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.

Short Answer

Expert verified
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 by step solution

01

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.
02

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.
03

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.

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