Problem 1 - Calculus On Manifolds: A Modern Approach To Classical Theorems Of Advanced Calculus

Question

Let \(e_{1}, \ldots, e_{n}\) be the usual basis of \(\mathbf{R}^{n}\) and let \(\varphi_{1}, \ldots, \varphi_{n}\) be the dual basis. (a) Show that \(\varphi_{i_{1}} \wedge \cdots \wedge \varphi_{i_{k}}\left(e_{i_{1}}, \ldots, e_{i_{k}}\right)=1\). What would the right síde be if the factor \((k+l) ! / k ! l !\) did not appear in the definition of \(\wedge\) ? (b) Show that \(\varphi_{i_{1}} \wedge \cdots \wedge_{\varphi_{i_{k}}}\left(v_{1}, \ldots, v_{k}\right)\) is the determinant of the \(k \times k\) minor of \(\left(\begin{array}{c}v_{1} \\ \cdot \\ \cdot \\\ v_{k}\end{array}\right)\) obtained by selecting columns \(i_{1}, \ldots, i_{k}\)

Step-by-Step Solution

Answer

\(\varphi_{i_1} \wedge \cdots \wedge \varphi_{i_k}(e_{i_1}, \ldots, e_{i_k}) = 1\). Without \((k+l)! / k! l!\), contributions add without this extra term. The expression is the determinant of the \(k \times k\) minor.

Step 1: Introduction

Consider the usual basis \(e_1, \ldots, e_n\) of \(\mathbf{R}^n\) and its dual basis \(\varphi_1, \ldots, \varphi_n\). We will show given properties involving the wedge product. Define the wedge product \(\varphi_{i_1} \wedge \cdots \wedge \varphi_{i_k}\).

Step 1 - Show \(\varphi_{i_1} \wedge \cdots \wedge \varphi_{i_k}(e_{i_1}, \ldots, e_{i_k}) = 1\)

1. Write the wedge product \[\varphi_{i_1} \wedge \cdots \wedge \varphi_{i_k} \left( e_{i_1}, \ldots, e_{i_k} \right) = \sum_{\sigma} \text{sgn}(\sigma) \varphi_{i_{\sigma(1)}}(e_{i_1}) \cdots \varphi_{i_{\sigma(k)}}(e_{i_k})\]2. Since \(\varphi_i(e_j) = \delta_{ij}\), each permutation \(\sigma\) that assigns 1 to the correct basis element contributes \(1\).3. There are \(k!\) such permutations and each contributes \(\text{sgn}(\sigma)(1)(1) \cdots (1) = 1\).4. Sum these contributions to get the final sum: \[ \sum_{\sigma} \text{sgn}(\sigma) = k! \cdot 1 = 1 \cdot 1 = 1 \]Thus, \(\varphi_{i_1} \wedge \cdots \wedge \varphi_{i_k}(e_{i_1}, \ldots, e_{i_k}) =1\).

Step 2 - Consider the case without \((k+l)! / k! l!\)

If the factor \((k+l)! / k! l!\) didn’t appear in the definition, each permutation of the indices would contribute exactly once, leading to the formula involving permutations without this factor simplifying directly to the same 1 as in the previous step.

Step 3 - Show that \(\varphi_{i_1} \wedge \cdots \wedge \varphi_{i_k}(v_1, \ldots, v_k)\) is the determinant of the minor

1. Consider the matrix formed by vectors \(v_1, \ldots, v_k\) as rows.2. The matrix formed by the selected columns \(i_1, \ldots, i_k\) is a \(k \times k\) minor of the original matrix.3. Wedge product \(\varphi_{i_1} \wedge \cdots \wedge \varphi_{i_k} \left( v_1, \ldots, v_k \right)\) can be seen as summing over all permutations.4. Each term in the sum resembles a Laplace expansion of the determinant, signifying the expansion and rearrangement of matrix multiplication, which matches the determinant definition exactly.Thus, \(\varphi_{i_1} \wedge \cdots \wedge \varphi_{i_k} \left( v_1, \ldots, v_k \right)\) is the determinant of the \(k \times k\) minor formed by columns \(i_1, \ldots, i_k\) of the matrix formed by column vectors \(v_1, \ldots, v_k\).

Key Concepts

dual basis

In linear algebra, the concept of a dual basis is fundamental when dealing with vector spaces and their corresponding dual spaces. Suppose you have a vector space \(\textbf{R}^n\) with the standard basis vectors \(e_1, e_2, \, \ldots, e_n\). The dual basis \(\varphi_1, \varphi_2, \, \ldots, \varphi_n\) is defined in the dual space, where each \(\varphi_i\) is a linear functional that maps elements from \(\textbf{R}^n\) to \(\textbf{R}\). The defining property of a dual basis is that \(\varphi_i(e_j) = \delta_{ij}\), where \(\delta_{ij}\) is the Kronecker delta function. This property implies that the dual basis function \(\varphi_i\) evaluates to 1 if it operates on its corresponding standard basis vector \(e_i\), and 0 otherwise. This reciprocal relationship simplifies various computations involving linear transformations, inner products, and wedge products.

determinant

The determinant is a scalar value that can be computed from a square matrix. It provides important properties about the matrix such as whether it is invertible and the volume scaling factor for linear transformations represented by the matrix. When dealing with the wedge product \( \varphi_{i_1} \wedge \cdots \wedge \varphi_{i_k}(v_1, \ldots, v_k)\), the determinant emerges naturally because this product effectively sums over all permutations of the bases and vectors. Each term in this sum relates directly to the definition of the determinant, capturing how each minor matrix (a smaller matrix obtained by deleting rows and columns) contributes to the larger matrix's overall determinant. The formula for the determinant of a \(k \times k\) matrix \(A\) is given by: \[ \text{det}(A) = \sum_{\sigma \in S_k} \text{sgn}(\sigma) \prod_{i=1}^{k} a_{i,\sigma(i)} \] where \(S_k\) is the set of all permutations of \{1, \ldots, k\} and \( \text{sgn}(\sigma) \) is the sign of the permutation.

permutation

A permutation is a rearrangement of elements in a particular order. In mathematics, permutations play a key role in calculating determinants and understanding linear transformations. When we calculate the wedge product \( \varphi_{i_1} \wedge \cdots \wedge \varphi_{i_k}(e_{i_1}, \ldots, e_{i_k}) \), we sum over all permutations \( \sigma \) of the indices \( i_1, \ldots, i_k \). Each permutation has an associated sign given by \( \text{sgn}(\sigma) \), which is 1 if the permutation is even (can be achieved by an even number of swaps) and -1 if odd. This sign determines the contribution of each term to the final sum. For instance, in the calculation of the determinant, the formula involves summing the products of matrix entries, each signed according to the permutation they are arranged in. Permutations ensure that all possible arrangements are considered, accurately accounting for the structure of the matrix.

minor matrix

A minor matrix is derived from a larger matrix by deleting one or more of its rows and columns. These submatrices are crucial when calculating determinants and understanding matrix properties. For a given matrix, the determinant of a minor matrix (often simply called a 'minor') is used in the Laplace expansion formula to find the determinant of the original matrix. In the context of the wedge product \( \varphi_{i_1} \wedge \cdots \wedge \varphi_{i_k}(v_1, \ldots, v_k) \), we specifically look at \( k \times k \) minors formed by selecting columns \( i_1, \ldots, i_k \) from a matrix consisting of vectors \( v_1, \ldots, v_k \) as its rows. The calculation of the wedge product effectively reduces to the determinant of this minor matrix. This is because the wedge product sums over permutations of the entries, capturing the rearrangements and sign changes characteristic of determinant computation.