Timeline for Topological degree and polynomial degree
Current License: CC BY-SA 3.0
8 events
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| Apr 9, 2014 at 16:29 | comment | added | UserX2017 | 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 $. | |
| Apr 7, 2014 at 18:57 | comment | added | Liviu Nicolaescu | @ 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. | |
| Apr 7, 2014 at 18:53 | history | edited | Liviu Nicolaescu | CC BY-SA 3.0 |
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| Apr 7, 2014 at 14:32 | comment | added | UserX2017 | 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? | |
| Apr 3, 2014 at 16:06 | history | edited | Liviu Nicolaescu | CC BY-SA 3.0 |
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| Apr 3, 2014 at 14:41 | history | edited | Liviu Nicolaescu | CC BY-SA 3.0 |
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| Apr 3, 2014 at 13:39 | history | edited | Liviu Nicolaescu | CC BY-SA 3.0 |
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| Apr 3, 2014 at 12:31 | history | answered | Liviu Nicolaescu | CC BY-SA 3.0 |