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I have some trouble to compute the homology of the torus with coefficients in $\Bbb F_p$ for $p$ a prime number. In particular I have a problem for $H_1$ :

1) The first way to compute it is to use the UCT $$H_1(S^1 \times S^1, \Bbb F_p) = H_1(S^1 \times S^1) \otimes \Bbb F_p \;\oplus\; \text{Tor}(H_0(S^1\times S^1); \Bbb F_p)\\ =\Bbb F_p ^2 \oplus \text{Tor}(\Bbb Z; \Bbb F_p) = \Bbb F_p ^2$$

2) The second way is to use the fact that $\Bbb F_p \otimes \Bbb F_p = \Bbb F_p$ so that $$H_1(S^1 \times S^1, \Bbb F_p) = H_1(C'_*(S^1, \Bbb F_p) \otimes C'_*(S^1, \Bbb F_p)) \\\cong \bigoplus_{r+s=1} H_r( S^1 ,\Bbb F_p) \otimes H_{s}( S^1 ,\Bbb F_p) \oplus \bigoplus_{r+s=0} \text{Tor}(H_r( S^1 ,\Bbb F_p) ; H_{s}( S^1 ,\Bbb F_p)) \\= H_0 \otimes H_1 \;\oplus\; H_1 \otimes H_0 \;\oplus\; \text{Tor}(H_0,H_0) \\= \Bbb F_p \otimes \Bbb F_p \;\oplus\; \Bbb F_p \otimes \Bbb F_p \;\oplus\; \Bbb F_p = \Bbb F_p^3 $$ where $C'_*(E,A) := C_*(E) \otimes A$, $C_*(E)=$ set of singular chains in $E$, and $A$ any abelian group. Here I used Eilenberg-Zilber and Künneth theorems.

Here is my question:

Why do I find $\Bbb F_p^2$ with the first method, and $\Bbb F_p^3$ with the second one?

Thank you!

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When you apply the Kunneth theorem, the Tor is being taken over $\mathbb{F}_p$, not over $\mathbb{Z}$, and so it vanishes. Over a field the Kunneth theorem just says that the homology of a product is a graded tensor product of homologies.

The assumption that $p > 3$ does not help you in any way.

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  • $\begingroup$ Thank you for your answer! Yes, but in the first line of 2) : $$H_1(\;C'_*(S^1, \Bbb F_p) \otimes C'_*(S^1, \Bbb F_p)\;)$$ I take the homology with coefficients in the integers. So then I should apply Künneth with $\text{Tor}_1^{\Bbb Z}$, in my opinion. $\endgroup$ – Watson Jul 3 '16 at 18:22
  • $\begingroup$ @Watson: the version of the Kunneth theorem I assume you're using (en.wikipedia.org/wiki/…) computes homology with coefficients in a PID $R$ in terms of Tor over $R$. Here $R = \mathbb{F}_p$. $\endgroup$ – Qiaochu Yuan Jul 4 '16 at 6:03

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