1
$\begingroup$

Why are diagonalizable elements of Lie algebra called "semi-simple"?
Is there a notion of "simple" elements? Is it related to "semi-simplicity" of the Lie algebra?

$\endgroup$
2
  • $\begingroup$ I don't know much about lie algebras but there is a general concept of simplicity and semi-simplicity for modules and algebras over modules (a lie algebra is an algebra over a field with additional properties). Check the chapter VI of the book Algèbres et Modules (I. Assem) for further information.. $\endgroup$
    – PtF
    Jul 25, 2015 at 12:34
  • 1
    $\begingroup$ The first question has been answered here. $\endgroup$ Jul 25, 2015 at 20:13

1 Answer 1

0
$\begingroup$

As far as I understand, this terminology is used as an analogy of what happens with matrices, and is related to the notion of semi-simple module mentioned in the comments.

Consider an $n\times n$ matrix $A$ over the complex numbers. Then $A$ defines a structure of $\mathbb{C}[X]$-module over $\mathbb{C}^n$, by letting $X$ act by $A$. Now, this module is semi-simple (that is, it is a direct sum of simple (or irreducible) modules) if and only if the matrix $A$ is diagonalizable. Indeed, $A$ is conjugate to a Jordan normal form, and each Jordan block defines an indecomposable direct summand of the module. It is not hard to see that a Jordan block defines a simple module if and only if it is one-dimensional. Hence the module is semi-simple if and only if all Jordan blocks have size 1, which means that $A$ is diagonalizable.

For abstract Lie algebras, an element is then semi-simple if its adjoint action is given by a diagonalizable matrix.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .