Problem 1 - A Course in Modern Mathematical Physics: Groups, Hilbert Space and Differential Geometry

Question

Let \(E_{f}^{i}\) be the matrix whose \((i, j)\) th component is 1 and all other components vanish. Show that these matrices form a basis of \(G \mathcal{C}(n, \mathbb{R})\), and have the commutator relations $$ \left[E_{t}^{\prime}, E_{t}^{k}\right]=\delta_{i}^{t} E_{t}^{k}-\delta^{k}, E_{t^{-}}^{\prime} $$ Write out the structure constants with respect to this algebra in this basis.

Step-by-Step Solution

Answer

The given matrices \(E_{f}^{i}\) form a basis of \(G \mathcal{C}(n, \mathbb{R})\) as they are linearly independent and there are exactly n² such matrices which equals to the dimension of the given vector space. The commutator relationships are verified for different combinations of i, j and k, with structure constants calculated to be -1 for i ≠ j and 0 otherwise.

Step 1: Understanding the given matrix

The matrix \(E_{f}^{i}\) is a square matrix of order 'n' with all entries being 0, except for the (i, j) element which is 1. Since i and j can take values between 1 and n, there will be n² such matrices. Hence there are n² matrices in the set of all \(E_{f}^{i}\). As the dimension of \(G \mathcal{C}(n, \mathbb{R})\) is n², the set of all \(E_{f}^{i}\) can form a basis. If two different \(E_{f}^{i}\) matrices are added or multiplied by a scalar, the resulting matrix will still have exactly one element being 1 and rest being 0, so they are linearly independent.

Step 2: Showing the commutator relations

The commutator of two matrices A and B, denoted as [A,B], is defined as AB - BA. For the given matrices, the (i, j) element in \(E_{f}^{i}\) is 1 and in \(E_{f}^{j}\) it is 0 and vice versa, therefore the commutator \([E_{f}^{i}, E_{f}^{j}]\) would be \(E_{f}^{i} E_{f}^{j} - E_{f}^{j} E_{f}^{i}\). As matrix multiplication is not commutative and for the case when i ≠ j the multiplication \(E_{f}^{i} E_{f}^{j}\) gives a zero matrix while multiplication \(E_{f}^{j} E_{f}^{i}\) gives a new matrix \(E_{f}^{k}\) where k ≠ i,j, so the commutator is \(-E_{f}^{k}\). When i = j, the commutator is a zero matrix because \(E_{f}^{i} E_{f}^{j}\) equals \(E_{f}^{j} E_{f}^{i}\). The relations are consistent with what is asked in the problem.

Step 3: Writing out the structure constants

The structure constants of the algebra with respect to this basis can be calculated using the commutation relations obtained in step 2. The structure constants \(C_{ij}^{k}\) are defined by the commutation relations \([E_{f}^{i}, E_{f}^{j}] = C_{ij}^{k} E_{f}^{k}\). From step 2, it is clear that when i ≠ j, \(C_{ij}^{k} = -1\) where k ≠ i,j and when i = j, all the structure constants are 0.

Key Concepts

Matrix Representation

In the realm of Lie algebras, matrices can serve as a powerful tool to represent elements of the algebraic structure. Let’s take a closer look at matrix representation using the matrix \(E_{f}^{i}\). This matrix is special in that all of its elements are zero except for one: the \(i, j\) component, which is set to 1.
A system of such matrices \(E_{f}^{i}\) forms a basis for the algebra \(G\mathcal{C}(n, \mathbb{R})\). Since \(i\) and \(j\) can each vary from 1 to \(n\), there are a total of \(n^2\) possible matrices, aligning perfectly with the dimension of the algebra \(G\mathcal{C}(n, \mathbb{R})\). This shared dimension confirms that these matrices can form a complete basis.
Additionally:

• They are linearly independent. Any scalar multiplication or addition of different \(E_{f}^{i}\) matrices produces another matrix within the set with only one non-zero entry.
• Each matrix contributes to spanning the vector space since it is constructed from unique position pairings \((i, j)\).

Thus, these matrices effectively capture the structure of the given algebra through their straightforward but powerful representation.

Commutator Relations

Commutator relations are central to understanding the interactions between elements in a Lie algebra. The commutator of two matrices \(A\) and \(B\) is expressed as \([A, B] = AB - BA\). This operation reveals the non-commutative nature of matrix multiplication in Lie algebras.
For the given matrices \(E_{f}^{i}\) and \(E_{f}^{j}\), their commutation depends on whether \(i\) and \(j\) are equal or not. Here's how it works:

• If \(i eq j\), the multiplication \(E_{f}^{i} E_{f}^{j}\) results in a zero matrix, whereas \(E_{f}^{j} E_{f}^{i}\) produces a new matrix \(E_{f}^{k}\), where \(k eq i, j\). Therefore, the commutator is \(-E_{f}^{k}\).
• If \(i = j\), both products \(E_{f}^{i} E_{f}^{j}\) and \(E_{f}^{j} E_{f}^{i}\) coincide, causing the commutator to be zero.

These subtleties in commutator relations illustrate the intricacies of matrix interactions within the algebra, providing a deeper comprehension of their structural behavior.

Structure Constants

Structure constants represent the fundamental constants connecting elements of a Lie algebra concerning their commutation. They offer a blueprint for how basis elements relate through the commutator operation. In our exercise, these constants are central to reconstituting the structure of the algebra.
The structure constants \( C_{ij}^{k} \) are defined by the relationship: \([E_{f}^{i}, E_{f}^{j}] = C_{ij}^{k} E_{f}^{k}\). By closely examining the commutation relations:

• When \(i eq j\), the constants \( C_{ij}^{k} = -1 \) for some \( k eq i, j \), signifying that the resultant from the commutator is the negative of a certain basis matrix \(E_{f}^{k}\).
• When \(i = j\), the commutator results in a zero matrix, hence all corresponding structure constants \( C_{ii}^{k} = 0 \).

Thus, these structure constants provide a concise yet comprehensive map of the algebra's properties, allowing mathematicians to evaluate the dynamics and interactions of its elements.