Integral over the Adjoint Map

Question

$\newcommand{\ad}{\operatorname{ad}}$ Let $\ad_X\in \operatorname{End}(\mathfrak{g})$ be the adjoint map for arbitrary but fixed $X\in\mathfrak{g}$. Denote with $C_{\mathfrak{g}}(X)$ the centraliser of $X$ in $\mathfrak{g}$.

Consider the following expression, with $\mathfrak{g}=su(N)$: $$\int_0^te^{sX}Ae^{-sX}\,ds=\int_0^t\exp(s\,\ad_X)A\,\mathrm ds,\quad A\in su(N).$$

I've been wondering if there is a way to perform the integral and noticed that $$[\exp(t \ad_X)\ad_X^{-1}-\ad_X^{-1}]A$$ should do the job when $\ad_X^{-1}$ exists. Clearly this is not the case for the whole algebra, but is it true if we restrict the domain of $\ad_X$ to $su(N)\setminus C_{su(N)}(X)$, and accordingly $A\in su(N)\setminus C_{su(N)}(X)$?

This is a standard result. In fact the inverse of the adjoint map you are looking at vanishes to yield a celebrated function of ad$_X$ related to Bernouli numbers. it is routinely used in physics. You want more? — Cosmas Zachos, Jun 09 '19 at 21:02
Cheers! I checked the BCH formula but must have overlooked this part somehow... — JDecou, Jun 10 '19 at 10:27
@CosmasZachos Actually, would you mind to show how the above formula emerges from the one you linked to? — JDecou, Jun 10 '19 at 13:36
You want something beyond the formal Taylor series expansion in t , s around 0? — Cosmas Zachos, Jun 10 '19 at 13:45
Well, if I expand and perform the integral, is there a way I can rewrite the infinite sum into sth like the above? — JDecou, Jun 10 '19 at 14:10

Cosmas Zachos · Accepted Answer · 2019-06-10T16:59:13.563

0

$\newcommand{\ad}{\operatorname{ad}}$ OK, just represent your function as the Taylor series in t around zero, the standard result, $$[\exp(t \ad_X)\ad_X^{-1}-\ad_X^{-1}]A =t \phi (t \ad_X) ,$$ where $$ \phi(x)\equiv \sum^\infty_{n=0} \frac{x^{n}}{(n+1)!} =\int_0^{t}\!\!ds ~~e^{sx}. $$ I'm not sure what it is that you are after.

edited Jun 10 '19 at 16:59

answered Jun 10 '19 at 14:34

Cosmas Zachos

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Integral over the Adjoint Map

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