Consider the path fibration: $K(\mathbb Z,2r-1)\rightarrow PK(\mathbb Z,2r)\rightarrow K(\mathbb Z,2r).$

Suppose that $H^*(K(\mathbb Z,2r-1);\mathbb Q)=H^*(S^{2r-1};\mathbb Q).$

We want to show that $H^*(K(\mathbb Z,2r);\mathbb Q)=\mathbb Q[a_{2r}]$.

The Gysin sequence gives that we have an isomorphism $$H^i(K(\mathbb Z,2r);\mathbb Q)\stackrel{\cup e}{\rightarrow}H^{i+2r}(K(\mathbb Z,2r);\mathbb Q)$$ where $\cup e$ is the cup product with the rational euler class.


(1) Why is it that $e$ and the fundamental class $a_{2r}\in H^{i+2r}(K(\mathbb Z,2r);\mathbb Q)\cong \mathbb Q$ are non-zero multiples of each other?

(2) What does the "fundamental class" mean in this context? and finally,

(3) I'm not clear as to how to deduce that $H^*(K(\mathbb Z,2r);\mathbb Q)=\mathbb Q[a_{2r}]$.


2) The defining property of $K(\mathbb Z,2r)$ is that for any $X$, we have $H^{2r}(X, \mathbb Z)= [X, K(\mathbb Z,2r)]$. In particular for $X=K(\mathbb Z,2r)$ we know that $H^{2r}(X, \mathbb Z)= \mathbb Z$ by Hurewitz and is generated by the element corrsponding to the identity map. This is the fundamental class. Similarly with $\mathbb Q$ coefficients.

1, 3) $H^i(K(\mathbb Z,2r);\mathbb Q)\stackrel{\cup e}{\rightarrow}H^{i+2r}(K(\mathbb Z,2r);\mathbb Q)$ means that $\cup e$ is an isomorphism from $H^0(K(\mathbb Z,2r);\mathbb Q)$ to $H^{2r}(K(\mathbb Z,2r);\mathbb Q)$. The later is generated by the fundamental class. Hence $e$ is plus or minus the fundamental class.

Now, by induction on $i$, all the $H^i$ with $i$ not multiple $2r$ are zero, while the others are all isomorpic to $\mathbb Q$ and are generated by $e$, $e^2$ etc. or equivalently by $a_{2r}$, $a_{2r}^2$ etc. That is, $H^*(K(\mathbb Z,2r);\mathbb Q)=\mathbb Q[a_{2r}]$.

  • $\begingroup$ thanks that explains it all.. $\endgroup$ – palio May 25 '11 at 16:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.