Derived Lie Algebra of $\mathfrak{so}_3 \mathbb{C}$

[Ted123]

Homework Statement

Find the derived lie algebra of $\mathfrak{so}_3 \mathbb{C}$, the 3x3 antisymmetric matrices with entries in $\mathbb{C}$ with Lie bracket the matrix commutator $[X,Y]=XY-YX$ for any $X,Y\in \mathfrak{so}_3 \mathbb{C}$.

The Attempt at a Solution

Since $\mathfrak{so}_3 \mathbb{C}$ is simple the derived lie algebra of $\mathfrak{so}_3 \mathbb{C}$ (an ideal) must equal $\mathfrak{so}_3 \mathbb{C}$. Is this right? If so, how do I justify this?

 

[mighty2000]

it is important to justify our conclusions with evidence and reasoning. In this case, your reasoning is correct. Since \mathfrak{so}_3 \mathbb{C} is a simple Lie algebra, it does not have any non-trivial ideals. Therefore, the derived Lie algebra must be equal to \mathfrak{so}_3 \mathbb{C} itself.

To further justify this, we can look at the definition of a derived Lie algebra. The derived Lie algebra of a Lie algebra \mathfrak{g} is the smallest ideal of \mathfrak{g} such that the quotient \mathfrak{g}/\mathfrak{g}' is abelian. In this case, since \mathfrak{so}_3 \mathbb{C} is already a simple Lie algebra, it cannot be decomposed into any smaller ideals. Therefore, the derived Lie algebra must be equal to \mathfrak{so}_3 \mathbb{C} itself.

Furthermore, we can also look at the Lie bracket [X,Y] for any X,Y\in \mathfrak{so}_3 \mathbb{C}. Since the Lie bracket is defined as the matrix commutator, it follows that [X,Y] is also an element of \mathfrak{so}_3 \mathbb{C}. This shows that the derived Lie algebra is closed under the Lie bracket operation, further supporting the fact that it must be equal to \mathfrak{so}_3 \mathbb{C} itself.

In conclusion, your reasoning is correct and justified. The derived Lie algebra of \mathfrak{so}_3 \mathbb{C} is equal to \mathfrak{so}_3 \mathbb{C} itself, as it is a simple Lie algebra with no non-trivial ideals.