The group $SU(2)$ consists of $2\times2$ unitary matrices with determinant $1$ can be put in the form:
$$U=\begin{pmatrix}a &b \\ c & d\end{pmatrix}$$
By invoking the conditions: $det(U)=1$ and $UU^\dagger=1$, we reduce the matrix $U$ to $$U=\begin{pmatrix}a &b \\ -b^* & a^*\end{pmatrix}$$
This is a 2 dimensional representation of $SU(2)$ group.
Now, consider a three dimensional representation:
$$U=\begin{pmatrix}a &b&c \\ d & e &f \\ g&h&i\end{pmatrix}$$
and then invoke the same conditions $det(U)=1$ and $UU^\dagger=1$.
Isn't this just the same as the 3 dimensional representation for $SU(3)$ group?
I know this in't correct, but I don't know what's wrong.
Best Answer
If you are a "physics lover" as your username suggests, then I suggest you think about it like this:
Any $SU(2)$ matrix can be written in the form $$ U = \exp \left(i \theta_x J_x + i \theta_y J_y + i \theta_z J_z \right),$$ where $\theta_x, \theta_y, \theta_z$ are real numbers and $$J_x = \frac 1 2 \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right), \ \ \ J_y = \frac 1 2 \left( \begin{array}{cc} 0 & -i \\ i & 0 \end{array} \right) \ \ \ J_z = \frac 1 2 \left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right).$$ Hopefully you recognise $J_x, J_y, J_z$ as the angular momentum operators for spin-$1/2$ particles.
To get the three-dimensional representation, you still write $U$ in the form$$ U = \exp \left(i \theta_x J_x + i \theta_y J_y + i \theta_z J_z \right),$$ using the same $\theta_x, \theta_y, \theta_z$. But now, you change $J_x, J_y, J_z$ to $$J_x = \left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & -i \\ 0 & i & 0 \end{array} \right), \ \ \ J_y = \left( \begin{array}{ccc} 0 & 0 & i \\ 0 & 0 & 0 \\ -i & 0 & 0 \end{array} \right), \ \ \ J_z = \left( \begin{array}{ccc} 0 & -i & 0 \\ i & 0 & 0 \\ 0 & 0 & 0 \end{array} \right).$$ These are the angular momentum operators for spin-$1$ particles.
To construct the $n$-dimensional irreducible representation of $SU(2)$, use the spin $j = (n-1)/2$ angular momentum matrices as $J_x, J_y, J_z$.