I have read in a book that the group $\mathrm{SU}(2)$ is one of the irreducible representations of the rotation group. The book begin saying that the rotation group has 3 generators $J_{1}, J_{2}$ and $J_{3}$, and the algebra is $$[J_{i},J_{j}] = i \epsilon _{ijk} J_{k}.$$ After, considering the Casimir operator $J^{2} = J_{1}^{2} + J_{2}^{2} + J_{3}^{2}$ the book finds the eigenvalues of $J^{2}$ and $J_{3}$ in the same basis: $$J^{2} |j,m \rangle = j(j + 1) |j,m \rangle ,$$ $$J_{3} |j,m \rangle = m |j,m \rangle ,$$ where $j = 0, 1/2, 1, 3/2, \ldots$ and $m = -j, -j+1,\ldots, 0, \ldots , j-1, j$. It says that for each value of $j$ corresponds one irreducible representation, for $j = 1/2$ this representation is the group $\mathrm{SU}(2)$, for $j = 1$ this is the $\mathrm{SO}(3)$, and so on...

But the book doesn't prove that this is in fact a representation. How can I do this?

And what is this rotation group? How is it defined?

Edit: I have read it in the book Matemática para físicos com aplicações by João Neto, vol. 1 (the book is in Portuguese). The book says "when two groups have the same algebra, we say that these are not two distinct groups, but different representations of the same group". So he concluded that $\mathrm{SU}(2)$ and are $\mathrm{SO}(3)$ are representations of the same group, which it call of rotation group.

  • $\begingroup$ Related: physics.stackexchange.com/q/96045/2451 , physics.stackexchange.com/q/106102/2451 , physics.stackexchange.com/q/96542/2451 and links therein. $\endgroup$ – Qmechanic May 19 '19 at 21:03
  • $\begingroup$ I wonder if the statement should be that the Lie algebra $\hbox{su}(2)$ is a representation of the rotation group $SO(3)$. $\endgroup$ – user52817 May 19 '19 at 21:39
  • $\begingroup$ @user52817 It should not. A Lie algebra and a group representation are two different things. As Emilio Pisanty explains, the relationship is that SU(2) is the “covering group” of SO(3). Both are groups. Neither is a Lie algebra nor a group representation. Both have Lie algebras and group representations. $\endgroup$ – G. Smith May 19 '19 at 22:50
  • $\begingroup$ @G.Smith My guess is that user52817 was referring to the fact that the structure constants do in fact give a representation, viz. the adjoint. (But this is a rep. of the algebra, not the group, so the claim is indeed wrong). $\endgroup$ – AccidentalFourierTransform May 20 '19 at 0:06
  • $\begingroup$ @AccidentalFourierTransform It isn’t kosher to consider the structure constants of a Lie algebra to “be” the Lie algebra itself, is it? $\endgroup$ – G. Smith May 20 '19 at 0:10

when two groups have the same algebra, we say that these are not two distinct groups, but different representations of the same group

This is bonkers. When two groups have the same algebra, then we say that they have the same algebra, i.e. that they are locally isomorphic. We do not say that they are the same group; we only say this when there is a group isomorphism between the two, and this is not the case with $\rm SU(2)$ and $\rm SO(3)$.

Moreover, that quote (if the translation is accurate) is abusing its notation to a pretty remarkable degree, particularly as regards the term "representation". Within this context, the term "representation" of a group $G$ should only be used in this specific technical sense, i.e. to refer to a vector space $V$ and to the linear action $R:G\to \mathrm{End}(V)$ on it.

The author seems to be getting at some form of "sameness" between $\rm SU(2)$ and $\rm SO(3)$, which does exist:

  • the two groups are locally isomorphic, and
  • more globally, $\rm SU(2)$ is the covering group for $\rm SO(3)$.

However, there is no standard sense in which the two can be considered "representations" of some other, "more abstract" group. If your translations are accurate, then the author is drawing outside the lines. It's not quite serious enough that I would tell you to drop the book immediately and find one which isn't wrong, but it's definitely a strong warning sign $-$ take everything on that book with a grain of salt, and start looking for a good replacement.

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