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Hello

This question somehow is related to a previous question I asked here. Let me set some notations. Let $\mathbb{F}_p$ be a finite field with $p$ element. Consider the following symmetric matrix

\begin{equation} J:=\begin{pmatrix} 0 & I_n \\\\ I_n & 0 \end{pmatrix}, \end{equation}

where $I_n$ is the identity matrix. Now the special orthogonal group is defined by

$$SO_{2n}(\mathbb{F}_p):=\{A\in SL_n(\mathbb{F}_p): AJA^T=J \}.$$

Obviously, when $\sigma,\tau\in M_n(\mathbb{F}_p)$ are skew-symmetric the following matrices belongs to the special orthogonal group. \begin{equation} \begin{pmatrix} I_n & \sigma\\\\ 0 & I_n \end{pmatrix}, \begin{pmatrix} I_n & 0\\\\ \tau & I_n \end{pmatrix}\in SO_{2n}(\mathbb{F}_p) \end{equation}

I have done some computations to show that these matrices generate $SO_{2n}(\mathbb{F}_p)$. But now, based on the question I have asked, I might have done some mistakes in my computations. Is it true that matrices mentioned above are a generating set?

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In even characteristic the set of invertible $A$ s.t. $AJA^\top=J$ would give you the symplectic group rather than an orthogonal group. –  Dima Pasechnik Jul 15 '12 at 12:12

1 Answer 1

When $n=1$ then your matrices $\sigma$ and $\tau$ must be zero (since they are skew-symmetric), and hence your two generators are equal to one. But $-id\in SO_{2n}(\mathbb F_p)$, so the group is not actually trivial.

But even if $n>1$ there is nothing that keeps you from choosing $\sigma=\tau=0$. So maybe you want to at least consider all matrices of the given form.

Edit: So, following the comments, I now assume that you let $\sigma$ and $\tau$ range over all skew-symmetric matrices (instead of just picking two; however it still suffices to let them range over a basis). Still, for $n=2$, the group generated by the matrices from the question is isomorphic to $SL_2(\mathbb F_p)$. Now one just has to compare orders to see that this isn't $SO_{4}(\mathbb F_p)$. Or just use GAP:

gap> A := One(GF(5))*[[1,0,0,0],[0,1,0,0],[0,-1,1,0],[1,0,0,1]];;
gap> B := One(GF(5))*[[1,0,0,1],[0,1,-1,0],[0,0,1,0],[0,0,0,1]];;
gap> G := Group(A, B);
<matrix group with 2 generators>
gap> Size(G);
120
gap> Size(SO(1,4,5)); 
14400

Edit 2: The matrices don't necessarily generate the group for $n=3$ either (checked for $p=5$ and $p=7$; for bigger $n$ the computation take longer than a few seconds so I didn't go there):

gap> n := 3;; p := 5;;
gap> f := sigma -> AsList(BlockMatrix([[1,1,IdentityMat(n)], [1,2,sigma],[2,1,NullMat(n,n)], [2,2,IdentityMat(n)]], 2, 2));
function( sigma ) ... end
gap> M := Concatenation(List([1..n], i -> List([i+1..n], j -> List([1..n], k -> List([1..n], function(l) if k = i and j = l then return 1; elif k = j and l = i then return -1; fi; return 0; end)))));;
gap> G := Group(One(GF(p))*Concatenation(List(M, f), List(M, v -> TransposedMat(f(v)))));
<matrix group with 6 generators>
gap> Size(G); Size(SO(1,2*n,p)); Size(SO(1,2*n,p)) mod Size(G);
14508000000
29016000000
0
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I meant for $n>1$. Would you please give more details by your second paragraph? I am not sure if I understand it entirely. –  M.B Jul 15 '12 at 5:20
    
@Florian: I think that M.B. does intend that $\sigma$ and $\tau$ should range over all skew-symmetric $n \times n$ matrices. –  Geoff Robinson Jul 15 '12 at 10:36
    
@Florian: I am so thankful to you for your comments. Indeed I did my computations for $n\geq 3$, and I thought this should work for $n=2$, which based on your comment is wrong. I don't know how to work with Gap. Would you please tell me how I can checked it for $n=3$? By the way, as Prof. Robinson mentioned here I take all skew-symmetric matrices. –  M.B Jul 15 '12 at 15:56
    
@M.B: You can in principle use the GAP code I just posted to check for arbitrary $n$ and $p$, but I see no reason why the matrices should generate the group for bigger choices of $n$ and $p$. –  Florian Eisele Jul 15 '12 at 16:47

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