2.5.3 Tensor Basis Independence Property

The Tensor Basis Independence Property ensures tensor components remain consistent regardless of the chosen basis, foundational in algebraic structures and tensor calculus.

Tensor Basis Independence Property is the guarantee that no nontrivial linear combination of the basis tensor products, formed from a chosen basis of the underlying vector space and its dual, produces the zero tensor, ensuring that every tensor has a unique representation in terms of these basis products. It is the second of the two properties, alongside the spanning property, required to establish that the induced tensor products form a genuine basis of a tensor space, and it is the property responsible for the well-definedness of tensor components.

Statement of the Property

Setting

Let $V$ be a vector space of dimension $n$ over a field $F$ , with basis $e_{1}, \dots, e_{n}$ and dual basis $e^{1}, \dots, e^{n}$ . For a tensor space $T_{s}^{r} (V)$ , the independence property asserts that

c_{j_{1} \dots j_{s}}^{i_{1} \dots i_{r}} e_{i_{1}} \otimes \dots \otimes e^{j_{s}} = 0

implies

c_{j_{1} \dots j_{s}}^{i_{1} \dots i_{r}} = 0

for every admissible index combination, so no nontrivial vanishing combination of the basis tensor products exists.

Proof of the Independence Property

Step One: Assuming a Vanishing Combination

Suppose the linear combination

Z = c_{j_{1} \dots j_{s}}^{i_{1} \dots i_{r}} e_{i_{1}} \otimes \dots \otimes e^{j_{s}}

equals the zero tensor, meaning it agrees with the zero multilinear map on every input.

Step Two: Evaluating on a Fixed Basis Input Tuple

Fix an arbitrary admissible index combination $k_{1}, \dots, k_{r}, l_{1}, \dots, l_{s}$ , and evaluate $Z$ on the input tuple $(e^{k_{1}}, \dots, e_{l_{1}}, \dots)$ :

0 = Z (e^{k_{1}}, \dots, e_{l_{1}}, \dots) = c_{j_{1} \dots}^{i_{1} \dots} e_{i_{1}} (e^{k_{1}}) \dots e^{j_{s}} (e_{l_{s}})

Step Three: Applying the Duality Relations

Using $e_{i} (e^{k}) = e^{k} (e_{i}) = δ_{i}^{k}$ for each factor, every term in the sum vanishes except the single term whose indices match $k_{1}, \dots, l_{1}, \dots$ exactly, leaving

0 = c_{l_{1} \dots}^{k_{1} \dots}

Step Four: Generalizing to Every Coefficient

Since the index combination $k_{1}, \dots, l_{1}, \dots$ was arbitrary, this argument applies to every coefficient in turn, showing that all $n^{r + s}$ coefficients must vanish, which establishes independence.

The Role of Duality in the Proof

Why the Dual Basis Is Essential

The independence proof relies entirely on the orthogonality relation between a basis and its dual basis; without a dual basis satisfying $e^{i} (e_{j}) = δ_{j}^{i}$ , there would be no mechanism for isolating a single coefficient by evaluation, since no natural pairing would distinguish one basis tensor product from another.

Contrast with the Underlying Vector Space Proof

Independence of the basis tensor products is not automatically inherited from independence of $e_{1}, \dots, e_{n}$ in $V$ alone; it requires the interplay between this basis and its dual, mediated by the pairing that defines evaluation of a tensor on its arguments.

Consequence: Uniqueness of Components

No Two Distinct Coefficient Arrays Represent the Same Tensor

Independence guarantees that if two coefficient arrays produce the same tensor,

a_{j_{1} \dots}^{i_{1} \dots} e_{i_{1}} \otimes \dots = b_{j_{1} \dots}^{i_{1} \dots} e_{i_{1}} \otimes \dots

then subtracting and applying independence to the resulting vanishing combination forces $a_{j_{1} \dots}^{i_{1} \dots} = b_{j_{1} \dots}^{i_{1} \dots}$ for every index combination. This is precisely what makes the components of a tensor a well-defined, unambiguous description of it.

Minimality of the Spanning Set

Independence also certifies that the spanning set of basis tensor products is minimal: no basis tensor product can be discarded and still leave a spanning set, since discarding one would eliminate the sole means of achieving a nonzero coefficient on it, contradicting the ability to represent tensors that genuinely require that term.

Interaction with Dimension

Exact Dimension Count

Combined with the spanning property, independence upgrades the bound $dim T_{s}^{r} (V) \leq n^{r + s}$ to an exact equality, since a spanning set of size $n^{r + s}$ that is also independent cannot be reduced further, and cannot be extended by any tensor not already in its span.

Preservation Under Change of Basis

Because the transformation relating basis tensor products across two bases of $V$ is given by an invertible matrix built from the change-of-basis matrix and its inverse, independence in one basis implies independence in every other basis, so the property does not depend on which particular basis of $V$ was used to construct the tensor basis products.