Element structure of special linear group of degree two over a field

Let $$K$$ be a field. Consider the special linear group of degree two $$SL(2,K)$$. The goal of this article is to describe the element structure of $$SL(2,K)$$.

Related descriptions

 * Element structure of special linear group of degree two over a finite field
 * Element structure of special linear group of degree two over a finite discrete valuation ring
 * Element structure of general linear group of degree two over a field
 * Element structure of general linear group of degree two over a finite field

Broad description
Note that in characteristic two, in the cases of scalar matrices and Jordan block, the eigenvalue of 1 subcase coincides with the eigenvalue of -1 subcase.

Identification between GL(2)-conjugacy classes and field elements
The trace is a mapping:

Conjugacy classes in $$SL(2,K)$$ $$\to$$ $$K$$

Further, if we are interested in the conjugacy classes of $$SL(2,K)$$ relative to $$GL(2,K)$$ (i.e., the relation of being conjugate in $$GL(2,K)$$, then this mapping is almost bijective: for each of the elements $$2,-2 \in K$$, there are two inverse images: the scalar matrix and the Jordan block.

Thus, if we are interested in conjugacy with respect to $$GL(2,K)$$, we can identify:


 * Case of characteristic not two: $$GL(2,K)$$-conjugacy classes in $$SL(2,K)$$ $$\leftrightarrow$$ $$K \bigsqcup$$ a two-point set
 * Case of characteristic two: $$GL(2,K)$$-conjugacy classes in $$SL(2,K)$$ $$\leftrightarrow$$ $$K \bigsqcup$$ a one-point set

In particular, for the size of $$K$$ a prime power $$q$$, we have:


 * Case of characteristic not two (i.e., odd $$q$$): There are $$q + 2$$ conjugacy classes in $$SL(2,q)$$ relative to $$GL(2,q)$$
 * Case of characteristic two: There are $$q + 1$$ conjugacy classes in $$SL(2,q)$$ relative to $$GL(2,q)$$

This, however, does not classify the conjugacy classes of $$SL(2,q)$$, because a single $$GL(2,q)$$-conjugacy class living in $$SL(2,q)$$ can split in $$SL(2,q)$$ into multiple conjugacy classes. As noted above, the splitting depends on the sizes of certain quotient groups of the multiplicative group of the field.

Splitting summary
Here, we provide a brief summary of the splitting of conjugacy classes from $$GL_2$$ to $$SL_2$$. As noted in the broad description, there are two types of conjugacy classes that (may) split. More details are in subsequent sections. The key justification in both cases is the splitting criterion for conjugacy classes in the special linear group:


 * Conjugacy classes of the Jordan block form: There are two such $$GL_2$$-conjugacy classes, with representatives:

$$\begin{pmatrix} 1 & 1 \\ 0 & 1 \\\end{pmatrix}, \begin{pmatrix} -1 & 1 \\ 0 & -1 \\\end{pmatrix}$$

Each of these conjugacy classes splits into $$|K^*/(K^*)^2|$$ conjugacy classes with respect to $$SL_2$$. Moreover, if we let $$S$$ be a transversal (i.e., a system of coset representatives) for $$(K^*)^2$$ in $$K^*$$, then the conjugacy class representatives are given by:

$$\begin{pmatrix} 1 & a \\ 0 & 1 \\\end{pmatrix}, a \in S$$

and:

$$\begin{pmatrix} -1 & a \\ 0 & -1 \\\end{pmatrix}, a \in S$$


 * Conjugacy classes whose characteristic polynomial is an irreducible quadratic with distinct roots in a separable quadratic extension $$L$$ of $$K$$. Each such conjugacy class splits into $$|K^*/N(L^*)|$$ where $$N(L^*)$$ denotes the image of $$L^*$$ under the norm map $$L \to K$$ for the quadratic extension $$L$$ of $$K$$.

Note that, for any separable quadratic extension $$L$$ of $$K$$, we have $$(K^*)^2 \subseteq N(L^*) \subseteq K^*$$, so $$K^*/N(L^*)$$ can be viewed as a quotient of $$K^*/(K^*)^2$$, viewed as a group.

Note, further, that when the characteristic of the field is not two, the set of possibilities for the extension $$L$$ of $$K$$ is itself given by the set of non-identity elements of $$K^*/(K^*)^2$$.

Below, we summarize some important cases for the field $$K$$:

Central elements
The center is a subgroup of order either 1 or 2, depending on whether the characteristic of the field is two or not two. In the case of characteristic not equal to two, the center is the subgroup:

$$\left\{ \begin{pmatrix} 1 & 0 \\ 0 & 1 \\\end{pmatrix}, \begin{pmatrix} -1 & 0 \\ 0 & -1 \\\end{pmatrix} \right\}$$

In the case of characteristic two, the center is the trivial subgroup.

Jordan block of size two
Over $$GL_2$$, there are two conjugacy classes of elements of $$SL_2$$ of this type in characteristic not two. In characteristic two, both of these may split further:


 * The conjugacy class of $$\begin{pmatrix} 1 & 1 \\ 0 & 1 \\\end{pmatrix}$$. In particular, this conjugacy class includes all matrices of the form $$\begin{pmatrix} 1 & \lambda \\ 0 & 1 \\\end{pmatrix}$$ and also all matrices of the form $$\begin{pmatrix} 1 & 0 \\ \lambda & 1 \\\end{pmatrix}$$ where $$\lambda \in K^\ast$$.
 * The conjugacy class of $$\begin{pmatrix} -1 & 1 \\ 0 & -1 \\\end{pmatrix}$$. In particular, this conjugacy class includes all matrices of the form $$\begin{pmatrix} -1 & \lambda \\ 0 & -1 \\\end{pmatrix}$$ and also all matrices of the form $$\begin{pmatrix} -1 & 0 \\ \lambda & -1 \\\end{pmatrix}$$ where $$\lambda \in K^\ast$$.

In characteristic two, both these conjugacy classes collapse into a single one with respect to $$GL_2$$ because $$1 = -1$$.

Splitting criterion
By the splitting criterion for conjugacy classes in the special linear group, we need to do the following:


 * First, find the centralizer of any element in the conjugacy class
 * Then determine its image under the determinant map to $$K^*$$
 * Then identify the coset space of that image in $$K^*$$ with the pices into which the $$GL_2$$-conjugacy class splits in $$SL_2$$.

The $$GL_2$$-centralizer of any element with a Jordan block of size two is the subgroup:

$$\{ \begin{pmatrix} a & b \\ 0 & a \\\end{pmatrix} \mid a \in K^\ast, b \in K \} \cong K^\ast \times K$$

The image of this under the determinant map is:

$$\{ a^2 \mid a \in K^\ast \}$$

In other words, it is the subgroup of squares in $$K^\ast$$, denoted $$(K^*)^2$$.

The cosets of this subgroup in $$K^*$$ are given by equivalence classes of elements of $$K^*$$ up to multiplication by squares. The collectin of cosets can be identified with the quotient group $$K^*/(K^*)^2$$, also called the multiplicative group modulo squares of $$K$$. Now, if $$S$$ is a collection of coset representatives of $$(K^*)^2$$ in $$K^*$$, consider the set of matrices:

$$\begin{pmatrix} c & 0 \\ 0 & 1 \\\end{pmatrix} \mid c \in S \}$$

Conjugating a $$GL_2$$-conjugacy class representative by each of these gives a collection of $$SL_2$$-conjugacy class representatives. The upshot is that the $$SL_2$$-conjugacy class representatives for the $$GL_2$$-conjugacy class of $$\begin{pmatrix}1 & 1 \\ 0 & 1 \\\end{pmatrix}$$ are:

$$\{ \begin{pmatrix} 1 & c \\ 0 & 1 \\\end{pmatrix} \mid c \in S \}$$

In particular, the number of such conjugacy classes is $$|K^*/(K^*)^2|$$.

In the case of characteristic not two, we have another $$GL_2$$-conjugacy class, the one with representative $$\begin{pmatrix}-1 & 1 \\ 0 & -1 \\\end{pmatrix}$$. The representatives for the $$GL_2$$-conjugacy class of $$\begin{pmatrix}-1 & 1 \\ 0 & -1 \\\end{pmatrix}$$ are:

$$\{ \begin{pmatrix} -1 & c \\ 0 & -1 \\\end{pmatrix} \mid c \in S \}$$.

In particular, the number of such conjugacy classes is $$|K^*/(K^*)^2|$$.

The upshot is that:


 * For characteristic not equal to two, the set of conjugacy classes with a size two Jordan block can be identified with two disjoint copies of $$K^*/(K^*)^2$$.
 * For characteristic two, the set of conjugacy classes with a size two Jordan block can be identified with a single copy of $$K^*/(K^*)^2$$.

Elements diagonalizable over a separable quadratic extension
These elements have pairs of distinct eigenvalues over a separable quadratic extension $$L$$ of $$K$$ but do not have eigenvalues over $$K$$. As we are working in $$SL_2$$, the eigenvalues must be inverses of each other, and the minimal polynomial is an irreducible polynomial of the form $$x^2 - ax + 1$$.

There are three parts to the program of understanding such elements:


 * Determine all the possible separable quadratic extensions of $$K$$.
 * Determine, for each extension $$L$$, what are the $$GL_2$$-conjugacy classes of elements of $$SL_2$$ whose minimal polynomial is irreducible over $$K$$ and splits over $$L$$. This is equivalent to finding all polynomials of the form $$x^2 - ax + 1$$ whose splitting field is $$L$$.
 * Determine, for each extension $$L$$, the number of pieces into which any $$GL_2$$-conjugacy class of the type identified above splits. As we shall see, the splitting behavior depends only on the field, and we do not need to know the precise original $$GL_2$$-conjugacy class.

Determination of separable quadratic extensions: characteristic not two
In case the characteristic of the field is not two, there is a correspondence:

Separable quadratic extensions of $$K$$ (up to equivalence as extensions) $$\leftrightarrow$$ Non-identity elements of the quotient group $$K^*/(K^*)^2$$, i.e., non-square elements up to multiplication by squares

The reason is as follows. Any quadratic extension of $$K$$ is obtained by adjoining a root of an irreducible quadratic. When the characteristic is not two, the quadratic formula shows that this is equivalent to adjoining the square root of the discriminant, so the extension is of the form $$K(\sqrt{D})$$ for $$D$$ a non-square in $$K^*$$. Further $$K(\sqrt{D_1})$$ and $$K(\sqrt{D_2})$$ are equivalent as quadratic extensions iff $$D_1/D_2$$ is a square in $$K^*$$. Thus, the separable quadratic extensions of $$K$$ up to equivalence are identified with non-identity cosets of $$(K^*)^2$$ in $$K^*$$.

If the characteristic of the field is two, the situation is different, because the separable quadratic extensions are precisely the ones that are not obtained by adjoining square roots. The general theory needs to be filled in.

Determination of minimal polynomials for a given separable quadratic extension: characteristic not two
Given a separable quadratic extension $$L = K(\sqrt{D})$$, we first note that there is at least one irreducible quadratic of the form $$x^2 - ax + 1 $$ that splits over this, and hence there exists at least one element of $$SL_2$$ whose minimal polynomial has splitting field $$L$$. First note that $$D \ne 1$$. Consider the polynomial:

$$x^2 - \frac{2(D + 1)}{(D - 1)}x + 1$$

The roots of this polynomial are:

$$\frac{D+1 \pm 2\sqrt{D}}{D - 1}$$

Having settled this, we turn to the question of determing the set:

$$\{ a \in K \mid x^2 - ax + 1 \mbox{ is irreducible, splits over } K(\sqrt{D})\}$$

Using the quadratic formula, we obtain that this is the set:

$$\{ a \in K \mid (a^2 - 4)/D \in (K^*)^2 \}$$

From a more global perspective, consider the set map:

$$K \setminus \{ -2,2\} \to K^\ast/(K^\ast)^2$$

given by:

$$a \mapsto a^2 - 4 \mbox{ taken modulo } (K^\ast)^2$$

We need to exclude $$\{ -2,2 \}$$ to make sure that the image is an invertible element.

The inverse image of the identity element of $$K^\ast/(K^\ast)^2$$ corresponds to the minimal polynomials that split over $$K$$. The inverse image of any non-identity element of $$K^\ast/(K^\ast)^2$$ corresponds to the minimal polynomials that split over the field extension $$L$$ corresponding to the non-identity element of $$K^\ast/(K^\ast)^2$$ by the correspondence detailed above:

Separable quadratic extensions of $$K$$ (up to equivalence as extensions) $$\leftrightarrow$$ Non-identity elements of the quotient group $$K^*/(K^*)^2$$, i.e., non-square elements up to multiplication by squares

In other words, we have a decomposition of $$K \setminus \{ -2,2\}$$ in terms of fibers above field extensions.

Determination of splitting criterion: all characteristics
We need to use the splitting criterion for conjugacy classes in the special linear group. In other words:


 * We first need to determine the centralizer of any element whose minimal polynomial splits over a separable quadratic extension $$L$$ of $$K$$.
 * We then need to determine the image of that centralizer under the determinant map to $$K^\ast$$.
 * We then need to identify the coset space of that image and use it to determine the $$SL_2$$-conjugacy classes into which the $$GL_2$$-conjugacy class splits.

The first step is easy. We note that the centralizer of this element can be identified with the image of $$L^*$$ in $$GL(2,K)$$ via the left action of $$L$$ on itself, viewing $$L$$ as a two-dimensional $$K$$-vector space.

Second, we need to understand the composite map:

$$L^* \to GL(2,K) \to K^*$$

It turns out that this composite is simply the norm map $$N$$ for the quadratic extension $$L$$ of $$K$$. The map sends $$\alpha \in L$$ to the product of $$\alpha$$ and its conjugate.

Thus, the image of the centralizer under the determinant map is the same as the image of $$L^*$$ under the algebraic norm $$N$$. Denote this as $$N(L^*)$$. Then, the coset space can be identified with the group $$K^*/N(L^*)$$. In particular, each $$GL_2$$-conjugacy class in $$SL_2$$ with minimal polynomial having splitting field $$L$$ must split into $$|K^*/N(L^*)|$$ many $$SL_2$$-conjugacy classes.

Further, for $$\alpha \in K^*$$, $$N(\alpha) = \alpha^2$$, so $$(K^*)^2 = N(K^*) \subseteq N(L^*)$$. Thus, the quotient $$K^*/N(L^*)$$ is a quotient of the quotient $$K^*/(K^*)^2$$.