Minimal Degree of map $S^2\times S^2\mapsto \mathbb{CP}^2$

Question

I am having troubles finding the minimal d such there is a map of such degree from $S^2\times S^2\mapsto \mathbb P^2$.

I know that the cohomology ring of $\mathbb P^2$ is thus I know that the degree will be given by $f^*(x^2)=(f^*(x))^2$ which will be the square of a class in $H^2(S^2\times S^2)$.

Thus $f^*(x)^2=(ha_1+ka_2)^2=2hk a_1a_2$ where $a_1$ and $a_2$ are the classes corresponding to $S^2\times pt$ and $pt\times S^2$.

How can I finish from here?

Thanks

Edit: It is complex projective space.

Answer 1

$a_1 a_2$ is a generator of $H^4(S^2\times S^2)$, so setting $h=k=1$ the lowest possible degree is $2$. Map $S^2\times S^2=\mathbb{C}P^1\times \mathbb{C} P^1\to \mathbb{C} P^2$ via $f([z:w],[z':w'])=[zz'+ww':zw':wz']$. This is surjective, checking directly that $\{z\neq 0\}\times \{z'\neq 0\}$ hits everything but $[0:0:1],[0:1:0]$ and finding specific preimages for the last two points. $h$ and $k$ are $f^*\phi(\eta)$, $f^*\phi(\kappa)$, where respectively $\eta$ and $\kappa$ are 2-cycles represented by $\mathbb{C}P^1\times \{[0:1]\}$ and $\{[0:1]\}\times \mathbb{C}P^1$. These are equal to $\phi(f_*(\eta))$ etc, which is computed as the intersection number of the Poincare dual of $\phi$ with $f_*(\eta)$ and $f_*(\kappa)$. $\phi$ is dual to $\mathbb{C}P^1\subset \mathbb{C}P^2$ as $\{[0:a:b]\}$.

We compute $f_*(\eta)=\{[w:z:0]\}$ and $f_*(\kappa)=\{[w':z':0]\}$, each of which is an embedded complex submanifold intersecting $PD(\phi)$ in the point $\{[0:1:0]\}$. Intersections of complex manifolds are either trivial or transverse and positive, so we get $h=k=1$. Note also that this is one of the maps you could get via apurv's technique, since each 2-cell of $S^2\times S^2$ is mapped onto a 2-cell of $\mathbb{C}P^2$ via a degree-1 map.

Answer 2

$S^2 \times S^2$ can be given a cell structure with 1 0-cell, 2 2-cells and 1 4-cell and $\mathbb{CP}^2$ can be given a cell structure with 1 0-cell, 1 2-cell and 1 4-cell.

Now map each of the 2-cells of $S^2 \times S^2$ to the 2-cell of $\mathbb{CP}^2$ using the identity map (this will make your $h,k$ both 1).

Because $\pi_3(\mathbb{CP}^2) = \pi_4(\mathbb{CP}^2) = 0$, there is no obstruction to extending this map to the entire $S^2 \times S^2$.

Remains to determine $h,k$. For this look at the map induced on the cellular chain complex to conclude that $h=k=1$.

Answer 3

$\mathbb{C}P^n$ is the $n$th symmetric product of $\mathbb{C}P^1$.

The map from the $n$th product of $\mathbb{C}P^1$ to $\mathbb{C}P^n$, which is the quotient map by the symmetric group, has $n!$ pre-images. All pre-images have degree $1$, since permutations are holomorphic maps, and holomorphic maps are always orientation preserving. So the quotient map has degree $n!$.

In this case, $\mathbb{C}P^1 \times \mathbb{C}P^1$ to $\mathbb{C}P^2$ by quotient map has degree $2$.

To see why $\mathbb{C}P^n$ is the nth symmetric product of $\mathbb{C}P^1$, note that a point in $\mathbb{C}P^1$ is the unique solution of a linear homogenous polynomial. e.g. $[a:b]$ is the solution to the homogenous polynomial $bx - ay$. Given $n$ points in $\mathbb{C}P^1$, multiply the corresponding $n$ linear homogenous polynomials together to get a degree $n$ homogenous polynomial. The $n+1$ coefficients of this polynomial become a single point in $\mathbb{C}P^n$, since global multiplication of a homoegenous polynomial doesn't change its roots. Finally, each degree $n$ homogenous polynomial has $n$ unordered roots, which is where the action of the symmetric group comes from.