# Is the physicist's notion of the generalized Dynkin index of a representation of a Lie-Algebra well defined?

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 $$$$d_R^{a_1...a_n}:=\frac{1}{n!} \sum_{\pi \in S_n} \mathrm{tr}\left[t^{a_{\pi(1)}}_Rt^{a_{\pi(1)}}_R...t^{a_{\pi(1)}}_R \right] =: I(R) d^{a_1...a_n} + \mathrm{products~of~lower~orders}$$$$ 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?

• 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. Sep 12, 2016 at 22:50