I was hoping for an explicit reference to the description of the mod 2 cohomology of a cyclic group $C_{2r}=\langle t \rangle$ of even order in terms of Stiefel-Whitney classes, i.e., that

$H^*(BZ_{2r};\mathbb{F}_2)=\mathbb{F}_2[x,y]/(x^2-ry)$, where $x=w_1(\chi)$ is the 1st Stiefel-Whitney class of the representation $\chi:Z_{2r} \rightarrow \mathbb{R}$ sending $t$ to -1, and $y=w_2(\rho)$ is the 2nd Stiefel-Whitney class (1st Chern class reduced mod 2) of the standard representation $\rho: \mathbb{Z}_{2r}\rightarrow \mathbb{C}$.


  • $\begingroup$ Are you sure of your statements? The 2nd Stiefel-Whiteney class of any 1-dimensional representation is trivial, and 1 st Chern class reduces to $w_1^2$. $\endgroup$
    – user43326
    Jan 14, 2015 at 8:06
  • $\begingroup$ @user43326: I don't think your second statement can be correct, since $w_1=0$ for any complex vector bundle. $\endgroup$
    – Mark Grant
    Jan 14, 2015 at 11:47
  • $\begingroup$ @user43326: I think so: $c_1(\rho)\in H^2(BZ_{2r};\mathbb{Z})=\mathbb{Z}_{2r}$ represents 1, so the mod 2 reduction (which is the 2nd Stiefel-Whitney class of the real 2-dimensional bundle) represents 1 in $H^2(BZ_{2r};\mathbb{F}_2)=\mathbb{F}_2$ as well. $\endgroup$
    – Fred
    Jan 14, 2015 at 13:16
  • $\begingroup$ @MarkGrant ,@Fred , you are right, sorry. $\endgroup$
    – user43326
    Jan 14, 2015 at 13:35

1 Answer 1


It is true that statements like this are completely classical and yet hard to find stated just like that in the literature! If noone has a reference, I would suggest to go as follows :

-- the cohomology groups are given in the book by Cartan and Eilenberg (oldest reference I can think of!)

-- the cohomology ring is obtained by looking at the exact sequence $G \to S^1 \to S^1$ where $G$ is your cyclic group of order $n$ and the map $S^1 \to S^1$ is multiplication by $n$; there is a fibration $G \to S^1 \to S^1 \to BG \to BS^1 \to BS^1$, and the "Gysin exact sequence" of the $S^1 \to BG \to BS^1$ part gives you the multiplicative structure. You have to use that $BS^1 = \mathbb{P}^1(\mathbb{C})$. In a nutshell: it is standard algebraic topology.

--this next step is probably more relevant to your question. It is always the case that $H^1(G, \mathbb{F}_2) = Hom(G, \mathbb{F}_2)$ (reference: any book on cohomology, such as Cartan-Eilenberg). This identifies the elements in degree $1$ as Stiefel-Whitney classes by definition (of course $\mathbb{F_2} = O_1(\mathbb{R})$...)

Then, the exponential exact sequence $0\to \mathbb{Z} \to \mathbb{C} \to \mathbb{C}^\times \to 1$ gives you $H^2(G, \mathbb{Z}) = Hom(G, S^1) ~~~(= Hom(G, GL_1(\mathbb{C})))$ when $G$ is finite. From there you can identify the elements in degree $2$.

OK this is way too long. I agree that a good reference would be better.


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