© | Dror Bar-Natan: Classes: 2004-05: Math 1300Y - Topology: (96) Next: A Genus 2 Surface by Mister Bailey
Previous: Homework Assignment 12

A Cup Product Example

This document in PDF: CupExample.pdf

A Fully Marked Surface Piece

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

$\displaystyle \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.
$

$\displaystyle \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.
$

$\displaystyle \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.
$

$\displaystyle \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.
$

$\displaystyle \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:

The Hopf Fibration

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Dror Bar-Natan 2005-04-28