Show, by generalising the method of question 1, that the cohomology ring of the \(n\)-torus \(T^n\) is an exterior algebra generated by \(H^1\). In other words, there is a basis \(\alpha_1, \ldots, \alpha_n\) for \(H^1(T^n;\mathbb{Z})\) such that \[ H^*(T^n;\mathbb{Z}) = \Lambda^*[\alpha_1, \ldots, \alpha_n]. \] (Recall: this is the free associative (but non-commutative) algebra in the \(\alpha_i\)’s, with relations \(\alpha_i\alpha_j = -\alpha_j\alpha_i\). We write \(\wedge\) for the product in this algebra. The dimension of \(H^k\) is then \(\binom{n}{k}\) with basis \(\{\alpha_{i_1} \wedge \alpha_{i_2} \wedge \cdots \wedge \alpha_{i_k} : i_1 < i_2 < \cdots < i_k\}\).)
“Question 1” refers to problem 1 of the same source sheet, which computes the cohomology ring of \(T = S^1 \times S^1\) from the definition of singular cohomology.
Solution
1. The integral cohomology ring of the circle is\[ H^*(S^1;\mathbb Z)\cong\Lambda(\alpha), \qquad |\alpha|=1. \]
There is one copy of \(\mathbb Z\) in degrees \(0\) and \(1\) and no higher cohomology, so the square of the degree-one generator is zero.
2. The cohomological Künneth theorem gives an isomorphism of graded rings\[ H^*(X\times S^1;\mathbb Z) \cong H^*(X;\mathbb Z)\otimes H^*(S^1;\mathbb Z) \]whenever \(H^*(X;\mathbb Z)\) is free abelian.
For the torus induction all relevant cohomology groups are free, so the Tor terms vanish. Compatibility of the cross product with cup products gives the tensor-product ring structure, with the Koszul sign rule.
3. Inductively,\[ H^*(T^n;\mathbb Z) \cong\bigotimes_{i=1}^n\Lambda(\alpha_i) \cong\boxed{\Lambda^*(\alpha_1,\ldots,\alpha_n)}, \]where \(|\alpha_i|=1\).
Use \(T^n=T^{n-1}\times S^1\) and apply 2 repeatedly. The graded tensor product of exterior algebras on degree-one generators is the exterior algebra on their union; distinct generators anticommute and each squares to zero.
4. Consequently\[ H^k(T^n;\mathbb Z)\cong\mathbb Z^{\binom nk}, \]with basis\[ \alpha_{i_1}\wedge\cdots\wedge\alpha_{i_k}, \qquad i_1<\cdots<i_k. \]
A degree-\(k\) exterior monomial is determined by choosing \(k\) distinct generators from the \(n\) available generators.