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The quantities $\delta_{ij}a_ib_j$ and $\epsilon_{ijk}a_ib_jc_k$ are invariant under the transformation of the $j=1$ (fundamental) representation of $SO(3)$. What would be the analogous expressions for the $j=1/2$ $SU(2)$ representation?

Attempt: The canonical definition of these groups are $$SO(3) = \left\{O \in GL(3, \mathbb{R}) : O^TO = 1, \det O = 1 \right\}\,\,\,\,\\SU(2) = \left\{U \in GL(2,\mathbb{C}) : U^{\dagger}U = 1, \det U = 1\right\}$$

Since the invariant in the rep $j=1$ in $SO(3)$, $\epsilon_{ijk}a_ib_jc_k$, only makes sense for $3D$ vectors, I am not really sure how to generalize this to $2D$. Can I say the invariant is the same in both cases, with the realization that I have simply a zero entry for the third component. (i.e just write $(a, b)$ in $2D$ is equivalent to $(a,b,0)$ in $3D$.

Thanks.

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The crucial difference between $\mathrm{SO(}N\mathrm{)}$ and $\mathrm{SU(}N\mathrm{)}$ is that for $\mathrm{SU(}N\mathrm{)}$ one can raise and lower indices, while for $\mathrm{SO(}N\mathrm{)}$ this is irrelevant. See also these notes for more information.

For $\mathrm{SU(}N\mathrm{)}$ we have: $$ U^\dagger U = 1 $$ which in component form is: $$ U^j{}_i U_j{}^k = \delta^k{}_i \;\;\; \text{and} \;\;\; U_i{}^j U^k{}_j = \delta^k{}_i $$ Thus: $$ \varphi_i \varphi^i \to \varphi'_i \varphi'^i = U_i{}^j U^i{}_k \varphi_j \varphi^k = \delta^j{}_k \varphi_j \varphi^k=\varphi_i \varphi^i $$ and: $$ \delta^{i}{}_{j} \to \delta'^{i}{}_{j} = U^i{}_k U_j{}^l \delta^k{}_l = U^i{}_k U_j{}^k = \delta^i{}_j $$ Furthermore: \begin{equation} \begin{aligned} \varepsilon^{i_1 i_2 \dots i_N} \to \varepsilon'^{i_1 i_2 \dots i_N} & = U^{i_1}{}_{j_1} U^{i_2}{}_{j_2} \cdots U^{i_N}{}_{j_N} \varepsilon^{j_1 j_2 \dots j_N} \\& = \mathrm{det}(U)\varepsilon^{i_1 i_2 \dots i_N} \\& = \varepsilon^{i_1 i_2 \dots i_N} \end{aligned} \end{equation} and using a similar argument one can show that: \begin{equation} \varepsilon_{i_1 i_2 \dots i_N} \to \varepsilon'_{i_1 i_2 \dots i_N} = \varepsilon_{i_1 i_2 \dots i_N} \end{equation}

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  • $\begingroup$ Do you mean to say that the invariant quantities in the $j=1/2$ rep are the same? (i.e the Kronecker delta and the Levi-Civita tensor are both invariant under any transformation), so the corresponding invariants for the $SU(2)$ rep is exactly $\delta_{ij}a_ib_j$ and $\epsilon_{ijk}a_ib_jc_k$? $\endgroup$ – CAF Mar 27 '14 at 8:53
  • $\begingroup$ In the line $...U^i_{\,k} U_j^{\,k} = \delta^i_{\, j}$, where are you using the fact that it is the condition $U^{\dagger}U = 1$ that holds rather than simply $U^TU=1$? $\endgroup$ – CAF Mar 27 '14 at 9:11
  • $\begingroup$ @CAF Look at the notes, page 14, I have attached (although it the conventions I've used in my message above are slightly different, you will get the idea). $\endgroup$ – Hunter Mar 27 '14 at 14:55
  • $\begingroup$ I did look at the notes, but it is not clear how the condition $U^{\dagger}U = 1$ is used. Seems to me it would work without this condition (i.e it would work with simply having $U^TU=1$) Also, the dimension of SU(2) is three, and given that the Kronecker and Levi-Civita tensors transform analogously under $U \in SU(2)$, are the invariants the same? Thanks for your help! $\endgroup$ – CAF Mar 27 '14 at 15:15
  • $\begingroup$ I am not sure about this last statement because an explicit representation of the generators in $SU(2)$ are scaled Pauli matrices which are $2x2$. So the matrix product $U a_i$ say, would only make sense if $a_i$ was 2D. And, in that case, I am not sure how to generalise the $SO(3)$ invariant $\epsilon_{ijk}a_ib_jc_k.$ Thanks for your help! $\endgroup$ – CAF Mar 27 '14 at 15:35

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