The Sign of the Linear Action

We compute the sign of \(\operatorname{GL}_n(\mathbb F_q)\) acting on \(\mathbb F_q^{\oplus n}\).

Proposition. Let \(q\) be an odd prime-power and \(n\) be a positive integer. The standard action of \(\operatorname{GL}_n(\mathbb F_q)\) on \(\mathbb F_q^{\oplus n}\) induces an embedding \[\iota\colon\operatorname{GL}_n(\mathbb F_q)\to\operatorname{Sym}\left(\mathbb F_q^{\oplus n}\right).\] Then \({\operatorname{sgn}}\circ\iota=\chi\circ\det\), where \(\chi\) is the quadratic character on \(\mathbb F_q^\times\).

We will handle the case of even \(q\) at the end because the argument is slightly different.

The main point is to compute \({\operatorname{sgn}}\circ\iota\) on some easy but generating elements of \(\operatorname{GL}_n(\mathbb F_q)\).

To state our supply of generating elements, we set up some notation. For the moment, we work over a general field \(k\), to be set to \(\mathbb F_q\) in a moment. For brevity, set \( G=\operatorname{GL} _ n(k) \), and let \(T\subseteq G\) be the subset of diagonal (but not necessarily scalar) matrices. For example, \(T\cong\left(k^\times\right)^n\). Continuing, we let \(E_{ij}\) be the \(n\times n\) matrix with zeroes everywhere except with a \(1\) in row \(i\) and column \(j\).

We are now ready to give our supply of generating elements.

Lemma 1. The group \(G\) is generated by \(T\) and the elements \[\{1+aE_{ij}:a\in\mathbb F_q^\times,i\ne j\}.\]

Proof of Lemma 1. This is essentially Gaussian elimination. Let’s begin by explaining by the relevance of Gaussian elimination. Given some \(n\times n\) matrix \(A\) with coefficients in \(k\), we can compute that \(A(1+aE_{ij})\) is the matrix \(A\) where we have added \(a\) times the \(i\)th column to the \(j\)th column. By taking the transpose, we also see that \((1+E_{ij})A\) is the matrix \(A\) where we do a similar row operation.

Now, the point is that such row and column operations can take any invertible \(n\times n\) matrix and eventually produce a diagonal matrix using the algorithm of Gaussian elimination. In short, one proceeds inductively, so it is enough to merely handle the bottom row and rightmost column. The given operations allow one to rearrange the rows and columns so that the bottom-right entry is nonzero, and then row and column operations can use this nonzero entry to make all other terms in the bottom row and rightmost column vanish.

In total, we have provided an algorithm to take any \(g\in G\) and multiply it on the left and right by matrices of the form \(1+aE_{ij}\) until we get a diagonal matrix in \(T\). This completes the proof. \(\blacksquare\)

It remains to compute \({\operatorname{sgn}}\circ\iota\) on our set of generating elements, which we do more or less directly.

Lemma 2. Let \(q\) be an odd prime power, and let \(\chi\) be the quadratic character on \(\mathbb F_q^\times\). For \(a\in\mathbb F_q^\times\), let \(\mu_a\colon\mathbb F_q\to\mathbb F_q\) be the multiplication-by-\(a\) map. Then \(\operatorname{sgn}\mu_a=\chi(a)\).

Proof of Lemma 2. This is essentially Zolotarev’s lemma. The map \(a\mapsto\operatorname{sgn}\mu_a\) defines a homomorphism \(\mathbb F_q^\times\to\{\pm1\}\). Because the group \(\mathbb F_q^\times\) is cyclic (of even order), there is a unique such nontrivial homomorphism, which we called \(\chi\), so it is enough to check that \(\operatorname{sgn}\mu_a=-1\) for some \(a\).

The point is to give a cycle decomposition of \(\mu_a\). Let \(r\) be the multiplicative order of \(a\). Then \(\mu_a\) decomposes into \(r\)-cycles, and we see that there must be \((q-1)/r\) such \(r\)-cycles. Thus, \[\operatorname{sgn}\mu_a=(-1)^{(r+1)\cdot(q-1)/r},\] which is \((-1)^{(q-1)/r}\) because \(q\) is odd. For example, if \(a\) is a generator, then this value is manifestly \(-1\). \(\blacksquare\)

Lemma 3. Let \(q\) be a power of the prime \(p\), and choose a positive integer \(n\ge2\). Given \(a\in\mathbb F_q^\times\), the cycle decomposition of \(1+aE_{ij}\) acting on \(\mathbb F_q^{\oplus n}\) has \(q^{n-1}/p\) total \(p\) cycles, and the remaining elements are fixed.

Proof of Lemma 3. By rearranging the rows and columns, we may assume that our element of interest is \[g=\operatorname{diag}\left(\begin{bmatrix}1 & a \\ 0 & 1\end{bmatrix},1,\ldots,1\right).\] As such, we see that \(g\) sends the \(n\)-tuple \(x=(x_1,x_2,\ldots,x_n)\) to \(gx=(x_1+ax_2,x_2,\ldots,x_n)\). As such, we see that the tuple \(x\) is fixed if and only if \(x_2=0\), and if \(x_2\) is nonzero, then \(x\) belongs to a \(p\)-cycle. Thus, we see that \(q^{n-1}\) elements belong to a \(p\)-cycle, whence the result follows. \(\blacksquare\)

Proof of Proposition. It is enough to check the equality \({\operatorname{sgn}}\circ\iota=\chi\circ\det\) on a set of elements generating \(G\), for which we use Lemma 1. There are two kinds of elements to work with.

• We show the equality for \(g\in T\). The element \(g\) can be decomposed into a product of diagonal matrices where all but one of the nonzero entries equals \(1\), so it suffices to work with such elements of \(T\). By rearranging the coordinates, we may as well assume that \[g=\operatorname{diag}(a,1,\ldots,1)\] for some \(a\in\mathbb F_q^\times\).

  We would like to show that \(\operatorname{sgn}\iota(g)=\chi(a)\), for which we use Lemma 2. If \(a\) is a square, then \(g\) is also a square, so both sides are \(1\). Otherwise, if \(a\) is not a square, we must show that \(\operatorname{sgn}\iota(g)=-1\). Well, \(\mu_a\colon\mathbb F_q\to\mathbb F_q\) has a cycle decomposition with an odd number of even-length cycles. Each such even-length cycle then produces \(q^{n-1}\) even cycles in \(\mathbb F_q^{\oplus n}\), so there is still an odd number of even-length cycles in the end.

• We show the equality for elements of the form \(1+aE_{ij}\). We show the equality for elements of the form \(1+aE_{ij}\), for which we use Lemma 3. Indeed, we see that \(\iota(g)\) decomposes only into cycles of odd length, all of which are even, so \(\operatorname{sgn}\iota(g)=+1\) follows. \(\blacksquare\)

Remark. Here is another way to handle the second case above, which avoids an explicit cycle decomposition. By symmetry, it will be enough to focus on the case where \(i>j\). In this case, we see that the subgroup \(B\subseteq G\) of upper-triangular elements with \(1\)s on the diagonal has odd order: the number of elements is \(q^{n(n-1)/2}\). Thus, \(B\) must be trivial under any homomorphism to \(\{\pm1\}\) because it has no index-\(2\) subgroups. The result follows.

As promised, we also handle the case where \(q\) is even separately.

Proposition. Let \(q\) be a power of \(2\). The standard action of \(\operatorname{GL}_n(\mathbb F_q)\) on \(\mathbb F_q^{\oplus n}\) induces an embedding \[\iota\colon\operatorname{GL}_n(\mathbb F_q)\to\operatorname{Sym}\left(\mathbb F_q^{\oplus n}\right).\] Then \({\operatorname{sgn}}\circ\iota\) is trivial unless \((q,n)=(2,2)\), in which case it is the unique quadratic character on \(\operatorname{GL}_2(\mathbb F_2)\).

Proof of Proposition. We imitate the previous proof. By Lemma 1, it is enough to handle elements of \(T\) and elements of the form \(1+aE_{ij}\), which we handle separately.

• Because \(\mathbb F_q^\times\) now has odd size, \({\operatorname{sgn}}\circ\iota\) is now forced to be trivial on \(T\) automatically.
• Consider \(g\) in the form \(1+aE_{ij}\), which we note is only relevant for \(n\ge2\). Here, Lemma 3 implies that the sign of \(\iota(g)\) is \((-1)^{q^{n-1}/2}\), which is \(+1\) unless \(q^{n-1}/2\) is odd. However, \(q^{n-1}/2\) is a power of \(2\), so it will only be odd when \(q^{n-1}=2\), which is equivalent to \(q=2\) and \(n=2\).

From the above casework, we see that it remains to show that there is a unique quadratic character on \(\operatorname{GL}_2(\mathbb F_2)\). Well, \(\operatorname{GL}_2(\mathbb F_2)\) is nonabelian group of order \(6\), so it must be isomorphic to \(S_3\), and a consideration of conjugacy classes shows that \(S_3\) has a unique normal subgroup of index \(2\), so there is a unique nontrivial character \(S_3\to\{\pm1\}\) (which is \(\operatorname{sgn}\)!). \(\blacksquare\)

Remark. It is perhaps worthwhile to have an explicit isomorphism \(\operatorname{GL}_2(\mathbb F_2)\to S_3\). Well, the natural action of \(\operatorname{GL}_2(\mathbb F_2)\) on \(\mathbb F_2^{\oplus2}\) has the fixed point \((0,0)\) and so descends to an action on the three-element set \(\mathbb F_2^{\oplus2}\setminus\{(0,0)\}\). Thus, we have a homomorphism \[\operatorname{GL}_2(\mathbb F_2)\to\operatorname{Sym}\left(\mathbb F_2^{\oplus2}\setminus\{(0,0)\}\right)\] We claim that this map is an isomorphism. Well, both sides have six elements, so it’s enough to check that the map is injective, which follows because an element \(g\in\operatorname{GL}_2(\mathbb F_2)\) is determined by \(g(1,0)\) and \(g(0,1)\). As an aside, we note that the previous sentence also provides an inverse map: send the permutation \(\sigma\) of \(\mathbb F_2^{\oplus2}\setminus\{(0,0)\}\) to the linear operator \(g\) with \(g(1,0)=\sigma(1,0)\) and \(g(0,1)=\sigma(0,1)\)—then \(g(1,1)=\sigma(1,1)\) follows automatically!

Remark. We take a moment to mention an alternate argument. One can show, though moderate amounts of pain, that the commutator subgroup of \(\operatorname{GL}_n(\mathbb F_q)\) is \(\operatorname{SL}_n(\mathbb F_q)\) unless \((q,n)=(2,2)\); the case \((q,n)=(2,2)\) can be handled separately. For \((q,n)\ne(2,2)\), this shows that \({\operatorname{sgn}}\circ\iota\) factors through \(\det\) except when \((q,n)=(2,2)\). It then remains to compute \({\operatorname{sgn}}\circ\iota\) on any matrix of given determinant, for which we can use the matrices in \(T\) as above. Anyway, an advantage of the argument presented in the above post is that we have provided explicit cycle decompositions for many interesting (but simple) elements of \(\operatorname{GL}_n(\mathbb F_q)\).