📂Representation Theory

Killing Form

Definition¹

The symmetric bilinear form $B$ defined as below on a finite-dimensional Lie algebra $\mathfrak{g}$ is called the Killing form.

$$ B(X, Y) = \trace(\ad_{X} \circ \ad_{Y}) \tag{1} $$

Here, $\trace$ is the trace of a linear transformation, and $\ad_{X} : \mathfrak{g} \to \mathfrak{g}$ is the linear map defined as follows.

$$ \ad_{X}(Y) = [X, Y]_{\mathfrak{g}} $$

Explanation

As the capital $K$ suggests, Killing is a person’s name. It is named after the German mathematician Wilhelm Karl Joseph Killing.

When an inner product is given on a vector space, it can be put to use in many ways, such as constructing a similarity function that compares how similar two vectors are, or expressing the geometric relationship between two vectors as a scalar value. However, a Lie algebra $\mathfrak{g}$ comes with no inner product; only the bracket $[\cdot,\cdot]$ is given. If we fix the first argument of the bracket, though, we get a linear transformation $\ad = [X, \cdot]$. Therefore, using the two linear transformations $\ad_{X}$ and $\ad_{Y}$ corresponding to elements $X, Y$ of the Lie algebra, we can consider the following mapping.

$$ (X, Y) \mapsto \trace(\ad_{X} \circ \ad_{Y}) $$

This is bilinear by the linearity of $\ad_{X}$ and $\trace$, and symmetric by the property of the trace.

$$ \trace(AB) = \trace(BA) $$

In this way, the Killing form is a symmetric bilinear form obtained naturally from the bracket of the Lie algebra. Moreover, as the theorem below shows, the Killing form can be used to determine whether or not $\mathfrak{g}$ is semisimple.

Properties

(a) For any $X$, $\ad_{X}$ is skew-symmetric with respect to $B$. $$ B(\ad_{X}(Y), Z) = - B(Y, \ad_{X}(Z)), \quad \forall Y, Z \in \mathfrak{g} $$

(b) It is invariant with respect to the bracket of $\mathfrak{g}$. $$ B([X, Y]_{\mathfrak{g}}, Z) = B(X, [Y, Z]_{\mathfrak{g}}) $$

Theorem

$B$ being nondegenerate means that for every $X \in \mathfrak{g}\setminus\left\{ 0 \right\}$ there exists $Y \in \mathfrak{g}$ satisfying the following.

$$ B(X, Y) \ne 0 $$

For a finite-dimensional Lie algebra $\mathfrak{g}$ over $\mathbb{R}$ or $\mathbb{C}$, $\mathfrak{g}$ being semisimple is equivalent to $B$ being nondegenerate.

$$ \text{$B$ is nondegenerate $\iff$ $\mathfrak{g}$ is semisimple} $$

This is called Cartan’s criterion.

Proof

For convenience, we omit the subscript of the bracket and write $[\cdot, \cdot] = [\cdot, \cdot]_{\mathfrak{g}}$.

Property (a)

We use the fact that $\ad$ is a Lie algebra homomorphism, that is, the following holds.

$$ \ad_{[X, Y]} = \ad_{X} \circ \ad_{Y} - \ad_{Y} \circ \ad_{X} $$

Expanding the left-hand side of the desired identity according to the definition $(1)$ of the Killing form gives the following.

$$ \begin{align*} B(\ad_{X}(Y), Z) &= B([X, Y], Z) \\ &= \trace\left( \ad_{[X, Y]} \circ \ad_{Z} \right) \\ &= \trace\left( \ad_{X} \circ \ad_{Y} \circ \ad_{Z} \right) - \trace\left( \ad_{Y} \circ \ad_{X} \circ \ad_{Z} \right) \end{align*} $$

The right-hand side is expanded in the same way.

$$ \begin{align*} - B(Y, \ad_{X}(Z)) &= - B(Y, [X, Z]) \\ &= - \trace\left( \ad_{Y} \circ \ad_{[X, Z]} \right) \\ &= - \trace\left( \ad_{Y} \circ \ad_{X} \circ \ad_{Z} \right) + \trace\left( \ad_{Y} \circ \ad_{Z} \circ \ad_{X} \right) \end{align*} $$

Cancelling the term $- \trace\left( \ad_{Y} \circ \ad_{X} \circ \ad_{Z} \right)$ that appears in both expansions, it remains only to check that the remaining terms are equal. By the cyclic property of the trace $\trace(ST) = \trace(TS)$, the following holds.

$$ \trace\left( \ad_{X} \circ \ad_{Y} \circ \ad_{Z} \right) = \trace\left( \ad_{Y} \circ \ad_{Z} \circ \ad_{X} \right) $$

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Property (b)

Substituting $Y$ for $X$ and $X$ for $Y$ in property (a) gives $B(\ad_{Y}(X), Z) = - B(X, \ad_{Y}(Z))$. Using this together with the antisymmetry of the bracket $[Y, X] = -[X, Y]$, we obtain the following.

$$ \begin{align*} B([X, Y], Z) &= - B([Y, X], Z) \\ &= - B(\ad_{Y}(X), Z) \\ &= B(X, \ad_{Y}(Z)) \\ &= B(X, [Y, Z]) \end{align*} $$

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