I'm pretty confused about the precise relation of the integral and the real cohomology of the classifying space $BG$ of a compact Lie group $G$. The natural map $H^n(BG;\mathbb{Z})\to H^n(BG;\mathbb{R})$ certainly kills all torsion, but can it have non-torsion elements in the kernel? For instance, if $G=SU_n$, then there is no torsion in $H^n(BG;\mathbb{Z})$ (see [Hatcher, Algebraic Topology, Theorem 4D.4]), so this map would be an injection for all $n$.

All expositions that I know (for instance [E. Brown, The cohomology of $BSO_n$ and $BO_n$ with integer coefficients]) somehow suggest that $H^n(BG;\mathbb{Z})\to H^n(BG;\mathbb{R})$ is injective, but this is certainly not the case for arbitrary spaces $X$ with torsion free $H^n(X;\mathbb{Z})$. The Corresponding statement for homology is true, since the singular chain complex is free, and so the universal coefficient theorem yiels $$ H_n(X;\mathbb{Z})\otimes \mathbb{R} \xrightarrow{\cong} H_n(X;\mathbb{R}). $$ However, this argument does not carry over to cohomology (or at least I don't see it).

What makes me suspicious is that if $H^n(BG;\mathbb{Z})\to H^n(BG;\mathbb{R})$ is injective for all $n$, then the long exact sequence induced from $\mathbb{Z}\to \mathbb{R} \to S^1$ would split, a thing I don't feel comfortable with.

  • 3
    $\begingroup$ You can use the naturality of the Universal Coefficient sequence for cohomology (Hatcher, Algebraic Topology, Theorem 3.2) to conclude that your map is injective whenever $H_{n-1}$ is torsion free. $\endgroup$ – Mark Grant Dec 16 '13 at 22:37

If $X$ is a space of finite type (meaning that the homology groups $H_i(X)$ are all finitely generated, a condition which applies in particular to $X=BG$ for $G$ a compact Lie group) then for each $n$ the map $H^n(X)\to H^n(X;\mathbb{R})$ is injective if and only if $H_{n-1}(X)$ is torsion free. Here and below integer coefficients are omitted from the notation.

To see this, note that the universal coefficient sequence $$ 0\to \operatorname{Ext}(H_{n-1}(X),A)\to H^n(X;A)\to \operatorname{Hom}(H_n(X),A)\to 0 $$ is natural with respect to homomorphisms $A\to A'$ of abelian coefficient groups. If $H_{n-1}(X)$ is torsion free, the ext groups $\operatorname{Ext}(H_{n-1}(X),\mathbb{Z})$ and $\operatorname{Ext}(H_{n-1}(X),\mathbb{R})$ both vanish, and the map $H^n(X)\to H^n(X;\mathbb{R})$ is injective if and only if $\operatorname{Hom}(H_n(X),\mathbb{Z})\to \operatorname{Hom}(H_n(X),\mathbb{R})$ is injective (which it clearly is).

Conversely, if $H_{n-1}(X)$ has torsion then so does $H^n(X)$, and this torsion is in the kernel of $H^n(X)\to H^n(X;\mathbb{R})$.

  • $\begingroup$ What is the definition of $H^n(X;G)$ if $G$ is non-abelian? $\endgroup$ – Qfwfq Dec 17 '13 at 11:15
  • $\begingroup$ @Qfwfq: Oops, there are too many $G$'s in may answer. I'll edit it, thanks. $\endgroup$ – Mark Grant Dec 17 '13 at 12:05

This is also equivalent to the corresponding statement for homology: Since $Hom(C,\mathbb{Z}) \otimes \mathbb{R} = Hom(C,\mathbb{R})$ for $C$ a free abelian group, the cochain complex with $\mathbb{R}$-coefficients is isomorphic to the cochain complex with $\mathbb{Z}$-coefficients tensored with $\mathbb{R}$. The same homological algebra as in the homology universal coefficient theorem gives you that $H^n(X) \rightarrow H^n(X; \mathbb{R})$ precisely kills torsion. (Which of course comes from $H_{n-1}$ if you invoke the universal coefficient sequence as in Marks answer.)

  • $\begingroup$ It is essential to restrict $C$ to also be finitely generated, otherwise the first map $\operatorname{Hom}(C,\mathbb{Z}) \otimes \mathbb{R} \to \operatorname{Hom}(C,\mathbb{R})$ will not be an isomorphism in general. $\endgroup$ – Ricardo Andrade Dec 17 '13 at 14:42
  • $\begingroup$ Oh, you're right. So this argument won't apply to singular cohomology, but for the spaces in question the cochain complex is quasi-isomorphic to a degreewise finitely-generated complex, so everything should work out. $\endgroup$ – Achim Krause Dec 17 '13 at 16:47
  • 1
    $\begingroup$ Dear @Achim Krause: Yes, any free chain complex whose homology has finite type (i.e. is finitely generated in each degree) is chain equivalent to a finite type, free chain complex. See, for example, lemma 5.5.9 of Spanier's book "Algebraic topology". In fact, we can directly apply theorem 5.5.10 of Spanier's book which gives a universal coefficient theorem for cohomology in terms of cohomology (as opposed to the usual universal coefficient theorem where cohomology is expressed in terms of homology), as long as the homology has finite type. $\endgroup$ – Ricardo Andrade Dec 17 '13 at 17:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.