0

Given an algebra $\mathfrak{g}$, the killing form is defined as $K(x, y) = \operatorname{Tr}(\operatorname{ad}(x) \circ \operatorname{ad}(y))$, but when $\mathfrak{g}=\mathfrak{gl}(n)$, we have that:

$\operatorname{Tr}([x, y])=\operatorname{Tr}(xy-yx)=0$

so is it true that $\operatorname{Tr}([\mathfrak{gl}(n), \mathfrak{gl}(n)])=0?$

kimchi lover
  • 24,277
Alessandro
  • 1,334
  • 3
    It is indeed true by definition that $Tr([x,y]) = 0$, but how does this have anything to do with the Killing form? – mi.f.zh Jan 31 '20 at 13:37
  • 1
    Remember $\text{ad}$ is not a commutator, but rather the commutator viewed as an operator on the basis of the algebra. Consider the Lie algebra formed from $x$, $y$ satisfying $[x,y] = y$, then $\text{ad}(x)$ is not traceless. – Cameron Williams Jan 31 '20 at 15:11

1 Answers1

1

The Killing form of $\mathfrak{gl}_n(K)$ can be expressed by the trace form, i.e., we have $$ B(X,Y)=2n \cdot \operatorname{tr}(XY) − 2 \operatorname{tr}(X)\operatorname{tr}(Y) $$ For $[\mathfrak{gl}_n(K), \mathfrak{gl}_n(K)]=\mathfrak{sl}_n(K)$ the Killing form therefore is given by $$ B(X,Y)=2n \cdot \operatorname{tr}(XY) $$ since $\operatorname{tr}([A,B])=\operatorname{tr}(AB-BA)=0$, as you said. Although this trace is zero, the Killing form of $\mathfrak{sl}_n(K)$ is not necessarily zero (but it can be for prime characteristic $p$ with $p\mid n$). I hope this helps with the misunderstanding.

Reference:

Uniqueness of the killing form

Dietrich Burde
  • 130,978