In Woit's: Quantum Theory, groups and representations, p109, 110, is proved the "Highest Weight Theorem":

Finite dimensional irreducible representations of $SU(2)$ have weights of the form $$-n,-n+2,\dots,n$$ for a non-negative integer, each with multiplicity 1, with $n$ a highest weight.

There is some part in the proof which I don't understand. One starts with the fact that there exists a highest weight $n$ and pick $v\in V_n$. Next we compute $$v_{n-2j}= \pi'(S_-)v_n\in V_{n-2j},$$

where $\pi'$ is the Lie algebra homomorphim between sl(2,$\mathbb{C}$) and $\mathrm{u}_{\mathbb{C}}(m)$ and $V_n$ is the "weight space". Now the proof says:

Consider the span of the $v_{n-2j}=\pi'(S_-)^jv_n \in V_{n-2j}$. To show that this is a representation one needs to show that the $\pi'(S_3)$ and $\pi'(S_+)$ leave it invariant.

What does it mean the author with "leave it invariant"? The proof goes on to find, as usual, that there is a minimum value of the weight equal to $-n$ but it is not clear what space is left invariant.

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    $\begingroup$ Is your question just about terminology? An operator $A$ leaves a space $B$ invariant, means $Ab\in B$ for all $b\in B$. i.e., if you apply $A$ to elements of $B$, you do not leave the space $B$, you get other elements of $B$. $\endgroup$ Jul 28, 2021 at 11:33
  • $\begingroup$ Yes and no... I know what invariant means, but here $\pi'(S_\pm)$ will change a vector from, say $v\in V_n$ to $V_{n\pm2}$, so I don't see what is what is left invariant, if I jump between subspaces instead of always staying in $V_n$ (as the $\pi'(S_z)$ does). $\endgroup$ Jul 28, 2021 at 11:55
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    $\begingroup$ leaves invariant the span, not the vector. $\endgroup$ Jul 28, 2021 at 12:53
  • $\begingroup$ Ok, I think this solves the issue.... (I thought $V_n$ was meant the invariant space, but it's something "bigger"). $\endgroup$ Jul 28, 2021 at 13:26

1 Answer 1


The Lie Algebra of $SU(2)$ is spanned by three elements: $S_x, S_y, S_z$. Alternatively, it is spanned by $S_+, S_-, S_z$ (this just a different basis). Note that by exponentiating Lie algebra elements, you get a group representation. (Because $SU(2)$ has a trivial fundamental group, you won't get any topological obstructions to this.) Now, if you act on any vector in a representation by a group (or Lie algebra) element, you get another vector in that representation. So, if you take any vector, and act on it by the orbit of every group element, that vector space you get will be the vector space of a representation of your group. Likewise, if you have a collection of vectors which are mapped into each other under the action of $S_+, S_-,$ and $S_z$, then any group element (we just get by exponentiation of linear combinations of $S_+, S_-, S_z$) will also map these vectors into each other.

We are essentially just saying that the vector space spanned by these vectors is closed under the group action.


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