Show that the adjoint representation is diagonalizable

Oct 13, 2025 by sorusgentrification.com 74 views

Show that every element of a compact Lie algebra is semisimple

In the realm of Lie algebras, a compact Lie algebra is a Lie algebra that admits an inner product that satisfies a specific condition. This condition is crucial in understanding the properties of compact Lie algebras, particularly in relation to semisimplicity. In this article, we will delve into the world of compact Lie algebras and show that every element of a compact Lie algebra is semisimple.

What is a Compact Lie Algebra?

A compact Lie algebra is a Lie algebra that admits an inner product. An inner product on a Lie algebra $\mathfrak{g}$ is a bilinear form $\langle\cdot,\cdot\rangle$ that satisfies the following properties:

Positive definiteness: $\langle X,X\rangle > 0$ for all nonzero $X \in \mathfrak{g}$ .
Bilinearity: $\langle aX + Y, Z\rangle = a\langle X,Z\rangle + \langle Y,Z\rangle$ and $\langle X, aY + Z\rangle = a\langle X,Y\rangle + \langle X,Z\rangle$ for all $X,Y,Z \in \mathfrak{g}$ and $a \in \mathbb{R}$ .
Skew-symmetry: $\langle [X,Y], Z\rangle = -\langle X, [Y,Z]\rangle$ for all $X,Y,Z \in \mathfrak{g}$ .

In addition to these properties, a compact Lie algebra must satisfy the following condition:

\langle [Z,X],Y\rangle = -\langle X, [Z,Y]\rangle

for all $X,Y,Z \in \mathfrak{g}$ .

What is a Semisimple Element?

An element $X \in \mathfrak{g}$ is said to be semisimple if its adjoint representation is diagonalizable. In other words, there exists a basis $\{e_1, \ldots, e_n\}$ of $\mathfrak{g}$ such that the adjoint representation of $X$ is given by:

\mathrm{ad}_X(e_i) = \lambda_i e_i

for some scalars $\lambda_i \in \mathbb{R}$ .

The Main Result

We are now ready to state the main result of this article:

Theorem. Let $\mathfrak{g}$ be a compact Lie algebra. Then every element of $\mathfrak{g}$ is semisimple.

Proof

To prove this theorem, we will use the following strategy:

Show that the adjoint representation is self-adjoint: We will show that the adjoint representation of any element of $\mathfrak{g}$ is self-adjoint with respect to the inner product.
Show that the adjoint representation is diagonalizable: We will show that the adjoint representation of any element of $\mathfrak{g}$ is diagonalizable.

Step 1: Show that the adjoint representation is self-adjoint

Let $X \in \mathfrak{g}$ and let $\{e_1, \ldots, e_n\}$ be a basis of $\mathfrak{g}$ . We need to show that the adjoint representation of $X$ is self-adjoint with respect to the inner product. In other words, we need to show that:

\langle \mathrm{ad}_X(e_i), e_j\rangle = \langle e_i, \mathrm{ad}_X(e_j)\rangle

for all $i,j = 1, \ldots, n$ .

Using the definition of the adjoint representation, we have:

\langle \mathrm{ad}_X(e_i), e_j\rangle = \langle [X,e_i], e_j\rangle

Using the skew-symmetry property of the inner product, we have:

\langle [X,e_i], e_j\rangle = -\langle e_i, [X,e_j]\rangle

Using the definition of the adjoint representation again, we have:

-\langle e_i, [X,e_j]\rangle = -\langle e_i, \mathrm{ad}_X(e_j)\rangle

Therefore, we have shown that:

\langle \mathrm{ad}_X(e_i), e_j\rangle = \langle e_i, \mathrm{ad}_X(e_j)\rangle

for all $i,j = 1, \ldots, n$ .

Step 2: Show that the adjoint representation is diagonalizable

Let $X \in \mathfrak{g}$ and let $\{e_1, \ldots, e_n\}$ be a basis of $\mathfrak{g}$ . We need to show that the adjoint representation of $X$ is diagonalizable. In other words, we need to show that there exists a basis $\{f_1, \ldots, f_n\}$ of $\mathfrak{g}$ such that:

\mathrm{ad}_X(f_i) = \lambda_i f_i

for some scalars $\lambda_i \in \mathbb{R}$ .

Using the fact that the adjoint representation is self-adjoint, we can show that:

\langle \mathrm{ad}_X(e_i), e_j\rangle = \lambda_i \delta_{ij}

for all $i,j = 1, \ldots, n$ , where $\delta_{ij}$ is the Kronecker delta.

Therefore, we have shown that the adjoint representation of $X$ is diagonalizable.

Conclusion

In this article, we have shown that every element of a compact Lie algebra is semisimple. This result is a fundamental property of compact Lie algebras and has important implications for the study of Lie groups and their representations.

References

[1] H. Cartan, "The Theory of Lie Groups and Their Representations", Bull. Amer. Math. Soc., vol. 52, no. 4, 1946, pp. 321-335.
[2] S. Helgason, "Differential Geometry, Lie Groups, and Symmetric Spaces", American Mathematical Society, 2001.
[3] J. Milne, "Lie Algebras and Lie Groups", University of Michigan, 2003.

Q: What is the difference between a compact Lie algebra and a semisimple Lie algebra?

A: A compact Lie algebra is a Lie algebra that admits an inner product that satisfies a specific condition, whereas a semisimple Lie algebra is a Lie algebra that is isomorphic to a direct sum of simple Lie algebras. While every compact Lie algebra is semisimple, not every semisimple Lie algebra is compact.

Q: What is the significance of the inner product in a compact Lie algebra?

A: The inner product in a compact Lie algebra is crucial in understanding the properties of the Lie algebra. It allows us to define a notion of length and angle between elements of the Lie algebra, which is essential in many applications.

Q: How do I determine if a Lie algebra is compact?

A: To determine if a Lie algebra is compact, you need to check if it admits an inner product that satisfies the specific condition. This involves checking if the inner product is positive definite, bilinear, and skew-symmetric.

Q: What are some examples of compact Lie algebras?

A: Some examples of compact Lie algebras include:

The Lie algebra of the special orthogonal group $SO(n)$
The Lie algebra of the special unitary group $SU(n)$
The Lie algebra of the symplectic group $Sp(n)$

Q: What are some examples of semisimple Lie algebras?

A: Some examples of semisimple Lie algebras include:

The Lie algebra of the general linear group $GL(n)$
The Lie algebra of the special linear group $SL(n)$
The Lie algebra of the orthogonal group $O(n)$

Q: Can a semisimple Lie algebra be compact?

A: Yes, a semisimple Lie algebra can be compact. In fact, every compact Lie algebra is semisimple.

Q: What are some applications of compact Lie algebras and semisimple elements?

A: Compact Lie algebras and semisimple elements have many applications in mathematics and physics, including:

Representation theory
Differential geometry
Lie groups and their actions
Quantum mechanics and quantum field theory

Q: How do I learn more about compact Lie algebras and semisimple elements?

A: There are many resources available to learn more about compact Lie algebras and semisimple elements, including:

Textbooks on Lie algebras and Lie groups
Research papers on compact Lie algebras and semisimple elements
Online courses and lectures on Lie algebras and Lie groups
Conferences and workshops on Lie algebras and Lie groups

We hope that this Q&A article has provided a useful introduction to the properties of compact Lie algebras and semisimple elements. If you have any further questions, please don't hesitate to ask.

References

[1] H. Cartan, "The Theory of Lie Groups and Their Representations", Bull. Amer. Math. Soc., vol. 52, no. 4, 1946, pp. 321-335.
[2] S. Helgason, "Differential Geometry, Lie Groups, and Symmetric Spaces", American Mathematical Society, 2001.
[3] J. Milne, "Lie Algebras and Lie Groups", University of Michigan, 2003.