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Let $F:\mathbb{C}^n\to \mathbb{C}^n$ be a homeomorphism homogeneous of degree 1 (i.e., $F(tx)=tF(x)$, $t>0$) and $g:\mathbb{C}^n\to \mathbb{C}$ a homogeneous polynomial of degree $k$. Let $L$ ($0\in L$) a complex line and $H=F(L)$ such that $g^{-1}(0)\cap H=\{0\}$. It is true that $h=g|_H\circ F|_L$ has topological degree less than or equal to k? This is true if F is linear map!

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What is the topological degree in the noncompact setting? –  Alex Degtyarev Apr 3 at 12:18
    
@AlexDegtyarev it can be defined as the value of the automorphism induced on $2n$-dimensional cohomology with compact support (or homology with infinity support). In that dimension the aforementioned cohomology is $\mathbb Z$. –  Fernando Muro Apr 3 at 14:46
    
@ Fernando: that still requires a properness assumption. Take the constant map from a vector space to itself. It's not proper and it does not induce morphism between cohomologies with compact supports. –  Liviu Nicolaescu Apr 3 at 15:01
    
@FernandoMuro Not any map induces a homomorphism in the homology with compact supports. As stated, the question just doesn't make sense (as the answer brlow points out). –  Alex Degtyarev Apr 3 at 15:03
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@LiviuNicolaescu and AlexDegtyarev sorry I misread the question and thought he was just talking about homeomorphisms. –  Fernando Muro Apr 3 at 15:30

1 Answer 1

Your question needs to be phrased more carefully.

Take $n=2$, $g(z_1,z_2)=z_1z_2$, $F=$ the identity map $\mathbb{C^2}\to\mathbb{C^2}$ and $L$ is the subspace

$$L=\lbrace(z_1,0),\;\;z_1\in\mathbb{C}\rbrace\subset \mathbb{C}^2. $$

Then $g\circ F|_L=0 $ and $g\circ F|_L$ does not have a degree.

In general for a map $g: \mathbb{C}\to \mathbb{C}$ to have a degree it has to be a proper map. The zero set of a homogeneous polynomial is a conical set in $\mathbb{C}^n$ and so is the image $F(L)$. If $F(L)$ intersects $Z_g:=g^{-1}(0)$ at a point $p_0\neq 0$, then the the intersection $Z_g\cap F(L)$ contains the whole ray $tp_0$, $t\geq 0$ and thus $g\circ F|_L$ cannot be proper and thus it does not even have a well defined degree.

It is not hard to see that $g\circ F|_L$ is proper if and only if $Z_g\cap F(L)=\lbrace 0\rbrace$. Assume that $L$ is the first coordinate line in $\mathbb{C}^n$. The degree of $g\circ F|_L$ is then the winding number of the loop

$$ [0,2\pi]\ni \theta \mapsto g\bigl(\, F(e^{i\theta},0,\dotsc, 0)\bigr)\in\mathbb{C}\setminus 0. $$

$\newcommand{\bC}{\mathbb{C}}$

Edit. The computation of the above winding number could be tricky. Here is an example that shows that the topological degree need not be equal to the algebraic degree. Assume $n=2$, $g(z_1,z_2)=z_1$, and $L=\{(z_1,0)\}\subset \bC^2$. Set

$$\Sigma=\bigl\{\, (z_1,z_2);\;\;|z_1|^2+|z_2|^2 =1\,\bigr\}\subset \bC^2. $$

We will construct degree $1$ homogeneous homeo $\bC^2\to\bC^2$ by fixing a homeo

$$\phi: \Sigma\to \Sigma. $$

We get a homeo $F_\phi: \bC^2\to\bC^2$ by setting

$$F_\phi(tp)=t\phi(p),\;\;\forall p\in\Sigma,\;\;t>0. $$

Take a tiny unknotted circle $K\subset \Sigma$ contained in a small neighborhood of the point $(1,0)\in \Sigma$. In particular, the projection of $K$ on the first coordinate axis $L$ is a small loop $K'$ with winding number $0$ because $K'$ is contained in a tiny disk of $L$ disjoint from the origin. The circle $\newcommand{\ve}{{\varepsilon}}$

$$ C =L\cap\Sigma=\bigl\{ (z_1,0),\;\;\;|z_1|=1\,\bigr\}\subset \Sigma. $$

describes another unknotted circle on $\Sigma$. In particular, it is isotopic to the circle $K$ and thus there exists a homeomorphism $\phi: \Sigma\to\Sigma$ such that $\phi(C)=K$.

The loop

$$ g\circ \phi: C\to \bC\setminus 0 $$

coincides with the loop $K'$ whose winding number is zero. This shows that

$$ \deg g\circ F_\phi|_L=0\neq \deg g=1. $$

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Thanks and sorry, suppose $ H \cap \{g = 0 \} = \{0 \} $. Then consider $ L \cong \mathbb {C} \cong \mathbb {R} ^2 $. Hence, define the topological degree of $ g |_H \circ F |_L $ to be the winding number of $ g|_{H \setminus \{0 \}} \circ F |_{L \setminus \{0 \}}: L \setminus \{0 \} \to \mathbb {C} \setminus \{0 \} $. Mr. Liviu Nicolaescu keeps going for my conjecture! Under the above conditions, with $ T $ the topological degree and $ k $ the algebraic degree, then $ | T | \leq k $. The Mr. can give a counter example? –  UserX10 Apr 7 at 14:32
    
@ Edson. I fixed an omission in my proof. You can get more interesting answers if in my proof you choose $K$ to be a knot disjoint from $C$ and having large linking number with $C$. I believe that $T$ as defined by you is equal to this likinking number. –  Liviu Nicolaescu Apr 7 at 18:57
    
Very hard to be true! For $\mathbb{C}^2$ in the above problem is equivalent to Zakiski's conjecture, that is true in this case! However I think your idea is very useful if I try to adapt for dimension $> 2 $. –  UserX10 Apr 9 at 16:29

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