Page 168 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Unit 15: Tensors in Cartesian Coordinates

15.2 Scalar Product of Vector and Covector Notes

Suppose we have a vector x and a covector a. Upon choosing some basis e , e , e , both of them
2
1
3
have three coordinates: x , x , x for vector x, and a , a , a for covector a. Let’s denote by a,x the
1
2
3
3
2
1
following sum:
3
a,x =  a x . ...(1)
i
i
i 1

The sum (1) is written in agreement with Einstein’s tensorial notation. It is a number depending
on the vector x and on the covector a. This number is called the scalar product of the vector x and
the covector a. We use angular brackets for this scalar product in order to distinguish it from the
scalar product of two vectors in E, which is also known as the dot product.
Defining the scalar product a,x by means of sum (1) we used the coordinates of vector x and
of covector a, which are basis-dependent. However, the value of sum (1) does not depend on any
basis. Such numeric quantities that do not depend on the choice of basis are called scalars or true
scalars.
Exercise 15.2: Consider two bases e , e , e and e ,e ,e , and consider the coordinates of vector
  
2
3
1
2
3
1
x and covector a in both of them. Prove the equality
3 3
 a x   a x .  i ...(2)
i

i
i
i 1 i 1


Thus, you are proving the self-consistence of formula (1) and showing that the scalar product
a,x given by this formula is a true scalar quantity.
Exercise 15.3: Let  be a real number, let a and b be two covectors, and let x and y be two vectors.
Prove the following properties of the scalar product:
(1) a + b; x = a, x + b, x ;
(2) a, x + y  a,x  a,y ;
(3)  a, x   a,x ;
(4) a, x   a,x .


Exercise 15.4: Explain why the scalar product a,x is sometimes called the bilinear function of
vectorial argument x and covectorial argument a. In this capacity, it can be denoted as f(a, x).
Remember our discussion about functions with non-numeric arguments.

Important note. The scalar product a,x is not symmetric. Moreover, the formula

a,x  x,a
is incorrect in its right hand side since the first argument of scalar product by definition should
be a covector. In a similar way, the second argument should be a vector. Therefore, we never can
swap them.

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