Let there be a Lie-Algebra $\mathfrak{g}$ (I am most interested in the cases, where $\mathfrak{g}$ is simple or semisimple) and a irreducible representation R. Let $t^a$ be a base of the Lie-Algebra and define $R(t^a) =: t^a_R$ for convenience.

Now physicists define the Index I(R) of the representation as \begin{equation} d_R^{a_1...a_n}:=\frac{1}{n!} \sum_{\pi \in S_n} \mathrm{tr}\left[t^{a_{\pi(1)}}_Rt^{a_{\pi(1)}}_R...t^{a_{\pi(1)}}_R \right] =: I(R) d^{a_1...a_n} + \mathrm{products~of~lower~orders} \end{equation} where d denotes a fundamental* symmetric, Ad-invariant tensor of the Lie-Algebra. (compare eq. (20) of Group theory factors for Feynman diagrams).

Now they claim, that d can be fixed in a representation-independent way. Why?

*As far as I understood a fundamental symmetric, Ad-invariant tensor of a Lie-Algebra is a Symmetric Ad-invariant tensor which is no Polynomial in Symmetric Ad-invariant tensors of lower rank.

Edit: No, as defined above it is a priori not clear, that it is well defined. In the following "unique" means always "unique up to a constant factor".

  • I assume, that the only thing we now about $d_R^{a_1...a_n}$ is, that it is symmetric and Ad-invariant.
  • There is a one-to-one correspondence between symmetric Ad-invariant tensors and Casimir Operators (compare Invariant tensors and Casimir operators for simple compact Lie groups)
  • The fundamental Casimir Operators are not unique (compare General Dynkin indices and their applications which can be found on p. 419 in Symmetries in Science)
  • Hence fundamental symmetric Ad-invariant tensors are not unique.
  • To simplify the issue consider the case n = 3. If $d$ could be fixed in a representation invariant way, then $d_R^{a_1a_2a_3}$ and $d_{R'}^{a_1a_2a_3}$ should be proportional. But since we've assumed, that the only thing we know about $d_R$ and $d_R'$ is, that it is symmetric, fundamental and Ad-invariant, there is by the preceding argumentation no reason why $d_R$ and $d_R'$ should be proportional.

If we could show, that $d_R$ and $d_R'$ satisfy further relations, respectively, which fix the invariant tensor, then the definition would be well-defined. One possibility of such relations would be orthogonality relations (compare again General Dynkin indices and their applications). But I can't see, why $d_R$ and $d_R'$ should satisfy such relations?

So I update my question: Are there relations which fix the form of $d_R$ uniquely as above and which are satisfied by $d_R$? In physics one generally assumes, that the $t^a$ are normalized as $k(t^a, t^b) = \delta^{ab}$, where $k$ is the killing form. Maybe this does define such relations also on higher order Dynkin Indices?

  • 1
    $\begingroup$ If I am not misunderstanding, I think mathematicians call this the "Dynkin Index". If so, the immediate answer is no: Starting with any non-trivial representation $R$ of $\mathfrak{g}$, one gets two reps of $\mathfrak{g}\oplus\mathfrak{g}$ with the same index: namely, the two projections $\mathfrak{g}\oplus\mathfrak{g}\rightarrow\mathfrak{g}$ followed by $R$. In some sense, this example is trivial; but I don't know of any non-trivial examples. $\endgroup$ Sep 12, 2016 at 22:50


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