How would I show that the Euler number for $ (S^1 × S^1 × S^1) $ is $0$? Would it be different if we considered $S^2 × S^1 $ or just $S^3$? If so, how?
Thanks for the help.
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The Euler characteristic of $S^1$ is zero and the Euler characteristic of a product of spaces is the product of their Euler characteristic since $\chi(S^1)=0$, $\chi(S^1\times S^1\times S^1)=\chi(S^2\times S^1)=0$.
Tsemo Aristide
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https://www.math.colostate.edu/~clayton/courses/601/601_7.pdf – Tsemo Aristide Mar 12 '19 at 15:32
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You could do that, by using the Künneth Theorem. You might want to take a deep dive into homology theory. – Lee Mosher Mar 12 '19 at 21:10
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The reason that talking about cell complexes is preferable in this context to talking about simplicial complexes is because defining a simplicial structure on a product of complexes is not a trivial task, but for a product of CW complexes it's actually pretty easy, and in particular it's not hard to count the number of cells and compute the Euler characteristic in terms of the factors. – William Mar 12 '19 at 22:34
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What are you allowed to use? – William Mar 12 '19 at 22:47
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For the torus, you can also draw the space and count the number of 0,1,2, and 3 cells. In particular, it can be constructed from a cube by identifying each pair of opposite square faces (see the picture below). This gives 1 zero cell, 3 one cells, 3 two cells, and 1 three cell, which gives an Euler characteristic of $$1-3+3-1 = 0.$$
Alvin Jin
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