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Killing form of nilpotent lie algebra

  1. Jun 30, 2008 #1
    The problem statement

    Let [tex]\mathfrak{g}[/tex] be a nilpotent Lie algebra. Prove that the Killing form of [tex]\mathfrak{g}[/tex] vanishes identically.

    The attempt 1

    [tex]\mathfrak{g}[/tex] itself is a solvable ideal, so [tex]\textrm{rad}(\mathfrak{g})=\mathfrak{g}[/tex] and [tex]\mathfrak{g}[/tex] is not semisimple. By Cartan's criterion the Killing form is degenerate, and there exists non-zero [tex]X\in\mathfrak{g}[/tex] so that

    \textrm{tr}(\textrm{ad}_X\textrm{ad}_Y)=0,\quad\forall Y\in\mathfrak{g}.

    This does not yet prove that the Killing form would vanish identically, but only that some rows vanish. The natural way to proceed seems to be by induction and using the quotient space [tex]\mathfrak{g}/\langle X\rangle[/tex], but this doesn't necessarily make sense because there doesn't seem to be any reason to believe that [tex]\langle X\rangle[/tex] would be an ideal in [tex]\mathfrak{g}[/tex].

    The attempt 2


    \mathfrak{h}=\{X\in\mathfrak{g}\;|\;\exists Y\in\mathfrak{g},\; \textrm{tr}(\textrm{ad}_X\textrm{ad}_Y)\neq 0\}.

    We want to prove [tex]\mathfrak{h}=\{0\}[/tex]. If it turned out that [tex]\mathfrak{h}[/tex] is a subalgebra, then the proof would be done. If it was a non-zero subalgebra, then it would be a non-zero nilpotent Lie algebra, and hence not semisimple, but it has a non-degenerate Killing form, in contradiction with the Cartan's criterion.

    Unfortunately I don't know if the [tex]\mathfrak{h}[/tex] defined like this is a subalgebra. If [tex]X_1,X_2\in\mathfrak{h}[/tex] are arbitrary, we would need to find [tex]Y\in\mathfrak{g}[/tex] so that

    \textrm{tr}(\textrm{ad}_{[X_1,X_2]}\textrm{ad}_Y)\neq 0.

    We know there exists [tex]Y_1,Y_2\in\mathfrak{g}[/tex] so that

    \textrm{tr}(\textrm{ad}_{X_k}\textrm{ad}_{Y_k})\neq 0,\quad k=1,2,

    but there still does not seem to be an obvious way to find [tex]Y[/tex].
  2. jcsd
  3. Mar 6, 2011 #2
    Probably too late for answer, but still i thought i should try.

    I think the solution is the first you wrote.

    Let L be the algebra and k(x,y) be the killing form, it holds k([x,y],z)=k(x,[y,z]). we define K={x in L s.t. k(x,y)=0 for any y in L}. Then K is an ideal of L because for any y,z in L an x in K we have k([x,y],z)=k(x,[y,z]), so [x,y] is in K.

    Then you can take L/K and continue by induction.
  4. Mar 7, 2011 #3


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    Doesn't this follow directly from the definition?

    [ad(x)ad(y)]^n maps into L^{2n} (or some power like that*). So if L is nilpotent, then so is ad(x)ad(y). (for any x,y)

    * L^k=[L,L^{k-1}], the lower central series.
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