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In the convention, where the Dynkin index $Tr(T_a T_b)$ of the lowest-dimensional representation is $\frac{1}{2} \delta_{ab}$, how can I normalize a given set of matrices properly?

For example, given the adjoint of $SU(2)$ written in terms of the Pauli matrices

$$ \sigma_1= \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \\ \end{array} \right) , \sigma_2= \left( \begin{array}{cc} 0 & -i \\ i & 0 \\ \end{array} \right), \sigma_3= \left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \\ \end{array} \right),$$

we have $Tr(\sigma_i \sigma_i)=2$. Does this mean the Pauli matrices are the correct normalized matrices for the adjoint representation, because the Dynkin index for the adjoint of $SU(2)$ is $2$?

Another example, would be the adjoint ($15$ dimensional) representation of $SU(4)$, which can be written in terms of $\lambda_i$ matrices as given in this paper (The Lie algebra su(4) Walter Pfeifer). These matrices satisfy $Tr(\lambda_i \lambda_i )=2$. Does this mean that these matrices must be multiplied by $\sqrt{2}$ in order to get the correctly normalized generators of $SU(4)$, because the Dynkin index is $4$?

A third, more different example, is the $10$-dimensional representatin of $SU(4)$, which consists of all symmetric, traceless $4\times 4$ matrices. A possible basis choice is

$$s(1)=\left( \begin{array}{cccc} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right);s(2)=\left( \begin{array}{cccc} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right);s(3)=\left( \begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ \end{array} \right); $$ $$s(4)=\left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right);s(5)=\left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ \end{array} \right);s(6)=\left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ \end{array} \right); $$ $$ s(7)= \sqrt{2}\left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right);s(8)= \sqrt{2}\left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right);s(9)=\sqrt{2}\left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right); $$ $$s(10)=\sqrt{2} \left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right), $$

which satisfy $Tr(s_i s_j)= 2 \delta_{ij} $, too. The Dynkin index for the $10$-dimensional representation of $SU(4)$ is $3$ and therefore, are the correctly normalized matrices here these matrices multiplied by $\frac{\sqrt{3}}{\sqrt{2}}$.

Any ideas would be much appreciated!

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There is no such a concept like "correct normalization" in Lie algebra representations. You just have to be consistent. By the way, the most used normalization for the adjoint of SU(2) is $$R(T_a)=\frac{1}{2}\sigma_a,$$ which gives $$[R(T_a),R(T_b)]=if_{abc}R(T_c)=i\epsilon_{abc}R(T_c),$$ and $$Tr(R(T_a)R(T_b))=\frac{\delta_{ab}}{2}.$$

Suppose now we use another normalization: $$R(T_a)\rightarrow \tilde R(T_a)=2R(T_a),$$ which is exactly the normalization you mentioned. The trace gives $$Tr(\tilde R(T_a)\tilde R(T_b))=2\delta_{ab}.$$ However we also need to renormalize the structure constants: $$[\tilde R(T_a),\tilde R(T_b)]=4i\epsilon_{abc}R(T_c)=i\tilde f_{abc}\tilde R(T_c).$$ Hence $$\tilde f_{abc}=2f_{abc}.$$

In general, the Dynkin index is given by $$I=\frac{dim(R)}{dim(g)}C_2(R),$$ where dim(R) and dim(g) are the dimensions of the representation and of the algebra, respectively, and C_2(R) is the eigenvalue of the second Casimir operator. It can be show this number depends on the normalization of the matrix. In our expample, it will change by the factor $2^2$.

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  • $\begingroup$ Thank you for this great answer. In both your examples the metric is not exactly Euclidean, and so the position of indicies should matter (raised and lowered should differ by a factor of 2). Do you know why every text on this topic overlooks this and treats the indicies as Euclidean? Thanks! $\endgroup$
    – Craig
    Nov 29, 2021 at 15:49
  • $\begingroup$ @Craig ?? the factor is a trivial convention normalization absorbable in the definition of the algebra elements in a representation R. The indices are Euclidean. $\endgroup$ Nov 29, 2021 at 16:35
  • $\begingroup$ @Cosmas Zachos : But people rarely actually do absorb the factor into the basis. Without doing so, the space is not Euclidean. Perhaps you could provide a more in depth answer to my question here: math.stackexchange.com/questions/4318891/… $\endgroup$
    – Craig
    Nov 29, 2021 at 17:04
  • $\begingroup$ @Craig Switching normalizations is the bread and butter of physics; you think this is a far-from-obvious point deserving serious discussion? $\endgroup$ Nov 29, 2021 at 17:12
  • $\begingroup$ @Cosmas Clearly I do not fully understand something so I'm inclined to say yes. I would like to see a justification as to why treating the structure constants as Euclidean does not lead to objects which are no longer invariant under a change of basis of the lie algebra, just as ignoring indices in special relativity leads to objects like $x_{\alpha}x_{\alpha} = t^2+x^2$ which are not actually scalars. $\endgroup$
    – Craig
    Nov 29, 2021 at 17:22

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