Allen Knutson said here in comments below the question that

I generally regard torsion in (co)homology as a sign that one should be computing K-theory instead, which has less of it.

I know one or two things about torsion groups, examples of cohomology groups that has non zero torsion part, some definitions $K$-theory.

Can some one help me to understand why torsion in cohomology should remind about computing $K$-theory? Any references are welcome.

Some not very well posed question is : what are (some of the) other signs that one should be computing K-theory?

  • $\begingroup$ Thank you @Todd Trimble.. :) $\endgroup$ Oct 2, 2019 at 11:30

1 Answer 1


Usually you are computing $H^*(X)$ or $K^*(X)$ for a reason; for example if $H^*(X)\not\simeq H^*(Y)$ then you know that $X$ and $Y$ are not homotopy equivalent, but if $H^*(X)\simeq H^*(Y)$ and this object has a complicated structure, then it is a reasonable guess that $X\simeq Y$. You can do this kind of argument with ordinary cohomology or complex $K$-theory or Morava $E$-theory or any of various other invariants; you just choose whichever one seems likely to be effectively calculable and sufficiently rich as to provide the information that you need. A common case is where $X$ is equivalent or closely related to a Borel construction $(Z\times EG)/G$ for some finite group $G$ and some $G$-space $Z$. In those cases, $H^*(X)$ will be related to $H^*(BG)$, and $H^*(BG)$ is typically an ugly sort of ring with many generators and random-looking relations and lots of torsion. (In fact, if $|G|=n$ and $a\in H^i(BG)$ with $i>0$ then $na=0$.) On the other hand, the Atiyah-Segal completion theorem says that $KU^0(BG)$ is a completion of the representation ring $R(G)$, and $R(G)$ is small, nicely structured and free of torsion. So $K$-theory is likely to be a more tractable invariant in these cases. The Morava $E$-theory ring $E^0(BG)$ is often free of torsion as well, and is a richer invariant than $KU^0(BG)$.

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    $\begingroup$ Thank you... :) I got most of it except the last four lines that starts with something about Atiyah-Segal completion theorem.. I did not heard about Atiyah-Segal completion theorem before , I got statement from wikipedia but did not understand much.. can you suggest some reference that help to understand what you said in last 4 lines.. $\endgroup$ Oct 2, 2019 at 12:08
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    $\begingroup$ The Atiyah-Segal completion theorem provides a clean statement about a forbidding space, $K(\mathbb RP^\infty)=\mathbb Z\oplus\mathbb Z_2$. But this is pretty much equivalent to a crude statement about manifolds: $K(\mathbb RP^{2n})=\mathbb Z\oplus \mathbb Z/2^n$, compared to $H(\mathbb RP^{2n})=\mathbb Z\oplus(\mathbb Z/2)^n$. Both have torsion, indeed groups of the same size, but the number of generators is a measure of "how much," a way in which $K$-theory has "less." $\endgroup$ Oct 3, 2019 at 0:47
  • $\begingroup$ @Ben can you suggest some references please.. $\endgroup$ Oct 3, 2019 at 13:37
  • $\begingroup$ Any textbook on topological $K$-theory should cover this example, because it's the first example of $K$-theory being different from homology. The upper bound comes from the Atiyah-Hirzebruch spectral sequence, or computing $K$-theory cell by cell. More interesting is the lower bound, that $K$-theory has torsion of high order. This is a Chern character exercise. The total Chern character $1+c_1+c_2+...$ is a homomorphism from $K$-theory to the multiplicative group of the units of the homology ring. Even though homology is additively 2-torsion, it can detect high torsion in $K$-theory. $\endgroup$ Oct 4, 2019 at 16:26
  • $\begingroup$ @BenWieland I do not know how I missed your comment. Thanks. I will ahve a look at some topological K theory book $\endgroup$ Oct 10, 2019 at 13:01

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