Proving the Killing Form of gl_n: A Simplified Approach

[lion8172]

Homework Statement

Does anybody know of a concise way to prove that the Killing form of gl_n is given by [tex] B(X,Y) = 2n Tr(X \cdot Y) - 2 Tr(X) Tr(Y)[/tex]?

Homework Equations

The Attempt at a Solution

I believe that this can be shown by noting that a basis for gl_n is given by the matrices E_{ij} whose only nonzero entries occur in the (i,j) entries. Using this fact, one could then explicitly work out the adjoint representation. Is there a simpler way, however?

 

[mighty2000]

Thank you for your question. The Killing form of gl_n can indeed be proven using the basis of matrices E_{ij} and the adjoint representation. However, there is a simpler way to prove this result.

First, we can rewrite the Killing form as B(X,Y) = \mathrm{Tr}(ad_X \circ ad_Y), where ad_X and ad_Y are the adjoint maps of X and Y, respectively. Using the fact that the adjoint representation of gl_n is given by ad_X(Y) = [X,Y], we can rewrite the Killing form as B(X,Y) = \mathrm{Tr}([X,Y]^2).

Next, we can use the Jacobi identity [X,[X,Y]] = [X,[Y,X]] + [Y,[X,X]] = [X,Y] + [Y,X] = 2[X,Y] to simplify the expression further. This gives us B(X,Y) = \mathrm{Tr}([X,Y]^2) = \mathrm{Tr}([X,[X,Y]] + [Y,[X,Y]]) = \mathrm{Tr}(2[X,Y]^2).

Finally, using the fact that \mathrm{Tr}(AB) = \mathrm{Tr}(BA) for any matrices A and B, we can rewrite the expression as B(X,Y) = \mathrm{Tr}(2[Y,X]^2) = 2\mathrm{Tr}([Y,X]\cdot [Y,X]) = 2\mathrm{Tr}(YX - XY)^2.

Since the trace is linear, we can expand this expression to get B(X,Y) = 2(\mathrm{Tr}(YX)^2 - \mathrm{Tr}(XY)^2). Using the property that \mathrm{Tr}(AB) = \mathrm{Tr}(BA), we can rewrite this as B(X,Y) = 2(\mathrm{Tr}(YX)^2 - \mathrm{Tr}(YX)^2) = 2\mathrm{Tr}(YX - XY)\mathrm{Tr}(XY - YX) = 2\mathrm{Tr}(XY - YX)^2.

Finally, using the fact that \mathrm{Tr}(XY) = \mathrm{Tr}(YX) for any matrices X and Y, we can simplify this expression to get B(X,Y) = 2\mathrm{Tr}(XY)^2 -