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This is kind of a question about Lie groups/algebras, but what is really hiding is some combinatorial work and some linear algebra

In the context of Matrix Lie groups we can define the ad $ad_x(Y)=[X,Y]=XY-YX$ and Ad maps $Ad_A(X)=AXA^{-1}$

I am trying show by direct calculation that $e^{ad_X}(Y) = Ad_{e^X}(Y)=e^X Ye^{-X}$

where $X$ and $Y$ are $n \times n$ matricies. (This is a question from Brian Hall's matrix Lie group book)

I have calculated that

$(ad_X)^m(Y) = \sum_{k=0}^m \binom{m}{k} X^kY(-X)^{m-k}$

where $(ad_X)^m(Y)=[X,\ldots,[X,[X,Y]]\cdots].$

Thus $e^{ad_X}(Y) = \sum_{p=0}^\infty \frac{(ad_X)^p(Y)}{p!}$ and so $e^{ad_X}(Y) = \sum_{p=0}^\infty \sum_{k=0}^p \binom{p}{k} \frac{1}{p!} X^kY(-X)^{p-k}$

I'm not really sure how to convert this into something that looks like the series expansion of $e^X Y e^{-X}$. Any tips?

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    Ok neat, that falls straight out now!2011-12-03

1 Answers 1

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Let $L_X(A)=XA$ and $R_X(A)=AX$ (left and right multiplication operators. Then $ad_X(A)=XA-AX=L_X(A)-R_X(A)=(L_X-R_X)(A)$. Note that the left and right multiplication operators commute with each other (because matrix multiplication is associative).

Therefore, $e^{ad_X}=e^{L_X-R_X}=e^{L_X}e^{-R_X}$ (since $e^{U+V}=e^Ue^V$ when $U$ and $V$ commute). So we find that

$e^{ad_X}A=e^{L_X}e^{-R_X}A=e^{L_X}\sum\limits_{n=0}^\infty \frac{(-1)^nR_X^n(A)}{n!} = e^{L_X}\sum\limits_{n=0}^\infty \frac{(-1)^nAX^n}{n!}=e^{L_X}Ae^{-X}$

$ = \sum\limits_{n=0}^\infty \frac{L_X^n}{n!} Ae^{-X} = \sum\limits_{n=0}^\infty \frac{X^nAe^{-X}}{n!} = e^XAe^{-X} = e^XA(e^X)^{-1}=Ad_{e^X}(A)$

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    I think if you compare the method you were trying to follow and what I have above, they are essentially the same. I've just hidden all of the difficult calculations under the guise of $e^{L_X-R_X}=e^{L_X}e^{-R_X}$. Proving this basic fact about exponentials requires the same kind of reindexing (and thus is just as hard to prove as your original statement).2011-12-03