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Problem of the Week #25 - November 19th, 2012

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Chris L T521

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Jan 26, 2012
995
Here's this week's problem.

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Problem: Any linear transformation $\delta$ of a Lie algebra $\mathfrak{g}$ with the property $\delta[Y,Z] = [\delta Y, Z] + [Y,\delta Z]$ is called a derivation of $\mathfrak{g}$. We denote the collection of all derivations of $\mathfrak{g}$ by the set $\text{Der}(\mathfrak{g})$. Show that $\text{Der}(\mathfrak{g})$ is the Lie algebra of the linear group $\text{Aut}(\mathfrak{g})$.

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You can use the following fact: The Lie algebra $\mathfrak{g}$ of a linear group $G$ consists of all matrices $X$ for which $\exp(\tau X)$ lies in $G$ for all $\tau\in\mathbb{R}$. Using this hint reduces the problem into showing that
\[\delta[Y,Z] = [\delta Y, Z] + [Y,\delta Z]\iff \exp(\tau\delta)[Y,Z] = [\exp(\tau\delta) Y, Z] + [Y,\exp(\tau\delta) Z].\]
(the second equation tells us that in this case, $\exp(\tau\delta)$ is an automorphism for all $\tau\in\mathbb{R}$).

Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

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Jan 26, 2012
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No one answered this week's question. I'd first like to apologize about my hint; as I was typing up the solution, I noticed I misread the hint and typed it out wrong for you guys. It should have said:

You can use the following fact: The Lie algebra $\mathfrak{g}$ of a linear group $G$ consists of all matrices $X$ for which $\exp(\tau X)$ lies in $G$ for all $\tau\in\mathbb{R}$. Using this hint reduces the problem into showing that
\[\delta[Y,Z] = [\delta Y, Z] + [Y,\delta Z]\iff \exp(\tau\delta)[Y,Z] = [\exp(\tau\delta) Y,\exp(\tau\delta) Z].\]
(the second equation tells us that in this case, $\exp(\tau\delta)$ is an automorphism for all $\tau\in\mathbb{R}$).
Sigh...anyways, here's my solution:

Proof: Let us assume the fact that the Lie algebra $\mathfrak{g}$ of a linear group $G$ consists of all matrices $A$ for which $\exp(\tau A)$ lies in $G$ for all $\tau\in\mathbb{R}$. Using this hint reduces the problem into showing that
\[\delta[X,Y] = [\delta X, Y] + [X,\delta Y]\iff \exp(\tau\delta)[X,Y] = [\exp(\tau\delta) X,\exp(\tau\delta) Y]\]
(i.e., $\exp(\tau \delta)$ is an automorphism for all $\tau\in\mathbb{R}$.

($\Leftarrow$): If $\exp(\tau\delta)[X,Y] = [\exp(\tau\delta) X,\exp(\tau\delta) Y]$ is true for all $\tau\in\mathbb{R}$, then
\[\begin{aligned} & \frac{d}{d\tau}\exp(\tau\delta)[X,Y]=\frac{d}{d\tau}[\exp(\tau\delta) X, \exp(\tau\delta)Y]\\ \implies & \delta\exp(\tau\delta)[X,Y]=[\delta\exp(\tau\delta)X,\exp(\tau\delta)Y]+[\exp(\tau\delta)X,\delta\exp(\tau\delta)Y]\end{aligned}\]
Setting $\tau=0$ gives us $\delta[X,Y] = [\delta X, Y] + [X,\delta Y]$.

($\Rightarrow$): Conversely, suppose that $\delta[X,Y] = [\delta X, Y] + [X,\delta Y]$. To see that $\exp(\tau\delta)[X,Y] = [\exp(\tau\delta) X,\exp(\tau\delta) Y]$ is true, rewrite this as $[X,Y]=\exp(-\tau\delta)[\exp(\tau\delta)X,\exp(\tau\delta)Y]$. Differentiating the RHS of this equation with respect to $\tau$ yields
\[\begin{aligned} &\exp(-\tau\delta)\left(-\delta[\exp(\tau\delta)X,\exp(\tau\delta)Y]+[\delta\exp(\tau\delta)X,\exp(\tau\delta)Y]+[\exp(\tau\delta)X,\delta\exp(\tau\delta)Y]\right)\\ = & \exp(-\tau\delta)\left(-\delta\exp(2\tau\delta)[X,Y]+\exp(2\tau\delta)\left([\delta X,Y]+[X,\delta Y]\right)\right)\\ = & \exp(-\tau\delta)\left(-\delta\exp(2\tau\delta)[X,Y]+\exp(2\tau\delta)(\delta[X,Y])\right)\qquad(\text{by our assumption})\\ = & 0\end{aligned}\]
Therefore, the RHS is independent of choice of $\tau$, and thus we have $[X,Y]=\exp(-\tau\delta)[\exp(\tau\delta)X,\exp(\tau\delta)Y]$ as seen (easily) when setting $\tau=0$. Q.E.D.
 
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