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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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    You are almost there! Replace the double summation $\sum\limits_{p=0}^{\infty}\sum\limits_{k=0}^p$ by the double summation $\sum\limits_{k=0}^{\infty}\sum\limits_{p=k}^{\infty}$ (these summations are equivalent), and substitute $\ell$ for $p-k$ in the second sum. Oh, and you made a mistake in computing $\left(\mathrm{ad}_X\right)^m \left(Y\right)$: there should be an $\binom{m}{k}$ in the sum. It will force you to replace the $\frac{1}{p!}$ in the last formula by something... closer to the goal.2011-12-03
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    Opps, that was a typo - I'll fix that up! And thanks for the tip - I'll try and see if I can figure it out2011-12-03
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    Ok neat, that falls straight out now!2011-12-03

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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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    Neat! A completely different approach to what I was doing2011-12-03
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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