1
$\begingroup$

I'm trying to get some intuition about the cup product for singular cohomology and was trying to check whether the graded commutativity $\varphi\smallsmile \psi = (-1)^{kl}\psi \smallsmile \varphi$ for $\varphi \in H^k(X)$ and $\psi \in H^l(X)$ is obvious for small $k,l$.

The most elementary case is when $k,l$ are both zero. Then $\varphi \smallsmile \psi$ is simply the pointwise product (where we may view the cocycles $\varphi$ and $\psi$ as functions on $X$) and the formula is immediate.

The next case is the product of a $0$-cocycle $\varphi$ with a $1$-cocycle $\psi$. The $1$-cocycle $\varphi \smallsmile \psi $ assigns to any $1$-simplex, which is to say curve, $f:\sigma_1 \to X$ the value $\varphi(f(1,0)) \psi(f)$. Similarly $\psi \smallsmile \varphi$ is the $1$-cocycle which assigns $\varphi(f(0,1))\psi(f)$ to any curve $f$. So we have that $\varphi \smallsmile \psi = \psi \smallsmile \varphi$ assigns the value $(\varphi(f(1)) - \varphi(f(0)))\psi(f)$ to $f$. That $\psi\smallsmile \varphi - \varphi \smallsmile \psi = 0$ in cohomology is now equivalent to this value only depending on the endpoints of the curve but I don't see why this is true.

$\endgroup$
1
$\begingroup$

The short answer is that $f(0)$ and $f(1)$ are homologous via $f$, so that $\varphi(f(0))-\varphi(f(1))$ is zero for any $f$.


The usual way to see that the cup-product is graded commutative is as follows:

Note that the value of a $1$-cocycle on a $1$-simplex $f$ is turned into its opposite if you reverse the orientation of $f$. Let's denote by $-f$ the simplex $f$ with opposite orientation.

$$\begin{eqnarray}(\varphi\smallsmile \psi)(f)&=&-(\varphi\smallsmile \psi)(-f)\\ \varphi(f(0))\psi(f)&=&-\varphi(f(1))\psi(-f)\end{eqnarray}$$ Now use again that $\psi(-f)=-\psi(f)$ to conclude.

$\endgroup$
  • $\begingroup$ I don't understand why it follows from $\psi$ being a cocycle that $\psi(f)$ only depends on the endpoints of $f$. This seems to assume all cocycles are coboundaries. $\endgroup$ – A. Van Werde Nov 30 '19 at 9:10
  • $\begingroup$ @A.VanWerde Sorry I was tired yesterday. $\endgroup$ – Arnaud Mortier Nov 30 '19 at 9:42

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.