TITLE: "a sign that one should be computing K-theory"
QUESTION [12 upvotes]: 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?

REPLY [14 votes]: 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 $KU^0(BG)$.