Tensor - 3 - Covector Introduction

epsilon covectors 𝜀^𝑖 are defined as:

• 𝜀¹(𝑒₁) = 1
• 𝜀¹(𝑒₂) = 0
• 𝜀²(𝑒₁) = 0
• 𝜀²(𝑒₂) = 1

Thus:

• 𝜀^𝑖(𝑒_𝑗) = 𝛿_𝑖𝑗 # the Kronecker delta function

In other words, let:

• 𝐸 = [𝑒₁|𝑒₂] a matrix whose columns are the basis vectors {𝑒₁, 𝑒₂}
• 𝐸ˆ = [𝜀¹|𝜀²] a matrix whose columns are the epsilon covectors {𝜀¹, 𝜀²}

The system of equations can be expressed as:

• ${\hat{E}}^{T}E=I$
• $\left[\begin{array}{cc}{𝜀}_1^1& {𝜀}_2^1\\ {𝜀}_1^2& {𝜀}_2^2\end{array}\right]\left[\begin{array}{cc}{e}_{11}& e_{21}\\ e_{12}& e_{22}\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

epsilon covector 𝜀^𝑖 consumes arbitrary vector 𝑣:

• 𝜀¹(𝑣) = 𝜀¹(𝑣[1]·𝑒₁ + 𝑣[2]·𝑒₂)
• 𝜀¹(𝑣) = 𝑣[1] · 𝜀¹(𝑒₁) + 𝑣[2] · 𝜀¹(𝑒₂)
• 𝜀¹(𝑣) = 𝑣[1]

• 𝜀²(𝑣) = 𝜀²(𝑣[1]·𝑒₁ + 𝑣[2]·𝑒₂)
• 𝜀²(𝑣) = 𝑣[1] · 𝜀²(𝑒₁) + 𝑣[2] · 𝜀²(𝑒₂)
• 𝜀²(𝑣) = 𝑣[2]

arbitrary covector 𝛼 consumes arbitrary vector 𝑣:

• 𝛼(𝑣) = 𝛼(𝑣[1]·𝑒₁ + 𝑣[2]·𝑒₂)
• 𝛼(𝑣) = 𝑣[1]·𝛼(𝑒₁) + 𝑣[2]·𝛼(𝑒₂)
• 𝛼(𝑣) = 𝜀¹(𝑣)·𝛼(𝑒₁) + 𝜀²(𝑣)·𝛼(𝑒₂)
• 𝛼(𝑣) = 𝜀¹(𝑣)·𝛼₁ + 𝜀²(𝑣)·𝛼₂
• 𝛼(𝑣) = 𝛼₁·𝜀¹(𝑣) + 𝛼₂·𝜀²(𝑣)
• 𝛼(𝑣) = (𝛼₁·𝜀¹ + 𝛼₂·𝜀²) (𝑣)

Thus

• 𝛼 = 𝛼₁·𝜀¹ + 𝛼₂·𝜀²

Thus

• any arbitrary covector 𝛼 can be expressed as a linear combination of basis epsilon covectors {𝜀¹, 𝜀²}

The epsilon covectors 𝜀¹ and 𝜀² form the basis vectors of the dual vector space 𝑉*:

1. 𝜀¹ and 𝜀² are linearly independent
   1.

   proof ¹, 𝜀²} from the dual vector space 𝑉* is linearly independent if:

   • scalars {𝑎₁, 𝑎₂} must be ALL zero such that:
   • 𝑎₁𝜀¹ + 𝑎₂𝜀² = 𝜀⁰ # where 𝜀⁰is the null functional (i.e. zero vector in 𝑉*)

   We need to prove the above statement.

   The expression 𝑎₁𝜀¹ + 𝑎₂𝜀² = 𝜀⁰ naturally leads to (𝑎₁𝜀¹ + 𝑎₂𝜀²)(𝑣) = 𝜀⁰(𝑣); for all 𝑣∊𝑉. Where we have both sides “consume” an arbitrary vector 𝑣∊𝑉:

   • (𝑎₁𝜀¹ + 𝑎₂𝜀²)(𝑣) = 𝜀⁰(𝑣); for all 𝑣∊𝑉
   • (𝑎₁𝜀¹ + 𝑎₂𝜀²)(𝑣) = 0; for all 𝑣∊𝑉 # by assumption that 𝜀⁰(𝑣) = 0 for all 𝑣∊𝑉
   • 𝑎₁𝜀¹(𝑣) + 𝑎₂𝜀²(𝑣) = 0; for all 𝑣∊𝑉 # because they are linear transformations

   Thus the original statement that we need to prove now becomes:

   • scalars {𝑎₁, 𝑎₂} must be ALL zero such that:
   • 𝑎₁𝜀¹(𝑣) + 𝑎₂𝜀²(𝑣) = 0; for all 𝑣∊𝑉

   Since the new statement is true for every vector 𝑣∊𝑉, it must also be true for the basis vectors {𝑒₁, 𝑒₂}:

   • that the scalars {𝑎₁, 𝑎₂} must be ALL zero such that:
   • 𝑎₁𝜀¹(𝑒_𝑖) + 𝑎₂𝜀²(𝑒_𝑖) = 0; for all 𝑒_𝑖∊{𝑒₁, 𝑒₂}

   Evaluating the above statement for each basis vector yields:

   • that the scalars {𝑎₁, 𝑎₂} must be ALL zero such that:
   • for 𝑒_𝑖=𝑒₁:
     • 𝑎₁𝜀¹(𝑒₁) + 𝑎₂𝜀²(𝑒₁) = 0
     • 𝑎₁*1 + 𝑎₂*0 = 0
     • 𝑎₁ = 0
   • for 𝑒_𝑖=𝑒₁:
     • 𝑎₁𝜀¹(𝑒₂) + 𝑎₂𝜀²(𝑒₂) = 0
     • 𝑎₁*0 + 𝑎₂*1 = 0
     • 𝑎₂ = 0

   Thus, 𝑎₁ and 𝑎₂ must equal to 0.

   A set of covectors {𝜀

2. every arbitrary covector 𝛼∊𝑉* can be expressed as a linear combination of {𝜀¹, 𝜀²}
   1. see above

The components of covector 𝛼 can be extracted as follows:

• 𝛼 = 𝛼₁·𝜀¹ + 𝛼₂·𝜀²
• 𝛼(𝑒₁) = 𝛼₁
• 𝛼(𝑒₂) = 𝛼₂

Thus:

• the components of an arbitrary covector (e.g. [2 1]) are measured by the number of covector/level-set lines that the basis vectors {𝑒₁, 𝑒₂} pierces