# Prove: Lie algebra generated by two $n\times n$ shift matrices is $\mathfrak{so}(n,\mathbb{C})$ ($n$ odd) or $\mathfrak{sp}(n,\mathbb{C})$ ($n$ even)

I wish to have a proof for the following result:

Let $$U_n$$ be an $$n\times n$$ upper shift matrix, and $$L_n = U_n^T$$ be a lower shift matrix. For example, $$U_5 = \begin{pmatrix} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}, \quad L_5 = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \end{pmatrix}.$$ Then the Lie algebra generated by $$U_n$$ and $$L_n$$ over $$\mathbb{C}$$ is isomorphic to $$\mathfrak{so}(n, \mathbb{C})$$ when $$n$$ is odd, and is isomorphic to $$\mathfrak{sp}(n, \mathbb{C})$$ when $$n$$ is even.

I tested this numerically with GAP over the field $$\mathbb{Q}$$: the Lie algebra generated by $$U_{2n+1}$$ and $$L_{2n+1}$$ over $$\mathbb{Q}$$ is of type $$B_n$$, while the Lie algebra generated by $$U_{2n}$$ and $$L_{2n}$$ over $$\mathbb{Q}$$ is of type $$C_n$$.

• For which bilinear forms? For the standard quadratic form $\mathfrak{so}(n,\mathbb{C})$ is the space of skew-symmetric matrices, which does not contain $L_n$ or $U_n$.
– abx
Sep 1, 2021 at 4:29
• @abx That should be part of the question. It should be some weird bilinear form which I could not determine. Sep 1, 2021 at 4:31
• Oh, I see. Maybe you could say that $L_n$ and $U_n$ generate a Lie algebra isomorphic to $\mathfrak{so}(n,\mathbb{C})$ or $\mathfrak{sp}(n,\mathbb{C})$.
– abx
Sep 1, 2021 at 5:13
• @abx Thanks for the suggestion! I've made the change in the text. Sep 1, 2021 at 5:16
• Here's a roadmap to a solution. (a) compute the space $B_n$ of invariant quadratic (for $n$ odd) or alternating forms by $U_n$ and $L_n$. Then $B_n$ should be 1-dimensional and generated by a non-degenerate form (otherwise the conclusion fails), say $q$. So, the Lie subalgebra $g_n$ generated by $\{U_n,L_n\}$ preserves $q$, i.e., is contained in so(q), resp sp(q). (b) Show the latter inclusion is an equality, e.g., computing enough iterated brackets to obtain a good lower bound on the dimension of $g_n$?
– YCor
Sep 1, 2021 at 6:36

Let $${\mathfrak g}$$ be the Lie algebra generated by $$U_n$$ and $$L_n$$. It's easy to check that $$U_n$$ and $$L_n$$ preserve the bilinear form determined by the matrix with $$1,-1,1,\ldots$$ down the antidiagonal and zero elsewhere. So $${\mathfrak g}$$ is contained in (a Lie algebra isomorphic to) $$\mathfrak{so} _n$$ for odd $$n$$, $$\mathfrak{sp} _n$$ for even $$n$$. Let's assume $$n$$ is odd; the argument is the same for even $$n$$. Clearly $${\mathfrak g}=\mathfrak{so} _3$$ if $$n=3$$. In the general case, $$[U_n, L_n]=H_n$$ is a diagonal matrix with just two non-zero entries $$\pm 1$$ and $$U_n-[H_n, U_n]$$, resp. $$L_n+[H_n, L_n]$$ equals $$U_{n-2}$$, resp. $$L_{n-2}$$ in the "middle" matrix subalgebra. By induction, $${\mathfrak g}$$ contains $$\mathfrak{so} _{n-2}$$. Now it's easy to see that $$[H_n, U_n]$$, resp. $$[H_n, L_n]$$ is a multiple of the "missing" positive, resp. negative root element, so in fact $${\mathfrak g}$$ equals $$\mathfrak{so} _n$$.