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A Cup Product Example

This document in PDF: CupExample.pdf

From this picture (drawn with help from Jacob Tsimerman) we can read the following:

$$\alpha_1^\dagger = a_1^\star + r_1^\star + r_2^\star, \qquad -\beta_1^\dagger = b_1^\star + r_3^\star + r_2^\star.$$

$$\partial_2A_{1+} = r_0, \quad \partial_0A_{1+} = a_1, \qquad \alp... ...(A_{1+}) = \alpha_1(\partial_2A_{1+})\beta_1(\partial_0A_{1+}) = 0\cdot 0 = 0.$$

$$\partial_2B_{1+} = r_1, \quad \partial_0B_{1+} = b_1, \qquad \alp... ...{1+}) = \alpha_1(\partial_2B_{1+})\beta_1(\partial_0B_{1+}) = 1\cdot(-1) = -1.$$

$$\partial_2A_{1-} = r_3, \quad \partial_0A_{1-} = a_1, \qquad \alp... ...(A_{1-}) = \alpha_1(\partial_2A_{1-})\beta_1(\partial_0A_{1-}) = 0\cdot 0 = 0.$$

$$\partial_2B_{1-} = r_4, \quad \partial_0B_{1-} = b_1, \qquad \alp... ..._{1-}) = \alpha_1(\partial_2B_{1-})\beta_1(\partial_0B_{1-}) = 0\cdot(-1) = 0.$$

So $\alpha_1^\dagger\cup\beta_1^\dagger=-B_{1+}^\star$ is a generator of $H^2$.

Exercise. Verify that $\beta_1^\dagger\cup\alpha_1^\dagger=-A_{1-}^\star$ is also a generator of $H^2$, but note that in $H^2$ we have $B_{1+}^\star=-A_{1-}^\star$ so the cup product is not commutative!

The Hopf Fibration as drawn by Penrose and Rindler:

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