Does interior multiplication come from an operation on $k$-vectors (instead of $k$-co vectors)?

Question

Usually, one sees the following definition of interior multiplication:

Let $V$ be an $n$-dimensional (real) vector space. $$ \iota_v :A_k(V) \to A_{k-1}(V) \, , \, (\iota_v \omega)(v_1, v_2, \dots, v_{k-1}) := \omega(v,v_1, v_2, \dots, v_{k-1}) ,$$

Where $A_k(V)$ is the vector space of all alternating multilinear maps $V^k \to \mathbb{R}$. Recall that $A_k(V) \cong \Lambda_k(V^*)$ where $\Lambda_k(V)$ is one of the direct summands of the Grassmann algebra (also known as the exterior algebra) of $V$.

My question: Does interior multiplication come from a suitable operation "lurking behind" on $\Lambda_k(V)$? i.e is there an opearation $j:\Lambda_k(V) \to \Lambda_{k-1}(V)$ s.t when aplplied to $V^*$ instead of $V$, it coincides with $i$.

A situation like this happens with the wedge product. One can define it explicitly on alternating maps (i.e $k$-covectors), but of course the wedge is defined on $k$-vectors, i.e $\alpha \wedge \beta = [\alpha \otimes \beta] $, and only when specializing to $V^*$ one gets the standard "wedge of forms" on $k$-covectors. (From this perspective, the wedge of the exterior algebra is a more "fundamental notion" then the wedge of forms).

Essentially, what I am asking is whether or not something analogous holds for the interior multiplication.

---

Comment: A possible approach to take is the following:

Given a vector space $V$, set $W=V^*$, and consider the interior multiplication of $W$, i.e given $\alpha \in W$, consider $i_{\alpha}:\Lambda_k(W^*) \to \Lambda_{k-1}(W^*)$. Since $W^*=V^{**}=V$ (the last equality is the natural identification between a space and its bi-dual) we consider the map $i_{\alpha}$ as a map $j^V_{\alpha}:\Lambda_k(V) \to \Lambda_{k-1}(V)$ (note we changed notation and gave a assigned a different letter for this "new" map to avoid confusion) .

In other words, we have for every $\alpha \in V^*$ a map $j^V_{\alpha}:\Lambda_k(V) \to \Lambda_{k-1}(V)$.

Considering this map as the "fundamental object", and taking $W=V^*$, we get that for every $v \in V^{**}=V$ there is a map $j^{V^*}_{v}:\Lambda_k(V^*) \to \Lambda_{k-1}(V^*)$. Now it remains to see if $j^{V^*}_{v}=i_v$. Actually, even if this is the case I am somehow unsatisfied from this solution. I would at least want to get an explicit expression ("formula") for the map $j^V_{\alpha}$.

Answer 1

What is behind your quesiton in my opinion is that in the area of tensor products and derived notions, there are two basic ways to describe things, either based on "decomposable elements" or based on multilinear maps. Some of the usual operations are much easier to describe in one picture than in the other, but I would not say that one picture is "more basic" than the other. In the most basic instance, starting from vector spaces $V$ and $W$, you can either take $V\otimes W$ defined as being spanned by elements of the form $v\otimes w$ as the basic object. Or you can say the basic object is $V^*\otimes W^*$, which is just the space of bilinear maps $V\times W\to\mathbb K$, where $\mathbb K$ is the ground field.

As you suspect, there is a basic insertion operator $i:V^*\times\Lambda^kV\to\Lambda^{k-1}V$. Most easily, this is described by identifying $\Lambda^kV$ with the space of $k$-linear alternating maps $(V^*)^k\to\mathbb K$ and then defining it as insertion as in the beginning of your question. But you can also write it out on decomposable elements, basically in the form $$ i_{\alpha}(v_1\wedge\dots\wedge v_k)=\sum_{i=1}^k(-1)^{i-1}\alpha(v_i)v_1\wedge\dots\wedge v_{i-1}\wedge v_{i+1}\wedge\dots\wedge v_k. $$ Exactly the same things happens (with the roles of $V$ and $V^*$ exchanged) if you look at the exterior multplication. This is easy to describe in the language of decomposable elements and a bit of work is needed to translate it into the picture of multilinear forms. But basically you can always switch freely between the two pictures.