Majer Sep 14 '10 at 18:45
Indeed, the size of the output can only give a lower bound on the complexity of the question... –  Thierry Zell Sep 14 '10 at 19:23
...unless the speaker is a politician ;-) –  Pietro Majer Sep 15 '10 at 15:58
To compute whether a polynomial map $F:R^2\to R^2$ is onto, I think you have to consider the closure of $Graph(F)\subset P^2\times P^2$, and project to the second factor (the map $F$ doesn't necessarily extend to $P^2$: consider $F(x,y)=(x,xy)$). If the image of $\overline(Graph(F))-Graph(F)$ in $P^2$ lies in $P^1$ at infinity, then the map should be proper. But I don't know enough algebraic geometry to make this precise (e.g. Harris only considers algebraically closed fields) books.google.com/… –  Ian Agol Sep 16 '10 at 3:46

1 Answer 1

up vote 3 down vote accepted

Here is an idea, I'm not 100% confident that it makes sense in all cases, but I'll try it anyway. Assuming that $f(\mathbb{R}^2)$ is 2-dimensional, the degree mod 2 of your map is the cardinality of the preimage of a generic point. If your components have degree respectively d and e, then Bezout gives you a preimage size of de.

That's projective solutions. In generic cases, you would expect no solutions at infinity. Unfortunately as David points out below, you may have an odd number of solutions at infinity.

So in the case when there's nothing at infinity, we have solutions over the complexes, but non-real solutions will come in complex conjugate pairs, so in that nice case the degree mod 2 is given by de.

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This is a very good answer which isn't 100% right. The trouble is that Bezout gives you $de$ roots in the projective plane, so you might not get the right count at $\infty$. For example, consider $(x,y) \mapsto (x^3+y^3+x, x^3+y^3+y)$. Bezout gives $9$ complex roots, but $3$ of them are on the line at $\infty$. Still, this approach will work in a lot of cases. –  David Speyer Sep 14 '10 at 19:45
Thierry -- in the affine case B\'ezout gives an inequality only: think of a map of the form $(x,y)\mapsto (x,y+f(x))$ where $deg(f)>1$. –  algori Sep 14 '10 at 20:01
My gut feeling was right: it was too good to be always true. I'll modify the answer accordingly. –  Thierry Zell Sep 14 '10 at 20:15
Thierry -- I started writing down basically the same answer when I thought about that example. It doesn't look like this has an easy soltion in terms of the degrees of the components. E.g. if the components are $(x,y^2+x)$ the map is even, but if the components are $(x,y+x^2)$, it is odd. –  algori Sep 14 '10 at 20:33
Correction to my above: the three roots on the line at infinity are double roots. If you look at $(x,y) \mapsto (x^3+y^3+p(x,y), x^3+y^3+q(x,y))$ for $p$ and $q$ generic quadrics, then there should only three roots at $\infty$. –  David Speyer Sep 14 '10 at 23:13

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