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Assume that $P\in \mathbb{C}[z]$ is a polynomial of degree $n$ with $n$ distinct roots $z_1,z_2,\ldots,z_n$.

We identify $\mathbb{R}^3$ with $\mathbb{C}\times \mathbb{R}$. Put $a_i=(z_i,0)$.

Then $\mathbb{R}^3 \setminus \{a_1,a_2,\ldots,a_n\}$ is the union of $\mathbb{R}^{3\geq0} \setminus \{a_1,a_2,\ldots,a_n\} $ and $\mathbb{R}^{3\leq0} \setminus \{a_1,a_2,\ldots,a_n\} $ where $\mathbb{R}^{3\geq0}=\mathbb{C} \times [0, +\infty)$ and $\mathbb{R}^{3\leq0}=\mathbb{C} \times (-\infty, 0]$.

This enable us to define a complex line bundle on $\mathbb{R}^3 \setminus \{a_1,a_2,\ldots,a_n\}$ using clutching function $1/P(z)$.

Since the wedge sum of $n$, $S^2$ is a deformation retract of $\mathbb{R}^3 \setminus \{a_1,a_2,\ldots,a_n\}$, we get a line bundle on the wedge sum of the spheres.

Does the isometric class of this line bundle depend on choosing initial polynomial $P(z)$? What is the explicit formulation of the (first) Chern class of this line bundle, in $\oplus_{i=1}^n \mathbb{Z}$, the cohomology of the base space?( In terms of the coefficients of $P(z)$)

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  • $\begingroup$ This should be an exercise in complex analysis. The coefficient of the Chern class on the sphere corresponding to $a_k$ should be computed as follows: take a small circle which winds once around $a_k$, take its image under the transformation $1/P(z)$, and compute the winding number of this curve around $a_k$. My guess is its the residue of the pole, or something lie that. $\endgroup$
    – Mark Grant
    Commented Aug 14, 2017 at 13:01
  • $\begingroup$ @MarkGrant Thank you. For $P(z)=z$ we get the tautological line bundle over $\mathbb{C}P^1\simeq S^2$. But may you elaborate your comment for arbitrary $P$. $\endgroup$
    – Ali Taghavi
    Commented Aug 14, 2017 at 13:06
  • $\begingroup$ if I remembered enough complex analysis to do this easily, I wouldn't have set it as an exercise ;) I believe it should follow from the residue theorem en.wikipedia.org/wiki/Residue_theorem and my comment above. If nobody more knowledgeable comes along, I might have a go later. $\endgroup$
    – Mark Grant
    Commented Aug 14, 2017 at 13:58
  • $\begingroup$ @MarkGrant thanks for this new comment. I do not think the residue effect on Chern classes . For example for P(z)=az , a in C, the line bundle is the Tautological bundle because az is homotopic to bz for $a\neq b$ . right? $\endgroup$
    – Ali Taghavi
    Commented Aug 14, 2017 at 14:38

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OK, here's how I think it goes. No need for the residue theorem or anything like that.

Let $S_k$ denote the sphere corresponding to $z_k$. To compute the Chern class of the restriction of your line bundle to $S_k$, it suffices to compute the degree of the map $$ f:S^1\to S^1,\qquad f(z) = \frac{|P(z)|}{P(z)} = \frac{\overline{P(z)}}{|P(z)|} $$ obtained by restricting your clutching function $1/P(z)$ to a small circle $S^1\subset\mathbb{C}$ which encloses $z_k$ and contains no other roots in its interior, and normalizing. But since this $f$ is the composition of the map $z\mapsto P(z)/|P(z)|$ of degree $n$ and the map $z\mapsto \bar{z}$ of degree $-1$, and since the degree of a composition is the product of the degrees, the degree of $f$ is $-n$.

Therefore I think that the Chern class of your bundle over $\vee_n S^2$ is $-n$ times the generator of each sphere.

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  • $\begingroup$ Thank you for your answer. Is $ \overline{P(z)} = P(\overline{z})$ essential in your argument? The coefficients are not necessarily real $\endgroup$
    – Ali Taghavi
    Commented Aug 14, 2017 at 19:05
  • $\begingroup$ @AliTaghavi: Oops, my mistake. No, I don't think so. I've corrected this and another mistake (I'd used $w^{-1}=\bar{w}/|w|$ instead of $w^{-1}=\bar{w}/|w|^2$ before). $\endgroup$
    – Mark Grant
    Commented Aug 14, 2017 at 19:45

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