Showing that $\mathbb{C}^2$ is irreducible as an $\mathfrak{sl}_2(\mathbb{C})$-module

Question

Let $\mathfrak{sl}_2(\mathbb{C}) = \{A \in \operatorname{M}_2(\mathbb{C}) \mid \operatorname{tr}(A)= 0\}$.

How to prove that $\mathbb{C}^2$ is an irreducible $\mathfrak{sl}_2(\mathbb{C})$-module?

Any hint will be appreciated.

Answer 1

You have a wrong definition of $\mathfrak{sl}_2(\Bbb C)$, it should be $\{A \in M_{2}(\Bbb C) : \text{tr}(A) = 0 \}$ .

Notice that $e = \pmatrix{0 & 1 \\ 0 & 0 }$ acts as follows : $e v_1 = 0$ and $ev_2 = v_1$. Similarly, for $f = \pmatrix{0 & 0 \\ 1 & 0 }$ we have $fv_1 = v_2$ and $fv_2 = 0$. Here $v_1 = (1,0)$ and $v_2 = (0,1)$.

From this it is easy to conclude that any $\mathfrak{sl}_2$-module spanned by a single vector $w \in \Bbb C^2 \backslash \{0\}$ should be everything.

Answer 2

Alternatively to the answer of @NicolasHemelsoet, there is an argument which you may view as simpler, but which does not generalize (say to the analogous question for $\mathfrak{sl}(n,\mathbb C)$).

Suppose that there is a non-trivial subspace $V$ in $\mathbb C^2$ that is invariant under the action of $\mathfrak{sl}(2,\mathbb C)$. Then $V$ has to have dimension one, so it is of the form $\mathbb C\cdot v$ for some nonzero vector $v\in\mathbb C^2$. But then for each $A\in\mathfrak{sl}(2,\mathbb C)$ we have to have $Av\in \mathbb C\cdot v$ by invariance, which says that $v$ is an eigenvector for each $A\in \mathfrak{sl}(2,\mathbb C)$. But already the elements $e_1$ and $e_2$ from above do not have a joint eigenvector.