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Charles Matthews
Charles Matthews
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homology Homology of a complex projective conic

Hi! LetLet $Q$ be a smooth conic (the zero set of a homogeneous degree 2 polynomial) in $\mathbb{P}^2(\mathbb{C})$ and let $j:Q\rightarrow\mathbb{P}^2(\mathbb{C})$ be the immersion in the projective plane. How can iI compute $j_*H_2(Q,\mathbb{Z})\subseteq H_2(\mathbb{P}^2(\mathbb{C}),\mathbb{Z})$? (within $H_i(X,\mathbb{Z})$, i meanmeans the i-th singular homology group).

I know that $Q$ is homeomorphic to $\mathbb{P}^1(\mathbb{C})$ so their homology groups are isomorphic, iI can restrict to the veroneseVeronese embedding $\phi:\mathbb{P}^1(\mathbb{C})\rightarrow\mathbb{P}^2(\mathbb{C})$, $\phi([x_0:x_1])=[{x_0}^2: x_0x_1 :{x_1}^2]$. I also notednote that the map $\phi$ is homotopic to $\psi: \mathbb{P}^1(\mathbb{C})\rightarrow\mathbb{P}^2(\mathbb{C})$, $\psi([x_0:x_1])=[{x_0}^2: 0 :{x_1}^2]$ that is a line "counted two times" so my guess is that $j_*H_2(Q,\mathbb{Z})$ can be the whole $H_2(\mathbb{P}^2(\mathbb{C}),\mathbb{Z})$ or it can be the subgroup of $H_2(\mathbb{P}^2(\mathbb{C}),\mathbb{Z})$ generated by two times its generator. How can iI proceed? Is there a simpler way to do it? Please tell me.

Thank you in advance.

homology of a complex projective conic

Hi! Let $Q$ be a smooth conic (the zero set of a homogeneous degree 2 polynomial) in $\mathbb{P}^2(\mathbb{C})$ let $j:Q\rightarrow\mathbb{P}^2(\mathbb{C})$ be the immersion in the projective plane. How can i compute $j_*H_2(Q,\mathbb{Z})\subseteq H_2(\mathbb{P}^2(\mathbb{C}),\mathbb{Z})$? (with $H_i(X,\mathbb{Z})$ i mean the i-th singular homology group).

I know that $Q$ is homeomorphic to $\mathbb{P}^1(\mathbb{C})$ so their homology groups are isomorphic, i can restrict to the veronese embedding $\phi:\mathbb{P}^1(\mathbb{C})\rightarrow\mathbb{P}^2(\mathbb{C})$ $\phi([x_0:x_1])=[{x_0}^2: x_0x_1 :{x_1}^2]$. I also noted that the map $\phi$ is homotopic to $\psi: \mathbb{P}^1(\mathbb{C})\rightarrow\mathbb{P}^2(\mathbb{C})$, $\psi([x_0:x_1])=[{x_0}^2: 0 :{x_1}^2]$ that is a line "counted two times" so my guess is that $j_*H_2(Q,\mathbb{Z})$ can be the whole $H_2(\mathbb{P}^2(\mathbb{C}),\mathbb{Z})$ or it can be the subgroup of $H_2(\mathbb{P}^2(\mathbb{C}),\mathbb{Z})$ generated by two times its generator. How can i proceed? Is there a simpler way to do it? Please tell me.

Thank you in advance.

Homology of a complex projective conic

Let $Q$ be a smooth conic (the zero set of a homogeneous degree 2 polynomial) in $\mathbb{P}^2(\mathbb{C})$ and let $j:Q\rightarrow\mathbb{P}^2(\mathbb{C})$ be the immersion in the projective plane. How can I compute $j_*H_2(Q,\mathbb{Z})\subseteq H_2(\mathbb{P}^2(\mathbb{C}),\mathbb{Z})$? (in $H_i(X,\mathbb{Z})$, i means the i-th singular homology group).

I know that $Q$ is homeomorphic to $\mathbb{P}^1(\mathbb{C})$ so their homology groups are isomorphic, I can restrict to the Veronese embedding $\phi:\mathbb{P}^1(\mathbb{C})\rightarrow\mathbb{P}^2(\mathbb{C})$, $\phi([x_0:x_1])=[{x_0}^2: x_0x_1 :{x_1}^2]$. I also note that the map $\phi$ is homotopic to $\psi: \mathbb{P}^1(\mathbb{C})\rightarrow\mathbb{P}^2(\mathbb{C})$, $\psi([x_0:x_1])=[{x_0}^2: 0 :{x_1}^2]$ that is a line "counted two times" so my guess is that $j_*H_2(Q,\mathbb{Z})$ can be the whole $H_2(\mathbb{P}^2(\mathbb{C}),\mathbb{Z})$ or it can be the subgroup of $H_2(\mathbb{P}^2(\mathbb{C}),\mathbb{Z})$ generated by two times its generator. How can I proceed? Is there a simpler way to do it?

Thank you in advance.

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Italo
Italo
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homology of a complex projective conic

Hi! Let $Q$ be a smooth conic (the zero set of a homogeneous degree 2 polynomial) in $\mathbb{P}^2(\mathbb{C})$ let $j:Q\rightarrow\mathbb{P}^2(\mathbb{C})$ be the immersion in the projective plane. How can i compute $j_*H_2(Q,\mathbb{Z})\subseteq H_2(\mathbb{P}^2(\mathbb{C}),\mathbb{Z})$? (with $H_i(X,\mathbb{Z})$ i mean the i-th singular homology group).

I know that $Q$ is homeomorphic to $\mathbb{P}^1(\mathbb{C})$ so their homology groups are isomorphic, i can restrict to the veronese embedding $\phi:\mathbb{P}^1(\mathbb{C})\rightarrow\mathbb{P}^2(\mathbb{C})$ $\phi([x_0:x_1])=[{x_0}^2: x_0x_1 :{x_1}^2]$. I also noted that the map $\phi$ is homotopic to $\psi: \mathbb{P}^1(\mathbb{C})\rightarrow\mathbb{P}^2(\mathbb{C})$, $\psi([x_0:x_1])=[{x_0}^2: 0 :{x_1}^2]$ that is a line "counted two times" so my guess is that $j_*H_2(Q,\mathbb{Z})$ can be the whole $H_2(\mathbb{P}^2(\mathbb{C}),\mathbb{Z})$ or it can be the subgroup of $H_2(\mathbb{P}^2(\mathbb{C}),\mathbb{Z})$ generated by two times its generator. How can i proceed? Is there a simpler way to do it? Please tell me.

Thank you in advance.