Computing (co)homology can be rather hard without the appropriate tools. The Serre spectral sequence is one such tool. We compute the important example of , and also provide a different proof of Proposition 3.22 in Hatcher’s Algebraic Topology, namely the computation of , where is the reduced James product (see section 4.J, for James). (Note that the cohomology rings are isomorphic since .) This example will illustrate the usefulness of the Serre spectral sequence. Everything here has coefficients in .
Recall that for a fibration such that is simply connected, if , then the Serre spectral sequence of says that . The universal coefficient theorem says that there is a natural short exact sequence that splits (albeit not naturally). Suppose or is torsionfree; then . This spectral sequence has a product, in the following sense. Suppose , and are Serre fibrations. Then a diagram:
Gives a product .
Our first goal is to study .
Theorem: , where the element is the image of under .
Proof. To prove this, consider the path-loop fibration . The path space is contractible. Hence the Serre spectral sequence says that . We know that if , and is zero otherwise. Consider the generator of ; then . Note that almost all differentials are zero, except . We therefore have:
We claim that all the nonvanishing maps are isomorphisms.
To see this, note that . But is contractible, so unless . Therefore if was not an isomorphism, we would get nontrivial elements in , which is weird. Therefore, if is a multiple of , and is zero otherwise. Say that generates . Then the generator of is , and thus the differential goes .
Now we will study the multiplicative structure on . We have to choose three Hopf fibrations; indeed, we can pick the most obvious ones: , , and . It can easily be checked that the desired diagram commutes, so that we have a product on spectral sequences. For the fibration , it is rather obvious that for all and for , and all differentials are zero.
Therefore, if now denotes the Serre spectral sequence for , the multiplication is . Since (under the UCT isomorphism ), and since we know that , it follows that because and . Thus, by induction, . The desired result follows.
Our next goal is to compute . Let denote the exterior algebra on one generator, i.e., , and let denote the divided polynomial algebra.
Theorem: If is odd, then where . If is even, then where and .
Proof. From the cohomological spectral sequence, we compute (like above) that is if , and is zero otherwise. The only nontrivial differential is . Arguing as above, we find that if is a multiple of , and is otherwise.
Let generate ; then generates . Consider a generator . Then sends , which is also (clearly) a generator of since generates .
Let’s consider first the case that is even. Then satisfies (by graded commutativity, since is odd). Therefore, can be written as for some integer . This means that . But from the Leibniz formula, we also know that . This is good, because (again because ). Thus is , i.e., if is even, else if is odd.
Also, commutes with everything, i.e., . Now, , so for some integer . It then follows that , so that . We determined that above (since ), so , i.e., by induction. To summarize:
Let’s now consider the case that is odd. Then commutes with everything since is even. Consequently, . Now, for some . This means that , i.e., . This therefore means that:
All of this is also at http://www.mit.edu/~sanathd/(co)homology-of-omega-S%5En.pdf.