What is the name for this relationship between a 1-form and a vector? all.
I have a question about Visual Differential Geometry and Forms - A Mathematical Drama in Five Acts (by Tristan Needham).
This book shows two relationships between Forms and vector. And I have a question on the second one.
The first one is a relationship between a 2-form $\Psi$ and a vector $\Psi$(underlined), which is shown as (34.10) on page p377. Their corresponding components are equal, that is, $\Psi^i$ = $\Psi$(underlined)$_i$. This relationship is called Hodge (star) duality operator.

The second one is a relationship between a 1-form $\phi$ and a vector $\phi$(underlined), which is shown as (34.14) on page p379. Their corresponding components are equal, that is, $\phi_i$ = $\phi$(underlined)$_i$.

My question is: What is the name for this second relationship?
P.S. This second relationship is so natural that it deserves a name, right?
 A: Given any finite dimensional vector space $V$ over a field $k$, and a choice of basis $e_1,\dots,e_n$, there is a canonical isomorphism $\alpha\colon V\cong V^*$, where $V^*$ is the dual vector space to $V$, i.e., the vector space of linear maps $V\to k$. (Linear maps $V\to k$ are also called one-forms.)
The isomorphism $\alpha$ is defined by sending the vector $e_i$ to the one-form $e^i$ (note the superscript) defined on a vector $v = v_1e_1 + \dots + v_ne_n$ by
$$
e^i(v) = e^i(v_1e_1 + \dots + v_ne_n) = v_i.
$$
In other words, $e^i$ just returns the $i$th component of the vector $v$. You should check that the one-forms $e^1,e^2,\dots,e^n$ form a basis for $V^*$. This basis is known as the corresponding dual basis for $V^*$. This is all linear algebra, but in differential geometry, this isomorphism is a simple version of the musical isomorphism, which is the more general term for the isomorphism between the tangent and cotangent bundles of a manifold.
In the example in Needham's book, if we choose the standard basis $\mathbf e_1,\mathbf e_2,\mathbf e_3$ for $\mathbb R^3$, then the one-forms $dx^1,dx^2,dx^3$ are precisely the corresponding dual basis for $(\mathbb R^3)^*$.
