Why is cohomology the direct product of the $H^n$?

Question

During a talk I mentioned in passing Borel's result that for $G$ a connected Lie group, $H^*(BG;\mathbb Q)$ is a polynomial ring. An audience member corrected me in very short order that no, it's in fact a power series ring. I responded that while that statement would follow from a common alternate definition of cohomology, for our limited computational purposes in the rest of the talk the standard Hatcher definition

$$H^* X := \bigoplus H^n X$$

would be unproblematic.

The response was that no, that was still wrong. As this question dragged on, and I ceased being able to imagine the other audience members were still listening, I panicked a bit and came to the somewhat credibility-damaging compromise that my interlocutor could mentally interpose a second pair of brackets whenever we faced a polynomial ring in the sequel. (He was quieted but unappeased.)

1. What does the definition $H^*X := \prod H^n X$ gain us? The examples I know are being able to define the Chern character $K^* \to H^*(-;\mathbb Q)$ and in general the first Chern class for complex-oriented cohomology theories, but I'm hoping for some overarching, systematic, moral reason.

2. What are some real errors produced by $H^* X := \bigoplus H^n X$, such that this gentleman would regard it not as a sometimes-less-convenient convention but a blatant falsehood?

3. Is there any advantage, on the other hand, to retaining the direct sum definition?

4. One encounters as well the convention that a graded ring $A$ is a sequence $(A_n)_{n \in \mathbb Z}$ of abelian groups and bilinear maps $A_m \times A_n \to A_{m+n}$ meeting a list of conditions. Is there a strong reason to prefer this over the direct sum definition, other than to eliminate the question "What is the degree of $0$?"?

I'm amazed at the temerity of your interlocutor to hold to this path when he apparently couldn't even suggest why your definition was wrong! — Kevin Carlson, Aug 23 '16 at 15:42
@MikeMiller: 1) I've seen that. Are there other reasons? Off the top of my head, it's the right target for the Chern character, but maybe there are others still. 2) That's what I'd thought before, but you can always do what people did in the old days and write the product version as $H^{**} X$ instead. 4) I mean, I think they're all interconvertible, but some people have a strong preference for the sequence-of-groups definition and I was trying to work out why. — jdc, Aug 23 '16 at 21:06
Sure, that's fair enough. I haven't had much reason to need to convert between them. There are the occasional situation in which a chain complex I have that presents cohomology is best written in terms of polynomials or power series and distinctly not as a series of groups, in which case the choice is obvious, but otherwise... eh. — , Aug 23 '16 at 21:11
@KevinCarlson: It's very possible he declined to expound further during my talk itself so as to avoid further derailing it. We had a bit of a chat afterward where he suggested that a. my definition is fine if I only care about the graded components and b. the actual wrongness of the direct sum definition comes in if one tries to discuss the cohomology of an infinite disjoint union. — jdc, Aug 23 '16 at 21:12

Thomas Rot · Answer 1 · 2016-08-23T21:57:52.213

3

This is not a full answer, but maybe gives a hint to the answer of question 1.

An important characteristic class is the Chern character. For (complex) line bundles it is defined as the the formal exponential of the first chern class, i.e. $ch(L)=e^{c_1(L)}:=1+c_1(L)+\frac{c_1(L)^2}{2!}+\ldots$. For higher dimensional vector bundles one defines the chern character by a formal splitting in chern roots. The chern character relates $K$-theory and cohomology, as the exponential has the wonderful property of turning sums into products.

If a cell complex is not finite, the sum, hence the chern character, naturally lives in $\prod_n H^{n}(X;\mathbb{Q})$ and not $\oplus_n H^{n}(X;\mathbb{Q})$. This occurs for the most important chern character of all, the chern character of the tautological bundle over $\mathbb{CP}^\infty$. Note that $\mathbb{CP}^\infty$ is the classifying space of $U(1)$.

edited Aug 23 '16 at 21:57

answered Aug 23 '16 at 20:39

Thomas Rot

10,399

2

To be obnoxiously pedantic, I don't think it lives in $\Bbb Z$ coefficients at all :) – Aug 23 '16 at 21:10
This is one of the reasons I had seen for 1 as well. Do you have any idea about 2? In the old days, the Chern character and other sorts of things expressed this way were seen as a reason to introduce a second notation $H^{*} X$ for the product version, with $H^ X$ being the sum version. Do you know why someone would want to expel the direct sum version from consideration altogether? – jdc Aug 23 '16 at 21:16
@jdc: No I don't, but we mathematicians are sometimes stubborn. If we learn a new way of looking at something we tend to view other ways as wrong. Obviously it is a well defined group with interesting properties... – Thomas Rot Aug 23 '16 at 22:00
@MikeMiller: Thanks, that was a bit to hasty. – Thomas Rot Aug 23 '16 at 22:00

Why is cohomology the direct product of the $H^n$?

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