5
$\begingroup$

Hatcher writes in his book "Homology calculations are often simplified by taking coefficients in a field, usually $\mathbb{Q}$ or $\mathbb{Z}_p$ for $p$ prime".

I wonder why one only consideres $\mathbb{Q}$ as field of char. $0$. I have seen proofs where one takes coefficients in $\mathbb{Q}$, but I have seen none where one takes coefficients in $\mathbb{R}$ or $\mathbb{C}$. Is there a reason for this? Does coefficients in $\mathbb{Q}$ give more information than coefficients in other fields of char. $0$? Or is it simply a convention to take $\mathbb{Q}$-coefficients?

$\endgroup$
5
  • $\begingroup$ It could have something to do with the fact that tensoring a $\mathbb{Z}$-module with $\otimes_{\mathbb{Z}}\mathbb{Q}$ is the ``simplest'' way to kill torsion? I think Hatcher does some rational homotopy theory so this is his preference.. The important properties of $\mathbb{R}$ (completeness) and $\mathbb{C}$ (algebraic closedness) aren't important for doing homology stuff. $\endgroup$
    – AnonymousCoward
    Jun 15, 2017 at 11:36
  • $\begingroup$ Why does this kill torsion? And when tensoring for example with $\mathbb{R}$ does this have torsion? $\endgroup$
    – user450093
    Jun 15, 2017 at 11:46
  • 1
    $\begingroup$ I'll just answer the second question: tensoring with $\mathbb{R}$ kills torsion just as well. One way of seeing that (assuming the truth of AnonymousCoward's comment) is that tensoring with $\mathbb{R}$ is the same as tensoring with $\mathbb{Q}$ and then tensoring the result with $\mathbb{R}$. So torsion gets killed in the first of these two steps and then neatly stays killed in the next step. However this means that one could consider the second step in this two-step process as overkill. The important word in A.C.'s answer is hence "simplest" $\endgroup$
    – Vincent
    Jun 15, 2017 at 11:52
  • 2
    $\begingroup$ $\Bbb R$ isn't entirely ignored. The whole deRham theory on smooth manifolds constructs a cohomology theory that looks like singular cohomology with real coefficients. $\endgroup$
    – John Hughes
    Jun 15, 2017 at 11:52
  • 2
    $\begingroup$ I might say it is because of the universal coefficient theorem. Loosely speaking the theorem states that if you know the homology with coefficient in $\mathbb{Q}$, then you know the homology with coefficient in any extension of $\mathbb{Q}$ (just take the tensor product). $\endgroup$
    – user10676
    Jun 15, 2017 at 13:45

1 Answer 1

Reset to default
5
$\begingroup$

If a cycle $\alpha$ represents a $p$-torsion class so that $\alpha+\ldots\alpha$ ($p$ times) represents the zero class, then working with $\mathbb Q$ coefficients one can form the cycle $\beta=\frac{1}{p}\alpha$ so that $\beta+\ldots+\beta$ ($p$ times) will also necessarily represent the zero class. But $\alpha=p\beta$ and therefore $[\alpha]=0$. Thus you can't have torsion when the coefficients are in $\mathbb Q$. The same goes for the other fields you mentioned but you don't gain any information, though in de Rham theory you have to work with real coefficients, as pointed out in the comments.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .