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.


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 $e_\alpha^{\;i}$.

  • 1
    $\begingroup$ $g^{ij}$ is an map from space of vectors to space of functionals. Since the two spaces are usually isomorphic, for finite dimensional vector spaces, one can have isomorphic maps. So no contradiction, see physics.stackexchange.com/q/603251 $\endgroup$
    – Cryo
    Dec 27, 2020 at 10:30
  • 2
    $\begingroup$ having said that, notation you start with is a bit misleading, yes $\endgroup$
    – Cryo
    Dec 27, 2020 at 10:38
  • $\begingroup$ @Cryo: thanks for your comments. In the list of steps, could you say which ones are erroneous ? $\endgroup$ Dec 27, 2020 at 11:09
  • $\begingroup$ I wouldn't call it erroneous, but 4. Raising/lowering of indices is not a trivial summation, it is mapping between two vector spaces. The mapping is not unique, and may not even be defined (singular metric). Perhaps if you gave an example of a real calculation you want to do, it would be easier to see how one can do it in a more careful way. I would define metric as a scalar product between vectors or co-vectors and then use it to induce a map between two spaces $\endgroup$
    – Cryo
    Dec 27, 2020 at 11:20
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    $\begingroup$ @pasabaporaqui As a general rule, don't trust wikipedia too much. Remember that what's written there may not be the result of agreement, but just of persistence of some particular user – who may be completely wrong. I've found several erroneous statements in Wikipedia. It's good to get a general idea maybe, but at the very least check its references. $\endgroup$
    – pglpm
    Dec 27, 2020 at 14:15

1 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.

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.

  • $\begingroup$ Great answer, much better and cleaner than mine. Thanks $\endgroup$
    – Cryo
    Dec 27, 2020 at 17:24
  • $\begingroup$ Great answer, I didn't know about this older point of view. Can you suggest some references? Another advantage of the separation between vector and covector spaces is that it allows us to introduce many notions (fluxes, differential, parallel connection, and so on) relying only on (differential) topological notions, without invoking a metric structure. $\endgroup$
    – pglpm
    Dec 27, 2020 at 17:56
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    $\begingroup$ @pglpm For examples of the older convention, you can consult basically any intro text in solid state physics which discusses crystal structures. David Tong has some solid state notes online (see p. 52 of these notes ), or you could find a copy of one of the standard texts like Ashcroft and Mermin or Kittel. I'm definitely in agreement with you; I think the main impediment to adopting this perspective in crystallography is that the field universally works in $\mathbb R^n$ for $n=1,2,3$ so the power and flexibility of a more general [...] $\endgroup$
    – J. Murray
    Dec 27, 2020 at 18:01
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    $\begingroup$ [...] manifold treatment probably isn't worth the hassle. $\endgroup$
    – J. Murray
    Dec 27, 2020 at 18:02
  • $\begingroup$ Thank you. If you happen to have time, please do add some of your remarks and references in that Wikipedia section, I think that'd be helpful for many people who stumble there. $\endgroup$
    – pglpm
    Dec 27, 2020 at 18:13

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