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I know that basis vector and basis of one-forms are related through

$$ \tilde{e}^\mu \cdot \vec{e}_\nu = \delta^\mu _\nu .\tag{1}$$

However, the metric has the property that allows to convert vectors to one-forms. So, can I try to say this: $$ \tilde{e}^\mu = g^{\mu \nu} \vec{e}_\nu , \tag{2}$$

if not explain me, please, maybe I don't clear with subscripts or can be some further.

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First I'll state three quick preliminaries so we're both on the same page, and then I'll answer the question. In the following, I'm going to use tildes to distinguish one-forms and their components from vectors and their components; it is traditional to drop this extra notation and simply write $X_\mu = g_{\mu \nu}X^\nu$ as though the $X$'s on either side are the same object. However, since they are emphatically not the same object, I use the notation $\tilde X_\mu = g_{\mu \nu} X^\nu$ to make this explicit.


Preliminary #1: The Metric

The metric $\mathbf{g}$ is an object which linearly eats two vectors and spits out a scalar (e.g. a real number). The components of $\mathbf{g}$ in a particular basis are what you get when you feed the metric the basis vectors:

$$g_{\mu \nu} \equiv \mathbf{g}(\hat e_\mu,\hat e_\nu)$$

Therefore, you often see the action of $\mathbf{g}$ on two vectors $\mathbf{X}=X^\mu \hat e_\mu$ and $\mathbf{Y} = Y^\nu \hat e_\nu$ written like this:

$$\mathbf{g}(\mathbf X,\mathbf Y)=\mathbf{g}(X^\mu \hat e_\mu,Y^\nu \hat e_\nu) = X^\mu Y^\nu \mathbf{g}(\hat e_\mu,\hat e_\nu) = X^\mu Y^\nu g_{\mu \nu}$$

We can pull the components $X^\mu,Y^\nu$ out front because $\mathbf g$ is linear.


Preliminary #2: One-forms

A one-form, or covector, is an object which linearly eats a vector and spits out a scalar. Note that the following is a one-form:

$$\tilde{\mathbf X} := \mathbf{g}(\mathbf{X},\bullet)$$

From a vector $\mathbf X$ and the metric $\mathbf g$, we can construct a one-form $\tilde{\mathbf X}$ by plugging $\mathbf{X}$ into the first slot of $\mathbf g$ and leaving the second slot empty. We say that the one-form $\tilde{\mathbf X}$ is dual to the vector $\mathbf X$.

$\tilde{\mathbf X}$ then acts on some vector $\mathbf Y$ in the obvious way:

$$\tilde{\mathbf X}(\mathbf Y) = \mathbf{g}(\mathbf X,\mathbf Y)$$

In particular, if we feed $\tilde{\mathbf X}$ a basis vector $\hat e_\nu$, we get

$$\tilde{\mathbf X}(\hat e_\nu) = \mathbf{g}(\mathbf X,\hat e_\nu) = X^\mu \mathbf{g}(\hat e_\mu, \hat e_\nu) = X^\mu g_{\mu \nu}$$ If we define the one-form basis $\hat \epsilon^\mu$ to have the property $\hat \epsilon^\mu (\hat e_\nu) = \delta^\mu_\nu$, then we can expand $\tilde{\mathbf X}$ in its components $\tilde X_\mu$. Performing the same action,

$$\tilde{\mathbf X}(\hat e_\nu) = \tilde X_\mu \hat \epsilon^\mu(\hat e_\nu) = \tilde X_\mu \delta^\mu_\nu = \tilde X_\nu$$

Comparing this to what we got before, we see that $\tilde X_\nu = X^\mu g_{\mu\nu}$.

Note that even though we say one forms eat vectors and spit out scalars, we can also say that vectors eat one-forms and spit out scalars. We simply define the action of a vector $\mathbf X$ on a one-form $\tilde{\mathbf Y}$ to be $$\mathbf X(\tilde{\mathbf Y}) := \tilde{\mathbf Y}(\mathbf X)$$ This will be relevant in a moment.


Preliminary #3: The Inverse Metric

We've seen that we can use the metric to associate a vector $\mathbf X$ to a dual one-form $\tilde{\mathbf X}$; we can also go the other direction and associate a covector $\tilde{\mathbf Y}$ to a dual vector $\mathbf Y$. We do this by defining the so-called inverse metric $\tilde{\mathbf g}$, which is a map which eats two one-forms and spits out a scalar.

Essentially this is just a metric on the space of one-forms in exactly the same way as $\mathbf g$ is a metric on the space of vectors. However, we link them together by demanding that if the vectors $\mathbf X$ and $\mathbf Y$ have one-form duals $\tilde{\mathbf X}$ and $\tilde{\mathbf Y}$, then

$$\mathbf{g}(\mathbf X,\mathbf Y) = \tilde{\mathbf g}(\tilde{\mathbf X},\tilde{\mathbf Y})$$

It's a straightforward exercise to show that this means that the components of the dual metric $\tilde g^{\mu \nu}$ satisfy the following relationship to the components of the metric:

$$\tilde g^{\mu \nu} g_{\nu \rho} = \delta^\mu_\rho$$

This means that if we express them in matrix form, the $\tilde g^{\mu\nu}$'s are the matrix inverse of the $g_{\mu\nu}$'s - hence the name "inverse metric". We can now use this to associate $\tilde{\mathbf Y}$ with a vector:

$$\mathbf Y = \tilde{\mathbf g}(\tilde{\mathbf{Y}},\bullet)$$ It's straightforward to demonstrate (by feeding the basis one-form $\hat\epsilon^\mu$ to $\mathbf Y$ as defined above) that the components of $\mathbf Y$ are, as expected, given by $$Y^\mu = \tilde{g}^{\mu \nu}\tilde Y_\nu$$


Now your question can be answered. It's true that the metric can "convert" vectors into one-forms in an abstract sense. However, when we talk about "raising" and "lowering" indices like you are doing, we are converting the components of a vector to the components of the corresponding one-form.

The expression that you wrote ($\tilde g^{\mu \nu} \hat e_\nu$, in my notation) is simply a linear combination of vectors, and is therefore not a one-form as defined here. It is, however, what you get when you convert the basis one-form $\hat \epsilon^\mu$ into a vector, which I will now show.

Note that the one-form dual to the unit vector $\hat e_\mu$ is not the basis one-form $\hat \epsilon^\mu$; it is the one-form

$$\tilde{\boldsymbol \omega} := \mathbf{g}(\hat e_\mu, \bullet)$$

which has components

$$\tilde \omega_{\nu} \equiv \tilde{\boldsymbol\omega}(\hat e_\nu) = \mathbf{g}(\hat e_\mu, \hat e_\nu) = g_{\mu \nu}$$

To be concrete, the dual to the basis vector $\hat e_0$ is the one form $$\tilde{\boldsymbol \omega} = g_{0\nu}\hat \epsilon^\nu = g_{00}\hat \epsilon^0 + g_{01}\hat \epsilon^1 + g_{02}\hat \epsilon^2 + g_{03} \hat \epsilon^3$$

Using the inverse metric in precisely the same way, the dual to the basis one-form $\hat \epsilon^0$ is the vector

$$\boldsymbol \omega = \tilde g^{0\nu}\hat e_\nu = \tilde g^{00}\hat e_0 + \tilde g^{01} \hat e_1 + \tilde g^{02} \hat e_2 + \tilde g^{03} \hat e_3$$


To conclude, we do not have that $$\hat\epsilon^\mu = g^{\mu \nu} \hat e_\nu$$ but rather that $$\underbrace{\tilde{\mathbf g}(\hat \epsilon^\mu,\bullet)}_{\text{The vector *dual* to the basis one-form }\hat\epsilon^\mu} = g^{\mu \nu}\hat e_\nu$$

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