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Ali Taghavi
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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 $n$, $2\_$ spheres$S^2$ 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 $P(z)$)

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 $n$, $2\_$ spheres 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 $P(z)$)

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 $P(z)$)

added 1 character in body; edited title
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Ali Taghavi
  • 356
  • 8
  • 31
  • 123

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 $n$, $2\_$ spheres 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 Chern the (first) Chern class of this line bundle, in terms of the coefficients of $P(z)$, in $\oplus_{i=1}^n \mathbb{Z}$, the cohomology of the base space?( In terms of the coefficients of $P(z)$)

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 $n$, $2\_$ spheres 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 Chern class of this line bundle, in terms of the coefficients of $P(z)$, in $\oplus_{i=1}^n \mathbb{Z}$, the cohomology of the base space?

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 $n$, $2\_$ spheres 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 $P(z)$)

added 1 character in body; edited title
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Ali Taghavi
  • 356
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  • 123

A line bundle on the directwedge sum of spheres associated to a polynomial $P(z)\in \mathbb{C}[z]$

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 direct sumwedge sum of $n$, $2\_$ spheres is a deformation retract of $\mathbb{R}^3 \setminus \{a_1,a_2,\ldots,a_n\}$, we get a line bundle on the direct sumwedge 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 Chern class of this line bundle, in terms of the coefficients of $P(z)$, in $\oplus_{i=1}^n \mathbb{Z}$, the cohomology of the base space?

A line bundle on the direct sum of spheres associated to a polynomial

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 direct sum of $n$ $2\_$ spheres is a deformation retract of $\mathbb{R}^3 \setminus \{a_1,a_2,\ldots,a_n\}$, we get a line bundle on the direct 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 Chern class of this line bundle, in terms of the coefficients of $P(z)$, in $\oplus_{i=1}^n \mathbb{Z}$, the cohomology of the base space?

A line bundle on the wedge sum of spheres associated to a polynomial $P(z)\in \mathbb{C}[z]$

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 $n$, $2\_$ spheres 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 Chern class of this line bundle, in terms of the coefficients of $P(z)$, in $\oplus_{i=1}^n \mathbb{Z}$, the cohomology of the base space?

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Ali Taghavi
  • 356
  • 8
  • 31
  • 123
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