# 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.