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Let the entities $\psi^i$ transform as the fundamental representation of $\mathrm{SU}(n)$, denoted by ${\bf n}$: $$ \psi^{\prime i}=U^{i}_{~j}\psi^j, $$ where, of course, $U$ represents $n\times n$ unitary matrices with unit determinant. Taking complex conjugates on both sides, we get, $$ \left(\psi^{\prime i}\right)^{*}=\left(U^{i}_{~j}\right)^{*}(\psi^j)^{*}. $$ Introducing the notation $$ \psi_i=\left(\psi^i\right)^{*}, \quad U_{i}^{~j}=\left(U^{i}_{~j}\right)^{*},\tag{1} $$ we have $$ \psi^\prime_i=U_{i}^{~j}\psi_j. $$ Therefore, by definition, $\psi_i=(\psi^i)^{*}$, which transform as $\bar{\bf n}$.


On the other hand, if I am not wrong, for the group $\mathrm{SU}(2)$, $\psi_i$ is also defined to be equal to $$\psi_i=\varepsilon_{ij}\psi^j.\tag{2}$$

How can we have two different definitions [in (1) and in (2)] for the same object $\psi_i$?

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OP's question is closely related to the fact that

are inequivalent representations, except for $n=2$ due to the existence of the invariant 2D Levi-Civita symbol $\epsilon$, cf. the following identity $$(\rho(g)^{-1})^T~=~\epsilon \rho(g)\epsilon^{-1},\qquad g~\in~SU(2).\tag{3}$$

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$^1$Note that the dual/contragredient/transposed representation (2) is equivalent to the complex conjugate representation $\bar{\rho}$ due to unitarity.

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  • $\begingroup$ Right. But my problem is somewhat more basic. It is true that for SU(2) group, $(\psi^i)^*=\varepsilon_{ij}\psi^j$. I don't think this is true because $\varepsilon_{ij}$ is just a real antisymmetric matrix and I am making a mistake. $\endgroup$ Commented Sep 25, 2023 at 16:19
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    $\begingroup$ Can you expand a bit more on the connections? I'm familiar with all the linked concepts, but I don't see the implications between them. Maybe the core claim is that the fundamental representation and its complex conjugate are generically not (and should not be expected to be) equivalent, but that SU(2) is an exception? If so, how does the Levi-Civita work in connecting the two? $\endgroup$ Commented Sep 25, 2023 at 16:22
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Commented Sep 26, 2023 at 7:21

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