# Introduction to Mathematical Physics/Dual of a vectorial space

## Dual of a vectorial space

### Definition

Definition:

Let ${\displaystyle E}$ be a vectorial space on a commutative field ${\displaystyle K}$. The vectorial space ${\displaystyle {\mathcal {L}}(E,K)}$of the linear forms on ${\displaystyle E}$ is called the dual of ${\displaystyle E}$ and is noted ${\displaystyle E^{*}}$.

When ${\displaystyle E}$ has a finite dimension, then ${\displaystyle E^{*}}$ has also a finite dimension and its dimension is gela to the dimension of ${\displaystyle E}$. If ${\displaystyle E}$ has an inifinite dimension, ${\displaystyle ^{*}}$ has also an inifinite dimension but the two spaces are not isomorphic.

chaptens

### Tensors

In this appendix, we introduce the fundamental notion of tensor\index{tensor} in physics. More information can be found in ([#References|references]) for instance. Let ${\displaystyle E}$ be a finite dimension vectorial space. Let ${\displaystyle e_{i}}$ be a basis of ${\displaystyle E}$. A vector ${\displaystyle X}$ of ${\displaystyle E}$ can be referenced by its components ${\displaystyle x^{i}}$ is the basis ${\displaystyle e_{i}}$:

${\displaystyle X=x^{i}e_{i}}$

In this chapter the repeated index convention (or {\bf Einstein summing convention}) will be used. It consists in considering that a product of two quantities with the same index correspond to a sum over this index. For instance:

${\displaystyle x^{i}e_{i}=\sum _{i}x^{i}e_{i}}$

or

${\displaystyle a_{ijk}b_{ikm}=\sum _{i}\sum _{k}a_{ijk}b_{ikm}}$

To the vectorial space ${\displaystyle E}$ correspond a space ${\displaystyle E^{*}}$ called the dual of ${\displaystyle E}$. A element of ${\displaystyle E^{*}}$ is a linear form on ${\displaystyle E}$: it is a linear mapping ${\displaystyle p}$ that maps any vector ${\displaystyle Y}$ of ${\displaystyle E}$ to a real. ${\displaystyle p}$ is defined by a set of number ${\displaystyle x_{i}}$ because the most general form of a linear form on ${\displaystyle E}$ is:

${\displaystyle p(Y)=x_{i}y^{i}}$

A basis ${\displaystyle e^{i}}$ of ${\displaystyle E^{*}}$ can be defined by the following linear form

${\displaystyle e^{j}e_{i}=\delta _{i}^{j}}$

where ${\displaystyle \delta _{i}^{j}}$ is one if ${\displaystyle i=j}$ and zero if not. Thus to each vector ${\displaystyle X}$ of ${\displaystyle E}$ of components ${\displaystyle x^{i}}$ can be associated a dual vector in ${\displaystyle E^{*}}$ of components ${\displaystyle x_{i}}$:

${\displaystyle X=x_{i}e^{i}}$

The quantity

${\displaystyle |X|=x_{i}x^{i}}$

is an invariant. It is independent on the basis chosen. On another hand, the expression of the components of vector ${\displaystyle X}$ depend on the basis chosen. If ${\displaystyle \omega _{k}^{i}}$ defines a transformation that maps basis ${\displaystyle e_{i}}$ to another basis ${\displaystyle e'_{i}}$

eqcov

${\displaystyle e'_{i}=\omega _{i}^{k}e_{k}}$

we have the following relation between components ${\displaystyle x_{i}}$ of ${\displaystyle X}$ in ${\displaystyle e_{i}}$ and ${\displaystyle x'_{i}}$ of ${\displaystyle X}$ in ${\displaystyle e'_{i}}$:

eqcontra

${\displaystyle x^{i}=\omega _{k}^{i}x'^{k}}$

This comes from the identification of

${\displaystyle X=x'^{i}e'_{i}=x'^{i}\omega _{i}^{k}e_{k}}$

and

${\displaystyle X=x^{i}e_{i}}$

Equations eqcov and eqcontra define two types of variables: covariant variables that are transformed like the vector basis. ${\displaystyle x_{i}}$ are such variables. Contravariant variables that are transformed like the components of a vector on this basis. Using a physicist vocabulary ${\displaystyle x_{i}}$ is called a covariant vector and ${\displaystyle x^{i}}$ a contravariant vector.

Covariant and contravariant components of a vector ${\displaystyle X}$.}
figcovcontra

Let ${\displaystyle x_{i}}$ and ${\displaystyle y_{j}}$ two vectors of two vectorial spaces ${\displaystyle E_{1}}$ and ${\displaystyle E_{2}}$. The tensorial product space ${\displaystyle E_{1}\otimes E_{2}}$ is the vectorial space such that there exist a unique isomorphism between the space of the bilinear forms of ${\displaystyle E_{1}\times E_{2}}$ and the linear forms of ${\displaystyle E_{1}\otimes E_{2}}$. A bilinear form of ${\displaystyle E_{1}\times E_{2}}$ is:

${\displaystyle b(x,y)=b^{ij}x_{i}x_{j}}$

It can be considered as a linear form of ${\displaystyle E_{1}\otimes E_{2}}$ using application ${\displaystyle \otimes }$ from ${\displaystyle E_{1}\times E_{2}}$ to ${\displaystyle E_{1}\otimes E_{2}}$ that is linear and distributive with respect to ${\displaystyle +}$. If ${\displaystyle e_{i}}$ is a basis of ${\displaystyle E_{1}}$ and ${\displaystyle f_{j}}$ a basis of ${\displaystyle E_{2}}$, then

${\displaystyle x\otimes y=x_{i}y_{j}e_{i}\otimes e_{j}.}$

${\displaystyle e_{i}\otimes e_{j}}$ is a basis of ${\displaystyle E_{1}\otimes E_{2}}$. Thus tensor ${\displaystyle x_{i}y_{j}=T_{ij}}$ is an element of ${\displaystyle E_{1}\otimes E_{2}}$. A second order covariant tensor is thus an element of ${\displaystyle E^{*}\otimes E^{*}}$. In a change of basis, its components ${\displaystyle a{ij}}$ are transformed according the following relation:

${\displaystyle a'_{ij}=\omega _{i}^{k}\omega _{j}^{l}a_{kl}}$

Now we can define a tensor on any rank of any variance. For instance a tensor of third order two times covariant and one time contravariant is an element ${\displaystyle a}$ of ${\displaystyle E^{*}\otimes E^{*}\otimes E}$ and noted ${\displaystyle a_{ij}^{k}}$.

A second order tensor is called symmetric if ${\displaystyle a_{ij}=a_{ji}}$. It is called antisymmetric is ${\displaystyle a_{ij}=-a_{ji}}$.

Pseudo tensors are transformed slightly differently from ordinary tensors. For instance a second order covariant pseudo tensor is transformed according to:

${\displaystyle a'_{ij}=det(\omega )\omega _{i}^{k}\omega _{j}^{l}a_{kl}}$

where ${\displaystyle det(\omega )}$ is the determinant of transformation ${\displaystyle \omega }$.

secformultens

Let us introduce two particular tensors.

• The Kronecker symbol ${\displaystyle \delta _{ij}}$ is defined by:

${\displaystyle \delta _{ij}=\left\{{\begin{array}{ll}1&{\mbox{ si }}i=j\\0&{\mbox{ si }}i\neq j\end{array}}\right.}$

It is the only second order tensor invariant in ${\displaystyle R^{3}}$ by rotations.

• The signature of permutations tensor ${\displaystyle e_{ijk}}$ is defined by:

${\displaystyle e_{ijk}=\left\{{\begin{array}{ll}1&{\mbox{if permutation }}ijk{\mbox{ of }}1,2,3{\mbox{ is even}}\\-1&{\mbox{if permutation }}ijk{\mbox{ of }}1,2,3{\mbox{ is odd}}\\0&{\mbox{if }}ijk{\mbox{ is not a permutation of }}1,2,3\end{array}}\right.}$

It is the only pseudo tensor of rank 3 invariant by rotations in ${\displaystyle R^{3}}$. It verifies the equality:

${\displaystyle e_{ijk}e_{imn}=\delta _{jm}\delta _{kn}-\delta _{jn}\delta _{km}}$

Let us introduce two tensor operations: scalar product, vectorial product.

• Scalar product ${\displaystyle a.b}$ is the contraction of vectors ${\displaystyle a}$ and ${\displaystyle b}$ :

${\displaystyle a.b=a_{i}b_{i}}$

• vectorial product of two vectors ${\displaystyle a}$ and ${\displaystyle b}$ is:

${\displaystyle (a\wedge b)_{i}=e_{ijk}a_{j}b_{k}}$

From those definitions, following formulas can be showed:

${\displaystyle {\begin{matrix}a.(b\wedge c)&=&a_{i}(b\wedge c)_{i}\\&=&a_{i}\epsilon _{ijk}b_{j}c_{k}\\&=&\epsilon _{ijk}a_{i}b_{j}c_{k}\\&=&\left|{\begin{array}{ccc}a_{1}&a_{2}&a_{3}\\b_{1}&b_{2}&b_{3}\\c_{1}&c_{2}&c_{3}\\\end{array}}\right|\end{matrix}}}$

Here is useful formula:

${\displaystyle a\wedge (b\wedge c)=d(a.c)-c(a.b)}$

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## Green's theorem

Green's theorem allows to transform a volume calculation integral into a surface calculation integral.

Theorem:

Let ${\displaystyle \omega }$ be a bounded domain of ${\displaystyle R^{p}}$ with a regular boundary. Let ${\displaystyle {\vec {n}}}$ be the unitary vector normal to hypersurface ${\displaystyle \partial \omega }$ (oriented towards the exterior of ${\displaystyle \omega }$). Let ${\displaystyle t_{ij\dots q}}$ be a tensor, continuously derivable in ${\displaystyle \omega }$, then:\index{Green's theorem}

${\displaystyle \int _{\omega }t_{ij\dots q,r}dv=\int _{\partial \omega }t_{ij\dots q}n_{r}ds}$

Here are some important Green's formulas obtained by applying Green's theorem:

${\displaystyle \int _{\omega }{\mbox{ grad }}\phi dv=\int _{\partial \omega }\phi {\vec {n}}{\vec {ds}}}$

${\displaystyle \int _{\omega }{\mbox{ rot }}{\vec {u}}dv=\int _{\partial \omega }({\vec {n}}\wedge {\vec {u}}){\vec {ds}}}$

${\displaystyle \int _{\omega }f_{,i}gdv+\int _{\omega }fg_{,i}dv=\int _{\partial \omega }fg{\vec {\nu }}_{i}{\vec {ds}}}$