I am interested in understanding the irreducible representations for the $\mathrm{SU}(3)$ group and, more generally, the $\mathrm{SU}(N)$ group. My initial question was as follows:
- Is there a $2$-dimensional irreducible representation for the $\mathrm{SU}(3)$ group?
The irreducible representation of $\mathrm{SU}(3)$ are indexed by $(n, m)$, with $n \geq m \geq 0$, and for an irreducible representation $V_{(n,m)}$, the dimension can be expressed as $$ \dim V_{(n,m)} = \frac{1}{2}(n+1)(m+1)(n+m+2). $$ The first few dimensions are $1,3,3,6,8,10,6,15,15, \ldots$.
I am curious if this result extends to other values of $N$. Specifically, is it true that:
- For any $N \geq 1$, there is no irreducible representation of $\mathrm{SU}(N)$ with a dimension between $1$ and $N$.
Additionally, I would like to know if this statement holds true for other matrix groups as well.
Your guess is correct for $\mathrm{SU}(n)$. This can be answered of course by using the (nontrivial) Weyl dimension formula. Below is a proof without using WDF but only uses more elementary facts in highest weight theory.
First let me list the relevant notations and facts we need. Let $G$ be a connected semisimple compact Lie group, let $\mathfrak g$ be the complexification of the Lie algebra of $G$, and let $\mathfrak h$ be a Cartan subalgebra. Write $\Sigma$ for the root system of $(\mathfrak g, \mathfrak h)$, write $W$ for the Weyl group of $\Sigma$, and choose a set of positive roots $\Sigma^+$. This defines the set of fundamental weights $\omega_1,\ldots, \omega_k$, the lattice of dominant weights, and its interior $$\bar P_+ = \Big\{\sum_{i=1}^n c_i \omega_i \mid c_i \in \mathbb Z_{\ge 0} \Big\},\quad P_+ = \Big\{\sum_{i=1}^n c_i \omega_i \mid c_i \in \mathbb Z_{>0} \Big\}.$$ For each $\lambda \in P_+$, its centralizer $W^\lambda$ in $W$ is trivial. For $\lambda \in \bar P_+$, $W^\lambda$ starts becoming nontrivial, and in general if you have more zero coefficients $c_i$, $W^\lambda$ is larger. In the extreme case $\lambda = 0$ all $c_i$'s are zero, so $W^\lambda = W$. The next extreme case is if you have only one nonzero $c_i$, i.e. $\lambda = \omega_i$. All other $\lambda$ have strictly smaller $W^\lambda$ compared to the first two cases.
Each finite dimensional irreducible representations of $G$ is a finite dimensional representation $U$ of $\mathfrak g$ (by connectedness of $G$, I think this rep of $\mathfrak g$ is itself irreducible, but we do not need this), and any irreducible subrepresentation $V$ of $U$ is unique determined by an element $\lambda$ in $\bar P_+$, called the highest weight of $V$ (in fact $V$ is a direct summand of $U$; again we don't need this). The set of weights (i.e. $\mathfrak h$-eigenvalues) in $V$ is stable under the action of $W$ on $\mathfrak h^*$.
In our case, $G = \mathrm{SU}(n)$, $\mathfrak g = \mathfrak{sl}(n,\mathbb C)$, $W \cong S_n$ the symmetric group. The stabilizer $W^{\omega_i}$ of $\omega_i$ is isomorphic to $S_i \times S_{n-i}$.
Now let $\lambda$ be in $\bar P_+$ and let $V$ be the corresponding finite dimensional representation. Since the set of weights in $V$ is stable under $W$, $W \cdot \lambda$ is a subset of weights, and so we have at least $|W \cdot \lambda|$-many distinct weights in $V$, and hence $V$ is at least $|W \cdot \lambda|$-dimensional. Now $$|W \cdot \lambda| = |W/ W^\lambda| = |W|/ |W^\lambda| = |S_n| / |W^\lambda| = n^!/|W^\lambda|.$$ The smallest number this can be is of course $1$ -- this is when $W^\lambda = W$ and $\lambda = 0$, corresponding to the trivial representation. The next smallest number is when $\lambda = \omega_i$ is a fundamental weight (because for other $\lambda$'s the stabilizer $W^\lambda$ is strictly smaller than the case $\lambda = \omega_i$ for some $i$), in which case $|W^\lambda| = |S_i \times S_{n-i}| = i!(n-i)!$, and $$|W \cdot \omega_i| = \frac{n!}{i!(n-i)!} = \begin{pmatrix} n\\i \end{pmatrix},$$ the binomial coefficient. The smallest number among these is $\begin{pmatrix} n\\1 \end{pmatrix} = n$. Therefore there are at least $n$-dimensions in a nontrivial representation $V$ and hence in $U$. This proves your claim.