What is the relationship between derivations and automorphisms of a Lie algebra?

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In summary, a derivation of a Lie algebra is a linear map that preserves the Lie bracket operation, while an automorphism is a bijective linear map that also preserves the Lie bracket operation. Every automorphism is a derivation, but not every derivation is an automorphism. Derivations and automorphisms play a crucial role in understanding the structure of a Lie algebra and can be applied to other mathematical structures as well.
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Chris L T521
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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 $\tau\in\mathbb{R}$).

 
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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.
 

FAQ: What is the relationship between derivations and automorphisms of a Lie algebra?

What is a derivation of a Lie algebra?

A derivation of a Lie algebra is a linear map that preserves the Lie bracket operation. In other words, if we have two elements X and Y in the Lie algebra, the derivation D satisfies the property that D([X,Y]) = [D(X),Y] + [X,D(Y)].

What is an automorphism of a Lie algebra?

An automorphism of a Lie algebra is a bijective linear map that preserves the Lie bracket operation. This means that it maps elements to elements with the same Lie bracket relationships, and the inverse map also preserves these relationships.

What is the relationship between derivations and automorphisms of a Lie algebra?

Every automorphism of a Lie algebra is also a derivation, but not every derivation is an automorphism. In addition, the set of all derivations of a Lie algebra forms a Lie subalgebra, while the set of all automorphisms forms a Lie group.

How do derivations and automorphisms relate to the structure of a Lie algebra?

Derivations and automorphisms are important in understanding the structure of a Lie algebra because they provide information about how the elements of the Lie algebra interact with each other. For example, derivations can be used to define the concept of a Lie subalgebra, while automorphisms can help identify symmetries and invariants of the Lie algebra.

Can the relationship between derivations and automorphisms be applied to other mathematical structures?

Yes, the concept of derivations and automorphisms can be generalized to other algebraic structures such as rings, algebras, and groups. In these cases, the definitions and properties may vary slightly, but the underlying idea of preserving the structure remains the same.

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