Let G be a Lie group and φ : G × G → G, (g, h) → ghg−1.
Let d(e,e) φ : T(e,e) (G × G) → TeG be the differential of φ , where Tx M designate the tangent space of a differential variety M.
My questions are:
1) how to calculate d(e,e) φ? (So we can see if it is linear or bilinear map)
2) How to prove that T(e,e)(G × G) is isomorphic to TeG × TeG ?
Dear Hanifa,
2) if M and N are two manifolds then T(x,y)(M \times N) =Tx(M)\times Ty(N). It is evident in local coordinates.
1) Let L=Te(G) be the Lie algebra of G. It is not hard to prove that
d(e,e)(\phi): L\times L ->L is just the (bi)linear mapping:
(X,Y) -> adX (Y)=[X,Y].
I think that the part 1.5 of
https://www.math.upenn.edu/~wziller/math650/LieGroupsReps.pdf
or (see especially Chapter 3)
http://www.math.stonybrook.edu/~kirillov/mat552/liegroups.pdf
will provide more details.
Best regards, Yuri
Dear Yuri, thank you for your answer, But it is theoretic, please if you can calculate the differential of φ : G × G → G, (g, h) → ghg−1, because it said in a theorem that d(e,e) φ is not bi linear. it is just a linear map.
Note: I haven't much knowledge in Lie Algebra and differential Geometry, that why I find difficulties. Of course the question is in Lie Algebra, not in Algebraic Number Theory as RG showed.
Again dear Yuri thanks.
Best
I ask if we derive for g and for h, i.e. dφ= ∂φ / ∂g + ∂φ / ∂h, and if we consider g and h as variables or as functions?
@Hanifa, Yes the differential is always a linear map between tangent spaces. For this example $d_{(e,e)}\varphi$ is the second projection \( (X,Y)\to Y \). You can find below the details.
The partial derivative with respect to h is easy since h \mapsto ghg^{-1} is linear, so its differential is the map itself, X \mapsto gXg^{-1} = Ad(g)X, where X \in L(G) = Lie algebra of G.
For the partial derivative with respect to g we put g(t) = e^{tX} for any X \in L(G) and differentiate g(t)hg(t)^{-1} = g(t)hg(-t) with respect to t at t=0. Since g(0) = e, this gives
g'(0)h - hg'(0) = Xh - hX = h(h^{-1}Xh - X) = h(Ad(h^{-1}X - X)
(I am using matrix notation, assuming that G \subset GL(V). The result is the same for abstract Lie groups where h(...) in the last term is replaced with dL_h(...) where L_h : G \to G denotes the left translation g \mapsto hg.)
Does this answer your question?
Best regards
Jost Eschenburg
Thank you Dear Jost your answer is sufficient.
Thank You all Dear participants for your precious answers. All your answers are exactly what I needed.
@Jost: If we denote $L:=d_{(e,e)}\varphi$ and $d_0:=\frac{d}{dt}|_{t=0}$ then
$L(X,Y)=L(X,0)+L(0,Y)=d_0\varphi(\exp tX,e)+d_0\varphi(e,\exp tY)=d_0(e)+d_0(\exp tY)=0+Y=Y$
Dear Mohamed, your computation is correct. I have computed the partial derivative of \varphi(g,h) with respect to g at (e,h) for an arbitrary h. For h=e this partial derivative is zero, hence only the partial with respect to h remains which gives Y as in your computation.
Best regards
Jost
Dear Hanifa
an inner automorphism is a certain type of automorphism of a group defined in terms of a fixed element of the group, called the conjugating element. Formally, if G is a group and a is an element of G, then the inner automorphism defined by a is the map f from G to itself defined for all x in G by the formula
f(x) = a−1xa.
Here we use the convention that group elements act on the right.
http://math.arizona.edu/~cais/594Page/soln/solnexam1.pdf
This is a very good question with many possible answers. In addition to excellent observations already made, there is a bit more to add.
A recent study of differentiating conjugation maps is given in
https://arxiv.org/pdf/1212.6180.pdf
A related study is given in
https://arxiv.org/pdf/hep-th/9902132.pdf
Thank you very much dear James and all who contributed in this question for the references and answers.
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