Quaternion Algebras

Given non-zero elements ${a}$ and ${b}$ in a field ${F}$ , the quaternion algebra ${\left(\frac{a, b}{F}\right)}$ over a field ${F}$ is the 4-dimensional vector space with the relations

$\displaystyle i^{2}=a, ~j^{2}=b, \mbox { and } j i=-i j \mbox{ if } char F \neq 2$

In characteristic ${2}$ , the quaternion algebra ${\left[\frac{a, b}{F}\right)}$ is defined by the relations

$\displaystyle i^{2}+i=a, j^{2}=b, \mbox { and } k=i j=j(i+1) \mbox{ if } char F =2 .$

Any algebra generated by non-zero elements ${i}$ , ${j}$ satisfying these relations with ${a}$ and ${b}$ non-zero has to be the quaternion algebra, that is it will be 4-dimensional and generated by ${i}$ , ${j}$ and ${ij}$ .

If ${a}$ or ${b}$ are zero, we wont get a quaternion algebra. For instance ${\mathbb F[i, j]}$ modulo the relations ${i^2=j^2=ij=ji=0}$ is just 3-dimensional.

We denote ${ij}$ by ${k}$ , then we have ${k^2= -ab}$ when ${char F \neq 2}$ .

We assume ${char F \neq 2}$ from now, in characteristic 2 we have to make some modifications to all the statements and arguments. (Basically because quadratic forms over characteristic ${2}$ behave differently)

Examples: a) Hamiton’s quaternions are ${\mathbb H = \left(\frac{-1, -1}{\mathbb R}\right)}$ . They are closely related to the rotations ${SO(3)}$ , ${SO(4)}$ and the group ${SU(2)}$ .

b) The algebra ${ \left(\frac{1, 1}{F}\right)}$ is isomorphic to the algebra of matrices ${M_2(F)}$ by the following identification

$\displaystyle 1 \mapsto\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right],~~i \mapsto\left[\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right], ~~j \mapsto\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right], ~~k \mapsto\left[\begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array}\right]$

c) ${\mathbf H (\mathbb Q)= \left(\frac{-1, -1}{\mathbb Q}\right)}$ is a division algebra. But ${\mathbf H (\mathbb C)= \left(\frac{-1, -1}{\mathbb C}\right)}$ is equivalent to ${\mathbf H (\mathbb C)= \left(\frac{1, 1}{\mathbb C}\right) \cong M_2(\mathbb C)}$ is the matrix algebra.

In general, ${\left(\frac{a, b}{F}\right)}$ can be viewed as a sub-algebra of a matrix algebra (possibly over a extension field) by the following identification

$\displaystyle x+ iy+ zj+w i j \longrightarrow \left[\begin{array}{cc} x+y \sqrt{a} & b(z+w \sqrt{a}) \\ z-w \sqrt{a} & x-y \sqrt{a} \end{array}\right]$

$\displaystyle 1 \mapsto\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right],~~i \mapsto\left[\begin{array}{cc} \sqrt{a} & 0 \\ 0 & - \sqrt{a} \end{array}\right], ~~j \mapsto\left[\begin{array}{ll} 0 & b \\ 1 & 0 \end{array}\right], ~~k \mapsto\left[\begin{array}{cc} 0 & b \sqrt{a} \\ - \sqrt{a} & 0 \end{array}\right]$

If ${a=1}$ , the algebra ${\left(\frac{1, b}{F}\right)}$ is isomorphic to ${M_2( F).}$

Note that we have the isomorphism

$\displaystyle \left(\frac{a, b}{F}\right)\cong \left(\frac{b, a}{F}\right)$

because we can simply interchange ${i}$ and ${j}$ .

We can also consider the basis ${1, i, k, ik}$ or ${1, j, k, jk}$ which gives us the equivalences

$\displaystyle {\left(\frac{a, b}{F}\right)\cong \left(\frac{a, -ab}{F}\right)}$ , ${\left(\frac{a, b}{F}\right)\cong \left(\frac{b, -ab}{F}\right)}$

We can also rescale ${i}$ and ${j}$ to ${r i}$ and ${sj}$ to see that

$\displaystyle \left(\frac{a, b}{F}\right)\cong \left(\frac{r^2a, s^2b}{F}\right)$

The equivalence ${\left(\frac{a, b}{F}\right)\cong \left(\frac{a, -ab}{F}\right)}$ allows us to see that

$\displaystyle \left(\frac{a, -a}{F}\right)\cong \left(\frac{a, a^2}{F}\right) \cong \left(\frac{a, 1}{F}\right) \cong M_2(F)$

So we can assume ${a, b}$ to be representatives of the square classes ${F^{\times}/F^{\times 2}}$ . Also if we extend the scalars to field ${K}$ which has square roots of at least one ${a}$ and ${b}$ , the algebra will be isomorphic to ${\left(\frac{1, b}{K}\right)}$ or ${\left(\frac{a, 1}{K}\right)}$ both of which are equivalent to the matrix algebra ${M_2 (K).}$ Such a field is called splitting field of algebra.

For example, everything is a square in ${\mathbb C}$ , so ${\left(\frac{1, 1}{\mathbb C}\right) \cong M_2(\mathbb C)}$ is the only quaternion algebra over ${\mathbb C}$ . Over reals, we have to just consider ${a, b = \pm 1}$ , and ${(a, b) =(-1, -1)}$ gives the Hamilton’s algebra ${\mathbb H}$ and every other choice is of the form ${\left(\frac{1, b}{\mathbb R}\right)}$ , so equivalent to ${M_2(\mathbb R).}$ But both the algebra when we extend the class become isomorphic to ${M_2 (\mathbb C)}$ . That is ${\mathbb H \otimes_{\mathbb R} \mathbb C \cong M_2(\mathbb C)}$ and ${M_2(\mathbb R) \otimes_{\mathbb R} \mathbb C \cong M_2(\mathbb C).}$

When ${p =1 \mod 4}$ , ${-1}$ is square ${\mod p}$ and hence ${\mathbf H (\mathbb F_p): = \left(\frac{-1, -1}{\mathbb F_p}\right) \cong M_2(\mathbb F_p).}$

Conjugates, Norm, Trace:

We have the conjugate map

$\displaystyle \alpha \rightarrow \bar \alpha: x+ iy+ zj+w k \longrightarrow x- iy- zj-w k$

And the norm

$\displaystyle N(\alpha) = \alpha \bar \alpha = \bar{\alpha} \alpha=x^{2}-a y^{2}-b z^{2}+a b w^{2}$

If the norm doesn’t represent zero non-trivially, then every non-zero element can be inverted.

$\displaystyle \alpha ^{-1} = \frac{\bar \alpha}{N(\alpha)} .$

Therefore every element satisfies a quadratic equation with coefficient in ${F}$ given by

$\displaystyle \alpha^2 - (\alpha + \bar \alpha) \alpha + \alpha \bar \alpha =0$

We call ${\alpha +\bar \alpha}$ the trace. So pure quaternions with ${x=0}$ are the trace zero elements. Trace zero elements have squares which are in ${F}$ , in fact only non-scalar elements with squares in the field.

Thus we can think quaternions as an noncommutative analog of quadratic field extensions. In fact, the only quadratic division algebras (all elements satisfy degree at most ${2}$ equations) are the quaternions and quadratic field extensions. Division property is important because otherwise a lot degree 2 algebras exist.

Question: When do we get algebras which are not split, that is not isomorphic to ${M_2(F)}$ ?

For instance if the norm form doesn’t represent zero, the algebra is a division algebra. So it cannot be the matrix algebra because the matrix algebra has zero divisors i.e., ${AB=0, A\neq 0, B \neq 0}$ .

The quadratic form ${x^{2}-a y^{2}-b z^{2}+a b w^{2}}$ is the norm form.

So if the form doesn’t represent zero, the algebra cannot be split. And It can also be seen that if this form represents zero non-trivially, the algebra is the matrix algebra.

Proof: If ${a}$ is square, the algebra is split. So assume ${a}$ is not a square and the norm represent zero, that is we have a non-trivial solution to ${x^{2}-a y^{2}-b z^{2}+a b w^{2}=0}$ which implies that

$\displaystyle b =\frac{x^{2}-a y^{2}}{z^{2}-a w^{2}}= r^2-as^2$

Now check that ${1, i, rj + sk, rk+asj}$ is a basis for the algebra with ${( rj + sk)^2=b^2}$ . This is a basis because the determinant ${\left|\begin{array}{cc} r & s \\ a r & r \end{array} \right|=b \neq 0.}$

So the algebra is equivalent to ${\left(\frac{a, b^2}{F}\right) \cong M_2(F).}$

So we have two cases: Either
a) the algebra is a division algebra, ${x^{2}-a y^{2}-b z^{2}+a b w^{2}}$ doesn’t represent zero non-trivially.
b) The algebra is the matrix algebra ${M_2(F)}$ .

We also consider the forms ${-a y^{2}-b z^{2}+a b w^{2}}$ and ${aX^2+bY^2}$

${x^{2}-a y^{2}-b z^{2}+a b w^{2}}$ represents zero non-trivially iff ${-a y^{2}-b z^{2}+a b w^{2}}$ represents zero non-trivially. And both the forms represent ${0}$ non-trivially iff ${aX^2+bY^2}$ represent ${1.}$

Proof: One side is obvious. So assume ${x^{2}-a y^{2}-b z^{2}+a b w^{2}}$ has a non-trivial solution with ${x\neq 0}$ . Let ${ \alpha =x+ iy+ zj+w i j }$ and consider ${\beta }$ orthogonal (with the bilinear form associated to the quadratic form) to the both ${1}$ and ${\alpha}$ . We have traces of ${\alpha \beta}$ and ${\bar \alpha \beta}$ both zero. Because ${\alpha +\bar \alpha =2x \neq 0}$ , one of ${\alpha \beta}$ and ${\bar \alpha \beta}$ is non-zero and we found a trace zero element with norm 0.

If we have a solution with ${w=1}$ for the second form ${-a y^{2}-b z^{2}+a b w^{2}}$ , we get

$\displaystyle a\left(\frac{z}{a w}\right)^{2}+b\left(\frac{y}{b w}\right)^{2}=1$ , and we found a solution to ${aX^2+bY^2=1}$ . So we consider the case ${w=0}$ , so that we have non-trivial solution to ${aX^2+bY^2=0}$ . In this case ${aX_0^2+bY_0^2=0}$ and hence ${aX^2+bY^2= aX^2 -a(Y \frac{X_0}{Y_0})^2}$ , which represents ${1}$ .

So either we have a split matrix algebra, or a division algebra. To check if it’s a division algebra, we have to understand the properties of these quadratic forms.

We finally see reason for the notation ${\left(\frac{a, b}{F}\right)}$ . The Hilbert symbol ${(a, b)_{F}}$ equals ${1}$ iff ${aX^2+bY^2}$ represent ${1}$ , which happens if and only if the algebra is split, that is equal to ${M_2(F)}$ .

Example: Consider ${\left(\frac{a, p}{\mathbb Q}\right)}$ where ${a}$ is not a square ${\bmod ~p}$ . By the above discussion we need to check if ${aX^2+pY^2=1}$ has rational solutions. But a rational solution will imply that ${a (r/s)^2+ p(r'/s)^2=1}$ , which gives ${ar^2=s^2 \bmod p}$ , and is not possible because ${a\neq \square \bmod p}$ For example if ${p = 1 \bmod 4}$ , then ${-1}$ is not square ${\bmod ~p}$ which implies that ${\left(\frac{-1, p}{\mathbb Q}\right) \cong M_2(\mathbb Q)}$

We saw the definining criterion for ${\left(\frac{a, b}{F}\right)}$ to be split matrix algebra. ${aX^2+bY^2=1}$ have solutions in ${F}$ . Some special cases where this happens are
1) Either of ${a}$ , ${b}$ is ${1}$ , or in general either of them is a square.
2) ${b =-a}$
3) If ${a \ne 0, 1}$ and ${b =1-a}$ .
4) ${b=x^2-ay^2}$ (This is equivalent to ${aX^2+bY^2=1}$ , and hence the most general case)

Alternatively if we find $a, b$ with ${ b \neq x^2-ay^2}$ , then we have a division algebra.

Example: 1) The algebra ${\mathbf H(\mathbb F) =\left(\frac{-1, -1}{F}\right)}$ is a division algebra iff there ${-1}$ is not sums of squares.

2) Any quaternion algebra over ${\mathbf F_p}$ for ${p}$ odd is split. This is because ${1-by^2}$ and ${ax^2}$ as ${x, y}$ vary have to intersect to give a solution to ${ax^2+by^2=1}$ . ( both of them are of size ${\frac{p+1}{2}}$ )

Question: When are the algebras different (or the same)?

We saw some conditions on ${(a, b)}$ for ${\left(\frac{a, b}{F}\right)}$ to be the matrix algebra, ${b}$ has to of the form ${x^2-ay^2}$ . Similarly we ${\left(\frac{a, b_1}{F}\right)}$ and ${\left(\frac{a, b_2}{F}\right)}$ is the same quaternion algebra iff ${\frac{b_1}{b_2} =x^2-ay^2}$ . The matrix algebra corresponds to taking ${b_2 =1}$ .

Thus scaling ${b}$ by an element of the form ${x^2-ay^2}$ doesn’t change the algebra. Also two elements ${b_1}$ and ${b_2}$ different modulo the subgroup of elements of the form ${x^2-ay^2}$ represent different quaternion algebras. To prove this we need to show there is a conjugation map $z \to qzq^{-1}$ on the ${\left(\frac{a, b_1}{F}\right)}$ to send $i^2 =a$ to the basic element of ${\left(\frac{a, b_2}{F}\right)}$ satisfying $i^2 =a$

For example ${\left(\frac{-1, p}{\mathbb Q}\right)}$ , ${\left(\frac{-1, q}{\mathbb Q}\right)}$ for ${p , q= 3 \bmod 4}$ represents different quaternion algebras as because ${\frac{q}{p}}$ cannot be written as ${X^2+Y^2}$ . So we have infinitely many distinct division algebras over ${\mathbb Q}$ !

The case of ${\mathbb Q}$ , we also other isomorphism between ${\left(\frac{a, b}{F}\right)}$ and ${\left(\frac{a', b'}{F}\right)}$ , and the quaternion algebra is determined by the discriminant.

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