most recent 30 from http://mathoverflow.net 2013-05-18T10:03:50Z http://mathoverflow.net/feeds/question/32276 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/32276/homology-of-a-complex-projective-conic Italo 2010-07-17T10:11:08Z 2010-07-17T10:46:40Z

<p>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). </p> <p>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? </p> <p>Thank you in advance.</p>

http://mathoverflow.net/questions/32276/homology-of-a-complex-projective-conic/32278#32278 Robin Chapman 2010-07-17T10:25:40Z 2010-07-17T10:25:40Z

<p>The subgroup $j_* H_2(Q)$ must be generated by twice the generator of $H_2(P^2(\mathbb{C}))$ (I'm dropping the coefficient group from my notation). To see this, your map $\psi$ decomposes as the embedding from $P^1$ into $P^2$ (which induces isomorphism on $H_2$) composed with the map $[x,y]\to[x^2,y^2]$ on $P^1$. This map has degree $2$ and so acts on $H_2$ by multiplication by $2$. To see this, think of it as the squaring map on the Riemann sphere. This is homotopic to the suspension of the double covering of $S^1$ onto itself. (That is multiplication by $2$ on $H_1(S^1)$ so suspends to multiplication by $2$ on $H^2(S^2)$).</p> <p>All this generalizes to degree $d$ rational curves in $P^2$.</p>