# Quail Entry: Cohomology of a Torus

The result I want is that if is the -torus, then is spanned by with . Where for . In other words

(1)

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

The desired (1) ultimately follows from the following isomorphism. Simply inductively apply it to etc.

Isomorphism (A): . The map sends .

here where , and is cross product which must like box product.

consider the l.e.s for the pair . Observe that it breaks up into **split** s.e.s as follows

(2)

pf: Note where (note only the cases are of concern, so no funny business with .

Fix a base point and identify . Then identity factors

in cohomology get

it follows that is injective. This shows how l.e.s breaks up into s.e.s; that 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): .

A similar argument (using that def. retract to ) shows the l.e.s for breaks up into s.e.s of the form

Restricting to one of the copies, it follows that is an isomorphism.

Elements of look like (this is cross product) where is nonzero only on , where it is .

It turns out .

pf: is verification of commutative diagram, it starts with and the legs are

(3)

(4)

both end at .

to do it take rep in the cochain complex and do stuff like extend to and that etc.

Finally, in the case and using (3),(4) it follows that is a generator. Hence

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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- January 18, 2010 / 2:33 pm

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