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Killing form of ${s l}_{2}$

Example

Let $k$ have characteristic zero and put $g = {s l}_{2} (k)$ . For

$e = (\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}), f = (\begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array}), h = (\begin{array}{cc} 1 & 0 \\ 0 & - 1 \end{array}),$

one has $K (h, h) = 8$ , $K (e, f) = K (f, e) = 4$ , and all other basis pairings are zero. Thus $K$ is nondegenerate.

Facts & Assumptions

Given: The displayed matrices over a characteristic-zero field.

[L1]

The Killing form is $K (x, y) = tr ({ad}_{x} {ad}_{y})$ (Killing form).

[L2]

A finite-dimensional characteristic-zero Lie algebra is semisimple exactly when its Killing form is nondegenerate (Cartan's semisimplicity criterion).

Verification

technique · direct matrix calculation

1.1givenalgebra

In the ordered basis $(e, f, h)$ , the relations $[h, e] = 2 e$ , $[h, f] = - 2 f$ , and $[e, f] = h$ give ${ad}_{e} = (\begin{array}{ccc} 0 & 0 & - 2 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \end{array}), {ad}_{f} = (\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 2 \\ - 1 & 0 & 0 \end{array}), {ad}_{h} = (\begin{array}{ccc} 2 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & 0 \end{array}) .$

2.1L1step 1.1algebra

Squaring the last matrix gives trace $8$ . Multiplying the first two matrices in either order gives trace $4$ . Their squares have trace zero, and multiplying either by ${ad}_{h}$ in either order also has trace zero. By [L1], these are precisely the claimed pairings.

3.1L2step 2.1algebra∎

The Gram matrix has determinant $\det (\begin{array}{ccc} 0 & 4 & 0 \\ 4 & 0 & 0 \\ 0 & 0 & 8 \end{array}) = - 128.$ Characteristic zero makes this scalar nonzero, so $K$ is nondegenerate. In particular [L2] recovers semisimplicity. Every basis pairing was calculated, including the zero pairings, and no choice principle is used.

Depends on

Used by

Same complexification with different killing form signatures Counterexample
Compact and split real forms of sl two c Example
All real forms of a complex semisimple lie algebra are isomorphic False statement
Every element of a complex semisimple Lie algebra is semisimple False statement
The root-space decomposition classifies real semisimple Lie algebras with no extra data False statement

Dependency tree · two levels

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Sources

Kirillov, An Introduction to Lie Groups and Lie Algebras, Example 5.51 (standard reference, not scraped)

Killing form of sl2sl2​