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Let $\mathfrak{g}$ be a semi-simple Lie algebra and let $\phi:\mathfrak{g}\rightarrow\mathfrak{gl}(V)$ be its finite-dimensional complex irreducible representation. You can define two non-degenerate symmetric forms on $\mathfrak{g}$:

  1. Standard Killing form: $K(X,Y)=tr(ad_X\circ ad_Y)$

  2. "Killing-like" form associated with $\phi$: $K_{\phi}(X,Y)= tr(\phi(X)\phi(Y))$

In general, is there any connection between $K_\phi$ and $K$? I know for instance that for defining representation of $\mathfrak{gl}(N,\mathbb{C})$ both forms are proportional. Is it true for general semi-simple Lie algebra? If not, is there a separate name for $K_{\phi}$?

I'm asking this question because in math-physical literature connected to research I'm doing people tend to confuse these two forms: $K_\phi$ is used to define second order Casimir invariant of a given representation. Yet, in some articles there is simply $K$ instead of $K_{\phi}$.

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    $\begingroup$ Small nitpick: $\mathfrak{gl}(N,\mathbb{C}) is *not* semisimple. $$ $$ No matter, over the complex numbers, if the Lie algebra is simple then they are proportional, since any two invariant bilinear forms are proportional. (Over the reals, for example, the latter statement is false: c.g., $\mathfrak{so}(3,1)$, but I don't recall whether the two forms in the answer are still proportional.) $$ $$ For semisimple Lie algebras which are not simple, I don't think they have to be proportional. $\endgroup$
    – José Figueroa-O'Farrill
    Commented Dec 14, 2010 at 22:53
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    $\begingroup$ This question comes up in both mathematics and physics literature. For example in section 22.1 of my 1972 Springer Graduate Text on Lie algebras I gave a purely mathematical treatment for a semisimple Lie algebra over a field such as $\mathbb{C}$. Here a "Casimir element" attached to a representation and trace function is contrasted with a "universal Casimir element" defined using the Killing form. To treat semisimple (or reductive) rather than just simple Lie algebras, you just need to be careful about the ideals acting as zero in a representation. $\endgroup$
    – Jim Humphreys
    Commented Dec 17, 2010 at 14:34
  • $\begingroup$ I cannot find any proof of this fact. In fact even with the schematic answer given by @BugsBunny below, I cannot prove that for simple $\mathfrak g$ the form $K_\phi$ is proportional to $K_{\text{Ad}}$. Could somebody direct me to one? $\endgroup$
    – Arturo don Juan
    Commented Dec 19, 2018 at 1:20

2 Answers 2

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They are proportional if $g$ is simple. The form $K_\phi$ defines a homomorphism from the adjoint to the coadjoint representation. If the adjoint representation is irreducible, i.e. $g$ is simple, you know all such homomorphisms are proportional by Schur's lemma.

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  • $\begingroup$ This is true over $\mathbb{C}$, but is it true over $\mathbb{R}$, say? Look at $\mathfrak{so}(3,1)$, say. It possesses a two-parameter family of invariant nondegenerate bilinear form, yet the adjoint action is irreducible. $\endgroup$
    – José Figueroa-O'Farrill
    Commented Dec 14, 2010 at 22:57
  • $\begingroup$ You need endomorphisms of the adjoint representation to be $R$. This means the adjoint representations is absolutely irreducible. I guess it is equivalent to the complexification of $g$ being simple... $\endgroup$
    – Bugs Bunny
    Commented Dec 14, 2010 at 23:06
  • $\begingroup$ OK thanks -- so the question remains whether for, say, $\mathfrak{so}(3,1)$ the two forms are always proportional. Is that obvious? Is that even true? $\endgroup$
    – José Figueroa-O'Farrill
    Commented Dec 14, 2010 at 23:42
  • $\begingroup$ Thanks a lot for your answers. I find them very inspiring. I'm a newbie in this field and I still have to learn A LOT :) Just a little question: What if (for semi-simple case) all those constants of proportionality that correspond to a different simple Lie algebras (on which $\mathfrak{g}$ decomposes) turn out to be the same? Is there a reason for which it should not be the case? $\endgroup$
    – Michał Oszmaniec
    Commented Dec 15, 2010 at 0:07
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    $\begingroup$ @Jose: The following seems an easy exercise to me: for a simple Lie algebra over non-algebraically closed field, every two symmetric invariant nondegenerate forms are proportional up to an element of the centroid. This is also proved, as far as I understand (but I haven't seen the paper) in: W.C. Waterhouse, Invariant bilinear forms on semisimple Lie algebras, Algebras Groups Geom. 9 (1992), No.1, 49-52. $\endgroup$
    – Pasha Zusmanovich
    Commented Jan 4, 2011 at 16:31
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These two forms are proportional provided $\mathfrak{g}$ is simple. For the semi-simple case, you can rescale on each of the simple factors.

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  • $\begingroup$ Is this still true over the reals when the complexification of the algebra is not semisimple? e.g., \mathfrak{so}(3,1) $\endgroup$
    – José Figueroa-O'Farrill
    Commented Dec 14, 2010 at 22:55
  • $\begingroup$ If they are both defined over the same field, then they can be proportional over a field extension if and only if they're already proportional over a base field. No? The issue is probably one of characteristic $p$ versus characteristic $0$. See <a href=mathoverflow.net/questions/46813/…> $\endgroup$
    – Keerthi Madapusi
    Commented Dec 14, 2010 at 23:00
  • $\begingroup$ You are right, I was assuming the base field was $\mathbb{C}$, and had Schur's lemma in mind. $\endgroup$
    – David Hill
    Commented Dec 14, 2010 at 23:02
  • $\begingroup$ I don't think this is a characteristic $p$ issue. The point is that I assummed we were working over an algebraically closed field. Schur's lemma doesn't work over the reals. $\endgroup$
    – David Hill
    Commented Dec 14, 2010 at 23:04
  • $\begingroup$ Yes, but formally speaking: the two invariant forms are elements of some vector space over the base field. It seems from your argument, that, once I tensor over $\mathbb{C}$, they are proportional. So shouldn't they already be proportional over $\mathbb{R}$? Or am I just being stupid? $\endgroup$
    – Keerthi Madapusi
    Commented Dec 14, 2010 at 23:07

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