$ \mathbf{e}^i = g^{ij} \mathbf{e}_j$ interpretation

Question

I've problems in the interpretation of the expression:

$$ \mathbf{e}^i = g^{ij} \mathbf{e}_j$$

that can be found, by example, in this wiki chapter. Also here.

Step by step of my erroneous logic:

1. The elements of the expression are a vector $\mathbf{e}_i$ belonging to the basis of the tangent space; $\mathbf{e}^i$ of the basis of cotangent space; and the metric tensor $g$.
2. Since $\mathbf{e}_i$ is a vector of the basis of the tangent space, it is a contravariant vector.
3. Since $\mathbf{e}_i$ is a contravariant vector, it can be expressed in index notation as $e^\alpha_{\;i}$.
4. By usual lowering/raising index $g^{ij}e^\alpha_{\;j} = e^{\alpha\,i}$
5. By paralelism between initial expression $ \mathbf{e}^i = g^{ij} \mathbf{e}_j$ and previous one $e^{\alpha\;i}=g^{ij}e^\alpha_j$, I can say that $\mathbf{e}^i$ corresponds to $e^{\alpha\,i}$
6. Since $\mathbf{e}^i$ vector is expressed as $e^{\alpha\,i}$, it is a contravariant vector.
7. But $\mathbf{e}^i$ can not be contravariant because it is a vector of the basis of the cotangent space. Contradiction.

Not found where is the error in previous sequence, all steps seems basic and true.

Addendum:

Another way to reach same contradiction:

1b. The set of all vectors that forms the basis of the tangent space $\{\mathbf{e}_1,\mathbf{e}_2,\dots\}$ is expressed in index form as $e^\alpha_{\;i}$.

2b. $e^\alpha_{\;i}$ expresses all the set of basis vectors of the tangent space. $e^\alpha_{\;i}$ is a tensor with two indexes, $\alpha$ contravariant (related to the space components) and $i$ covariant (related to the index in the basis set).

3b. $g^{ij}$ is a tensor that, given two covariant tensors produces an scalar. In other words, given a covariant vector/tensor, produces a contravariant vector/tensor. Or, more generically, maps from a (n+m)-tensor with n contravariant dimensions and m covariant ones to another (n+m)-tensor with (n+1) contravariant dimensions and (m-1) covariant ones.

4b. Applying $g^{ij}$ over $e^\alpha_{\;j}$ we map the $j$ covariant dimension of $e$ to contravariant, obtaining a tensor twice contravariant $e^{\alpha\;i}$

5b. Since $e^{\alpha\;i}$ has two contravariant indexes, it can not be the set of basic vectors of the cotangent space. The basis of cotangent space is expected in the form $e_\alpha^{\;i}$.

Answer 1

I think your confusion comes from the fact that there are several different ways of looking at covariance and contravariance.

The old-school way to treat this issue is to say that given a basis $\mathbf e_i$ for a vector space $V$, we can define a dual basis for $V$ which we write as $\mathbf e^i = g^{ij}\mathbf e_j$. In this framework, both $\mathbf e_i$ and $\mathbf e^i$ belong to $V$. Correspondingly, a vector $\mathbf v\in V$ can be expanded in terms of the original basis or the dual basis, i.e. $\mathbf v = v^i\mathbf e_i = v_i \mathbf e^i$. The $v^i$ are called the contravariant components of $\mathbf v$, while the $v_i$ are the covariant components of $\mathbf v$. In order for this equality to hold, we must have that $v_i = g_{ij} v^j$, where $g_{ij}$ and $g^{ij}$ are matrix inverses of one another.

The inner product between vectors is given by $\mathbf e_i \cdot \mathbf e_j = g_{ij}$. As a result, $\mathbf e^i \cdot \mathbf e_j = g^{ik}\mathbf e_k \mathbf e_j = g^{ik}g_{kj} = \delta ^i_j$. Therefore, we can write the inner product between two vectors in any of the following equivalent ways:

$$\mathbf v\cdot \mathbf w = v^i w^j\mathbf e_i \cdot \mathbf e_j = v_i w^j = v^i w^j g_{ij}$$

Note that at no point have we left the vector space $V$. There is no notion of a dual space here; everything takes place in a single vector space, and contravariance and covariance of vector or tensor components is purely a property of which basis you elect to expand the vector or tensor in. This convention is still in use in fields like crystallography, where $\mathbf e_i$ might represent the lattice vectors of some crystal and the $\mathbf e^i$ are the reciprocal lattice vectors.

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The more modern treatment is to say that given a vector space $ V$ and a basis $\mathbf e_i$, we can define a basis $\boldsymbol \epsilon^i$ for the (algebraic) dual space $V^*$ by the condition that $\boldsymbol \epsilon^i(\mathbf e_j) = \delta^i_j$. Any non-degenerate bilinear form (such as a metric) defines an isomorphism between $V$ and $V^*$. Any vector $\mathbf v\in V$ has a covector partner $\mathbf v^\flat\in V^*$ given by $$\mathbf v^\flat = \mathbf g(\mathbf v,\bullet)$$ whose action on a vector $\mathbf w\in V$ is then $$\mathbf v^\flat(\mathbf w) = \mathbf g(\mathbf v,\mathbf w) = g_{ij} v^i w^j$$

This approach is ultimately much cleaner in my opinion. Vectors and covectors become clearly different geometrical objects with different transformation properties, and the differences can be manifested in clearly basis-independent ways. However, it should be noted that the older and newer perspectives are ultimately equivalent.