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Consider a matrix Lie group equipped with a left &/or right invariant metric.

The adjoint of linear transformation $A$ with respect to the inner product is denoted as $A^*$.

Here what is actually meant by $A^*$? How to compute an expression like $(ad(X))^*(Y)$ where X,Y are lie algebra elements.

Edited:

Note the definition: $ < (ad(X))^*(Y) , Z > = < Y, [X,Z] >$ for any lie algebra element $Z$.

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If $A:\mathbb{C}^n\to\mathbb{C}^n$ is linear and $\mathbb{C}^n$ has the (hermitian) inner prouduct $\langle x,y\rangle$ then the adjoint of $A$ (which is given by the conjugate transpose of the matrix for $A$) satisfies $\langle Ax,y\rangle=\langle x,A^*y\rangle$.

in a lie algebra, $\text{ad}(A):\frak{a}\to\frak{a}$ is given by $\text{ad}(A)(X)=[A,X]$. then $\text{ad}(A)^*$ is the adjoint of the linear transformation $\text{ad}(A)$ (which probably just means the conjugate transpose if youre working directly on a matrix space)

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thanks yoyo, however i'm unsure whether it goes with < (ad(X))^*(Y) , Z > = < Y, [X,Z] > –  sam Apr 15 '11 at 13:42
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For it to be the conjugate transpose, you have to be sure that the matrix is with respect to an orthonormal basis. –  Jonas Meyer May 10 '11 at 4:36
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