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Does the Lie-Algebra of $SU(2)$ always have 3 generators? Because I'm reading about different representations of $SU(2)$ as $2\times2$, $3\times3$, $4\times4$ matrices etc. But I guess that even if we are looking at $7\times7$ matrix representation of $SU(2)$ we only need $3$ such matrices corresponding to the $3$ generators right?

And if we have a representation by say $7\times 7$ matrices, is it not possible by using similarity transformations to find an $3-$dimensional invariant subspace corresponding to these $3$ linear operators? Then what point is it to look for higher representations of $SU(2)$ then just $3\times 3$ real matrices?

Torsten Schoeneberg
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Higgsino
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    If I understand you correctly, you are correct in your first guess: the image of the representation is (unless it's trivial) isomorphic to $SU(2)$ (and the analogous fact would hold true for any simple Lie algebra) and hence has three generators. However, with your final question, you seem to be basically asking what is the point of representation theory, if the represented group or Lie algebra is already given as a matrix group or Lie algebra?. That's a valid but very broad question which has nothing to do with $SU(2)$ in particular. – Torsten Schoeneberg Jan 05 '19 at 05:34
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    And let me add to my above comment that that exact question was asked, and got some very insightful answers, on MathOverflow: https://mathoverflow.net/q/153740/27465 – Torsten Schoeneberg Jan 05 '19 at 19:06
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    It is not true that every $7$-dimensional representation of SU$(2)$ has a $3$-dimensional invariant subspace. SU$(2)$ has an irreducible $7$-dimensional representation. – Andreas Blass Jan 06 '19 at 02:58
  • ok, so given any set of 3 linearly independent 7x7 matrices, it is not true that you by a change of coordinates can find an invariant subspace with a lower dimension then 7? I find this interesting because then it makes a big difference by how you choose your set of 3 linearly independent matrices. Because I could obviously choose them such that there is an invariant 3 dimensional subspace which they act on. – Higgsino Jan 06 '19 at 22:38
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    Higgsino, to find an invariant subspace for the generators of a representation means to find a subrepresentation. Given that representations of semisimple Lie algebras are completely reducible, saying that one is irreducible is equivalent to it has no proper subrepresentation / there is no proper invariant subspace. – Torsten Schoeneberg Jan 07 '19 at 21:10

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