Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I am interested in clearing up a confusion of mine. I will try to make my question as clear as I can but I apologize in advance if this is not the case.

Given a unitary group of some unital involutive algebra, $U(A) =$ {$u \in A| uu^* = u^*u = I $},

I can define the adjoint action (and right action shortly) of u as $(Ad u)\xi:=u\xi u^*$, Where $\xi \in H$, the Hilbert space on which I am representing the algebra $A$.

Define right action on elements of the hilbert space using Connes Real Structure operator J. $\xi u^* = JuJ^*\xi$. At the level of the algebra the adjoint is then given as $(ad B)\xi:= B\xi - \xi B$.

My Question: How does this generalize in the case where the algebra A is non-associative and H is a general 'bimodule' over A? is conjugation always well defined?

For example. Say I am interested in the group G2 which is generated by the derivation algebra of the octonions (ie it is the automorphism group of the octonions). I can write out the root diagram for G2 so I know there should be an adjoint representation. The octonions O however are are not-associative. As further example if I take A = H = O, how is the adjoint action and conjugation defined (explicitly)?

I imagine there is a very simple answer to this question. Any direct help you could offer or an appropriate reference would be greatly appreciated.

share|improve this question
    
I should say that in the associative case, the the adjoint is given by the derivation $(adB)\xi = D_B \xi = B\xi - \xi B$, so I would imagine at the level of the algebra the adjoint might be given by a similar derivation, but I am not sure. I don't know how this guess would work at the level of the group. –  SMF Aug 20 '12 at 18:32
add comment

1 Answer 1

I hope this clears up some of your confusion:

  1. The group $U(A)$ acts on $A$ itself by conjugation: $\mathrm{Ad}_u(a) = uau^\ast$.

  2. If $A$ acts on a Hilbert space $H$ then the group $U(A)$ acts on $H$ also just by restricting the action of $A$ on $H$. What you have written, $\mathrm{Ad}_u(\xi) = u \xi u^\ast$, doesn't make sense. What does it mean to multiply a vector on the right by an operator?

  3. What is true is that there is a certain intertwining relation between the conjugation action of $U(A)$ on $A$ and the action of $U(A)$ on $H$. Specifically: $$ u (a \xi) = u a u^\ast u \xi = \mathrm{Ad}_u(a)(u \xi) $$

  4. The other adjoint action that you have written, $\mathrm{ad}_B(\xi) = B\xi - \xi B$, also does not make sense as stated, for the same reason as above: it doesn't make sense to multiply a vector on the right by an operator. This type of adjoint action is usually associated with Lie algebras. Since $A$ is an associative algebra, it is a Lie algebra with the bracket $[a,b] = ab - ba$. For any Lie algebra, the adjoint action is defined as $\mathrm{ad}_X(Y) = [X,Y]$.

  5. For Lie groups and Lie algebras, the relation between these two notions of adjoint is standard, and not really at the appropriate level for this site. But it works like this: let $G$ be a Lie group with Lie algebra $\mathfrak{g}$. For $g \in G$, define $C_g : G \to G$ by $C_g(x) = gxg^{-1}$. This is a smooth map which takes the identity to the identity. The differential of this map at the identity is then a linear map from $\mathfrak{g}$ to $\mathfrak{g}$, which is called $\mathrm{Ad}_g$. Then the map $Ad : g \mapsto Ad_g$ is a smooth group homomorphism from $G$ to $GL(\mathfrak{g})$; the differential of this map at the identity is a linear map $ad : \mathfrak{g} \to \mathfrak{gl(g)}$. You can check that this agrees with the notion of adjoint defined in item 4 above.

I would suggest thinking about these different notions of adjoint for a while and then coming back to ask a more focused question.

share|improve this answer
    
Hi thank for responding. I could have been better with my definitions - although I was trying to keep my post short. What I mean by $Ad_u(\xi) = u\xi u^*$ is the following. Define right operation on elements of H as: $\xi u^* = (u^*)^0\xi$, where $b^0 = Ja^* J^*$. J is a real structure operator that acts on the hilbert space H. It is a generalization of the tomita operator. So at the level of the group (Ad u) = uJuJ^*. Exponentiating you can see that at the level of the algebra the adjoint is given as $(ad B) = B - JB^*J^*$. Focused: Is there an analogous J in the non-associative case? –  SMF Aug 21 '12 at 0:31
1  
The existence of the modular conjugation operator requires a cyclic and separating vector for $A$ in the Hilbert space $H$. In general I don't think you'll have that. You will probably need to examine the specific representation of whatever non-associative algebra you are working with to see if you can define the $J$ in a similar way as is done for von Neumann algebras. –  MTS Aug 21 '12 at 19:34
add comment

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.