Ytukyg

In other words, how to show that $mathbb{RP}^2$ is not contractible in $mathbb{CP}^2$.

Any hints would be appreciated.

I will accept Georges' challenge.

1) Cohomological. The fundamental class of the real projective plane defines a class $[Bbb{RP}^2] in H_2(Bbb{CP}^2;Bbb Z/2)$. To see that it is non-trivial, recall that the cup product in cohomology is Poincare dual to the intersection product in homology. If we can show that $[Bbb{RP}^2] frown [Bbb{RP}^2] = [pt]$, then we have what we want. Take one of these classes to be represented by ${[x_0 : x_1 : x_2] mid x_i in Bbb R}$ and another to be ${[x_0 : omega x_1 : omega^2 x_2] mid x_i in Bbb R}$ where $omega$ is your favorite nontrivial cube root of unity. (These two embeddings are homotopic, hence homologous.) For $[x_0 : x_1 : x_2] = [y_0 : omega y_1 : omega^2 y_2]$, where all $x_i$ and $y_i$ are real, we must have $x_1 = y_1 = x_2 = y_2 = 0$, and hence the only intersection point is $[1:0:0]$. The two tangent spaces of these embeddings are $Bbb R^2 subset Bbb C^2$ and $Bbb R omega oplus Bbb R omega^2 subset Bbb C^2$, which sum to the whole $Bbb C^2$, so this is a transverse intersection, and the product is indeed one point (and hence nontrivial, as desired).

2) An immediate route to the above computation comes from symplectic geometry: $Bbb{RP}^2$ is Lagrangian and hence the normal bundle is isomorphic to its tangent bundle; so the self-intersection number is $chi(Bbb{RP}^2) = 1$.

3) Or observe that the restriction of the tautological line bundle of $Bbb{CP}^2$ is the complexification of the tautological line bundle on $Bbb{RP}^2$, and hence has $c_1(ell_{Bbb C}) = beta w_1(ell_{Bbb R}) neq 0 in H^2(Bbb{RP}^2;Bbb Z/2)$. Here we use either the general formula for a real vector bundle, $c_{2i+1}(E_{Bbb C}) = beta w_{4i+1}(E)$, or think of the Bockstein $beta$ as the composite $H^1(X;Bbb Z/2) to H^1(X; U(1)) to H^2(X;Bbb Z)$, sending a flat real line bundle to a flat complex line bundle with the same transition functions in $pm 1 subset U(1)$ and then forgetting the flat structure.

4) Amusingly, because $pi_2(Bbb{CP}^2) to H_2(Bbb{CP}^2)$ is an isomorphism, we see that we may compute $pi_2(Bbb{RP}^2) to pi_2(Bbb{CP}^2)$ as the map $pi_2(Bbb{RP}^2) to H_2(Bbb{RP}^2) to H_2(Bbb{CP}^2)$; of course, $H_2(Bbb{RP}^2;Bbb Z) = 0$, so the map is not detected on $pi_2$. Further, by considering the diagram comparing the fiber sequences $S^1 to S^5 to Bbb{CP}^2$ and $O(2) to S^3 to Bbb{RP}^2$ and the fact that the inclusion $S^3 to S^5$ is null-homotopic, we see that the map is zero on all homotopy groups.

There is still something the homotopy theory can say, though. If $Bbb{RP}^2 to Bbb{CP}^2$ was null-homotopic, there would in particular be a lift of the initial map to a map to $S^5$ (using the homotopy lifting lemma), and in particular by restriction a section $Bbb{RP}^2 to S^3$ of the original fibration. There are a number of reasons this is impossible: one is that there are no embedded $Bbb{RP}^2$s in $S^3$.

On the relation $c_1(ell otimes Bbb C) = beta w_1(ell)$ for real line bundles $ell$.

We may define a map $H^1(X; Bbb Z/2) to H^1(X; U(1)) to H^2(X;Bbb Z)$, the first by inclusion of coefficients $pm 1 to U(1)$ and the second by the boundary map in the long exact sequence $H^1(X; Bbb R) to H^1(X; U(1)) to H^2(X; Bbb Z)$. In terms of calculability, that long exact sequence helps you identify $H^1(X; U(1))$ as an extension of $H^2_{text{tors}}(X; Bbb Z)$ by $U(1)^{b_1}$, and the boundary map is just projection to the group of components $H^2_{text{tors}}(X;Bbb Z)$, which then includes into $H^2(X;Bbb Z)$. In any case, all you need to know to calculate the kernel of this map is which classes of $H^1(X; Bbb Z/2)$ lift to classes of $H^1(X;Bbb Z)$, and both of these have nice definitions in terms of homomorphisms out of the fundamental group.

At the level of cocycles, if $sigma$ is a $Bbb Z/2$-cocycle and $tilde sigma$ is a lift to a $Bbb Z$-cochain, then just as we considered $sigma$ as a $U(1)$-cocycle above we may consider $tilde sigma$ as a $Bbb R$-cochain. The boundary map $H^1(X; U(1)) to H^2(X; Bbb Z)$ takes a lift of $sigma$ as a $U(1)$-cocycle to a $Bbb R$-cochain $tilde sigma$, and restricts $partial tilde sigma$ to a $Bbb Z$-valued cocycle. By definition, $beta[sigma] = [partial tilde sigma]$, and so we see that the procedure outlined above gives another definition of the Bockstein.

Now interpreting the three groups in terms of line bundles, $H^1(X;Bbb Z/2)$ is isomorphic to the group of real line bundles, $H^1(X; U(1))$ is isomorphic to the group of flat complex line bundles (those equipped with some system of local trivializations with constant transition maps), and $H^2(X; Bbb Z)$ is the group of complex line bundles. Writing $mathcal U(1)$ for the sheaf of continuous maps to $U(1)$ and similarly $mathcal R$ for continuous maps to $Bbb R$, the long exact sequence of cohomology for $Bbb Z to mathcal R to mathcal U(1)$ identifies $H^1(X; mathcal U(1)) = H^2(X; Bbb Z)$, and comparing this sequence to that of $Bbb Z to Bbb R to U(1)$ identifies the map $H^1(X; U(1)) to H^1(X; mathcal U(1)) cong H^2(X; Bbb Z)$ as the boundary map we described at the beginning.

In particular, $H^1(X;Bbb Z/2) to H^1(X; U(1))$ sends a real line bundle, perhaps described as a collection of local trivializations with transition maps in $Bbb Z/2$, to a complex line bundle with the same local trivializations and the same transition maps $Bbb Z/2 subset U(1)$; we consider this as a flat line bundle. (This is sending $ell$ to $ell otimes Bbb C$ equipped with a particular flat structure induced by that of $ell$.) The above paragraph identifies the next map $H^1(X; U(1)) to H^2(X; Bbb Z)$ as forgetting the flat structure. Thus the composite $H^1(X; Bbb Z/2) to H^2(X;Bbb Z)$ sends $ell mapsto ell otimes Bbb C$, considered as real and complex line bundles, respectively. Because $w_1$ and $c_1$ give the isomorphisms between groups of line bundles and these cohomology groups, and we identified that composite above as the Bockstein $beta$, we have seen what we wanted.