Question

Let \(A_{i}\) and \(B_{i j}\) be alternating covariant tensors in an open region of \(E^{4}\). Show that the exterior product of \(A_{i} d x^{i}\) and \(\frac{1}{2} B_{i j} d x^{i} d x^{j}\) equals \(\frac{1}{3 !} C_{i j k} d x^{i} d x^{j} d x^{k}\), where \(C_{i j k}\) is the exterior product of the tensors \(A_{i}\) and \(B_{i j}\).

Short answer

Question: Show that the exterior product of 𝐴𝑖 𝑑x^𝑖 and (1/2) 𝐵𝑖𝑗 𝑑x^𝑖 𝑑x^𝑗 results in a form that involves their exterior product 𝐶𝑖𝑗𝑘 in an open region of 𝐸^4. Answer: The exterior product of 𝐴_𝑖 𝑑x^𝑖 and (1/2) 𝐵_𝑖𝑗 𝑑x^𝑖 𝑑x^𝑗 equals (1/3!) 𝐶_𝑖𝑗𝑘 𝑑x^𝑖 𝑑x^𝑗 𝑑x^𝑘, where 𝐶_𝑖𝑗𝑘 = 𝐴_𝑖 𝐵_𝑗𝑘 - 𝐴_𝑗 𝐵_𝑖𝑘.

Step-by-step solution

Compute the exterior product of A_i dx^i and B_ij dx^i dx^j

Here, we are given two alternating covariant tensors: A_i dx^i and (1/2) B_ij dx^i dx^j. We need to find the exterior product of these two tensors. The exterior product of two tensors is defined as: (P ∧ Q)_ij...kl = P_iq...r Q_j...klr - P_jq...r Q_i...klr Where P and Q are the two tensors and the indices i, j, k, and l are distinct. Now, let's compute the exterior product of A_i dx^i and (1/2) B_ij dx^i dx^j: (A_i dx^i ∧ (1/2) B_ij dx^i dx^j)_ijk = (1/3!) A_i B_jk - A_j B_ik (since indices are distinct) Here, C_ijk is the exterior product of the tensors A_i and B_ij.

Compare the exterior product with the given form

Now, we need to show that the exterior product obtained in the previous step is equal to the given form: (1/3!) C_ijk dx^i dx^j dx^k We already have: C_ijk = A_i B_jk - A_j B_ik The exterior product of A_i dx^i and (1/2) B_ij dx^i dx^j that we have obtained is: (1/3!) A_i B_jk - A_j B_ik Comparing this with the given form, we can conclude that: (1/3!) C_ijk dx^i dx^j dx^k = (1/3!) C_ijk dx^i dx^j dx^k (since indices are distinct) Thus, the exterior product of 𝐴_𝑖 𝑑x^𝑖 and (1/2) 𝐵_𝑖𝑗 𝑑x^𝑖 𝑑x^𝑗 equals (1/3!) 𝐶_𝑖𝑗𝑘 𝑑x^𝑖 𝑑x^𝑗 𝑑x^𝑘.

Key concepts

Alternating Covariant Tensors

In the realm of tensor algebra, an alternating covariant tensor is one that changes sign whenever two of its indices are swapped. This type of tensor represents quantities that are inherently related to orientation in space. For example, consider a tensor \( A_{ij} \) with two lower indices in a space \( E^n \). If \( A_{ij} \) is alternating, then swapping the indices results in \( A_{ji} = -A_{ij} \).

Now, focusing on our exercise, \( A_i \) and \( B_{ij} \) are given as alternating covariant tensors, which implies that for any pair of indices \( i \) and \( j \) in \( B_{ij} \) we have \( B_{ij} = -B_{ji} \) and for \( A_{i} \) being a 1-index tensor, no swapping is possible.

These properties are crucial while calculating the exterior product, as seen in the step-by-step solution, because the antisymmetry leads to certain terms vanishing, simplifying the computation. In the context of multivariable calculus, alternating covariant tensors correspond to what are known as differential forms, which leads us to our next concept.

Differential Forms

Continuing our exploration, differential forms provide a powerful way to abstract and generalize the concepts of calculus to multiple dimensions. A differential form is a kind of function that you can integrate over a manifold, a term for a space that is locally like a piece of \(\mathbb{R}^n\). They are directly related to covariant tensors; in fact, a differential form of degree \( k \) can be thought of as an alternating covariant tensor of rank \( k \).

The exterior product mentioned in the original exercise relates directly to differential forms. It's an operation that combines two forms into a new one in a way that extends the cross product concept in three dimensions. For instance, \( A_{i} dx^{i} \) and \( \frac{1}{2} B_{ij} dx^{i} dx^{j} \) are differential forms of degree 1 and 2, respectively. When we take their exterior product, we obtain a new differential form of degree 3, as shown in the solution.

Students should appreciate that when performing calculations with differential forms, the ordering of the \( dx^{i} \) terms in the product matters due to their antisymmetric nature, reinforcing what we understand as orientation in these higher-dimensional spaces.

Tensor Algebra

In the universe of mathematical structures, tensor algebra is a framework that extends the concepts of vectors and matrices to higher-dimensional arrays of numbers that transform in specific ways under coordinate transformations. Tensors are geometric entities that can encode physical quantities and their relationships in any number of dimensions. They are the backbone of fields such as differential geometry and relativity.

The exterior product we have been discussing is an operation within tensor algebra, expressive of the interplay between different tensors. In our exercise, tensor algebra rules enabled us to amalgamate the alternating covariant tensors \( A_i \) and \( B_{ij} \) to obtain a new tensor \( C_{ijk} \) that encapsulates combinations of the components of the initial tensors while respecting their antisymmetric properties.

Mastering tensor algebra is crucial for students interested in advanced physics and engineering because it provides the language to describe complex systems within a robust mathematical framework. In particular, understanding how exterior products work helps to unravel the complex relationships between geometric quantities in a way that can be further generalized to even more intricate structures within the field.