# What is this Lie algebra?

Consider two matrices $A,B \in \mathfrak{su}(N)$ which are both diagonal in the standard basis and non-zero.

If we consider the new matrix $\tilde{B} := FBF^{\dagger}$ where $F$ is the `quantum' fourier transform matrix:

$$F_N = \frac{1}{\sqrt{N}} \begin{bmatrix} 1&1&1&1&\cdots &1 \\ 1&\omega&\omega^2&\omega^3&\cdots&\omega^{N-1} \\ 1&\omega^2&\omega^4&\omega^6&\cdots&\omega^{2(N-1)}\\ 1&\omega^3&\omega^6&\omega^9&\cdots&\omega^{3(N-1)}\\ \vdots&\vdots&\vdots&\vdots&&\vdots\\ 1&\omega^{N-1}&\omega^{2(N-1)}&\omega^{3(N-1)}&\cdots&\omega^{(N-1)(N-1)} \end{bmatrix}$$ (where $\omega=\exp(2i\pi/N)$, see the wikipedia page on Quantum Fourier Transform)

what are the possibilities for the Lie algebra generated (i.e. all linear combinations of all bracket expressions) as $\left<A, \tilde{B}\right>_{Lie}$? I.e. what algebras can be generated this way by varying $B$ while keeping it diagonal.

Specifically, will the Lie group associated to whatever algebra is generated contain a subgroup isomorphic to any non-trivial Clifford group?

• I'm confused as $\mathfrak{su}(N)$ refers to some specific basis, as well as the matrix $F_N$, while "diagonal in the same basis" possibly refers to some possibly other (and possibly not orthogonal) basis. The question already makes sense when $A,B$ are diagonal in the standard basis, is this what you mean? Second, do you mean the real Lie subalgebra generated by $A,\tilde{B}$? – YCor Apr 14 '16 at 18:27
• Amended accordingly. – Benjamin Apr 14 '16 at 18:34

I can't see your question having a completely general answer. In the 'generic' case, I think the subalgebra generated by $A$ and $\tilde{B}$ is just $\mathfrak{su}(N)$.
To see this, first of all note that $\tilde{B}$ is of the form $$\begin{bmatrix} 0 & b_1 & b_2 & \dots & b_{N-1} \\ b_{N-1} & 0 & b_1 & \dots & b_{N-2} \\ b_{N-2} & b_{N-1} & 0 & \dots & b_{N-3} \\ \vdots & \vdots & \vdots & \dots & \vdots \\ b_{1} & b_{2} & b_{3} & \dots & 0 \end{bmatrix}$$ with $b_{N-i}=-\overline{b_i}$ for $1\leq i\leq N-1$, and all such matrices are possible. Let $A$ have diagonal entries $a_1{\rm i},\ldots ,a_N{\rm i}$ (where the $a_i$ are real and sum to zero). In the generic case, the square differences $(a_i-a_j)^2$ for $i<j$ are all distinct. Then the subspace of elements $x$ of $\mathfrak{su}(N)$ satisfying $[A,[A,x]]=-(a_i-a_j)^2 x$ is two-dimensional, spanned by $e_{ij}-e_{ji}$ and ${\rm i}(e_{ij}+e_{ji})$, where $e_{ij}$ denotes the matrix with $1$ in the $(i,j)$ position, and $0$ elsewhere.
Generically, the $b_i$ are all non-zero, and hence $\tilde{B}$ is supported on this subspace of elements satisfying $[A,[A,x]]=-(a_i-a_j)^2 x$, specifically the component of $x$ is $$b_{j-i}e_{ij}-\overline{b_{j-i}}e_{ji}=\beta(e_{ij}-e_{ji})+\gamma{\rm i}(e_{ij}+e_{ji})$$ where $\beta,\gamma$ are real (and at least one is non-zero). Since the $(a_i-a_j)^2$ are distinct, it follows that $\beta(e_{ij}-e{ji})+\gamma{\rm i}(e_{ij}+e_{ji})$ belongs to $\langle A,\tilde{B}\rangle$. Now since $$[A,\beta(e_{ij}-e_{ji})+\gamma{\rm i}(e_{ij}+e_{ji})]=(a_i-a_j)(-\gamma(e_{ij}-e_{ji})+\beta{\rm i}(e_{ij}+e_{ji}))$$ and since we cannot have $\gamma/\beta=-\beta/\gamma$, then $\langle A,\tilde{B}\rangle$ contains both $e_{ij}-e_{ji}$ and ${\rm i}(e_{ij}+e_{ji})$.
Finally, $[e_{ij}-e_{ji},{\rm i}(e_{ij}+e_{ji})]=2{\rm i}(e_{ii}-e_{jj})$ is also in $\langle A,\tilde{B}\rangle$.