If \(T^n\) is the \(n\)-dimensional torus, the product of \(n\) circles, show that the Pontryagin ring \(H_*(T^n; \mathbb{Z})\) is the exterior algebra \(\Lambda_\mathbb{Z}[x_1, \dots, x_n]\) with \(|x_i| = 1\).
Solution
1. \(H_*(S^1; \ZZ) = \ZZ \oplus \ZZ x\) with \(|x| = 1\), and the Pontryagin product (induced by the group structure of \(S^1\)) satisfies \(x^2 = 0\).
\(H_1(S^1) = \ZZ\) generated by \(x\), and \(x \cdot x = 0\) since \(H_2(S^1) = 0\).
2. By the Künneth theorem, \(H_*(T^n) \cong H_*(S^1)^{\otimes n}\).
\(T^n = (S^1)^n\), and the Künneth theorem computes the homology of a product.
3. The Pontryagin product on \(H_*(T^n)\) is the tensor product of the Pontryagin products on each factor (the group structure of \(T^n\) is the product of the group structures).
the multiplication on \(T^n\) is coordinatewise.
4. Hence \(H_*(T^n) \cong \Lambda_\ZZ[x_1, \ldots, x_n]\), the exterior algebra on generators \(x_1, \ldots, x_n\) of degree \(1\) (where \(x_i\) is the generator of \(H_1\) of the \(i\)-th circle factor).
<1>1–<1>3 (the tensor product of exterior algebras on one generator each is the exterior algebra on all the generators).
5. Q.E.D.
4.