Dual basis

The dual base is a term from linear algebra that has two different meanings:

• For a given basis of a finite-dimensional vector space , an associated dual basis of the dual space is constructed. ${\displaystyle V}$ ${\displaystyle V^{*}}$
• For a given basis of a Euclidean vector space , another, the first dual basis of, is constructed. ${\displaystyle V}$ ${\displaystyle V}$

Dual basis in the dual space V *

definition

Let it be a -dimensional vector space over a body . (In applications, the body is often or .) Next is a base of . ${\displaystyle V}$ ${\displaystyle n}$ ${\displaystyle K}$ ${\displaystyle \mathbb {R} }$ ${\displaystyle \mathbb {C} }$ ${\displaystyle \{e_{1},\dotsc ,e_{n}\}}$ ${\displaystyle V}$

Then there is exactly one linear mapping for each with and for , because a linear mapping is uniquely determined by the images on a basis. The so defined form a basis of the dual space , the basis of the dual basis. With the Kronecker delta notation, is the defining property of the dual base . ${\displaystyle i\in \{1,\dots ,n\}}$ ${\displaystyle e_{i}^{*}:V\rightarrow K}$ ${\displaystyle e_{i}^{*}(e_{i})=1}$ ${\displaystyle e_{i}^{*}(e_{j})=0}$ ${\displaystyle j\neq i}$ ${\displaystyle e_{i}^{*}}$ ${\displaystyle V^{*}}$ ${\displaystyle \{e_{1},\dotsc ,e_{n}\}}$ ${\displaystyle V}$ ${\displaystyle e_{i}^{*}(e_{j})=\delta _{ij}}$

Behavior when changing base

Be a basis of and the corresponding dual basis. Next is a second basis of with . ${\displaystyle \{e_{1},\dotsc ,e_{n}\}}$ ${\displaystyle V}$ ${\displaystyle \{e_{1}^{*},\dotsc ,e_{n}^{*}\}}$ ${\displaystyle \{a_{1},\dotsc ,a_{n}\}}$ ${\displaystyle V}$ ${\displaystyle a_{j}=\sum _{k}A_{kj}e_{k}}$

As a matrix of a base change is invertible . The components of the inverse are denoted by. A comparison of ${\displaystyle A}$ ${\displaystyle A^{-1}}$ ${\displaystyle A_{ik}^{-1}}$

${\displaystyle \sum _{k}A_{ik}^{-1}{e}_{k}^{*}(a_{j})=\sum _{k}A_{ik}^{-1}A_{kj}=\delta _{ij}}$

with the defining property immediately results in the transformation behavior of the dual basis: ${\displaystyle a_{i}^{*}(a_{j})=\delta _{ij}}$

${\displaystyle a_{i}^{*}=\sum _{k}A_{ik}^{-1}e_{k}^{*}}$.

Calculation on a fixed basis

A finite-dimensional vector space of the dimension above the body is always isomorphic to the coordinate space of the column vectors with entries from . If you choose isomorphism ${\displaystyle n}$ ${\displaystyle K}$ ${\displaystyle K^{n}}$ ${\displaystyle K}$

${\displaystyle e_{1}\mapsto {\begin{pmatrix}1\\0\\0\\\vdots \end{pmatrix}}}$, etc., ${\displaystyle e_{1}^{*}\mapsto \left(1,0,0,\dotsc \right)}$

is mapped to the i-th row of as above . ${\displaystyle a_{i}^{*}}$ ${\displaystyle A^{-1}}$

Tensor notation

In the tensor formalism of the theory of relativity , one writes the basis of a vector space (such as a tangent space ) with upper indices , calls these vectors contravariant and understands them as column vectors. The associated covariant basis is then exactly the dual basis presented above in the form of row vectors. This one writes with lower indices . The defining condition is then . ${\displaystyle (e^{i})_{i}}$ ${\displaystyle (e_{i})_{i}}$ ${\displaystyle e^{j}e_{i}=\delta _{i}^{j}}$

The reason for this notation is the different transformation behavior of the vectors when changing the base . If the linear transformation, which maps one basis to another , then applies: ${\displaystyle L}$ ${\displaystyle (e^{i})_{i}}$ ${\displaystyle (e'^{i})_{i}}$

${\displaystyle \delta _{i}^{j}=e^{j}e_{i}=e^{j}L^{-1}Le_{i}=e^{j}L^{-1}e'^{i}}$

and one reads that the dual basis is transformed by means of . If one looks at coordinates with respect to the bases, one finds similar relationships. If is and is , then if Einstein's summation convention is observed for a vector : ${\displaystyle L^{-1}}$ ${\displaystyle L=(l_{\ j}^{i})}$ ${\displaystyle L^{-1}=({\tilde {l}}_{\ j}^{i})}$ ${\displaystyle v=\lambda _{i}e^{i}}$

${\displaystyle v=\lambda _{i}e^{i}=\lambda _{i}\delta _{j}^{i}e^{j}=\lambda _{i}{\tilde {l}}_{\ k}^{i}l_{\ j}^{k}e^{j}=\lambda _{i}{\tilde {l}}_{\ k}^{i}e'^{k}}$.

The coefficient from to the basis vector is , that is, the coefficients are also transformed using the inverse transformation matrix. In general, all (contravariant) quantities that transform using are written with upper indices and all (covariant) quantities that are opposite, i.e. using transform, with lower indices. ${\displaystyle v}$ ${\displaystyle e'^{k}}$ ${\displaystyle \lambda _{i}{\tilde {l}}_{\ k}^{i}}$ ${\displaystyle L}$ ${\displaystyle L^{-1}}$

Dual basis in the Euclidean vector space V

Definition and calculation

Let be any basis of a Euclidean vector space . The dual basis for this is defined by the property ${\displaystyle \{{\vec {a}}_{1},\dotsc ,{\vec {a}}_{n}\}}$ ${\displaystyle V}$ ${\displaystyle \{{\vec {a}}_{1}^{*},\dotsc ,{\vec {a}}_{n}^{*}\}}$ ${\displaystyle V}$

${\displaystyle {\vec {a}}_{i}^{*}\cdot {\vec {a}}_{j}=\delta _{ij}}$,

Here is the scalar product . ${\displaystyle \cdot }$

Furthermore, let an orthonormal basis in ,   describe the change of basis with the invertible matrix . By comparing ${\displaystyle \{{\hat {e}}_{1},\dotsc ,{\hat {e}}_{n}\}}$ ${\displaystyle V}$ ${\displaystyle \textstyle {\vec {a}}_{j}=\sum _{k}A_{kj}{\hat {e}}_{k}}$ ${\displaystyle A}$

${\displaystyle (\sum _{k}A_{ik}^{-1}{\hat {e}}_{k})\cdot {\vec {a}}_{j}=\sum _{k}A_{ik}^{-1}A_{kj}=\delta _{ij}}$

with results ${\displaystyle {\vec {a}}_{i}^{*}\cdot {\vec {a}}_{j}=\delta _{ij}}$

${\displaystyle {\vec {a}}_{i}^{*}=\sum _{k}A_{ik}^{-1}{\hat {e}}_{k}}$.

With the dyadic product this is written: ${\displaystyle \otimes }$

${\displaystyle A=\sum _{k}{\vec {a}}_{k}\otimes {\hat {e}}_{k}\quad \Leftrightarrow \quad A^{-1}=\sum _{k}{\hat {e}}_{k}\otimes {\vec {a}}_{k}^{*}}$

The vectors here form the columns of the matrix (or the second level tensor) and the dual basis is found in the rows of the inverse ${\displaystyle \{{\vec {a}}_{1},\dotsc ,{\vec {a}}_{n}\}}$ ${\displaystyle A}$ ${\displaystyle A^{-1}.}$

Special case R ³

In the vector space with standard scalar product and cross product , with the above equation and the formula for matrix inversion , we find : ${\displaystyle \mathbb {R} ^{3}}$ ${\displaystyle \cdot }$ ${\displaystyle \times }$

${\displaystyle {\begin{aligned}{\vec {a}}_{1}^{*}=&{\frac {{\vec {a}}_{2}\times {\vec {a}}_{3}}{{\vec {a}}_{1}\cdot ({\vec {a}}_{2}\times {\vec {a}}_{3})}}={\frac {{\vec {a}}_{2}\times {\vec {a}}_{3}}{\begin{vmatrix}{\vec {a}}_{1}&{\vec {a}}_{2}&{\vec {a}}_{3}\end{vmatrix}}}\\{\vec {a}}_{2}^{*}=&{\frac {{\vec {a}}_{3}\times {\vec {a}}_{1}}{{\vec {a}}_{2}\cdot ({\vec {a}}_{3}\times {\vec {a}}_{1})}}={\frac {{\vec {a}}_{3}\times {\vec {a}}_{1}}{\begin{vmatrix}{\vec {a}}_{1}&{\vec {a}}_{2}&{\vec {a}}_{3}\end{vmatrix}}}\\{\vec {a}}_{3}^{*}=&{\frac {{\vec {a}}_{1}\times {\vec {a}}_{2}}{{\vec {a}}_{3}\cdot ({\vec {a}}_{1}\times {\vec {a}}_{2})}}={\frac {{\vec {a}}_{1}\times {\vec {a}}_{2}}{\begin{vmatrix}{\vec {a}}_{1}&{\vec {a}}_{2}&{\vec {a}}_{3}\end{vmatrix}}}\end{aligned}}}$

The denominator of the fractions contains the late product formed with the basis vectors , which is invariant to a cyclical exchange of its arguments, and which is equal to the determinant of the matrix which is formed from the basis vectors. The defining property is immediately apparent here.

Application from crystallography

The determination of this dual basis in is important when describing crystal lattices . There the primitive lattice vectors form a (generally not orthonormal) basis of the . The scalar product between basis vectors of the reciprocal basis and primitive lattice vectors is in the crystallographic convention: ${\displaystyle \mathbb {R} ^{3}}$ ${\displaystyle \{{\vec {a}}_{1},{\vec {a}}_{2},{\vec {a}}_{3}\}}$ ${\displaystyle \mathbb {R} ^{3}}$ ${\displaystyle {\vec {b}}_{i}}$ ${\displaystyle {\vec {a}}_{j}}$

${\displaystyle {\vec {b}}_{i}\cdot {\vec {a}}_{j}=\delta _{ij}}$,

${\displaystyle \{{\vec {b}}_{1},{\vec {b}}_{2},{\vec {b}}_{3}\}}$is therefore the too dual basis in . ${\displaystyle \{{\vec {a}}_{1},{\vec {a}}_{2},{\vec {a}}_{3}\}}$ ${\displaystyle \mathbb {R} ^{3}}$

Example: The primitive lattice vectors of the face-centered cubic (fcc) lattice are:

${\displaystyle {\vec {a}}_{1}={\frac {a}{2}}\left({\hat {e}}_{y}+{\hat {e}}_{z}\right)}$

${\displaystyle {\vec {a}}_{2}={\frac {a}{2}}\left({\hat {e}}_{x}+{\hat {e}}_{z}\right)}$

${\displaystyle {\vec {a}}_{3}={\frac {a}{2}}\left({\hat {e}}_{x}+{\hat {e}}_{y}\right)}$

The above equations for the result: ${\displaystyle \mathbb {R} ^{3}}$

${\displaystyle {\vec {b}}_{1}={\frac {1}{a}}\left(-{\hat {e}}_{x}+{\hat {e}}_{y}+{\hat {e}}_{z}\right)}$

${\displaystyle {\vec {b}}_{2}={\frac {1}{a}}\left({\hat {e}}_{x}-{\hat {e}}_{y}+{\hat {e}}_{z}\right)}$

${\displaystyle {\vec {b}}_{3}={\frac {1}{a}}\left({\hat {e}}_{x}+{\hat {e}}_{y}-{\hat {e}}_{z}\right)}$

These form a body-centered cubic (bcc) grid.

Generalization to pseudo-Riemannian metrics

In the finite-dimensional vector space with pseudo-Riemannian metric and a basis consider the dual vector defined by ${\displaystyle V}$ ${\displaystyle g}$ ${\displaystyle \{{\vec {e}}_{1},\dotsc ,{\vec {e}}_{n}\}}$ ${\displaystyle \alpha _{i}}$

${\displaystyle \alpha _{i}({\vec {v}}):=e_{1}^{*}\wedge e_{2}^{*}\wedge \dotsc \wedge e_{n}^{*}({\vec {e}}_{1},{\vec {e}}_{2},\dotsc ,{\vec {e}}_{i-1},{\vec {v}},{\vec {e}}_{i+1},\dotsc ,{\vec {e}}_{n})}$.

Then applies

${\displaystyle g({\vec {e}}_{i}^{*},{\vec {e}}_{j})=\delta _{ij}}$   with . ${\displaystyle {\vec {e}}_{i}^{*}:=\sharp \alpha _{i}}$

Here, the dual vector in the dual space from the first meaning, the outer product , and by the pseudo-Riemannian metric induced isomorphism between and . ${\displaystyle e_{i}^{*}}$ ${\displaystyle \wedge }$ ${\displaystyle \sharp }$ ${\displaystyle V^{*}}$ ${\displaystyle V}$

See also

• Dual space

swell

• Gerd Fischer: Lineare Algebra , Vieweg-Verlag, ISBN 3-528-97217-3 .
• Hans Stephani: General theory of relativity . Deutscher Verlag der Wissenschaften, Berlin 1991, ISBN 3-326-00083-9 .