3.6.2 Tensor Covector Scalar Output

Tensor Covector Scalar Output is a mathematical operation that evaluates a covector acting on a tensor to produce a scalar value.

Tensor Covector Scalar Output is the property that the result of pairing a covector with a vector always belongs to the underlying field itself, rather than to any vector space, dual space, or higher tensor space, making the evaluation operation the unique rank-reducing step that terminates a chain of tensor contractions. Whenever a covector f in V* is applied to a vector v in V, written <f, v> or f(v), the returned quantity f(v) is a single number in the field F, carrying no direction, no basis dependence in its value, and no further index structure.

Nature of the Scalar Output

Membership in the Field

For a vector space V defined over a field F, such as the real numbers or the complex numbers, the output of f(v) for f in V* and v in V is always an element of F. It is not a vector, not a covector, and not an object requiring a basis to be expressed, since elements of F are the same regardless of any basis chosen for V.

Zero Free Indices

In index notation, a covector has one lower index and a vector has one upper index. The evaluation operation contracts these two indices against each other,

f (v) = f_{i} v^{i}

using the Einstein summation convention. Because the upper index on v^i and the lower index on f_i are identical and summed away, the resulting expression f_i v^i has zero free indices. An object with zero free indices is, by definition, a scalar: a (0, 0) tensor.

Basis Independence of the Scalar Value

Why the Number Does Not Depend on Coordinates

Although the intermediate components f_i and v^i are computed relative to a chosen basis, and different bases give different component values, the final summed scalar f_i v^i is identical no matter which basis was used to compute it. This happens because a change of basis transforms f_i by a matrix A and v^i by the inverse matrix A^{-1} (or the reverse, depending on convention), and these two transformations exactly cancel when the contracted sum is recomputed.

{\tilde{f}}_{k} {\tilde{v}}^{k} = (A_{k}^{i} f_{i}) ({(A^{- 1})}_{j}^{k} v^{j}) = f_{i} v^{i}

Contrast with Vector-Valued or Tensor-Valued Results

This scalar output is what distinguishes the evaluation operation from other tensor operations such as the outer product, which produces a higher-rank tensor with free indices remaining, or a linear map applied to a vector, which produces another vector still carrying one free upper index. Only when every upper index is matched with a corresponding lower index, as happens in the covector-vector pairing, does the result collapse entirely to a scalar.

Role as the Terminal Step of Contraction Chains

Reducing Rank to Zero

In longer tensor expressions, repeated contractions progressively lower the total rank of an expression by two at each step, one upper index and one lower index at a time. The covector-vector evaluation is the smallest possible instance of this process: it starts from a total rank of 1 + 1 = 2 and reduces it to 0, leaving a pure scalar with no further contraction possible.

Example: Extracting a Component

A common use of the scalar output is component extraction. If e^i is a dual basis covector and v is any vector, then e^i(v) yields the scalar equal to the i-th component of v in that basis,

e^{i} (v) = v^{i}

demonstrating that scalar outputs of the evaluation operation are precisely how individual numerical coordinates of a vector are recovered from the abstract vector itself.

Diagrammatic Summary

The diagram shows the covector contributing one index and the vector contributing one index, both consumed by the contraction, leaving a rank-zero scalar with no remaining structure.