4
$\begingroup$

As we introduced a cohomology group of a complex $K$ (for example over $\mathbb{Z}$) as the set $Hom(C,\mathbb{Z})$, we also talked about the cup product to give the cohomology group a ring structure.

In particular, the cup product is defined as follows: We define the cup product on the basis elements of the simplex as

$$\cup : C^n(K) \times C^m(K) \to C^{n+m}(K)$$

$$(\varphi \cup \psi)(\sigma_{n+m}) := \varphi(\sigma_{\leq n}) \cdot \psi(\sigma_{\geq n})$$

and then take the linear map corresponding to the image of the basis elements.

I was wondering about the following observation: Since $\varphi \cup \psi, \varphi,\psi$ are linear maps, we get for a basis element $\sigma$

$2(\varphi \cup \psi)(\sigma)=(\varphi \cup \psi)(\sigma)+(\varphi \cup \psi)(\sigma)=(\varphi \cup \psi)( \sigma + \sigma)=\varphi(\sigma_{\leq n}+\sigma_{\leq n}) \cdot \psi(\sigma_{\geq n} + \sigma_{\geq n})=2\varphi(\sigma_{\leq n}) \cdot 2\psi(\sigma_{\geq n})=4(\varphi \cup \psi)(\sigma)$

So the map cannot be linear anymore. Where is my mistake in this observation?

$\endgroup$
5
  • $\begingroup$ I recommend you have a more precise title. Unfortunately I can’t confidently suggest one as I don’t know the subject, maybe: “why are the maps linear in the cup product (cohomology)?” $\endgroup$
    – FShrike
    Oct 15, 2022 at 10:17
  • $\begingroup$ Thanks for your comments! $\endgroup$
    – mkfrnk
    Oct 15, 2022 at 10:49
  • $\begingroup$ "we introduced a cohomology group of a complex $K$ as the set $Hom(C,\mathbb{Z})$". What is the relation betweeen $K$ and $C$? Is $K$ a chain complex or something else? And the cohomology groups are certainly not the groups $Hom(C,\mathbb Z)$, but the cohomology groups of the cochain complex $Hom(C, \mathbb Z)$. $\endgroup$
    – Paul Frost
    Oct 15, 2022 at 15:40
  • $\begingroup$ Sorry, this was a typo. I meant $Hom(K,\mathbb{Z})$ $\endgroup$
    – mkfrnk
    Oct 15, 2022 at 16:00
  • $\begingroup$ @mkfrnk Then you should correct it in the question. $\endgroup$
    – Paul Frost
    Oct 17, 2022 at 9:44

1 Answer 1

4
$\begingroup$

Your third step is incorrect. It is not true that $(\phi \cup \psi)(2\sigma) = \phi(2\sigma_{\le n}) \psi(2\sigma_{\ge n})$.

The problem is that $2\sigma$ is not a basis element. The formula you gave defined $\phi \cup \psi$ on basis elements and extended linearly, so by definition, $(\phi \cup \psi)(2\sigma) = 2(\phi \cup \psi)(\sigma) = 2\phi(\sigma_{\le n}) \psi(\sigma_{\ge n})$.

Your formula would only make sense if $2\sigma$ was a basis element with $(2\sigma)_{\le n} = 2\sigma_{\le n}$ and $(2\sigma)_{\ge n} = 2\sigma_{\ge n}$, but this is simply not true.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .