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I'm trying to prove the Leibniz Rule: $$\partial(u \smile v) = \partial(u) \smile v + (-1)^{|u|} \ u \smile \partial(v) $$ and I'm having some difficulties.

Here's my attempt: Let $|u| = p$, $|v|=q$ and $\sigma: \Delta^{p+q+1} \longrightarrow X$.

  1. $(\partial u \smile v)([\sigma]) = \displaystyle \sum_{i=0}^{p} (-1)^{i} u([\sigma d^{i} \lambda_{p}]) \cdot v([\sigma \rho_{q}]) + (-1)^{p+1} \ u([\sigma \lambda_{p}]) \cdot v([\sigma \rho_{q}])$. Ok, so far.

  2. Here's the problem with the proof I found: $(u \smile \partial v)([\sigma]) = u([\sigma \lambda_{p}]) \cdot \partial v ([\sigma \rho_{q+1}])$. But $\partial v ([\sigma \rho_{q+1}]) = \displaystyle \sum_{j=p}^{p+q+1} (-1)^{j-p} v([\sigma \rho_{q+1} d^{j-p}])$. We split the sum: $j = p: (-1)^{p-p} v([\sigma \rho_{q+1} d^{j}]) = v([\sigma \rho_{q}])$ and $j \geq p+1: (-1)^{p} \displaystyle \sum_{j=p+1}^{p+q+1} (-1)^{j} v([\sigma\rho_{q+1} d^{j}])= (-1)^{p}\displaystyle \sum_{j=p+1}^{p+q+1} (-1)^{j} v([\sigma d^{j} \rho_{q}])$.

I don't understand the sum when $j \geq p+1$. Shouldn't there be a $-(-1)^{p}$ and why do we have the $d^{j}$ instead of $d^{j-p}$?

Any help would be greatly appreciated.

  • Pleas recall what is the cup product... – Jean Marie Jun 08 '17 at 15:01
  • Given $u \in C^{p}(X, \Bbbk)$, $v \in C^{q}(X, \Bbbk)$ and $\sigma: \Delta^{p+q} \longrightarrow X$, the cup product of $u$ and $v$ is defined as: $(u \smile v)([\sigma]) := u([\sigma \circ \lambda_{p}]) \cdot v([\sigma \circ \rho_{q}])$ where $\lambda_{p}: \Delta^{p} \longrightarrow \Delta^{p+q}$ is defined by $\lambda_{p}(e_{i}) = e_{i}$ and $\rho_{q}: \Delta^{q} \longrightarrow \Delta{p+q}$ is given by $\rho_{q}(e_{i}) = e_{i+p}$. –  Jun 08 '17 at 15:09

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