2009-10-13 07:45:47 8 Comments

Standard algebraic topology defines the cup product which defines a ring structure on the cohomology of a topological space. This ring structure arises because cohomology is a contravariant functor and the pullback of the diagonal map induces the product (using the Kunneth formula for full generality, I think.)

I've always been mystified about why a dual structure, perhaps an analogous (but less conventional) "co-product", is never presented for homology. Does such a thing exist? If not, why not, and if so, is it such that the cohomology ring structure can be derived from it?

I am aware of the intersection products defined using Poincare duality, but I'm seeking a true dual to the general cup product, defined via homological algebra and valid for the all spaces with a cohomology ring.

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## 5 comments

## @David E Speyer 2016-04-02 16:04:25

$\def\ZZ{\mathbb{Z}}\def\RR{\mathbb{R}}\def\PP{\mathbb{P}}$I noticed these excellent answers were missing an explicit example of a space $X$ and a class $\eta$ in $H_{\ast}(X, \ZZ)$ such that the image of $\eta$ in $H_{\ast}(X \times X)$ (under the diagonal map) is not in the image of $H_{\ast}(X) \otimes H_{\ast}(X) \to H_{\ast}(X \times X)$. So, for the record, this occurs for the fundamental class of $\mathbb{RP}^3$.

VerificationWe write points of $\RR \PP^3$ in homogenous coordinates as $(w:x:y:z)$. For $t \in [0,1]$, define $$D_t := {\Big \{} ((w_1:x_1:y_1:z_1), (w_2:x_2:y_2:z_2)) \in \RR \PP^3 \times \RR \PP^3 : $$ $$w_2 x_1 = t w_1 x_2,\ x_2 y_1 = t x_1 y_2,\ y_2 z_1 = t y_1 z_2,$$ $$w_2 y_1 = t^2 w_1 y_2,\ x_2 z_1 = t^2 x_1 z_2,\ w_2 z_1 = t^3 z_1 w_2 {\Big \}}$$ and set $D = \bigcup_{t \in [0,1]} D_t$.$D_1$ is the diagonal and, for $t \in (0,1]$, we have $D_t \cong \RR \PP^3$, the equations defining $D_t$ simply say that $(w_2:x_2:y_2:z_2) = (t^3 w_1: t^2 x_1: t y_1: z_1)$. The space $D_0$ is a bit more interesting: the equations are $w_2 x_1 = x_2 y_1 = y_2 z_1 = w_2 y_1 = z_2 x_1 = z_2 w_1=0$, and considering cases shows that $$D_0 = (\RR \PP^3 \times \RR \PP^0) \cup (\RR \PP^2 \times \RR \PP^1) \cup (\RR \PP^1 \times \RR \PP^2) \cup (\RR \PP^0 \times \RR \PP^3).$$ Here $\RR \PP^2 \times \RR \PP^1$, for example, is all points of the form $(\ast: \ast: \ast : 0) \times (0 : 0 : \ast : \ast)$.

$D$ is an oriented manifold with corners, whose boundary is $D_1 - D_0$. So $D_1$ is homologous to $D_0$.

Put the standard CW-structure on $\RR \PP^3$, writing $C_i$ for the cell of dimension $i$, and write $C_{ij}$ for $C_i \times C_j$. It is easy to compute in this structure that $$H_3 = \ZZ \cdot [C_{30}] \oplus \ZZ \cdot [C_{03}] \oplus (\ZZ/2\ZZ) \cdot ([C_{21}]+[C_{12}]).$$ The image of $\bigoplus_{i+j =3 } H_i(\RR \PP^3) \otimes H_j(\RR \PP^3)$ is the first two summands, and $[D_0] = [C_{30}] + [C_{21}] + [C_{12}] + [C_{03}]$, so $[D_0]$ is not in the image of $\bigoplus_{i+j =3 } H_i \otimes H_j$. (Recall that $H_1(\RR \PP^3) = H_2(\RR \PP^3) = 0$. )

I'll point out two things which confused me at first. First of all, why isn't $[D_0]$ the image of $\sum_{i+j=3} [\RR \PP^i] \otimes [\RR \PP^j]$? The answer is that $\RR\PP^2$ is not orientable, so it does not define a class in $H_2(\RR \PP^3)$.

Once we realize that, what does $[D_0]$ mean, since the $\RR \PP^1 \times \RR \PP^2$ and $\RR \PP^2 \times \RR \PP^1$ components are also not orientable? The answer is that the orientation on the interior of $D$ (which is $(0,1) \times \RR \PP^3$, so orientable) gives rise to orientations on the interiors of the components of $D_0$, but these orientations are discontinuous across the codimension $2$ strata where these components meet. Nonetheless, this is enough to give us a map from an oriented closed cell to each component, whose boundaries cancel out, so we can speak of the $H_3$ class of $[D_0]$ even though we can't speak of the classes of its individual components.

## @john mangual 2016-04-02 17:46:51

Can you explain why $C_{ij}\equiv C_i \times C_j$ is such a natural piece of notation? I have never had intuition for co-multiplication. To me it looks just likr factoring. Is $6=2\times 3$ a co-multiplication ?

## @David E Speyer 2016-04-02 17:58:52

Just double subscripts to save space on the page.

## @Qiaochu Yuan 2014-07-04 05:11:52

Here is a situation where you really use this coalgebra structure (which, as other answers have mentioned, exists over a field in particular).

If $X$ is a homotopy associative $H$-space, then $H_{\bullet}(X, \mathbb{Q})$ has the structure of an algebra with the multiplication given by the Pontryagin product induced from the multiplication $X \times X \to X$. But $H_{\bullet}(X, \mathbb{Q})$ also has a coalgebra structure induced from the diagonal map $X \to X \times X$, and these structures are compatible, making $H_{\bullet}(X, \mathbb{Q})$ a Hopf algebra. The same is true of the cohomology, and these Hopf algebras are dual in an appropriate sense. This is the reason Hopf invented Hopf algebras.

The Hopf algebra structure on $H_{\bullet}(X, \mathbb{Q})$ is a surprisingly strong invariant of $X$; in particular, it determines the rational homotopy of $X$ as follows. Recall that if $X$ is an $H$-space then $\pi_{\bullet}(X)$ has the structure of a graded Lie algebra with bracket given by the Samelson bracket.

Example.$H_{\bullet}(\Omega S^{n+1}, \mathbb{Q})$ is known to be isomorphic to $\mathbb{Q}[x]$, a polynomial algebra on a generator $x$ of degree $n$, which can be identified with a generator of $H_n(S^n, \mathbb{Q})$. The only possible coalgebra structure is $\Delta x = x \otimes 1 + 1 \otimes x$. This gives$$\Delta x^2 = (x \otimes 1 + 1 \otimes x)^2 = x^2 \otimes 1 + \left( 1 + (-1)^{n^2} \right) x \otimes x + 1 \otimes x^2$$

from which we conclude that if $n$ is even then $x$ is primitive but not $x^2$, but if $n$ is odd then $x$ and $x^2$ are both primitive. These are the only possible primitive elements, and this computes the rational homotopy groups of the spheres; we get

$$\pi_{n-1}(\Omega S^n) \otimes \mathbb{Q} \cong \pi_n(S^n) \otimes \mathbb{Q} \cong \mathbb{Q}$$

and

$$\pi_{4n-2}(\Omega S^{2n}) \otimes \mathbb{Q} \cong \pi_{4n-1}(S^{2n}) \otimes \mathbb{Q} \cong \mathbb{Q}$$

and everything else vanishes rationally.

## @Jim Conant 2014-07-04 06:08:49

This is really nice. I've not seen this particular presentation before.

## @Tyler Lawson 2009-10-15 11:58:14

Homology is not naturally a coalgebra unless you take field coefficients or unless your object has torsion-free homology groups over the integers. The basic issue, as mentioned above, is that even though you have a split exact universal coefficient sequence for the homology of a product, the splitting isn't natural. You don't actually need homology to be dual to cohomology because that would involve some additional finiteness properties.

However, if your space has torsion-free homology with integer coefficients, then $H_*(X;R) = H_*(X) \otimes R$ for all $R$, and so you get a coalgebra structure on the homology of $X$ with coefficients in $R$ simply as the base change of the one over the integers. If $R$ is an algebra over a field then you get a coalgebra structure with no assumptions on $X$ by base-change from said field.

I should probably point out that the Kunneth formula is more complicated than stated in a previous answer. There's an exact sequence

$0 \to H_*(C;\mathbb{Z}) \otimes H_*(D;M) \to H_*(C \otimes D;M) \to \operatorname{Tor}(H_*(C;\mathbb{Z}), H_*(D;M)) \to 0$

but notice that one side involves integer coefficients and the other coefficients in a general module. If you want the universal coefficient theorem with the same coefficients on both sides it takes the form of a spectral sequence with $E_2$-term

$\operatorname{Tor}^R_{p,q} (H_*(X;R), H_*(Y;R))$

converging to $H_*(X \times Y;R)$. (The bigrading on $\operatorname{Tor}$ comes because we're taking $\operatorname{Tor}$ of graded modules.)

In general if $E$ is a generalized homology-cohomology theory then flatness of $E_* X$ over the ground ring $E_*$ guarantees a coalgebra structure on the $E$-homology of $X$. This also may or may not have anything to do with duality, because flatness and projectiveness are not the same.

As mentioned, you still do have a coalgebra structure on the chains of $X$ (or the "$E$-homology object of $X$" in the stable homotopy category), and this is really just some kind of failure of the homology groups to mimic what's going on behind the scenes.

## @Loop Space 2009-10-16 08:21:55

The universal coefficient theorem wasn't stated in a previous answer.

## @Tyler Lawson 2009-10-16 12:01:51

yargh, I meant the Kunneth formula. fixed.

## @Loop Space 2009-10-21 09:02:14

Also, although this is the most general form for chains, for singular cohomology then it's a little elaborate, isn't it? After all, singular chains are free (by definition!) so the complication of coefficients doesn't arise. Or am I missing something?

## @Tyler Lawson 2009-10-21 12:40:07

You need that singular chains are free to get the conclusion about the Tor spectral sequence in the first place; the spectral sequence is a general one computing H_

(C ⊗_R D) from H_C and H_* D when C and D are (nonnegative) chain complexes of R-modules with one of them levelwise free. One example is to look at the mod-4 homology of RP^2 x RP^2 from the mod-4 homology of its factors. Having said that, the spectral sequences you get always collapse at E_3 because everything is arising from integral coefficients, but if you leave the higher Tors out it doesn't work.## @Loop Space 2009-10-13 13:08:03

The Eilenberg-Zilber theorem says that for singular homology there is a natural chain homotopy equivalence:

S

_{*}(X) ⊗ S_{*}(Y) ≅ S_{*}(X × Y)The map in the

reversedirection is the Alexander-Whitney map. Therefore we obtain a mapS

_{*}(X) → S_{*}(X × X) → S_{*}(X) ⊗ S_{*}(X)which makes S

_{*}(X) into a coalgebra.My source (Selick's

Introduction to Homotopy Theory) thenstatesthat this gives H_{*}(X) the structure of a coalgebra. However, I think that the Kunneth formula goes the wrong way. The Kunneth formula says that there is a short exact sequence of abelian groups:0 → H

_{*}(C) ⊗ H_{*}(D) → H_{*}(C ⊗ D) → Tor(H_{*}(C), H_{*}(D)) → 0(the astute will complain about a lack of coefficients. Add them in if that bothers you)

This is split, but not naturally, and when it is split it may not be split as modules over the coefficient ring. To make H

_{*}(X) into a coalgebra we need that splitting map. That requires H_{*}(X) to be flat (in which case, I believe, it's an isomorphism).That's quite a strong condition. In particular, it implies that cohomology is dual to homology.

Of course, if one works over a field then everything's fine, but then integral homology is

somuch more interesting than homology over a field.In the situation for cohomology, only

someof the directions are reversed, which means that the natural map is still from the tensor product of the cohomology groups to the cohomology of the product. Since the diagonal map now gets flipped, this is enough to define the ring structure on H^{*}(X).There are deeper reasons, though. Cohomology is a

representablefunctor, and its representing object is a ring object (okay, graded ring object) in the homotopy category. That's the real reason why H^{*}(X) is a ring (the Kunneth formula has nothing to do with defining this ring structure, by the way). It also means that cohomology operations (aka natural transformations) are, by the Yoneda lemma, much more accessible than the corresponding homology operations (I don't know of any detailed study of such).Rings and algebras, being varieties of algebras (in the sense of universal or general algebra) are generally much easier to study than coalgebras. Whether this is more because we have a greater history and more experience, or whether they are inherently simpler is something I shall leave for another to answer. Certainly, I feel that I have a better idea of what a ring looks like than a coalgebra. One thing that makes life easier is that often spectral sequences are spectral sequences of rings, which makes them simpler to deal with - the more structure, the less room there is for things to get out of hand.

Added Later:One interesting thing about the coalgebra structure - when it exists - is that it is genuinely a coalgebra. There's no funny completions of the tensor product required. The comultiplication of a homology element is always afinitesum.Two particularly good papers that are worth reading are the ones by Boardman, and Boardman, Johnson, and Wilson in the Handbook of Algebraic Topology. Although the focus is on operations of cohomology theories, the build-up is quite detailed and there's a lot about general properties of homology and cohomology theories there.

Added Even Later:One place where the coalgebra structure has been extremely successfully exploited is in the theory of cohomology cooperations. For a reasonably cohomology theory, the cooperations (which are homology groups of the representing spaces) are Hopf rings, which are algebra objects in the category of coalgebras.## @Ben Webster 2009-10-13 15:36:11

Is there any reason one couldn't have spectral sequences of coalgebras?

## @Loop Space 2009-10-13 17:58:36

The short answer is that I don't see why not, but you'd need every term in the spectral sequence to be flat in order to get this. I'm also not so sure how much help it would be. The point about rings is that once you know where x goes to then you know where x^2 goes to. But knowing where x goes to doesn't obviously tell you where everything in the comultiplication of x goes to.

## @Jim Conant 2014-07-04 06:05:38

Nice answer! I am amused by the term cooperation.

## @Ben Webster 2009-10-13 12:56:22

Yes, homology is a coalgebra, at least with coefficients over a field, with the coproduct given by the pushforward X -> X x X. This is dual to the ring structure on cohomology, so they determine each other.

## @Loop Space 2009-10-13 17:59:55

I know this is included in my answer above, but so that answers are self-contained: for this to be always correct everything needs to be over a field right from the start.

## @Mikael Vejdemo-Johansson 2009-10-13 18:05:51

This is the answer I'd have brought - with the added comment that the reason I hear most often /why/ we don't study the coalgebra of homology is that coalgebras are so much less studied than algebras, and we don't have a firm intuition formed for coalgebras in the same way.

## @Mikael Vejdemo-Johansson 2009-10-13 18:06:40

@ Andrew Does it have to be a field? Or is it enough that all Tor vanish? And for that matter, does 'is a field' follow from 'all Tor vanish'?

## @Ben Webster 2009-10-13 18:39:37

Well, semi-simple assures that all modules are flat. As this question of Anton's shows, any other examples would be pretty pathological. mathoverflow.net/questions/208/…

## @Loop Space 2009-10-14 06:45:54

I'd need to look it up to be sure, but the slogan that I've absorbed from algebraic topology is that to have "good" behaviour for all spaces then you need the coefficient ring to be a (graded) field. But "good" generally means Kunneth formula and cohomology dual to homology, rather than just one of them. I don't know enough about homological algebra to state the conditions precisely, though, without looking them up (the Boardman et al papers are a good reference, btw). But that's why the Morava K-theories are so popular: they are the only ones where the coefficient ring is a graded field.