The general construction of classifying spaces says that for any group homomorphism $H \rightarrow G$ there is a associated map between the classifying spaces $BH \rightarrow BG$.

I am trying to see how it works in explicit cases. It is known that $BS^1 = \mathbb{C}P^\infty$, $B\mathbb{Z}/2 = \mathbb{R}P^\infty$ . Also the cohomology with field coefficients of those spaces can be found in many sources. In general, the inclusion of $\mathbb{Z}/2 \rightarrow S^1$ induces a map in cohomology

$$ H^*(\mathbb{C}P^\infty) \cong k[x] \rightarrow H^*(\mathbb{R}P^\infty) \cong k[y]$$

where $deg(x) = 2$ and $deg(y) =1$. My guess is that the map should send $x$ to $y^2$, but I don't really know how to prove it.


Recall that $\mathbb{C}P^n$ is the quotient of $S^{2n+1}\subseteq\mathbb{C}^{2(n+1)}$ by the free action of $S^1\subseteq\mathbb{C}$. At the same time $\mathbb{R}P^{2n+1}$ is the quotient of $S^{2n+1}$ by the free $\mathbb{Z}_2$-action. Since $\mathbb{Z}_2\cong\{\pm 1\}\subseteq S^1$, the quotient $\gamma_\mathbb{R}:S^{2n+1}\rightarrow\mathbb{R}P^{2n+1}$ factors over $\gamma_\mathbb{C}:S^{2n+1}\rightarrow\mathbb{C}P^n$ to give a map $\varphi:\mathbb{R}P^{2n+1}\rightarrow\mathbb{C}P^n$, which has the structure of a principal $S^1/\mathbb{Z}_2\cong S^1$ bundle.

If we write the cosets in $\mathbb{R}P^{2n+1}$ as $[x_0,y_0,x_1,y_1,\dots, x_n,y_n]$, with each $x_i,y_i\in\mathbb{R}$, then

$\varphi([x_0,y_0,x_1,y_1,\dots, x_n,y_n])=[x_0+iy_0,\dots,x_n+iy_n]$.

Stabilising this map we get $\varphi:\mathbb{R}P^\infty\rightarrow\mathbb{C}P^\infty$, which acts by representing a point in $\mathbb{R}P^\infty$ by a point in $\mathbb{R}P^{2N+1}$ for suitably large $N$, and applying the previous map to get a point in $\mathbb{C}P^N\subseteq\mathbb{C}P^\infty$.

The map $\varphi:\mathbb{R}P^\infty\rightarrow\mathbb{C}P^\infty$ is a principal $S^1$-bundle, and so one may use, for instance, a Gysin sequence to calculate its action on cohomology. However its even easier than this. Since $H^*(\mathbb{R}P^\infty;\mathbb{Z}_2)\cong\mathbb{Z}_2[y_1]$ and $H^*(\mathbb{C}P^\infty;\mathbb{Z}_2)\cong\mathbb{Z}_2[x_2]$ are polynomial algebras, it suffices to determine the action of $\varphi^*$ on the generator $x_2$.

Now the inclusions of $S^2\cong\mathbb{C}P^1\subseteq\mathbb{C}P^\infty$ and $SO(3)\cong\mathbb{R}P^3\subseteq\mathbb{R}P^\infty$ induce isomorphisms on cohomology in degree $2$, so it suffices to determine what happens with $\varphi:SO(3)\rightarrow S^2$. This last map has many nice descriptions. For instance it is the projection of the fibre sequence $SO(2)\rightarrow SO(3)\rightarrow S^2$ given by the canonical action of $SO(3)$ on the unit sphere in $\mathbb{R}^3$. Alternatively it is the unit sphere bundle of the tangent bundle $TS^2$ of $S^2$.

Since the Euler characteristic may be given by evaluating the Euler class on the fundamental class, we get $\chi(S^2)=<e(S^2),[S^2]>=2$. However, the Euler class is also the top Chern class (which is the only Chern class in this case). This tells us that unit sphere bundle of the tangent bundle is classified by a degree 2 map $2:S^2\rightarrow BS^1$, and we conclude that there is a homotopy fibration


If we study the Serre spectral sequence with mod 2 coefficients for this fibration we find an exact sequence

$0\rightarrow H^2(S^2;\mathbb{Z}_2)\xrightarrow{\varphi^*} H^2(SO(3);\mathbb{Z}_2)$

since $2^*=0:H^2(BS^1;\mathbb{Z}_2)\rightarrow H^2(S^2;\mathbb{Z}_2)$ is the zero homomorphism. This tells us that $\varphi^*$ in injective in degree 2. Since there is only one option for its image we have


and so your conjecture was correct.

If you don't want to invoke the Serre spectral sequence, consider another method. There are cofibration sequences

$S^1\xrightarrow{2}S^1\rightarrow \mathbb{R}P^2\xrightarrow{q}S^2$

$S^2\xrightarrow{\gamma_\mathbb{R}}\mathbb{R}P^2\xrightarrow{j} \mathbb{R}P^3$.

By an argument in integral cohomology the composition $q\circ\gamma_\mathbb{R}\simeq\ast$ is homotopically trivial. Since $\gamma_\mathbb{R}$ is a cofibration we get an extension over the cofiber $\mathbb{R}P^3\rightarrow S^2$. It's a little trickier to show rigorously, but this extension may be chosen to be $\varphi$.

Now $q^*:H^2(S^2;\mathbb{Z}_2)\xrightarrow{\cong}H^2(\mathbb{R}P^2;\mathbb{Z}_2)$ is an isomorphism. However we have just seen that

$q^*=(\varphi\circ j)^*=j^*\circ\varphi^*$

so in particular $\varphi^*$ cannot be trivial, and thus for algebraic reasons must be an isomorphism. Hence


  • $\begingroup$ Thanks @Tyrone everything is pretty much clear now. $\endgroup$ – C. Zhihao Mar 15 '18 at 16:39

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