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I'm studying the adjoint representation of $GL_n(\mathbb{C})$, and I'm encountering some confusion about the denominator that appears in the matrix entries of the adjoint action.

The adjoint representation of $GL_n(\mathbb{C})$ is given by the map $\rho: GL_n(\mathbb{C}) \to GL_n(\mathbb{C})$ defined by $$ \rho(A)(M) = A M A^{-1}. $$ for all $M \in GL_n(\mathbb{C})$.

The book that I'm reading tells:

Note that $\det(A)^{n-1}$ is a common denominator for the rational functions $\rho_{kl}(A)$.

Now, I don't get why the matrix entries $\rho_{kl}(A)$ have $\det(A)^{n-1}$ as a common denominator, rather than just $\det(A)$. Probably I've been missing something elementary. Could anyone clarify this point?

Thanks in advance for your insights!

1 Answers1

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This is a mistake/typo. The common denominator is indeed $\det A$. Also, the image of $\rho$ is not in $\operatorname{GL}_n(\mathbb{C})$ but $\operatorname{GL}_{n^2}(\mathbb{C})$ (invertible maps on the $n^2$-dimensional space of matrices $\mathbb{C}^{n \times n}$).

Vladimir Lysikov
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  • Ah yes, writing $GL_n(\mathbb{C})$ instead of $GL_{n^2}(\mathbb{C})$ for the image of $\rho$ was my typo. Thank you! –  Dec 28 '24 at 17:59