Covariance and contravariance of vectors

Definition

The general formulation of covariance and contravariance refers to how the components of a coordinate vector transform under a change of basis (passive transformation). Thus let V be a vector space of dimension n over a field of scalars S, and let each of f = (X₁, ..., X_n) and f′ = (Y₁, ..., Y_n) be a basis of V. Also, let the change of basis from f to f′ be given by

for some invertible n×n matrix A with entries ${\displaystyle a_j^i}$. Here, each vector Y_j of the f′ basis is a linear combination of the vectors X_i of the f basis, so that

$${\displaystyle Y_j=\sum _i{a}_j^i{X}_i,}$$

which are the columns of the matrix product ${\displaystyle \mathbf{f}A}$.

Contravariant transformation

A vector ${\displaystyle v}$ in V is expressed uniquely as a linear combination of the elements ${\displaystyle X_i}$ of the f basis as

where vⁱ[f] are elements of the field S known as the components of v in the f basis. Denote the column vector of components of v by v[f]:

$${\displaystyle \mathbf{v}[\mathbf{f}]=\left[\begin{array}{c}{v}^1[\mathbf{f}]\\ v^2[\mathbf{f}]\\ ⋮\\ v^n[\mathbf{f}]\end{array}\right]}$$

so that (2) can be rewritten as a matrix product

$${\displaystyle v=\mathbf{f}\mathbf{v}[\mathbf{f}].}$$

The vector v may also be expressed in terms of the f′ basis, so that

$${\displaystyle v={\mathbf{f}}^{\prime }\mathbf{v}[{\mathbf{f}}^{\prime }].}$$

However, since the vector v itself is invariant under the choice of basis,

$${\displaystyle \mathbf{f}\mathbf{v}[\mathbf{f}]=v={\mathbf{f}}^{\prime }\mathbf{v}[{\mathbf{f}}^{\prime }].}$$

The invariance of v combined with the relationship (1) between f and f′ implies that

$${\displaystyle \mathbf{f}\mathbf{v}[\mathbf{f}]=\mathbf{f}A\mathbf{v}[\mathbf{f}A],}$$

giving the transformation rule

$${\displaystyle \mathbf{v}[{\mathbf{f}}^{\prime }]=\mathbf{v}[\mathbf{f}A]=A^{-1}\mathbf{v}[\mathbf{f}].}$$

In terms of components,

$${\displaystyle v^i[\mathbf{f}A]=\sum _j{\tilde{a}}_j^i{v}^j[\mathbf{f}]}$$

where the coefficients ${\displaystyle {\tilde{a}}_j^i}$ are the entries of the inverse matrix of A.

Because the components of the vector v transform with the inverse of the matrix A, these components are said to transform contravariantly under a change of basis.

The way A relates the two pairs is depicted in the following informal diagram using an arrow. The reversal of the arrow indicates a contravariant change: $${\displaystyle \begin{array}{cc}\mathbf{f}& ⟶{\mathbf{f}}^{\prime }\\ v[\mathbf{f}]& ⟵v[{\mathbf{f}}^{\prime }]\end{array}}$$

Covariant transformation

A linear functional α on V is expressed uniquely in terms of its components (elements in S) in the f basis as

$${\displaystyle \alpha (X_i)={\alpha }_i[\mathbf{f}],i=1,2,\dots ,n.}$$

These components are the action of α on the basis vectors X_i of the f basis.

Under the change of basis from f to f′ (via 1), the components transform so that

Denote the row vector of components of α by α[f]:

$${\displaystyle \alpha [\mathbf{f}]=\left[\begin{array}{c}{\alpha }_1[\mathbf{f}],{\alpha }_2[\mathbf{f}],\dots ,{\alpha }_n[\mathbf{f}]\end{array}\right]}$$

so that (3) can be rewritten as the matrix product

$${\displaystyle \alpha [\mathbf{f}A]=\alpha [\mathbf{f}]A.}$$

Because the components of the linear functional α transform with the matrix A, these components are said to transform covariantly under a change of basis.

The way A relates the two pairs is depicted in the following informal diagram using an arrow. A covariant relationship is indicated since the arrows travel in the same direction: $${\displaystyle \begin{array}{cc}\mathbf{f}& ⟶{\mathbf{f}}^{\prime }\\ \alpha [\mathbf{f}]& ⟶\alpha [{\mathbf{f}}^{\prime }]\end{array}}$$

Had a column vector representation been used instead, the transformation law would be the transpose $${\displaystyle {\alpha }^{\mathrm{T}}[\mathbf{f}A]=A^{\mathrm{T}}{\alpha }^{\mathrm{T}}[\mathbf{f}].}$$

Coordinates

The choice of basis f on the vector space V defines uniquely a set of coordinate functions on V, by means of $${\displaystyle x^i[\mathbf{f}](v)=v^i[\mathbf{f}].}$$ The coordinates on V are therefore contravariant in the sense that $${\displaystyle x^i[\mathbf{f}A]=\sum _{k=1}^n{\tilde{a}}_k^i{x}^k[\mathbf{f}].}$$ Conversely, a system of n quantities vⁱ that transform like the coordinates xⁱ on V defines a contravariant vector (or simply vector). A system of n quantities that transform oppositely to the coordinates is then a covariant vector (or covector).

This formulation of contravariance and covariance is often more natural in applications in which there is a coordinate space (a manifold) on which vectors live as tangent vectors or cotangent vectors. Given a local coordinate system xⁱ on the manifold, the reference axes for the coordinate system are the vector fields $${\displaystyle X_1=\frac{\mathrm{\partial }}{\mathrm{\partial }{x}^1},\dots ,X_n=\frac{\mathrm{\partial }}{\mathrm{\partial }{x}^n}.}$$ This gives rise to the frame f = (X₁, ..., X_n) at every point of the coordinate patch.

If yⁱ is a different coordinate system and $${\displaystyle Y_1=\frac{\mathrm{\partial }}{\mathrm{\partial }{y}^1},\dots ,Y_n=\frac{\mathrm{\partial }}{\mathrm{\partial }{y}^n},}$$ then the frame f' is related to the frame f by the inverse of the Jacobian matrix of the coordinate transition: $${\displaystyle {\mathbf{f}}^{\prime }=\mathbf{f}{J}^{-1},J={\left(\frac{\mathrm{\partial }{y}^i}{\mathrm{\partial }{x}^j}\right)}_{i,j=1}^n.}$$ Or, in indices, $${\displaystyle \frac{\mathrm{\partial }}{\mathrm{\partial }{y}^i}=\sum _{j=1}^n\frac{\mathrm{\partial }{x}^j}{\mathrm{\partial }{y}^i}\frac{\mathrm{\partial }}{\mathrm{\partial }{x}^j}.}$$

A tangent vector is by definition a vector that is a linear combination of the coordinate partials ${\displaystyle \mathrm{\partial }/\mathrm{\partial }{x}^i}$. Thus a tangent vector is defined by $${\displaystyle v=\sum _{i=1}^n{v}^i[\mathbf{f}]X_i=\mathbf{f}\mathbf{v}[\mathbf{f}].}$$

Such a vector is contravariant with respect to change of frame. Under changes in the coordinate system, one has $${\displaystyle \mathbf{v}\left[{\mathbf{f}}^{\prime }\right]=\mathbf{v}\left[\mathbf{f}{J}^{-1}\right]=J\mathbf{v}[\mathbf{f}].}$$

Therefore, the components of a tangent vector transform via $${\displaystyle v^i\left[{\mathbf{f}}^{\prime }\right]=\sum _{j=1}^n\frac{\mathrm{\partial }{y}^i}{\mathrm{\partial }{x}^j}{v}^j[\mathbf{f}].}$$

Accordingly, a system of n quantities vⁱ depending on the coordinates that transform in this way on passing from one coordinate system to another is called a contravariant vector.

Covariant and contravariant components of a vector with a metric

In a finite-dimensional vector space V over a field K with a non-degenerate symmetric bilinear form g : V × V → K (which may be referred to as the metric tensor), there is little distinction between covariant and contravariant vectors, because the bilinear form allows covectors to be identified with vectors. That is, a vector v uniquely determines a covector α via $${\displaystyle \alpha (w)=g(v,w)}$$ for all vectors w. Conversely, each covector α determines a unique vector v by this equation. Because of this identification of vectors with covectors, one may speak of the covariant components or contravariant components of a vector, that is, they are just representations of the same vector using the reciprocal basis.

Given a basis f = (X₁, ..., X_n) of V, there is a unique reciprocal basis f^# = (Y¹, ..., Yⁿ) of V determined by requiring that $${\displaystyle g(Y^i,X_j)={\delta }_j^i,}$$ the Kronecker delta. In terms of these bases, any vector v can be written in two ways: $${\displaystyle \begin{array}{cc}v& =\sum _i{v}^i[\mathbf{f}]X_i=\mathbf{f}\mathbf{v}[\mathbf{f}]\\ & =\sum _i{v}_i[{\mathbf{f}}^{\mathrm{♯}}]Y^i={\mathbf{f}}^{\mathrm{♯}}{\mathbf{v}}^{\mathrm{♯}}[\mathbf{f}].\end{array}}$$ The components vⁱ[f] are the contravariant components of the vector v in the basis f, and the components v_i[f] are the covariant components of v in the basis f. The terminology is justified because under a change of basis,

$${\displaystyle \mathbf{v}[\mathbf{f}A]=A^{-1}\mathbf{v}[\mathbf{f}],{\mathbf{v}}^{\mathrm{♯}}[\mathbf{f}A]=A^T{\mathbf{v}}^{\mathrm{♯}}[\mathbf{f}]}$$ where ${\displaystyle A}$ is an invertible ${\displaystyle n\times n}$ matrix, and the matrix transpose has its usual meaning.

Euclidean plane

In the Euclidean plane, the dot product allows for vectors to be identified with covectors. If ${\displaystyle {\mathbf{e}}_1,{\mathbf{e}}_2}$ is a basis, then the dual basis ${\displaystyle {\mathbf{e}}^1,{\mathbf{e}}^2}$ satisfies $${\displaystyle \begin{array}{cc}{\mathbf{e}}^1\cdot {\mathbf{e}}_1=1,& {\mathbf{e}}^1\cdot {\mathbf{e}}_2=0\\ {\mathbf{e}}^2\cdot {\mathbf{e}}_1=0,& {\mathbf{e}}^2\cdot {\mathbf{e}}_2=1.\end{array}}$$

Thus, e¹ and e₂ are perpendicular to each other, as are e² and e₁, and the lengths of e¹ and e² normalized against e₁ and e₂, respectively.

Example

For example, suppose that we are given a basis e₁, e₂ consisting of a pair of vectors making a 45° angle with one another, such that e₁ has length 2 and e₂ has length 1. Then the dual basis vectors are given as follows:

• e² is the result of rotating e₁ through an angle of 90° (where the sense is measured by assuming the pair e₁, e₂ to be positively oriented), and then rescaling so that e² ⋅ e₂ = 1 holds.
• e¹ is the result of rotating e₂ through an angle of 90°, and then rescaling so that e¹ ⋅ e₁ = 1 holds.

Applying these rules, we find $${\displaystyle {\mathbf{e}}^1=\frac{1}{2}{\mathbf{e}}_1-\frac{1}{\sqrt{2}}{\mathbf{e}}_2}$$ and $${\displaystyle {\mathbf{e}}^2=-\frac{1}{\sqrt{2}}{\mathbf{e}}_1+2{\mathbf{e}}_2.}$$

Thus the change of basis matrix in going from the original basis to the reciprocal basis is $${\displaystyle R=\left[\begin{array}{cc}\frac{1}{2}& -\frac{1}{\sqrt{2}}\\ -\frac{1}{\sqrt{2}}& 2\end{array}\right],}$$ since $${\displaystyle [{\mathbf{e}}^1{\mathbf{e}}^2]=[{\mathbf{e}}_1{\mathbf{e}}_2]\left[\begin{array}{cc}\frac{1}{2}& -\frac{1}{\sqrt{2}}\\ -\frac{1}{\sqrt{2}}& 2\end{array}\right].}$$

For instance, the vector $${\displaystyle v=\frac{3}{2}{\mathbf{e}}_1+2{\mathbf{e}}_2}$$ is a vector with contravariant components $${\displaystyle v^1=\frac{3}{2},v^2=2.}$$

The covariant components are obtained by equating the two expressions for the vector v: $${\displaystyle v=v_1{\mathbf{e}}^1+v_2{\mathbf{e}}^2=v^1{\mathbf{e}}_1+v^2{\mathbf{e}}_2}$$ so $${\displaystyle \begin{array}{cc}\left[\begin{array}{c}{v}_1\\ v_2\end{array}\right]& =R^{-1}\left[\begin{array}{c}{v}^1\\ v^2\end{array}\right]\\ & =\left[\begin{array}{cc}4& \sqrt{2}\\ \sqrt{2}& 1\end{array}\right]\left[\begin{array}{c}{v}^1\\ v^2\end{array}\right]\\ & =\left[\begin{array}{c}6+2\sqrt{2}\\ 2+\frac{3}{\sqrt{2}}\end{array}\right]\end{array}.}$$

Three-dimensional Euclidean space

In the three-dimensional Euclidean space, one can also determine explicitly the dual basis to a given set of basis vectors e₁, e₂, e₃ of E₃ that are not necessarily assumed to be orthogonal nor of unit norm. The dual basis vectors are:

$${\displaystyle {\mathbf{e}}^1=\frac{{\mathbf{e}}_2\times {\mathbf{e}}_3}{{\mathbf{e}}_1\cdot ({\mathbf{e}}_2\times {\mathbf{e}}_3)};{\mathbf{e}}^2=\frac{{\mathbf{e}}_3\times {\mathbf{e}}_1}{{\mathbf{e}}_2\cdot ({\mathbf{e}}_3\times {\mathbf{e}}_1)};{\mathbf{e}}^3=\frac{{\mathbf{e}}_1\times {\mathbf{e}}_2}{{\mathbf{e}}_3\cdot ({\mathbf{e}}_1\times {\mathbf{e}}_2)}.}$$

Even when the e_i and eⁱ are not orthonormal, they are still mutually reciprocal: $${\displaystyle {\mathbf{e}}^i\cdot {\mathbf{e}}_j={\delta }_j^i,}$$

Then the contravariant components of any vector v can be obtained by the dot product of v with the dual basis vectors:

$${\displaystyle q^1=\mathbf{v}\cdot {\mathbf{e}}^1;q^2=\mathbf{v}\cdot {\mathbf{e}}^2;q^3=\mathbf{v}\cdot {\mathbf{e}}^3.}$$

Likewise, the covariant components of v can be obtained from the dot product of v with basis vectors, viz.

$${\displaystyle q_1=\mathbf{v}\cdot {\mathbf{e}}_1;q_2=\mathbf{v}\cdot {\mathbf{e}}_2;q_3=\mathbf{v}\cdot {\mathbf{e}}_3.}$$

Then v can be expressed in two (reciprocal) ways, viz. $${\displaystyle \mathbf{v}=q^i{\mathbf{e}}_i=q^1{\mathbf{e}}_1+q^2{\mathbf{e}}_2+q^3{\mathbf{e}}_3.}$$ or $${\displaystyle \mathbf{v}=q_i{\mathbf{e}}^i=q_1{\mathbf{e}}^1+q_2{\mathbf{e}}^2+q_3{\mathbf{e}}^3}$$ Combining the above relations, we have $${\displaystyle \mathbf{v}=(\mathbf{v}\cdot {\mathbf{e}}^i){\mathbf{e}}_i=(\mathbf{v}\cdot {\mathbf{e}}_i){\mathbf{e}}^i}$$ and we can convert between the basis and dual basis with $${\displaystyle q_i=\mathbf{v}\cdot {\mathbf{e}}_i=(q^j{\mathbf{e}}_j)\cdot {\mathbf{e}}_i=({\mathbf{e}}_j\cdot {\mathbf{e}}_i)q^j}$$ and $${\displaystyle q^i=\mathbf{v}\cdot {\mathbf{e}}^i=(q_j{\mathbf{e}}^j)\cdot {\mathbf{e}}^i=({\mathbf{e}}^j\cdot {\mathbf{e}}^i)q_j.}$$

If the basis vectors are orthonormal, then they are the same as the dual basis vectors.

Vector spaces of any dimension

The following applies to any vector space of dimension n equipped with a non-degenerate commutative and distributive dot product, and thus also to the Euclidean spaces of any dimension.

All indices in the formulas run from 1 to n. The Einstein notation for the implicit summation of the terms with the same upstairs (contravariant) and downstairs (covariant) indices is followed.

The historical and geometrical meaning of the terms contravariant and covariant will be explained at the end of this section.

Definitions

1. Covariant basis of a vector space of dimension n: ${\displaystyle {\mathbf{e}}_{\mathbf{j}}\triangleq }$ {any linearly independent basis for which in general is ${\displaystyle {\mathbf{e}}_{\mathbf{i}}\cdot {\mathbf{e}}_{\mathbf{j}}\ne {\delta }_{ij}}$}, i.e. not necessarily orthonormal (D.1).
2. Contravariant components of a vector ${\displaystyle \mathbf{v}}$: ${\displaystyle v^i\triangleq \{v^i\mid \mathbf{v}=v^i{\mathbf{e}}_{\mathbf{i}}\}}$ (D.2).
3. Dual (contravariant) basis of a vector space of dimension n: ${\displaystyle {\mathbf{e}}^{\mathbf{i}}\triangleq \{{\mathbf{e}}^{\mathbf{i}}:{\mathbf{e}}^{\mathbf{i}}\cdot {\mathbf{e}}_{\mathbf{j}}={\delta }_j^i\}}$ (D.3).
4. Covariant components of a vector ${\displaystyle \mathbf{v}}$: ${\displaystyle v_i\triangleq \{v_i\mid \mathbf{v}=v_i{\mathbf{e}}^{\mathbf{i}}\}}$ (D.4).
5. Components of the covariant metric tensor: ${\displaystyle g_{ij}\triangleq {\mathbf{e}}_{\mathbf{i}}\cdot {\mathbf{e}}_{\mathbf{j}}}$; the metric tensor can be considered a square matrix, since it only has two covariant indices: ${\displaystyle G\triangleq \{g_{ij}\}}$; for the commutative property of the dot product, the ${\displaystyle g_{ij}}$ are symmetric (D.5).
6. Components of the contravariant metric tensor: ${\displaystyle g^{ij}\triangleq \{h_{ij}:G^{-1}=\{h_{ij}\}\}}$; these are the elements of the inverse of the covariant metric tensor/matrix ${\displaystyle G^{-1}}$, and for the properties of the inverse of a symmetric matrix, they're also symmetric (D.6).

Corollaries

• ${\displaystyle g^{ij}{g}_{jk}={\delta }_k^i}$ (1).
  Proof: from the properties of the inverse matrix (D.6).
• ${\displaystyle {\mathbf{e}}^i=g^{ij}{\mathbf{e}}_j}$ (2).
  Proof: let's suppose that ${\displaystyle \{A^{ij}\mid {\mathbf{e}}^i=A^{ij}{\mathbf{e}}_j\}}$; we will show that ${\displaystyle A^{ij}=g^{ij}}$. Taking the dot product of both sides with ${\displaystyle {\mathbf{e}}_k}$:$${\displaystyle \begin{array}{cc}& {\mathbf{e}}^i\cdot {\mathbf{e}}_k=(A^{ij}{\mathbf{e}}_j)\cdot {\mathbf{e}}_k\\ \stackrel{\text{(D.3,D.5)}}{\to }& A^{ij}{g}_{jk}={\delta }_k^i;\end{array}}$$multiplying both sides by ${\displaystyle g^{mk}}$:$${\displaystyle \begin{array}{cc}& g^{mk}{A}^{ij}{g}_{jk}=g^{mk}{\delta }_k^i\\ \stackrel{\text{(D.5)}}{\to }& g^{mk}{g}_{kj}{A}^{ij}=g^{mi}\\ \stackrel{\text{(1)}}{\to }& {\delta }_j^m{A}^{ij}=g^{mi}\\ ⟶& A^{im}=g^{mi}\stackrel{\text{(D.6)}}{=}{g}^{im}.◼\end{array}}$$
• ${\displaystyle {\mathbf{e}}^i\cdot {\mathbf{e}}^j=g^{ij}}$ (3).
  Proof: $${\displaystyle \begin{array}{cc}& {\mathbf{e}}^i\stackrel{\text{(2)}}{=}{g}^{ik}{\mathbf{e}}_k;\\ & {\mathbf{e}}^j\stackrel{\text{(2)}}{=}{g}^{jm}{\mathbf{e}}_m\\ ⟶& {\mathbf{e}}^i\cdot {\mathbf{e}}^j=g^{ik}{g}^{jm}({\mathbf{e}}_k\cdot {\mathbf{e}}_m)\stackrel{\text{(D.5)}}{=}{g}^{ik}{g}^{jm}{g}_{km}\stackrel{\text{(1)}}{=}{\delta }_m^i{g}^{jm}=g^{ji}\stackrel{\text{(D.6)}}{=}{g}^{ij}.◼\end{array}}$$
• ${\displaystyle {\mathbf{e}}_i=g_{ij}{\mathbf{e}}^j}$ (4).
  Proof: let's suppose that ${\displaystyle \{B_{ij}\mid {\mathbf{e}}_i=B_{ij}{\mathbf{e}}^j\}}$; we will show that ${\displaystyle B_{ij}=g_{ij}}$. Taking the dot product of both sides with ${\displaystyle {\mathbf{e}}^k}$: ${\displaystyle {\mathbf{e}}_i\cdot {\mathbf{e}}^k=(B_{ij}{\mathbf{e}}^j)\cdot {\mathbf{e}}^k\stackrel{\text{(D.3,3)}}{\to }{B}_{ij}{g}^{jk}={\delta }_i^k}$; multiplying both sides by ${\displaystyle g_{mk}}$:$${\displaystyle \begin{array}{cc}& g_{mk}{B}_{ij}{g}^{jk}=g_{mk}{\delta }_i^k\\ \stackrel{\text{(D.5)}}{\to }& g_{mk}{g}^{kj}{B}_{ij}=g_{mi}\\ \stackrel{\text{(1)}}{\to }& {\delta }_m^j{B}_{ij}=g_{mi}\\ ⟶& B_{im}=g_{mi}\stackrel{\text{(D.5)}}{=}{g}_{im}.◼\end{array}}$$
• ${\displaystyle v^i=g^{ij}{v}_j}$ (5).
  Proof: $${\displaystyle \begin{array}{cc}& \mathbf{v}\stackrel{\text{(D.2)}}{=}{v}^i{\mathbf{e}}_i;\\ & \mathbf{v}\stackrel{\text{(D.4)}}{=}{v}_j{\mathbf{e}}^j\stackrel{\text{(2)}}{=}{v}_j(g^{ji}{\mathbf{e}}_i)\\ ⟶& v^i{\mathbf{e}}_i=v_j{g}^{ji}{\mathbf{e}}_i,\\ \mathrm{\forall }i& :v^i=g^{ji}{v}_j\stackrel{\text{(D.6)}}{=}{g}^{ij}{v}_j.◼\end{array}}$$
• ${\displaystyle v_i=g_{ij}{v}^j}$ (6).
  Proof: specular to (5).
• ${\displaystyle v_i=\mathbf{v}\cdot {\mathbf{e}}_i}$ (7).
  Proof: $${\displaystyle \begin{array}{c}\mathbf{v}\cdot {\mathbf{e}}_i\stackrel{\text{(D.4)}}{=}(v_j{\mathbf{e}}^j)\cdot {\mathbf{e}}_i\stackrel{\text{(D.3)}}{=}{v}_j{\delta }_i^j=v_i.◼\end{array}}$$
• ${\displaystyle v^i=\mathbf{v}\cdot {\mathbf{e}}^i}$ (8).
  Proof: specular to (7).
• ${\displaystyle \mathbf{u}\cdot \mathbf{v}=g_{ij}{u}^i{v}^j}$ (9).
  Proof: $${\displaystyle \mathbf{u}\cdot \mathbf{v}\stackrel{\text{(D.2)}}{=}(u^i{\mathbf{e}}_i)\cdot (v^j{\mathbf{e}}_j)=({\mathbf{e}}_i\cdot {\mathbf{e}}_j)u^i{v}^j\stackrel{\text{(D.5)}}{=}{g}_{ij}{u}^i{v}^j.◼}$$
• ${\displaystyle \mathbf{u}\cdot \mathbf{v}=g^{ij}{u}_i{v}_j}$ (10).
  Proof: specular to (9).

Historical and geometrical meaning

Considering this figure for the case of an Euclidean space with ${\displaystyle n=2}$, since ${\displaystyle \mathbf{v}=\mathbf{O}\mathbf{A}+\mathbf{O}\mathbf{B}}$, if we want to express ${\displaystyle \mathbf{v}}$ in terms of the covariant basis, we have to multiply the basis vectors by the coefficients ${\displaystyle v^1=\frac{|\mathbf{O}\mathbf{A}|}{|{\mathbf{e}}_{\mathbf{1}}|}}$, ${\displaystyle v^2=\frac{|\mathbf{O}\mathbf{B}|}{|{\mathbf{e}}_{\mathbf{2}}|}}$.

With ${\displaystyle \mathbf{v}}$ and thus ${\displaystyle \mathbf{O}\mathbf{A}}$ and ${\displaystyle \mathbf{O}\mathbf{B}}$ fixed, if the module of ${\displaystyle {\mathbf{e}}_{\mathbf{i}}}$ increases, the value of the ${\displaystyle v^i}$ component decreases, and that's why they're called contra-variant (with respect to the variation of the basis vectors module).

Symmetrically, corollary (7) states that the ${\displaystyle v_i}$ components equal the dot product ${\displaystyle \mathbf{v}\cdot {\mathbf{e}}_{\mathbf{i}}}$ between the vector and the covariant basis vectors, and since this is directly proportional to the basis vectors module, they're called co-variant.

If we consider the dual (contravariant) basis, the situation is perfectly specular: the covariant components are contra-variant with respect to the module of the dual basis vectors, while the contravariant components are co-variant.

So in the end it all boils down to a matter of convention: historically the first non-orthonormal basis of the vector space of choice was called "covariant", its dual basis "contravariant", and the corresponding components named specularly.

If the covariant basis becomes orthonormal, the dual contravariant basis aligns with it and the covariant components collapse into the contravariant ones, the most familiar situation when dealing with geometrical Euclidean vectors. ${\displaystyle G}$ and ${\displaystyle G^{-1}}$ become the identity matrix ${\displaystyle I}$, and:

$${\displaystyle \begin{array}{cc}{g}_{ij}& ={\delta }_{ij},\\ g^{ij}& ={\delta }^{ij},\\ \mathbf{u}\cdot \mathbf{v}& ={\delta }_{ij}{u}^i{v}^j=\sum _i{u}^i{v}^i\\ & ={\delta }^{ij}{u}_i{v}_j=\sum _i{u}_i{v}_i.\end{array}}$$

If the metric is non-Euclidean, but for instance Minkowskian like in the special relativity and general relativity theories, the basis are never orthonormal, even in the case of special relativity where ${\displaystyle G}$ and ${\displaystyle G^{-1}}$ become, for ${\displaystyle n=4,\eta \triangleq \mathrm{diag}(1,-1,-1,-1)}$. In this scenario, the covariant and contravariant components always differ.

Use in tensor analysis

The distinction between covariance and contravariance is particularly important for computations with tensors, which often have mixed variance. This means that they have both covariant and contravariant components, or both vector and covector components. The valence of a tensor is the number of covariant and contravariant terms, and in Einstein notation, covariant components have lower indices, while contravariant components have upper indices. The duality between covariance and contravariance intervenes whenever a vector or tensor quantity is represented by its components, although modern differential geometry uses more sophisticated index-free methods to represent tensors.

In tensor analysis, a covariant vector varies more or less reciprocally to a corresponding contravariant vector. Expressions for lengths, areas and volumes of objects in the vector space can then be given in terms of tensors with covariant and contravariant indices. Under simple expansions and contractions of the coordinates, the reciprocity is exact; under affine transformations the components of a vector intermingle on going between covariant and contravariant expression.

On a manifold, a tensor field will typically have multiple, upper and lower indices, where Einstein notation is widely used. When the manifold is equipped with a metric, covariant and contravariant indices become very closely related to one another. Contravariant indices can be turned into covariant indices by contracting with the metric tensor. The reverse is possible by contracting with the (matrix) inverse of the metric tensor. Note that in general, no such relation exists in spaces not endowed with a metric tensor. Furthermore, from a more abstract standpoint, a tensor is simply "there" and its components of either kind are only calculational artifacts whose values depend on the chosen coordinates.

The explanation in geometric terms is that a general tensor will have contravariant indices as well as covariant indices, because it has parts that live in the tangent bundle as well as the cotangent bundle.

A contravariant vector is one which transforms like ${\displaystyle \frac{d{x}^{\mu }}{d\tau }}$, where ${\displaystyle x^{\mu }}$ are the coordinates of a particle at its proper time ${\displaystyle \tau }$. A covariant vector is one which transforms like ${\displaystyle \frac{\mathrm{\partial }\phi }{\mathrm{\partial }{x}^{\mu }}}$, where ${\displaystyle \phi }$ is a scalar field.

Algebra and geometry

In category theory, there are covariant functors and contravariant functors. The assignment of the dual space to a vector space is a standard example of a contravariant functor. Contravariant (resp. covariant) vectors are contravariant (resp. covariant) functors from a ${\displaystyle \text{GL}(n)}$-torsor to the fundamental representation of ${\displaystyle \text{GL}(n)}$. Similarly, tensors of higher degree are functors with values in other representations of ${\displaystyle \text{GL}(n)}$. However, some constructions of multilinear algebra are of "mixed" variance, which prevents them from being functors.

In differential geometry, the components of a vector relative to a basis of the tangent bundle are covariant if they change with the same linear transformation as a change of basis. They are contravariant if they change by the inverse transformation. This is sometimes a source of confusion for two distinct but related reasons. The first is that vectors whose components are covariant (called covectors or 1-forms) actually pull back under smooth functions, meaning that the operation assigning the space of covectors to a smooth manifold is actually a contravariant functor. Likewise, vectors whose components are contravariant push forward under smooth mappings, so the operation assigning the space of (contravariant) vectors to a smooth manifold is a covariant functor. Secondly, in the classical approach to differential geometry, it is not bases of the tangent bundle that are the most primitive object, but rather changes in the coordinate system. Vectors with contravariant components transform in the same way as changes in the coordinates (because these actually change oppositely to the induced change of basis). Likewise, vectors with covariant components transform in the opposite way as changes in the coordinates.