Introduction to Mathematical Physics/Dual of a vectorial space

Dual of a vectorial space

Definition

Definition:

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

When $E$ has a finite dimension, then $E^{\ast }$ has also a finite dimension and its dimension is equal to the dimension of $E$. If $E$ has an infinite dimension, $E^{\ast }$ has also an infinite 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 $E$ be a finite dimension vectorial space. Let $e_i$ be a basis of $E$. A vector $X$ of $E$ can be referenced by its components $x^i$ is the basis $e_i$:

$X=x^i{e}_i$

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:

$x^i{e}_i=\sum _i{x}^i{e}_i$

or

$a_{ijk}{b}_{ikm}=\sum _i\sum _k{a}_{ijk}{b}_{ikm}$

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

$p(Y)=x_i{y}^i$

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

$e^j{e}_i={\delta }_i^j$

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

$X=x_i{e}^i$

The quantity

$|X|=x_i{x}^i$

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

eqcov

$e{\text{'}}_i={\omega }_i^k{e}_k$

we have the following relation between components $x_i$ of $X$ in $e_i$ and $x{\text{'}}_i$ of $X$ in $e{\text{'}}_i$:

eqcontra

$x^i={\omega }_k^{i}x{\text{'}}^k$

This comes from the identification of

$X=x{\text{'}}^{i}e{\text{'}}_i=x{\text{'}}^i{\omega }_i^k{e}_k$

and

$X=x^i{e}_i$

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

Covariant and contravariant components of a vector $X$.}

figcovcontra

Let $x_i$ and $y_j$ two vectors of two vectorial spaces $E_1$ and $E_2$. The tensorial product space $E_1\otimes E_2$ is the vectorial space such that there exist a unique isomorphism between the space of the bilinear forms of $E_1\times E_2$ and the linear forms of $E_1\otimes E_2$. A bilinear form of $E_1\times E_2$ is:

$b(x,y)=b^{ij}{x}_i{x}_j$

It can be considered as a linear form of $E_1\otimes E_2$ using application $\otimes$ from $E_1\times E_2$ to $E_1\otimes E_2$ that is linear and distributive with respect to $+$. If $e_i$ is a basis of $E_1$ and $f_j$ a basis of $E_2$, then

$x\otimes y=x_i{y}_j{e}_i\otimes e_j.$

$e_i\otimes e_j$ is a basis of $E_1\otimes E_2$. Thus tensor $x_i{y}_j=T_{ij}$ is an element of $E_1\otimes E_2$. A second order covariant tensor is thus an element of $E^{\ast }\otimes E^{\ast }$. In a change of basis, its components $aij$ are transformed according the following relation:

$a{\text{'}}_{ij}={\omega }_i^k{\omega }_j^l{a}_{kl}$

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 $a$ of $E^{\ast }\otimes E^{\ast }\otimes E$ and noted $a_{ij}^k$.

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

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

$a{\text{'}}_{ij}=det(\omega ){\omega }_i^k{\omega }_j^l{a}_{kl}$

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

secformultens

Let us introduce two particular tensors.

• The Kronecker symbol ${\delta }_{ij}$ is defined by:
  ${\delta }_{ij}=\{\begin{array}{cc}1& \text{ if }i=j\\ 0& \text{ if }i\ne j\end{array}$
  It is the only second order tensor invariant in $R^3$ by rotations.
• The signature of permutations tensor $e_{ijk}$ is defined by:

$e_{ijk}=\{\begin{array}{cc}1& \text{if permutation }ijk\text{ of }1,2,3\text{ is even}\\ -1& \text{if permutation }ijk\text{ of }1,2,3\text{ is odd}\\ 0& \text{if }ijk\text{ is not a permutation of }1,2,3\end{array}$

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

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

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

• Scalar product $a.b$ is the contraction of vectors $a$ and $b$ :
  $a.b=a_i{b}_i$
• vectorial product of two vectors $a$ and $b$ is:
  $(a\wedge b{)}_i=e_{ijk}{a}_j{b}_k$

From those definitions, following formulas can be showed:

$\begin{array}{ccc}a.(b\wedge c)& =& a_i(b\wedge c{)}_i\\ & =& a_i{ϵ}_{ijk}{b}_j{c}_k\\ & =& {ϵ}_{ijk}{a}_i{b}_j{c}_k\\ & =& \left|\begin{array}{ccc}{a}_1& a_2& a_3\\ b_1& b_2& b_3\\ c_1& c_2& c_3\end{array}\right|\end{array}$

Here is useful formula:

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

secappendgreeneq

Green's theorem

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

Theorem:

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

${\int }_{\omega }{t}_{ij\dots q,r}dv={\int }_{\partial \omega }{t}_{ij\dots q}{n}_{r}ds$

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

${\int }_{\omega }\text{ grad }\varphi dv={\int }_{\partial \omega }\varphi \overrightarrow{n}\overrightarrow{ds}$

${\int }_{\omega }\text{ rot }\overrightarrow{u}dv={\int }_{\partial \omega }(\overrightarrow{n}\wedge \overrightarrow{u})\overrightarrow{ds}$

${\int }_{\omega }{f}_{,i}gdv+{\int }_{\omega }f{g}_{,i}dv={\int }_{\partial \omega }fg{\overrightarrow{\nu }}_i\overrightarrow{ds}$