Quail Entry: Cohomology of a Torus

The result I want is that if T^n = (S^1)^n is the n-torus, then H^k(T^n) is spanned by a_{i_1} \smallsmile ... \smallsmile a_{i_k} with i_1 < ... < i_k.  Where a_i = p_i^*a for a \in H^1(S^1) \cong \mathbb{Z}.  In other words

H^*(T^n) = \wedge H^1(T^n) (1)

In what follows I’m suppressing the \smallsmile notation and the coefficient ring for cohomology.

The desired (1) ultimately follows from the following isomorphism.  Simply inductively apply it to Y = T^{n-1} etc.

Isomorphism (A): H^{n+1}Y \times H^nY \to H^{n+1}(S^1 \times Y).  The map sends (c,b) \mapsto 1\times c + a \times b.

here a = p^*a' where a' \in H^1S^1, and a \times b is cross product which must like box product.

consider the l.e.s for the pair (S^1 \times Y, s \times Y). Observe that it breaks up into split s.e.s as follows

...\xrightarrow{0} H^{n+1}(S^1\times Y, s\times Y) \to H^{n+1}(S^1 \times Y) \to H^{n+1} \xrightarrow{o} ... (2)

pf: Note H^{n+1}(S^1\times Y, s\times Y) \cong H^{n+1}(Y') where Y' = \frac{S^1 \times Y}{s \times Y} (note only the cases n\ge 1 are of concern, so no funny business with H^0.

Fix a base point y \in Y and identify Y' = (S^1 - s) \times Y \sqcup s \times y.  Then identity factors

Y' \subset S^1 \times Y \to Y'

in cohomology get

H^{n+1}(Y') \leftarrow H^{n+1}(S^1 \times Y) \xleftarrow{m} H^{n+1}(Y')

it follows that m is injective. This shows how l.e.s breaks up into s.e.s; that m has a section shows the s.e.s are split.

Because (2) is split, the middle term is isomorphic to the product of the left and right, so (A) ultimately follows from

Isomorphism (B): H^nY \cong H^{n+1}(I \times Y, \partial I \times Y) \cong H^{n+1}Y' \cong H^{n+1}(S^1 \times Y, s\times Y).

A similar argument (using that I \times Y def. retract to Y) shows the l.e.s for (I \times Y, \partial I \times Y) breaks up into s.e.s of the form

\xrightarrow{o} H^n(I \times Y) \to \begin{smallmatrix} H^n(1 \times Y) \\ \oplus \\ H^n(0 \times Y) \end{smallmatrix} \xrightarrow{\delta} H^{n+1}(I \times Y, \partial I \times Y) \xrightarrow{0}

Restricting \delta to one of the copies, it follows that \delta \colon H^n(o \times Y) \to H^{n+1}(I\times Y, \partial I \times Y) is an isomorphism.

Elements of H^n(o \times Y) look like 1_0 \times \beta (this is cross product) where 1_0 \in H^n(\partial I) is nonzero only on 0 \in \partial I, where it is 1.

It turns out \delta (1_0 \times \beta) = \delta(1_0) \times \beta.

pf: is verification of commutative diagram, it starts with H^k(A) \times H^l(Y) and the legs are

\xrightarrow{\delta \times 1} H^k(X,A) \times H^l(Y) \xrightarrow{\times} (3)

\xrightarrow{\times} H^{k+1}(A \times Y) \xrightarrow{\delta} (4)

both end at H^{k+l+1}(X\times Y, A \times Y).

to do it take rep in the cochain complex and do stuff like extend \phi \in C^k(A) to \bar \phi \in C^k(X) and that p^*(\delta(\psi)) = \delta(p^*(\psi)) etc.

Finally, in the case k = l = 0 and Y = pt using (3),(4) it follows that \delta(1_0) \in H^1(I, \partial I) is a generator. Hence

H^n(Y) \cong H^n(0 \times Y) \to H^{n+1}(I \times Y, \partial I \times Y)

\beta \mapsto 1_0\times \beta \mapsto \delta(1_0) \times \beta

is an iso and proves (B).

Note the argument puts in extra effort to actually determine what the various isomorphisms are and this is necessary to determine the cup product structure.

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