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Every 1-dimensional or simple complex Lie algebra admits an invariant, symmetric and non-degenerate bilinear form. This form is unique up to multiplication by a nonzero constant (which in Yang-Mills theories plays the role of the coupling constant).

Are there other Lie algebras with such a unique scalar product?

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    $\begingroup$ A reductive Lie algebra splits as the (commuting) direct sum of a commutative Lie algebra and many simple Lie algebras, so it supports a unique such form if and only if it is 1-dimensional commutative, or is actually simple. A general Lie algebra splits as the direct sum of its radical and a reductive Lie algebra. Obviously the reductive part must be as above. I'm not sure about the rest. $\endgroup$
    – LSpice
    Commented Jun 20, 2019 at 16:08
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    $\begingroup$ It does not split as a Lie algebra. In general you found yourself with a semi direct product. $\endgroup$
    – InfiniteLooper
    Commented Jun 20, 2019 at 16:13
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    $\begingroup$ If you have an invariant, symmetric and non-degenerate bilinear form as stated, and, if $\mathfrak i$ is an ideal in $\mathfrak g$. Then, the orthogonal $\mathfrak i ^{\perp}$ of $\mathfrak i$ is also an ideal and your algebra $\mathfrak g$ splits as $\mathfrak i + \mathfrak i ^\perp$. So, for a non reductive algebra, you don't even have one such a bilinear form. $\endgroup$
    – InfiniteLooper
    Commented Jun 20, 2019 at 17:04
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    $\begingroup$ @InfiniteLooper, yes, sorry, I am in the bad habit of writing "commuting direct sum" and "direct sum" in place of "direct sum" and "semi-direct product" for Lie algebras. Why couldn't we have an isotropic ideal ($\mathfrak i = \mathfrak i^\perp$)? $\endgroup$
    – LSpice
    Commented Jun 20, 2019 at 21:17
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    $\begingroup$ @InfiniteLooper It's indeed far from true: for instance for every Lie algebra $\mathfrak{g}$, the semidirect product $\mathfrak{g}\ltimes\mathfrak{g}^*$ has such a nondegenerate symmetric bilinear form (easy to figure out). Here where $\mathfrak{g}^*$ is viewed as abelian ideal and endowed with the coadjoint representation of $\mathfrak{g}$. $\endgroup$
    – YCor
    Commented Jun 26, 2019 at 15:37

2 Answers 2

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Let $\mathfrak g$ be a finite dimensional Lie algebra with an invariant, symmetric and non-degenerate bilinear form $\langle \cdot , \cdot \rangle $.

Claim : $\mathfrak g$ admits only one such a bilinear form if and only if $\mathfrak g $ is either simple or abelian and one dimensional.

Proof : Denote the Levi decomposition of $\mathfrak g$ by $\mathfrak r + \mathfrak s$, then for any $\lambda \in \mathbb R $ the form defined by $$\langle r_1 + s_1 , r_2 + s_2 \rangle_\lambda = \langle r_1 , r_2 \rangle + \langle s_1 , r_2 \rangle + \langle r_1 , s_2 \rangle +\lambda \langle s_1 , s_2 \rangle $$ Is invariant. For some $\lambda$ around 1, the form is also definite because it is an open property.

Having only one such form implies that $\mathfrak g$ is either solvable and equal to $\mathfrak r$ or semi-simple and equal to $\mathfrak s$. We know that in the semi-simple case, only simple ones have only one such form. Assume from now on, that $\mathfrak g = \mathfrak r$ is solvable. We write $\mathfrak g = [\mathfrak g , \mathfrak g] + A$ as a direct sum of vector spaces and as before for some $\lambda$ the bilinear form defined by $$\langle g_1 + a_1 , g_2 + a_2 \rangle_\lambda = \langle g_1 , g_2 \rangle + \langle g_1 , a_2 \rangle + \langle a_1 , g_2 \rangle +\lambda \langle a_1 , a_2 \rangle $$ is symmetric and definite.

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    $\begingroup$ Here's a slight variant of the proof, not using Levi. If $\mathfrak{g}$ is not simple or $\le 1$-dimensional, then it has a quotient that is simple or $\le 1$-dimensional. Since $\mathfrak{g}$ has a non-degenerate invariant bilinear form $b_0$ and this quotient too, it also admits a degenerate non-zero invariant bilinear form $b_1$. Hence the space of invariant bilinear forms on $\mathfrak{g}$ has dimension $\ge 2$. So it contains 2 non-proportional non-degenerate elements (say $b_0+tb_1$ for all but finitely many scalars $t$). $\endgroup$
    – YCor
    Commented Jun 26, 2019 at 14:05
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    $\begingroup$ Also the result (and proof) works over an arbitrary field of characteristic zero, when in the statement "simple" is replaced with "absolutely simple". (For a simple, non-absolutely-simple Lie algebra, the dimension of the space of invariant symmetric bilinear forms is $\ge 2$.) $\endgroup$
    – YCor
    Commented Jun 26, 2019 at 14:07
  • $\begingroup$ Using this approach you show that the dimension of such bilinear form is greater than the length of maximal chains of composition in the Lie algebra. Can you show the equality ? $\endgroup$
    – InfiniteLooper
    Commented Jun 26, 2019 at 15:06
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    $\begingroup$ If the abelianization has dimension $n$ this dimension is $\ge n^2$ ($\ge n(n+1)/2$ for the symmetric part). $\endgroup$
    – YCor
    Commented Jun 26, 2019 at 15:13
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    $\begingroup$ In general, for given $\mathfrak{g}$ let $K$ be the intersection of all kernels of all [symmetric] invariant bilinear forms. Then these are those for which $\mathfrak{g}/K$ is simple or 1-dimensional. Remains to understand $K$, and this goes along with understanding Lie algebras admitting a non-degenerate symmetric bilinear form. $\endgroup$
    – YCor
    Commented Jun 26, 2019 at 21:07
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I. Bajo and S. Benayadi already gave a complete answer for your question in their paper entitled "Lie algebras admitting a unique quadratic structure" in 1997, there you can find much more detail.

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  • $\begingroup$ Actually all details were given in the proofs above, e.g. in my 5-line comment. Anyway it's useful to have a published reference. $\endgroup$
    – YCor
    Commented Jun 11, 2021 at 13:47

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