Compute the Pontryagin ring structure in \(H_*(SO(5); \mathbb{Z})\).
Solution
1. \(SO(5)\) has \(H_*(SO(5); \mathbb{Z}) \cong \Lambda_\mathbb{Z}[x_1, x_3]\) (exterior algebra on generators of degrees \(1\) and \(3\)), as an abelian group.
the homology of \(SO(5)\) is the exterior algebra on generators in degrees \(1\) and \(3\) (from the standard computation of the homology of \(SO(n)\)).
2. The Pontryagin product (induced by the group multiplication) makes \(H_*(SO(5))\) into a graded-commutative Hopf algebra.
the group structure gives a Hopf algebra structure on homology.
3. Since \(H_*(SO(5))\) is an exterior algebra on primitive generators \(x_1\) (degree \(1\)) and \(x_3\) (degree \(3\)), the Pontryagin ring is exactly the exterior algebra \(\Lambda_\mathbb{Z}[x_1, x_3]\).
the generators are primitive (they come from the sphere factors in the CW structure), and the Pontryagin product of primitives in an exterior algebra is the exterior product.
4. Hence the Pontryagin ring is \(H_*(SO(5)) \cong \Lambda_\mathbb{Z}[x_1, x_3]\) with \(|x_1| = 1\), \(|x_3| = 3\), and \(x_1^2 = x_3^2 = 0\), \(x_1 x_3 = -x_3 x_1\).
<1>1–<1>3.
5. Q.E.D.
4.