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 ${\mathcal {L}}(E,K)$of the linear forms on $E$ is called the dual of $E$ and is noted $E^{*}$.

When $E$ has a finite dimension, then $E^{*}$ has also a finite dimension and its dimension is equal to the dimension of $E$. If $E$ has an infinite dimension, $E^{*}$ 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 $E^{*}$ called the dual of $E$. A element of $E^{*}$ 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 $E^{*}$ 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 $E^{*}$ 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 $e'_{i}$

eqcov

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

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

eqcontra

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

This comes from the identification of

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

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 $E^{*}\otimes E^{*}$. In a change of basis, its components $a{ij}$ are transformed according the following relation:

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

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 $a$ of $E^{*}\otimes E^{*}\otimes E$ 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:

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

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

secformultens

Let us introduce two particular tensors.

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

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

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

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

Here is useful formula:

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

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 ${\vec {n}}$ 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:

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

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

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