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How can I show that the the action of Lie algebra $\mathfrak{so}(2n+1)$ on $\mathbb{C}^{2n+1}$ is irreducible? Is there a simple way?

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up vote 1 down vote accepted

Hint: Show that the action of the Lie algebra is enough to move between the standard basis vectors of $\mathbb C^{2n+1}$; is this enough to show the only submodule is the whole space?

Hint+: think in terms of matrix units

Complete Answer:

Let $E_{ij}$ be the matrix unit with $1$ in the $(i,j)$ entry and $0$ elsewhere. Let $v_0, v_1,\ldots, v_n, v_{-1}, \ldots, v_{-n}$ be the standard basis of $\mathbb C^{2n+1}$ and consider the matrix description of $\mathfrak{so}(2n+1)$, i.e. the Lie algebra of matrices solving the equation $A^tM+MA=0$ for $$M=\begin{bmatrix}1&0&0\\ 0&0&I_n\\ 0&I_n&0 \end{bmatrix} $$. One can show the Lie algebra contains the matrices $E_i=E_{i,i+1}-E_{-i-1,-i}$ for $0<i<n$, $E_0=E_{n,0}-E_{0,-n}$, and as well as their transposes $F_i, F_0$. But then by inspection, we see that by acting on a basis element $v_i$ by some sequence of these matrices we can obtain a multiple of any other basis element $v_j$. (Ex. for $n=2$, $E_1E_2E_0v_0=v_1$). Then given any nonzero vector $v$, we can systematically act by these $E$ matrix elements until we obtain $v_1$, from which we can generate all basis vectors using the $F$ 's, hence every cyclic submodule is the whole space; i.e., it is irreducible.

PS. This argument is just the $\mathfrak{so}$-analogue of the same (but perhaps easier to parse) argument for why $\mathbb C^n$ is irreducible for $\mathfrak{sl}_n$, and indeed this sort of argument works for all the "natural" representations of the matrix Lie algebras.

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