We define the cap product in the following way :

$$a \frown z := \epsilon \otimes 1(a \otimes Ez \circ \Delta_\star(z))$$

Where :

$1. \hspace{0.3cm} \Delta_\star$ is the induced map on chain complex by the diagonal map $\Delta : X \longrightarrow X \times X$

$2. \hspace{0.3cm}Ez$ is a choince of an Eilenberg-Zielber map $Ez : C_\bullet(X \times X) \longrightarrow C_\bullet(X) \otimes C_\bullet(X)$

$3.\hspace{0.3cm} \epsilon$ is the map defined only in degree $0$, more accurately $\epsilon_n = 0$ if $n \ne 0$ and $\epsilon_0 : (C^\bullet(X)\otimes C_\bullet(X))_0 \longrightarrow \mathbb{Z}$ where $\langle f,x \rangle = f(x), f \in C^i(X),x \in C_i(X)$.

With those convention it should be clear that we call cap product the following map :

$$C^q(X) \otimes C_{p+q}(X) \overset{\frown}{\longrightarrow} C_p(X)$$

Now, since the Eilenberg-Zielber theorem due to acyclic models theorem we have that any choice (dependant on chain level) that extends the diagonal will be as good as the defined since they will induce the same map in homology.

Given the premise I'd like to substitute the $Ez \circ \Delta_\star(z)$ with $AW(z) = \sum\limits_{p+q = n} \sigma_p^1 \otimes \sigma_q^2$ the Alexander Whitney map, see the definition in a previous question here. What I don't understand is the following :

If I compute $\varphi \frown \sigma$ with $\varphi \in C^p$ and $\sigma : \Delta^{p+q} \longrightarrow X$ I end up having the formula $\varphi(\sigma_{|[e_q,\cdots,e_{p+q}]})\cdot \sigma_{|[e_0,\cdots e_q]}$ which is indeed the formula of the cap product here.

What I should get according to my professor is $(-1)^{pq}\varphi(\sigma_{|[e_q,\cdots,e_{p+q}]})\cdot \sigma_{|[e_0,\cdots e_q]}$. Where the sign on $(-1)$ comes from ? Reading "Tammo Tom Dieck, Algebraic Topology" at page $290$, I found the following

enter image description here

Is this somehow related to my problem? I thought $f = \epsilon$ and $g = \text{id}$ could work. I'm not sure since if the degree of a map is defined as the integer $p$ such that $f : C_q(X) \longrightarrow C_{q+p}(X)$ the identity should have degree $0$, which doesn't solve the problem.

Addendum : Is the sign of the $(1)^{|g||a|}$ there for particular reasons ? I thought this could be since if we define the differential of two chain complex with a factor $(-1)^{\bullet}$ then that is the right sign to achieve $d(f \otimes g) = (f\otimes g)d$, but computing the two of them, that doesn't seem relevant.

Any help in understanding this would de be appreciated since the construction of cap product and cup product is strictly related to understanding duality.


1 Answer 1


The issue here is common, and happens if one is not careful in distinguishing left and right actions and does not follow Koszul's sign rule. In particular, note that the formula in your link is a right action and evaluates $\psi$ on the first face of $\sigma$, which is not what your wrote. Everything is commutative in homology so it "doesn't really matter" but if one is a bit careful these mix-ups can be avoided.

In your case, you have an algebra of cochains $(C^*(X),\smile,\delta)$ and you want to make it act on the complex of chains $(C_*(X),d)$. By their very definition, $C_*(X)$ and $C^*(X)$ are paired, by a map $$\langle -,-\rangle : C^*(X)\otimes C_*(X)\longrightarrow \mathbb Z$$

such that $\langle\varphi,\sigma \rangle= \varphi(\sigma)$. Note that I am not assuming that these are homogeneous, and the "degree-by-degree" formula just follows from this. In particular, if you wanted to define the pairing to go in the other direction, this would introduce signs: if now $\psi$ is and $\tau$ are homogenous, then $\langle \tau,\psi\rangle = (-1)^{|\psi||\tau|} \psi(\tau)$, owing to the fact we use the braiding $x\otimes y \longmapsto (-1)^{|x||y|} y\otimes x$ before composition.

With this at hand, you can either act on the left or on the right, by using the Alexander coproduct in your post which I will write more simply as $\Delta(\sigma) = \sigma_{(1)}\otimes \sigma_{(2)}$ (a la Sweedler). In the case we act on the right, we obtain the following:

$$ \sigma\frown\varphi= (-1)^{|\varphi||\sigma|}\langle \varphi, \sigma_{(1)}\rangle \sigma_{(2)}$$

and the sign arises as follows: we begin with $\sigma\otimes \varphi$, apply $\Delta$ to obtain $\sigma_{(1)}\otimes \sigma_{(2)}\otimes \varphi$, interchange $\sigma_{(2)}$ and $\varphi$ introducing the sign $(-1)^{|\varphi||\sigma_{(2)}|}$ and then pair $\sigma_{(1)}$ and $\varphi$ with $\varphi$ on the "wrong side" introducing the sign $(-1)^{|\varphi||\sigma_{(1)}|}$ for a total of $(-1)^{|\varphi||\sigma_{(2)}|}(-1)^{|\varphi||\sigma_{(1)}|} = (-1)^{|\varphi||\sigma|}$ as $|\sigma_{(1)}|+|\sigma_{(2)}| = |\sigma|$.

This sounds a bit farfetched at first but it works: you can check that if we define $$\langle f\smile g, \sigma \rangle =(-1)^{|g||\sigma_{(1)}|} \langle f,\sigma_{(1)}\rangle \langle g, \sigma_{(2)}\rangle $$ where again I am being careful in introducing the sign because $g$ and $\sigma_{(1)}$ have been flipped, then we have that $\sigma \frown (\varphi\smile \psi) = (\sigma \frown \varphi) \frown \psi$. Indeed, both terms evaluate to $(-1)^{|\phi||\sigma|+|\psi||\sigma_{(2)}|}\varphi(\sigma_{(1)})\psi(\sigma_{(2)})\sigma_{(3)}$.

This suggests that the left action should be defined in a similar way: $$ \varphi \frown \sigma = (-1)^{|\varphi||\sigma_{(1)}|}\langle \varphi, \sigma_{(2)}\rangle \sigma_{(1)}$$ owing to the fact that we moved $\varphi$ past $\sigma_{(1)}$ to pair it with $\sigma_{(2)}$: this is the formula your professor gave to you (and it is correct!). Indeed, if you compute $(\varphi\smile\psi)\frown \sigma$ you will get the same as $\varphi\frown (\psi\frown \sigma)$, which would not be true with the formula you propose (up to signs). In this case you get

$$(-1)^{|\varphi||\sigma_{(1)}|+ |\psi|(|\sigma_{(1)}|+|\sigma_{(2)}|)} \sigma_{(1)} \varphi(\sigma_{(2)}) \psi(\sigma_{(3)}). $$

It turns out that because $\Delta$ is cocommutative in homology (and hence so is $\smile$), these two actions are also compatible, in the sense that one homology we do have $\varphi\frown \sigma = (-1)^{|\varphi||\sigma|}\sigma\frown \varphi$, and in particular many of these issues can be "swept under the rug" by carefully tweaking the formulas above with signs (this is usually harmless, but gives us all a headache when sign conventions do not match.)

  • 1
    $\begingroup$ (I tried to write this in a way that is not specific to $C_*(X)$, and in fact the idea above works if say $C$ is a graded dg coalgebra paired to its graded dual $A$, and you want to define right and left actions through the coproduct of $C$. Then all the formulas work with the signs given. Sweedler's notation is particularly handy to handle the computations!) $\endgroup$
    – Pedro
    May 9, 2021 at 18:22
  • $\begingroup$ Why do we require the Koszul's sign rule? $\endgroup$ Aug 29, 2021 at 18:26
  • $\begingroup$ Otherwise said, why do we need to require that the two action are compatible? $\endgroup$ Aug 31, 2021 at 8:29
  • $\begingroup$ @jacopoburelli The fact that the cohomology algebra of a space is commutative is something we like! And the compatibility means the bimodule is actually a symmetric one (i.e. a module for this commutative algebra). $\endgroup$
    – Pedro
    Aug 31, 2021 at 10:20
  • $\begingroup$ Okay thanks, just last question in order to understand this properly. In order to obtain the formula my professor gave to me, I have to make a switch which involes a braiding with the sign you defined. So my question is: Why do I know that the two (i.e $\varphi \otimes \sigma_{(1)} \otimes \sigma_{(2)}$ and $(-1)^{\lvert \varphi \rvert \lvert \sigma_{(1}\rvert} \sigma_{(1)} \otimes \varphi \otimes \sigma_{(2)}$) are the "same element"? Because it seems to me that they are the same under the "isomorphism of Koszul" and I don't understand why should they have same image. Thanks $\endgroup$ Aug 31, 2021 at 10:28

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