Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have often seen, in the context of operator theory and operator algebras, the notation $\mathrm{Ad}(U)a=UaU^*$, where $U$ is a unitary operator on a Hilbert space $H$ and $a$ is a bounded linear operator on $H$. I have no idea what "Ad" stands for, where/how this notation came into common use, nor whether it fits into a more general context (e.g., for similarities or other automorphisms outside of the context of operator theory). Some Google searching revealed a use of "$\mathrm{Ad}$" in the theory of Lie groups that doesn't quite match with the above, but might have a common origin.

Where does "$\mathrm{Ad}$" come from, especially in the context of $\mathrm{Ad}(U)a=UaU^*$?

share|improve this question
@Qiaochu: Thanks! So, "Adjoint". I still don't see how it all fits together, and I only speculate that there was some abuse of notation when it was borrowed in operator theory, but that certainly helps. –  Jonas Meyer Apr 23 '11 at 18:31
I'm pretty sure that @Qiaochu's right. I'd even go so far as to say that the motivation stems from the finite-dimensional case $A = M_{n}\mathbb{C}$, where the unitary group $U(n) = \mathcal{U}(A)$ leaves the subspace of self-adjoint matrices (= its Lie algebra) invariant. –  t.b. Apr 23 '11 at 18:47
Sorry, I should have said anti-self-adjoint ($a^{\ast} = - a$) before. –  t.b. Apr 23 '11 at 19:01
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.