De Rham and Koszul complexes Consider the algebraic de Rham complex of the $n$-dimensional plane: this is merely
$$\ldots\rightarrow Sym(V^*)\otimes\bigwedge^{k}V^*\rightarrow Sym(V^*)\otimes\bigwedge^{k+1}V^*\rightarrow\ldots
$$
with the standard de Rham differential $fdx_I\rightarrow\sum_{i=1}^n\frac{df}{dx_i}dx_i\wedge dx_I$.
On the other hand, there is a Koszul resolution of the ideal $(x_1,\ldots,x_n)\subset Sym(V^*)$ is defined  on a chain complex which is termwise the same as above, but with differential going in the opposite direction: e.g. it sends $dx_{i_1}\wedge\ldots\wedge dx_{i_k}$ to $x_{i_1}dx_{i_2}\wedge\ldots\wedge dx_{i_k}-x_{i_2}dx_{i_1}\wedge dx_{i_3}\wedge\ldots\wedge dx_{i_n}+\ldots+(-1)^{k-1}x_{i_k}dx_{i_1}\wedge\ldots dx_{i_{k-1}}$.
If we call the de Rham differential $d$, and the Koszul differential $e$, then $de+ed$ is scaling by the total degree (exterior and symmetric parts) of the form.
Having some familiarity with different types of cohomology groups that one encounters in algebraic geometry, it seems unusual to me that there would exist two different differentials on the same "complex", and that they would be related in such a way. Thus, I am curious whether this is an instance of some more general phenomenon, for example whether this occurs for certain complexes of sheaves on other spaces. Though I am not an expert in this language, it seems one way to phrase what happens in this example is that there is a single graded algebra with two different DGA structures which are related (by this scaling formula). Are there any/many other examples of this happening?
 A: What you have here is a graded version of an $\mathfrak{sl}_2$-triple. Compare for example the proof of the Lefschetz decomposition and hard Lefschetz theorem in Hodge theory, which is essentially the observation that $L$ and $\Lambda$ form the $e$ and $f$ of an $\mathfrak{sl}_2$-triple, because
$$[L,\Lambda] = (k-n)\operatorname{id}$$
on degree $k$ forms (so let $h$ act by $k-n$ in degree $k$).
I will first give a definition of a 'graded $\mathfrak{sl}_2$-triple', and then give a conceptual explanation for the example you gave.
Definition. Let $k$ be a field (algebraically closed of characteristic $0$, for simplicity), and let $C$ be the complex
$$\ldots \to 0 \to k \stackrel 0\to k \to 0 \to \ldots$$
with $k$ in (cohomological) degrees $0$ and $1$. Let $\mathfrak{gl}(C)$ be the graded Lie algebra defined by the graded algebra $\operatorname{Hom}^*(C,C)$ (all endomorphisms of $C = C^0 \oplus C^1$, with its natural grading), and let $\mathfrak{sl}(C)$ be the graded Lie subalgebra obtained by replacing $\operatorname{Hom}^0(C,C)$ by its trace $0$ elements, recalling that
$$\operatorname{tr}\big(f \colon C \to C\big) = \sum_i (-1)^i\operatorname{tr}\big(f^i \colon C^i \to C^i\big).$$
Explicitly, $\mathfrak{sl}(C)$ has a basis $\{e,f,h\}$ where $h \colon C \to C$ is the identity (which indeed has trace $0$), $e \colon C \to C[1]$ is the map
$$\begin{array}{ccccccccc}\ldots & \to &0 & \to & k & \to & k & \to & \ldots \\ & & \downarrow & & || & & \downarrow & & \\ \ldots & \to & k & \to & k & \to & 0 & \to & \ldots,\! \end{array}$$
and $f \colon C \to C[-1]$ is the map
$$\begin{array}{ccccccccc}\ldots & \to &k & \to & k & \to & 0 & \to & \ldots \\ & & \downarrow & & || & & \downarrow & & \\ \ldots & \to & 0 & \to & k & \to & k & \to & \ldots.\! \end{array}$$
With the graded Lie bracket, we get
\begin{align*}
[e,f] &= ef + fe = h,\\
[h,e] &= 0,\\
[h,f] &= 0.
\end{align*}
Viewing the complex $C$ with zero differentials as a graded object $C^0 \oplus C^1$, the map $e$ can be represented by the map $(a,b) \mapsto (b,0)$ and $f$ by $(a,b) \mapsto (0,a)$.
Definition. A graded $\mathfrak{sl}_2$-triple on a graded vector space $V$ is a triple $(e,f,h)$ of elements in $\operatorname{Hom}^*(V,V)$ of degrees $1$, $-1$, and $0$ respectively satisfying the identities
\begin{align*}
[e,f] &= ef + fe = h,\\
[h,e] &= 0,\\
[h,f] &= 0.
\end{align*}
Example. Let $V$ be a finite dimensional vector space, and let $V \otimes C$ be the two term complex $V \oplus V[-1]$ with zero differential. Then
$$\operatorname{Sym}(V \otimes C) = \operatorname{Sym}(V) \otimes \operatorname{Sym}(V[-1]) = \bigoplus_i \operatorname{Sym}(V) \otimes  \left(\bigwedge\nolimits^i V\right)[-i],$$
since the sign in the graded swap $K \otimes L \stackrel\sim\to L \otimes K$ replaces the symmetriser on $K^{\otimes n}$ by the antisymmetriser in odd degree. This is the complex you're studying, except it has differentials all $0$ for now.
But $\operatorname{Hom}^*(C,C)$ acts on $V \otimes C$, hence so does $\mathfrak{sl}_2(C)$. Then the latter also acts¹ on $T^*(V \otimes C)$ and its quotient $\operatorname{Sym}(V \otimes C)$.

Lemma. This graded $\mathfrak{sl}_2$-triple is the one described by the OP, with $e$ the Koszul differential, $f$ the de Rham differential, and $h$ multiplication by the total degree.

Proof. Writing
$$T^n(V \otimes C) = \big(V \oplus V[-1]\big)^{\otimes n}$$
with elements $(v_1,w_1) \otimes \ldots \otimes (v_n,w_n)$, the action of $\mathfrak{sl}_2(C)$ is given by
\begin{align*}
e \colon T^n(V \otimes C) \to& T^n(V \otimes C)[1]\\
(v_1,w_1) \otimes \ldots \otimes (v_n,w_n) \mapsto & \sum_{i=1}^n (v_1,-w_1) \otimes \ldots \otimes (v_{i-1},-w_{i-1}) \\ & \otimes (w_i,0) \otimes (v_{i+1},w_{i+1}) \otimes \ldots \otimes (v_n,w_n),\\\\
f \colon T^n(V \otimes C) \to& T^n(V \otimes C)[-1]\\
(v_1,w_1) \otimes \ldots \otimes (v_n,w_n) \mapsto & \sum_{i=1}^n (v_1,-w_1) \otimes \ldots \otimes (v_{i-1},-w_{i-1}) \\ & \otimes (0,v_i) \otimes (v_{i+1},w_{i+1}) \otimes \ldots \otimes (v_n,w_n),
\end{align*}
and $h$ is multiplication by $n$. In the above notation, the quotient map
$$T^n(V \otimes C) \to \operatorname{Sym}^n(V \otimes C) = \bigoplus_{i+j=n} \operatorname{Sym}^i V \otimes \left(\bigwedge\nolimits^j V\right)[-j]$$
is given by
$$(v_1,w_1) \otimes \ldots \otimes (v_n,w_n) \mapsto \sum_{I \amalg J = \{1,\ldots,n\}} v_I \otimes w_J,$$
where $v_I = \prod_{i \in I} v_i$ and $w_J = w_{j_1} \wedge \ldots \wedge w_{j_s}$ if $\{j_1 < \ldots < j_s\} = J$. Then $f$ descends to the de Rham differential $d$, and $e$ descends to the Koszul differential $e$. Indeed, consider an element
$$(v_1,0) \otimes \ldots \otimes (v_r,0) \otimes (0,w_1) \otimes \ldots \otimes (0,w_s)$$
lifting $v_I \otimes w_J$ where $I = \{1,\ldots,r\}$. Then $f$ maps this to
$$\sum_{i=1}^r (v_1,0) \otimes \ldots \otimes (v_{i-1},0) \otimes (0,v_i) \otimes (v_{i+1},0) \otimes \ldots \otimes (v_r,0) \otimes (0,w_1) \otimes \ldots \otimes (0,w_s),$$
which under the quotient $T^n(V \otimes C) \to \operatorname{Sym}^n(V \otimes C)$ maps to
$$\sum_{i=1}^r \left(\prod_{j \neq i} v_j\right) \otimes \big(v_i \wedge w_J\big),$$
which is the de Rham differential of $v_I \otimes w_J$. On the other hand $e$ takes it to
$$\sum_{j=1}^s (v_1,0) \otimes \ldots \otimes (v_r,0) \otimes (0,-w_1) \otimes \ldots (0,-w_{j-1}) \otimes (w_j,0) \otimes (0,w_{j+1}) \otimes \ldots \otimes (0,w_n),$$
which under the quotient $T^n(V \otimes C) \to \operatorname{Sym}^n(V \otimes C)$ maps to
$$\sum_{j=1}^s (-1)^{j-1} v_Iw_j \otimes w_{J \setminus\{j\}},$$
which is the Koszul differential of $v_I \otimes w_J$. Finally, just like on $T^n(V \otimes C)$, on $\operatorname{Sym}^n(V \otimes C)$ the map $h$ is just multiplication by $n$, the total degree of $v_I \otimes w_J$. $\square$

¹ The map $\operatorname{Hom}^*(C,C) \to \operatorname{Hom}^*(C \otimes C, C \otimes C)$ by $f \mapsto f \otimes f$ is not linear (nor does it preserve the grading). So we do not get a natural action of $\operatorname{Hom}^*(C,C)$ on $\operatorname{Sym}(V \otimes C)$. However for a graded Lie algebra $L$ acting on complexes $C$ and $D$, there is an action on $C \otimes D$ by
$$\rho_{C \otimes D}(x)(c \otimes d) = \Big(\rho_C(x) \otimes 1 + (-1)^{\deg(x)\deg(c)} \otimes \rho_D(x)\Big)(c \otimes d)$$
for $x \in L$, where $\rho_C(x)$ and $\rho_D(x)$ are the actions of $x$ on $C$ and $D$ respectively. Similarly one gets actions on $\operatorname{Sym}(C)$, etcetera.
