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I've been trying to understand the geometrical meaning of $k$-vectors and $k$-forms on some vector space $V$ of finite dimension $n$ over a field $\Bbb K$. Indeed, as I understood, a $k$-form $\omega \in \Lambda^k(V)$ can be thought of as a measurement object on $k$-vectors. So, it takes a $k$-vector and says how much of it goes through it. In that sense, a $1$-form picks vectors and tells how much of that vector goes throught it in a scale defined by the form itself. In that setting, we represent $\omega$ by a subspace of dimension $n-k$.

On the other hand, $k$-vectors as I know represents pieces of $k$-dimensional subspaces. This means that they represent line segments, plane segments, cube segments and so on. In that case, we represent a $k$-vector $v$ by a piece of a $k$-dimensional subspace.

We look that the thing representing $\omega$ has dimension $n-k$ and the thing representing $v$ has dimension $k$. The sum of these dimensions is always $n$ and what $\omega$ is supposed to do is to pick $v$ and return the scalar $\omega(v)$.

What's the meaning of this relationship between the dimensions? In all of that intuitive view, why should we expect this kind of behavior?

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You should go hurt whoever told you those things because they didn't talk about the inner product they'd silently fixed in the background. What is happening here is that fixing an inner product induces an isomorphism between $\bigwedge^k V^*$ and $\bigwedge^{n-k} V$, which permets us to identity a $k$-form with an $(n-k)$-vector.

To fix ideas, let $V$ be an $n$-dimensional real vector space. A $k$-vector is an element of $\bigwedge^k V$ and a $k$-form is an element of $\bigwedge^k V^*$. These spaces are dual to one another, either by explicit computation or abstract nonsense.

Now, the top exterior power of a finite-dimensional vector space is one-dimensional, so both $\bigwedge^n V$ and $\bigwedge^n V^*$ are of dimension one. We also see (this time by direct computation) that there are nondegenerate bilinear pairings $$ \bigwedge^k V \times \bigwedge^{n-k} V \to \bigwedge^n V \quad\hbox{and}\quad \bigwedge^k V^* \times \bigwedge^{n-k} V^* \to \bigwedge^n V^* $$ for all $k$.

The problem is that while the top exterior powers are of dimension one, we do not have a canonical isomorphism between them. To get one, we fix an inner product $g$ on $V$. Its determinant (or volume form) is a nonzero element of $\bigwedge^n V^*$ and thus gives an isomorphism of that space with the ground field $\mathbb R$. We get a compatible isomorphism of $\bigwedge^n V$ with $\mathbb R$ by considering the dual metric. Composing these isomorphisms with the above nondegenerate pairings induces isomorphisms $$ \bigwedge^k V \cong \bigwedge^{n-k} V^* \quad\hbox{and}\quad \bigwedge^k V^* \cong \bigwedge^{n-k} V. $$ We finally get the correspondence between $k$-forms and $(n-k)$-vectors by following this last noncanonical isomorphism: $$ \bigwedge^k V^* \to \bigwedge^{n-k} V. $$ Technically speaking we only need to fix a volume form on the space $V$ to get these isomorphisms, because such a form will induce the necessary isomorphisms of the top exterior powers with $\mathbb R$, which in turn will identify $\bigwedge^k V^*$ and $\bigwedge^{n-k} V$ via the (now non-) canonical nondegenerate pairings. Sloppy, awful people sometimes do this silently by fixing a basis of $V$, which they then either define to be orthonormal or use to define a nonzero element of $\bigwedge^n V$, which then brings us to this kind of confusion later on.

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Given a metric, which allows for inner products represented by $\cdot$, you know that any 1-vector can be identified with a corresponding 1-form in the following sense: let $v, w$ be vectors. The linear functional

$$\omega: v \mapsto v \cdot w$$

is linear in $v$, and therefore a 1-form. $\omega$ can be identified with $w$, in that all such 1-forms exhibit a 1:1 correspondence with vectors with vectors when there is an inner product of vectors.

There is a natural generalization of $\cdot$ for inner products of $k$-vectors with other $k$-vectors (which I won't go into here). Let $i$ be some $n$-vector. Let $W$ be some $(n-1)$-vector. Then consider the linear functional

$$\Omega: v \mapsto (v \wedge W) \cdot \frac{i}{i \cdot i}$$

This map is insensitive to the choice of $i$. Crucially, $v \wedge W = \alpha i$ for some scalar $\alpha$, so we get $i \cdot i/(i \cdot i) = 1$. $i$ could be chosen to be unit in this sense with no loss of generality. $\Omega$ as a 1-form could then be identified equally well with $W$; there is a one-to-one correspondence.

So what if $\omega(v) = \Omega(v)$? What is the relationship between $w$ and $W$?

Again, this relies upon generalizing the inner product, now to $k$-vectors of unequal grade. As before, I will not elaborate on this; I will only submit that, if $A = A_p$ is a $p$-vector and $B = B_q$ is a $q$-vector such that $p \leq q$, then $A_p \cdot B_q$ is a $(q-p)$-vector.

I will also submit that $(v \wedge W) \cdot i = v \cdot (W \cdot i)$. Thus, we can arrive at the conclusion that

$$\omega(v) = \Omega(v) \implies w = \frac{W \cdot i}{i \cdot i}$$

The inner product can be thought of as taking parallel vectors and annihilating them to scalars. This is at the heart of "Hodge duality," and this inner product with an $n$-vector $i$ should be interpreted as performing Hodge duality, or finding the orthogonal $k$-vector to whatever it's multiplying.

In other words, $W \cdot i$ is a vector wholly orthogonal to the $(n-1)$-vector $W$. That is, $w$ is just the normal vector to the $n-1$-dimensional subspace that $W$ corresponds to.

I've glossed over some important aspects (inner products of arbitrary $p$-vectors with arbitrary $q$-vectors); for this kind of idea, clifford algebra is quite appropriate. It has a "geometric product" of that is well-suited to these problems, combining the inner product and the wedge product into a single operation, and it makes some of the identities I've invoked out of thin air rather simple to prove.

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