# Determining a differential on an exterior algebra

Let $V$ be a graded vector space (over $\mathbb{N}$). It is claimed here 1 , page 5, that to determine a derivation on the free, graded-commutative algebra $\Lambda V$ it suffices to define it on $V$. I don't understand why this is so. Presumably it's referring to the fact that $\Lambda$ is left adjoint to the forgetful functor from commutative graded algebras to graded vector spaces, that's ok, but the problem is that the differential on a graded vector space is not supposed to be a morphism of graded commutative algebras, but a derivation!

What's going on?

• But being a derivation still means that it will be uniquely determined by what it does to $V$, via the Leibniz rule. – Tobias Kildetoft Jan 27 '17 at 9:55

First, the functor $\Lambda$ is left adjoint to the forgetful functor from commutative graded algebras to graded modules. $\Lambda V$ has a concrete description as a quotient of the tensor algebra $TV$.

The claim is that the forgetful functor from commutative differential graded algebras to graded differential modules (chain complexes) has a left adjoint, also denoted $\Lambda$, such that, when forgetting the differentials, it coincides with the previous $\Lambda$.

The differential $d_\Lambda$ in $\Lambda V$ where $V$ is a differential graded module can be defined explicitely.

We have $TV=k \oplus V \oplus V\otimes V\oplus \dots$ where $k$ is the base commutative ring.

Define $d_\Lambda$ to be zero on $k$, to be $d_V$ on $V$, and on $V\otimes V$ what it has to be is forced upon us by the Leibniz rule:

$$d_\Lambda(v\wedge w)=(d_\Lambda v)\wedge w + (-1)^{|v|} v \wedge d_\Lambda w.$$

On the higher tensors one proceeds by induction and the Leibniz rule, analogously. One quickly checks that $d_\Lambda$ is a differential, a derivation, such that $\Lambda$ is a functor and a left adjoint to the forgetful functor.