# Simplicial Complex Induced by a Morphism

Let $$(C, \otimes, I)$$ be a symmetric monoidal abelian category. For an object $$M$$ in $$C$$ with a map $$\rho : M \rightarrow I$$, we can form the chain complex $$M_*$$ where $$M_n = \otimes_{i = 1}^n M$$ and $$d : M_n \rightarrow M_{n-1}$$ is $$\sum_{i = 1}^n (-1)^i 1 \otimes \cdots \otimes \rho \otimes \cdots \otimes 1$$ where $$\rho$$ is in the $$i$$th slot (this is a less-than-ideal of writing the map, but I think it should be clear what I mean).

This is sort of like a Bar construction, but we got less information to start with than a typical module over a monad. My questions are

1. What is the simplicial object over $$C$$ corresponding to $$M_*$$ under the Dold-Khan correspondence?

2. Is there a name for this construction?

3. Can we view this as an instance of the Bar construction?

Example: Note that, if we take $$C$$ to be the symmetric monoidal category of $$R$$-modules over a commutative ring $$R$$ and modify the twist map $$\tau$$ so that $$\tau_{M} : M \otimes_R M \rightarrow M \otimes_R M$$ sends $$a \otimes b$$ to $$-b \otimes a$$, then we get $$\Lambda (M)$$ above. If $$M = R^n$$ and we are given a map $$\rho : R^n \rightarrow R$$, then $$M_*$$ above is the Koszul complex. So this means that the Koszul complex is an example of the kind of resolution above.

Another one along the same vein is De Rham cohomology.

Thanks very much!

• Is $M_0 = I$? I can't think of anything else it could be. – David Roberts Jul 21 at 22:27
• that's right, sorry – Dean Young Jul 21 at 22:54