# How many permutations do we need before we're in $SU\left( n\right)$?

Let $\mathcal{L}\subseteq \mathfrak{su}\left( n\right)$ be a Lie algebra for $n \geq 2$ with Lie group $G = e^{\mathcal L}$, and let $X \in G$ be represented by an $n\times n$ matrix (I prefer fixing a basis to work with something tangible).

Suppose $P_1, P_2, \ldots , P_k \in G$ are permutation matrices. What are the necessary conditions on $\left\{ P_k\right\}$ to guarantee that $G = SU\left( n\right)$? Do such conditions even exist?

This question arises in the context of a larger control problem as follows:

Let $n = 4$ and $G \subsetneq SU\left( n\right)$ by assumption. Let the group action on $X_1, X_2 \in G$ be the matrix product $X_1 X_2$.

Suppose we can generate any element of a group $G_1 \subset G$: $${G_1} = SU\left( 3 \right) \oplus 1 = \left\{ {X = \left[ {\begin{array}{*{20}{c}} U&{} \\ {}&1 \end{array}} \right],\,\,U \in SU\left( 3 \right)} \right\}$$

and let us consider the group $G_2 \subset G$: $${G_2} = \left\{ {Y = \left[ {\begin{array}{*{20}{c}} {{y_{11}}}&{{y_{12}}}&{}&{{y_{13}}} \\ {{y_{21}}}&{{y_{22}}}&{}&{{y_{23}}} \\ {}&{}&1&{} \\ {{y_{31}}}&{{y_{32}}}&{}&{{y_{33}}} \end{array}} \right],\,\,\left[ {\begin{array}{*{20}{c}} {{y_{11}}}&{{y_{12}}}&{{y_{13}}} \\ {{y_{21}}}&{{y_{22}}}&{{y_{23}}} \\ {{y_{31}}}&{{y_{32}}}&{{y_{33}}} \end{array}} \right] \in V} \right\},\,\,V \subseteq SU\left( 3 \right)$$

It is easily seen that if we can generate the permutation matrix $${P_{\left( {3,4} \right)}} = \left[ {\begin{array}{*{20}{c}} 1&{}&{}&{} \\ {}&1&{}&{} \\ {}&{}&{}&1 \\ {}&{}&1&{} \end{array}} \right]$$

then we can generate elements of $G_2$ such that $V = SU\left( 3\right)$.

In general, the converse doesn't hold. That is, $V = SU\left( 3\right) \not\Rightarrow P_{\left( {3,4} \right)} \in G$.

We see that $G_1$ "comprises" rows $1, 2, 3$, and $G_2$ comprises rows $1,2,4$. This concepts extends to general $n$, for groups comprising $2 \leq m < n$ rows, where we call such groups "$m$-subgroups" of $G$.

An identical derivation shows that if any such $m$-subgroup $G_a \sim SU\left( m\right)$ can be generated for row tuple $a = \left( a_1, a_2, \ldots , a_m\right)$, along with the permutation matrix $P_{a\rightarrow b}$, this implies that $m$-subgroup $G_b$ for row tuple $b = \left( b_1, b_2, \ldots , b_m\right)$ is also isomorphic to $SU\left( m\right)$. In this way, the $SU$-isomorphism of an $m$-subgroup extends to all $m$-subgroups we can reach via the set of permutations we can generate in $G$. Thus if we know the group $P$ of permutations that can be generated in $G$ (or in particular, a set of permutations $\left\{ P_k\right\}$ that generates this group), this saves us a considerable bit of work in checking each $m$-subgroup for isomorphism.

In the process of systematically determining $\left\{ P_k\right\}$, the issue then arises that we've assumed $G \subsetneq SU\left( n\right)$, hence if the set of $P_1, P_2, \ldots , P_k \in G$ implies $G = SU\left( n\right)$ at any point during the construction, we've violated our assumption. In the context of this control problem, this means that determining $SU$-isomorphism of the $m$-groups is moot and we needn't bother checking.

## 1 Answer

I am confused by your question in ways that don't fit in a comment.

"How many" is a strange question: If $G$ contains two permutations which generate $S_n$, then it contains all of $S_n$. For example, $S_n$ is generated by $(12)$ and $(123\cdots n)$.

Also, $SU(n)$ doesn't contain $S_n$: All elements of $SU(n)$ have determinant $1$, so $SU(n)$ only contains the permutation matrices with even sign. If $Z$ denotes the center of $SU(n)$ (a cyclic group of order $n$), then $SU(n)/Z$ contains $S_n$, so $SU(n)$ contains a $\mathbb{Z}/n \mathbb{Z}$ central extension of $S_n$. Also, $SU(n)$ contains a $(\mathbb{Z}/2\mathbb{Z})^{n-1}$ extension of $S_n$: The signed permutation matrices with determinant $1$.

Moreover, $SO(n)$ contains the even permutation matrices, and contains the signed permutation matrices with determinant $1$.

So, my basic question is this: Whatever you mean by $G$ containing $S_n$, how does it rule out the possibility that $G = SO(n)$?

• The permutation matrices $P_1, P_2, \ldots$ don't necessarily generate $S_n$. There may in fact be only one $P_k$ for a given system. I'm looking for the necessary conditions on the set $\left\{ P_k\right\}$, which is some set of permutations we've determined are members of the Lie group $G$, that allow us to conclude that $G = SU\left( n\right)$. Now I sense from your explanation that $SO\left( n\right)$ is a subgroup of $SU\left( n\right)$, and $S_n$ is a subgroup of $SO\left( n\right)$, hence you're saying that even if the set generates $S_n$, I still can't conclude... – COTO Nov 30 '14 at 15:23
• ...with certainty that $G = SU\left( n\right)$? If so, you've answered my question for me and the bounty is yours. Thank you. (ETA: With the respect to the above, replace $S_n$ with "set of permutations with even sign".) – COTO Nov 30 '14 at 15:25
• Right, $A_n$ (which is the standard notation for the even permutations in $S_n$) is contained in $SO(n)$, which in turn lies in $SU(n)$. Specifically, $SO(n)$ is real matrices with determinant $1$ whose inverse is equal to their transpose, and $SU(n)$ is complex matrices with determinant $1$ whose inverse is their conjugate transpose. – David E Speyer Nov 30 '14 at 19:13
• That's that then. Thank you. The bounty is yours. :) – COTO Dec 1 '14 at 16:36