Why is the Chevalley-Eilenberg differential a coderivation? For every Lie algebra $\mathfrak{g}$ we can consider the Chevalley-Eilenberg complex given by the exterior powers $\bigwedge^n \mathfrak{g}$ together with the differential $d_{\mathrm{CE}} \colon \bigwedge^n \mathfrak{g} \to \bigwedge^{n-1} \mathfrak{g}$ that is given by
$$
  d_{\mathrm{CE}}(x_1 \wedge \dotsb \wedge x_n)
  =
  \sum_{1 \leq i &lt; j \leq n}
  (-1)^{i+j-1}
  [x_i, x_j]
  \wedge x_1
  \wedge \dotsb
  \wedge \widehat{x_i}
  \wedge \dotsb
  \wedge \widehat{x_j} \wedge \dotsb \wedge x_n \,.
$$
We may regard the Lie algebra $\mathfrak{g}$ as a graded vector space concentrated in degree $1$.
Then the tensor algebra $\operatorname{T}(\mathfrak{g})$ becomes a graded Hopf algebra (such that $\mathfrak{g}$ consists of primitive elements) and its quotient $\bigwedge \mathfrak{g}$ inherits the structure of a graded Hopf algebra.
If I understand the nlab correctly then the Chevalley-Eilenberg differential $d_{\mathrm{CE}}$ is the unique extension of the Lie bracket $[-,-] \colon \bigwedge^2 \mathfrak{g} \to \mathfrak{g}$ to a graded coderivation for the underlying graded coalgebra structure of $\bigwedge \mathfrak{g}$.


  How can we show that $d_{\mathrm{CE}}$ is a graded coderivation of $\bigwedge \mathfrak{g}$, and that is it the unique one that extends the Lie bracket?


I have put my (so far unsuccessful) attempt below.
Coderivation

The differential $d_{\mathrm{CE}}$ is of degree $-1$ so we need to show that
\begin{equation}
  \Delta( d_{\mathrm{CE}}(c) )
  =
  \sum_{(c)}
  d_{\mathrm{CE}}( c_{(1)} ) \otimes c_{(2)}
  +
  (-1)^{|c_{(1)}|} c_{(1)} \otimes d_{\mathrm{CE}}(c_{(2)})
  \tag{1}
\end{equation}
for all $c = x_1 \wedge \dotsb \wedge x_n$ with $x_i \in \mathfrak{g}$ (i.e. that $\Delta$ is a homomorphism of chain complexes with respect to $d_{\mathrm{CE}}$).
I started explicitely calculating the right hand side:
By writing $x_1 \dotsm x_n$ instead of $x_1 \wedge \dotsb \wedge x_n$ we have
\begin{align*}
  {}&amp;
  \Delta(x_1 \dotsm x_n)
  \\
  ={}&amp;
  (x_1 \otimes 1 + 1 \otimes x_1)
  \dotsm
  (x_n \otimes 1 + 1 \otimes x_n)
  \\
  ={}&amp;
  \sum_{k=0}^n
  \;
  \sum_{\sigma \in S(k,n-k)}
  (-1)^{n_k(\sigma)}
  x_{\sigma(1)} \dotsm x_{\sigma(k)}
  \otimes
  x_{\sigma(k+1)} \dotsm x_{\sigma(n)}
\end{align*}
where $S(k,n-k) \subseteq S_n$ denotes the set of $k$-$(n-k)$-shuffles and
$$
  n_k(\sigma)
  =
  \#
  \{
    (i,j)
  \mid
    1 \leq i \leq k, \,
    k+1 \leq j \leq n, \,
    \sigma(i) &gt; \sigma(j)
  \} \,.
$$
For the right hand side of $(1)$ we hence get
\begin{align*}
  {}&amp;
  \sum_{k=0}^n
  \;
  \sum_{\sigma \in S(k,n-k)}
  (-1)^{n_k(\sigma)}
  \biggl[
    d_{\mathrm{CE}}(x_{\sigma(1)} \dotsm x_{\sigma(k)})
    \otimes
    x_{\sigma(k+1)} \dotsm x_{\sigma(n)}
  \\
  {}&amp;
  \phantom{
  \sum_{k=0}^n
  \;
  \sum_{\sigma \in S(k,n-k)}
  (-1)^{n_k(\sigma)}
  \biggl[
  }
    + (-1)^k
    x_{\sigma(1)} \dotsm x_{\sigma(k)}
    \otimes
    d_{\mathrm{CE}}(x_{\sigma(k+1)} \dotsm x_{\sigma(n)})
  \biggr]
  \\
  ={}&amp;
  \sum_{k=0}^n
  \;
  \sum_{\sigma \in S(k,n-k)}
  (-1)^{n_k(\sigma)}
  \\
  {}&amp;
  \biggl[
    \biggl(
      \sum_{1 \leq i &lt; j \leq k}
      (-1)^{i+j-1}
      [x_{\sigma(i)}, x_{\sigma(j)}]
      x_{\sigma(1)}
      \dotsb
      \widehat{x_{\sigma(i)}}
      \dotsb
      \widehat{x_{\sigma(j)}}
      \dotsb
      x_{\sigma(k)}
    \biggr)
    \otimes
    x_{\sigma(k+1)} \dotsm x_{\sigma(n)}
  \\
  {}&amp;
    +
    (-1)^k
    x_{\sigma(1)} \dotsm x_{\sigma(k)}
  \\
  {}&amp;
    \otimes
    \biggl(
      \sum_{1 \leq i &lt; j \leq n-k}
      (-1)^{i+j-1}
      [x_{\sigma(k+i)}, x_{\sigma(k+j)}]
      x_{\sigma(k+1)}
      \dotsb
      \widehat{x_{\sigma(k+i)}}
      \dotsb
      \widehat{x_{\sigma(k+j)}}
      \dotsb
      x_{\sigma(n)}
    \biggr)
  \biggr] \,.
\end{align*}
For the left side of $(1)$ one similarly find that
\begin{align*}
  {}&amp;
  \Delta( d( x_1 \dotsm x_n) )
  \\
  ={}&amp;
  \sum_{1 \leq i &lt; j \leq n}
  (-1)^{i+j-1}
  \Delta( [x_i, x_j] x_1 \dotsm \widehat{x_i} \dotsm \widehat{x_j} \dotsm x_n )
  \\
  ={}&amp;
  \sum_{1 \leq i &lt; j \leq n}
  (-1)^{i+j-1}
  ([x_i, x_j] \otimes 1 + 1 \otimes [x_i, x_j])
  (x_1 \otimes 1 + 1 \otimes x_1)
  \\
  {}&amp;
  \phantom{
    \sum_{1 \leq i &lt; j \leq n}
    (-1)^{i+j-1}
  }
  \dotsm
  (x_i \otimes 1 + 1 \otimes x_i)^{\wedge}
  \dotsm
  (x_j \otimes 1 + 1 \otimes x_j)^{\wedge}
  \dotsm
  (x_n \otimes 1 + 1 \otimes x_n)
\end{align*}
I guess that it could make sense that after multiplying out these terms and rearranging the resulting summands both sides of (1) give the same result
But I don’t really see this (and even less so for the signs).

Uniqueness

I think I have checked that if $C$ is any graded coalgebra and $d_1, d_2 \colon C \to C$ are graded derivations of the same degree $k$ that their equalizer
$$
  E
  :=
  \{
    x \in C
  \mid
    d_1(x) = d_2(x)
  \}
$$
is then a graded subcoalgebra of $C$.
We would like that any coderivation of degree $-1$ of $\bigwedge \mathfrak{g}$ is uniquely determined by the restriction $\bigwedge^2 \mathfrak{g} \to \mathfrak{g}$, so I suspect that $\bigwedge \mathfrak{g}$ is generated as a (graded) coalgebra by its homogeneous parts $\bigwedge^2 \mathfrak{g}$.
But I don’t know how to show this, or if this is correct.

I guess that for finite dimensional $\mathfrak{g}$ one could also dualize the problem and consider instead the cochain complex $\bigwedge \mathfrak{g}^*$ together with the (co)differential $d_{\mathrm{CE}}^*$ induced by $d_{\mathrm{CE}}$.
Then $\bigwedge \mathfrak{g}^*$ should be a graded algebra with derivation $d_{\mathrm{CE}}^*$, so $d_{\mathrm{CE}}^*$ should be uniquely determined by the restriction $\mathfrak{g}^* \to \bigwedge^2 \mathfrak{g}^*$ because $\bigwedge \mathfrak{g}^*$ is generated as an algebra by $\mathfrak{g}$.
But I’m not sure if this actually works.
Coderivation

The differential $d_{\mathrm{CE}}$ is of degree $-1$ so we need to show that
\begin{equation}
  \Delta( d_{\mathrm{CE}}(c) )
  =
  \sum_{(c)}
  d_{\mathrm{CE}}( c_{(1)} ) \otimes c_{(2)}
  +
  (-1)^{|c_{(1)}|} c_{(1)} \otimes d_{\mathrm{CE}}(c_{(2)})
  \tag{1}
\end{equation}
for all $c = x_1 \wedge \dotsb \wedge x_n$ with $x_i \in \mathfrak{g}$ (i.e. that $\Delta$ is a homomorphism of chain complexes with respect to $d_{\mathrm{CE}}$).
I started explicitely calculating the right hand side:
By writing $x_1 \dotsm x_n$ instead of $x_1 \wedge \dotsb \wedge x_n$ we have
\begin{align*}
  {}&amp;
  \Delta(x_1 \dotsm x_n)
  \\
  ={}&amp;
  (x_1 \otimes 1 + 1 \otimes x_1)
  \dotsm
  (x_n \otimes 1 + 1 \otimes x_n)
  \\
  ={}&amp;
  \sum_{k=0}^n
  \;
  \sum_{\sigma \in S(k,n-k)}
  (-1)^{n_k(\sigma)}
  x_{\sigma(1)} \dotsm x_{\sigma(k)}
  \otimes
  x_{\sigma(k+1)} \dotsm x_{\sigma(n)}
\end{align*}
where $S(k,n-k) \subseteq S_n$ denotes the set of $k$-$(n-k)$-shuffles and
$$
  n_k(\sigma)
  =
  \#
  \{
    (i,j)
  \mid
    1 \leq i \leq k, \,
    k+1 \leq j \leq n, \,
    \sigma(i) &gt; \sigma(j)
  \} \,.
$$
For the right hand side of $(1)$ we hence get
\begin{align*}
  {}&amp;
  \sum_{k=0}^n
  \;
  \sum_{\sigma \in S(k,n-k)}
  (-1)^{n_k(\sigma)}
  \biggl[
    d_{\mathrm{CE}}(x_{\sigma(1)} \dotsm x_{\sigma(k)})
    \otimes
    x_{\sigma(k+1)} \dotsm x_{\sigma(n)}
  \\
  {}&amp;
  \phantom{
  \sum_{k=0}^n
  \;
  \sum_{\sigma \in S(k,n-k)}
  (-1)^{n_k(\sigma)}
  \biggl[
  }
    + (-1)^k
    x_{\sigma(1)} \dotsm x_{\sigma(k)}
    \otimes
    d_{\mathrm{CE}}(x_{\sigma(k+1)} \dotsm x_{\sigma(n)})
  \biggr]
  \\
  ={}&amp;
  \sum_{k=0}^n
  \;
  \sum_{\sigma \in S(k,n-k)}
  (-1)^{n_k(\sigma)}
  \\
  {}&amp;
  \biggl[
    \biggl(
      \sum_{1 \leq i &lt; j \leq k}
      (-1)^{i+j-1}
      [x_{\sigma(i)}, x_{\sigma(j)}]
      x_{\sigma(1)}
      \dotsb
      \widehat{x_{\sigma(i)}}
      \dotsb
      \widehat{x_{\sigma(j)}}
      \dotsb
      x_{\sigma(k)}
    \biggr)
    \otimes
    x_{\sigma(k+1)} \dotsm x_{\sigma(n)}
  \\
  {}&amp;
    +
    (-1)^k
    x_{\sigma(1)} \dotsm x_{\sigma(k)}
  \\
  {}&amp;
    \otimes
    \biggl(
      \sum_{1 \leq i &lt; j \leq n-k}
      (-1)^{i+j-1}
      [x_{\sigma(k+i)}, x_{\sigma(k+j)}]
      x_{\sigma(k+1)}
      \dotsb
      \widehat{x_{\sigma(k+i)}}
      \dotsb
      \widehat{x_{\sigma(k+j)}}
      \dotsb
      x_{\sigma(n)}
    \biggr)
  \biggr] \,.
\end{align*}
For the left side of $(1)$ one similarly find that
\begin{align*}
  {}&amp;
  \Delta( d( x_1 \dotsm x_n) )
  \\
  ={}&amp;
  \sum_{1 \leq i &lt; j \leq n}
  (-1)^{i+j-1}
  \Delta( [x_i, x_j] x_1 \dotsm \widehat{x_i} \dotsm \widehat{x_j} \dotsm x_n )
  \\
  ={}&amp;
  \sum_{1 \leq i &lt; j \leq n}
  (-1)^{i+j-1}
  ([x_i, x_j] \otimes 1 + 1 \otimes [x_i, x_j])
  (x_1 \otimes 1 + 1 \otimes x_1)
  \\
  {}&amp;
  \phantom{
    \sum_{1 \leq i &lt; j \leq n}
    (-1)^{i+j-1}
  }
  \dotsm
  (x_i \otimes 1 + 1 \otimes x_i)^{\wedge}
  \dotsm
  (x_j \otimes 1 + 1 \otimes x_j)^{\wedge}
  \dotsm
  (x_n \otimes 1 + 1 \otimes x_n)
\end{align*}
I guess that it could make sense that after multiplying out these terms and rearranging the resulting summands both sides of (1) give the same result
But I don’t really see this (and even less so for the signs).

Uniqueness

I think I have checked that if $C$ is any graded coalgebra and $d_1, d_2 \colon C \to C$ are graded derivations of the same degree $k$ that their equalizer
$$
  E
  :=
  \{
    x \in C
  \mid
    d_1(x) = d_2(x)
  \}
$$
is then a graded subcoalgebra of $C$.
We would like that any coderivation of degree $-1$ of $\bigwedge \mathfrak{g}$ is uniquely determined by the restriction $\bigwedge^2 \mathfrak{g} \to \mathfrak{g}$, so I suspect that $\bigwedge \mathfrak{g}$ is generated as a (graded) coalgebra by its homogeneous parts $\bigwedge^2 \mathfrak{g}$.
But I don’t know how to show this, or if this is correct.

I guess that for finite dimensional $\mathfrak{g}$ one could also dualize the problem and consider instead the cochain complex $\bigwedge \mathfrak{g}^*$ together with the (co)differential $d_{\mathrm{CE}}^*$ induced by $d_{\mathrm{CE}}$.
Then $\bigwedge \mathfrak{g}^*$ should be a graded algebra with derivation $d_{\mathrm{CE}}^*$, so $d_{\mathrm{CE}}^*$ should be uniquely determined by the restriction $\mathfrak{g}^* \to \bigwedge^2 \mathfrak{g}^*$ because $\bigwedge \mathfrak{g}^*$ is generated as an algebra by $\mathfrak{g}$.
But I’m not sure if this actually works.