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(I have already done a similar question, but I did not express myself very well and the question was a bit confusing, so let me try again. If you find the question repetitive, I apologize and you can remove it). Consider a Lie group with 3 generators $J_{1}, J_{2}$ and $J_{3}$ that satisfy $$[J_{a},J_{b}] = i\epsilon_{abc} J_{c}.$$ We can show that $J^{2}$ commutes with all these generators, so we can diagonalize $J^{2}$ and one of them togheter (we choose $J_{3}$. If we call $|j,m \rangle$ a eigenvector of $J_{3}$ and $J^{2}$, we find that $$J^{2}|j,m \rangle = j(j+1)|j,m \rangle,$$ $$J_{3}|j,m \rangle = m |j,m \rangle,$$ where $j = 0, 1/2, 1, 3/2, ...$ and $m = -j, -j+1, ..., j$. I have read that for each value of $j$ is associated a representtion of this group, given by the matrices with elements $\langle j, m | J_{a} |j,m \rangle$. In fact, for $j = 1/2$ we obtein the Pauli matrices, which are related to the generators of $SU(2)$. But how can I show that this matrices form a representation? It's not obvious to me.

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  • $\begingroup$ Possible duplicate by OP: physics.stackexchange.com/q/481141/2451 $\endgroup$ – Qmechanic May 24 at 2:40
  • $\begingroup$ yes, it was to this question that I was referring to when I said that I had already asked a similar question $\endgroup$ – AlfredV May 24 at 12:12
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Simply show that the matrices satisfy the commutation relations that define the algebra. The commutator is defined in terms of matrix multiplication. For matrices $X,Y$, we define $[X,Y]=XY-YX$.

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  • $\begingroup$ Why? To be a representation, shouldn't these matrices have the same multiplification rule of the elements of the group? That is, if A and B are elements of the group, and D(A) and D(B) the matrices associated with these elements, then we should have D(A)D(B) = D(AB). Show that these matrices satisfy the commutation relation is equivalent to this? $\endgroup$ – AlfredV May 24 at 12:10
  • $\begingroup$ These are elements of the algebra, not the group. In any case for $x,y$ in the algebra there is usually no meaning for $xy$ so the requirement that $D(A)D(B)=D(AB)$ makes no sense. Our notion of multiplication on the algebra is commutation. $\endgroup$ – user26872 May 24 at 12:19
  • $\begingroup$ I'm sorry, I am confused. What's the definition of representation? What's the relation of the algebra with the group? $\endgroup$ – AlfredV May 24 at 13:40
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    $\begingroup$ Roughly, the algebra and group are related through exponentiation. $\endgroup$ – user26872 May 24 at 19:12

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