Let $K$ be a field, we can define the special orthogonal group $\text{SO}_2(K)$ as a matrix group

\[\text{SO}_2(K)=\left\{\begin{pmatrix}a & b \\ -b & a\end{pmatrix}\in \text{GL}_2(K):a^2+b^2=1\right\}.\]

One natural question to ask here is whether there is an alternative description of the group structure of $\text{SO}_2(K)$ under multiplication that is perhaps easier to grasp. This turns out to have an answer but to say it we must establish some terminology: We will say that $i\in K$ if the polynomial equation $x^2+1=0$ has a solution in $K$, and $i\not \in K$ if it does not. Moreover, $i$ will always denote a solution to this polynomial equation. With this in mind, we can state the structure theorem for $\text{SO}_2(K)$.

Theorem 1: Let $K$ be a field. Then if $i\in K$, $\text{SO}_2(K)\cong K^\times$, the multiplicative group of $K$. If $i\not \in K$, then $\text{SO}_2(K)\cong K(i)^\times/K^\times$.

The proof of this theorem that I present will look slightly ad-hoc. However, it is all a reflection of Galois cohomology, and in the second half of this blog post, I will use the solution to this problem to introduce some basic concepts in Galois cohomology.

Proof of Theorem 1: We will handle the case of $i\in K$ and $i\not \in K$ separately, as they are quite different.

Case 1: $i\in K$. In this case, the condition $a^2+b^2=1$ factors as $(a+bi)(a-bi)=1$. Inspired by this, we will define a map $\varphi:K^\times\to \text{SO}_2(K)$ by letting $\varphi(x)$ be the solution to the system equations

\[\begin{cases} a+bi=x, \\ a-bi=x^{-1}. \end{cases}\]

If we work out the solution to this system of equations, we get the following explicit form of $\varphi$

\[\varphi(x)=\left(\frac{x+x^{-1}}{2}, \frac{x-x^{-1}}{2i}\right)\]

This map also has an inverse, given by $(a,b)\mapsto a+bi$. These are clearly seen as inverses, so we just have to verify that $\varphi$ is a homomorphism. But this can be done by directly: for $x,y\in K^\times$, we have by computation

\[\begin{pmatrix}\frac{x+x^{-1}}{2} & \frac{x-x^{-1}}{2i} \\ -\frac{x-x^{-1}}{2i} & \frac{x+x^{-1}}{2} \end{pmatrix}\begin{pmatrix}\frac{y+y^{-1}}{2} & \frac{y-y^{-1}}{2i} \\ -\frac{y-y^{-1}}{2i} & \frac{y+y^{-1}}{2} \end{pmatrix}=\begin{pmatrix}\frac{xy+(xy)^{-1}}{2} & \frac{xy-(xy)^{-1}}{2i} \\ -\frac{xy-(xy)^{-1}}{2i} & \frac{xy+(xy)^{-1}}{2} \end{pmatrix}.\]

Case 2: $i\not \in K$. This case will take a bit more work, but we will also take a slightly more conceptual approach. We make the following observation: define a $2$-dimensional $K$-algebra $A$ by

\[A=\left\{\begin{pmatrix} a & b \\ -b & a\end{pmatrix}:a,b\in K\right\}\]

Note that we have a map $\text{det}:A\to K$, and $\text{SO}_2(K)=\text{Ker}(\text{det}:A^\times\to K^\times)$. Now, $A$ is isomorphic to a different object that might be more familiar: $K(i)$. We have an isomorphism $\varphi:K(i)\to A$ given by

\[a+bi\mapsto \begin{pmatrix}a & b \\ -b & a\end{pmatrix}.\]

This is an example of a general principle: if you have a degree $n$ field extension $L/K$, then for any $K$-basis of $L$ you can construct an embedding $L\hookrightarrow M_{n\times n}(K)$ by sending $\alpha$ to the matrix corresponding to the linear transformation $[\alpha]:L\cong K^n\to K^n$ given by multiplication by $\alpha$.

What does the determinant correspond to under this isomorphism? It turns out to be equal to the norm $N:K(i)\to K$, which is a map that takes $a+bi$ to $a^2+b^2$. This can also be defined by $N(x)=x\bar{x}$, where $\bar{x}=a-bi$ is the Galois conjugate of $x$ over $K$.

All this information can be combined in the following commutative diagram

All this tells us that we have an isomorphism $\text{Ker}(N:K(i)^\times\to K^\times)\cong \text{SO}_2(K)$. The upshot of this is that instead of studying matrices, we can just work within the field $K(i)$. Now that we have this perspective, I want to introduce a group homomorphism that I will call $\delta$. This will map $K(i)^\times$ to $K(i)^\times$, and it is given by the formula

\[\delta(x)=\frac{x}{\bar{x}}.\]

I will leave verifying that $\delta$ is a group homomorphism as an exercise. It turns out that $\delta$ is the key to proving case 2.

Lemma 1: Via the first isomorphism theorem, $\delta$ induces an isomorphism from $K(i)^\times/K^\times$ to $\text{Ker}(N:K(i)^\times\to K^\times)$.

Proof There are three steps to this proof. First, we need to show that $\text{Ker}(\delta)=K^\times$, second we need to show that $\text{Im}(\delta)\subseteq \text{Ker}(N:K(i)^\times\to K^\times)$, and finally we need to show that $\text{Im}(\delta)\supseteq \text{Ker}(N:K(i)^\times\to K^\times)$.

For step one, we see that by definition $x\in \text{Ker}(\delta)$ if and only if $1=\delta(x)=x/\bar{x}$. But this implies that $x=\bar{x}$, which by the Galois correspondence implies $x\in K$. If you don’t know Galois theory, we can also see this by writing $x=a+bi$ for $a,b\in K$, then $a+bi=x=\bar{x}=a-bi$ if and only if $b=0$, or $x\in K$.

Now for step two, we can do this by direct computation, verifying that $N(\delta(x))=1$:

\[N(\delta(x))=N\left(\frac{x}{\bar{x}}\right)=\frac{x}{\bar{x}}\overline{\frac{x}{\bar{x}}}=\frac{x}{\bar{x}}\frac{\bar{x}}{x}=1.\]

Now we have step three, our challenge is as follows: Suppose $y\in K(i)^\times$ such that $N(y)=1$. Can we construct an $x\in K(i)^\times$ such that $y=\delta(x)$? To do this, we will set $x=1+y$, then

\[\delta(x)=\frac{x}{\bar{x}}=\frac{1+y}{1+\bar{y}}=\frac{1+y}{1+y^{-1}}=\frac{1+y}{y^{-1}(y+1)}=y,\]

note that we are explicitly using the fact that $\bar{y}=y^{-1}$, which follows from $y\in \text{Ker}(N)$. There’s actually one minor issue here, which is that if $y=-1$, then $x=0\not\in K(i)^\times$. But this is just a single case, so we can verify that $y=\delta(i)$ in this case.

Now we collect the properties that we have proven about $\delta$: it is a group homomorphism from $K(i)^\times$ to itself, with kernel $K^\times$ and image $\text{Ker}(N:K(i)^\times\to K^\times)$. By the first isomorphism theorem, this means that $K(i)^\times/K^\times$ is isomorphic to $\text{Ker}(N:K(i)^\times\to K^\times)$. By since we previously obtained an isomorphism $\varphi$ between $\text{Ker}(N:K(i)^\times$ and $K^\times)\cong \text{SO}_2(K)$. Putting this all together, we get $K(i)^\times/K^\times \cong \text{SO}_2(K)$. $\quad \square$

If you like, we can encode all of this information in an exact sequence

Here, the map $\iota$ is the inclusion map.

Exercise: Try proving case 1 in the style of case 2. Hint1

Addendum: Galois Cohomology

This section is completely unnecessary to a basic understanding of everything that proceeded it in the post. However, it does put everything done on a more conceptual ground and introduces a very important tool in number theory: Galois cohomology. See, what was actually done in the proof of case 2 was a very special case of a foundational result in Galois cohomology, called Hilbert’s theorem 90, that I hope to introduce here. The following will assume familiarity with the Galois correspondence and modules.

Here’s the setup: let $L/K$ be a finite Galois extension of fields, and let $G=\text{Gal}(L/K)$ be the Galois group of this extension. The first thing to realize is that objects defined in terms of $L$ should be understood as having an action of $G$. This is a very general principle, but all we need here is that $L^\times$ is not just an abelian group, but a module over a (noncommutative) ring called $\mathbb{Z}[G]$.

What is $\mathbb{Z}[G]$? As a set, it is

\[\mathbb{Z}[G]=\left\{\sum_{g\in G}a_g[g]:a_g\in \mathbb{Z}\right\}\]

The $[g]$ are just symbols that satisfy a certain multiplication law, which is that $[g] [h]=[gh]$. The multiplication for all of $\mathbb{Z}[G]$ simply extends linearly.

How is $L^\times$ a $\mathbb{Z}[G]$ module? Well, the Galois group $G$ basically definitionally acts on $L$, so the $\mathbb{Z}[G]$-module structure comes from combining the action of $G$ and the natural $\mathbb{Z}$-module structure of any abelian group. More specifically, if $x\in L$ and $a=\sum_{g\in G}a_g[g]$, Then

\[x^a=\prod_{g\in G}g(x)^{a_g},\]

n.b. that we are writing $L^\times$ multiplicatively, so we write the action of $\mathbb{Z}[G]$ is as an exponential. In general, we call modules over $\mathbb{Z}[G]$ Galois modules.

$\mathbb{Z}[G]$ has a certain two-sided ideal $I_G$, called the augmentation ideal. We define a ring homomorphism $\epsilon:\mathbb{Z}[G]\to \mathbb{Z}$ by

\[\epsilon \left(\sum_{g\in G}a_g[g]\right)=\sum_{g\in G}a_g,\]

then we define $I_G=\text{Ker}(\epsilon)$. Next, let $A$ be a multiplicative right $\mathbb{Z}[G]$-module. Define a map $N:A\to A$ by $N(a)=\prod_{g\in G}a^g$. Note that if $A=L^\times$, $N$ recovers the regular norm map of fields. Now we are ready to introduce our first Galois cohomology group. Let $A$ be a right $\mathbb{Z}[G]$-module, define

\[\hat{H}^{-1}(L/K,A)=\text{ker}(N)/(AI_G),\]

this is called the $-1$st Tate cohomology group of $A$ (Exercise: show that this is well defined, i.e. $AI_G\subseteq \text{Ker}(N)$.) I claim that we can restate Lemma 1, in particular the third step of the proof, as $\hat{H}^{-1}(K(i)/K,K(i)^\times)\cong 0$. If $c\in \text{Gal}(K(i)/K)$ is complex conjugation, then one can show that $I_G=(1-[c])$ and $\delta$ is exponentiation by $1-[c]$. Putting these facts together, it turns out that $K(i)^\times I_G=\text{im}(\delta)$. Since Lemma 1 says that $\text{im}(\delta)=\text{Ker}(N)$, we get that $\hat{H}^{-1}(K(i)/K,K(i)^\times)=\text{ker}(N)/(K(i)^\times I_G)\cong 0$. This admits the following generalization:

Theorem 2: (Hilbert $90$) Let $L/K$ be a cyclic Galois extension of fields. Then $\hat{H}^{-1}(L/K,L^\times)\cong 0$.

This theorem was proven by Kummer in the $1850$s. (Hilbert’s name is only on it because this was the $90$th theorem in his Zahlbericht.) This is not what will often be stated as Hilbert $90$ in many modern texts, instead, they will state a generalization by Emmy Noether in the $1930$s, but this generalization is surprisingly unimportant for our story. (That does not, however, mean Noether’s version is unimportant. It is in fact very important and is the foundational result in Galois cohomology.)

Proof: Let $y\in \text{Ker}(N)$ and let $G=\text{Gal}(L/K)=\langle \sigma \rangle\cong \mathbb{Z}/n\mathbb{Z}$. We wish to show that $y\in AI_G$. In this case, we can actually write $I_G$ as the principal ideal $(1-\sigma)$, so this is equivalent to showing that $y=x^{1-\sigma}$ for some $x$. I will introduce a rather unusual function $x(t)$ defined on $L$.

\[x(t)=\sum_{j=0}^{n-1}t^{[\sigma^j]} y^{1+[\sigma] +\cdots+[\sigma^{j-1}]}\]

Now, we want to find a formula for $x(t)^{[\sigma]}$.

\[x(t)^{[\sigma]}=\sum_{j=0}^{n-1}t^{[\sigma^j][\sigma]} y^{(1+[\sigma] +\cdots+[\sigma^{j-1}])[\sigma]}=\sum_{j=0}^{n-1}t^{[\sigma]^{j+1}}y^{[\sigma]+[\sigma^2]+ \cdots+[\sigma^{j}]}=y^{-1}\sum_{j=0}^{n-1}t^{[\sigma^{j+1}]}y^{1+[\sigma] +\cdots+[\sigma^{j}]}\]

Now, if we inspect this sum, we notice that it is very similar to $x(t)$. In particular, its $j=0,1,\cdots,{n-2}$ terms correspond to the $j=1,2,\cdots,n-1$ terms of $x(t)$. If we look at the $j=n-1$ term, we get $t^{[\sigma^n]}y^{1+[\sigma]+\cdots+[\sigma^{n-1}]}$. But $[\sigma^n]=1$, so $x^{[\sigma^n]}=x$. Furthermore, $a^{1+[\sigma]+\cdots+[\sigma^{n-1}]}=N(a)=1$ by assumption, so the $j=n-1$ term of the sum on the right is equal to $y$, which is precisely the $j=0$ term of $x(t)$. The upshot of this is that

\[x(t)=\sum_{j=0}^{n-1}t^{[\sigma^{j+1}]}y^{1+[\sigma] +\cdots+[\sigma^{j}]}\]

so

\[x(t)^{[\sigma]}=y^{-1}x(t) \Longrightarrow y=x(t)/x(t)^{[\sigma]}.\]

This almost implies that $y\in L^\times I_G$, except there is one technicality: the possibility that $x(t)$ is $0$. However, there is always some value of $t$ such that $x(t)\neq 0$. To show this, note that $x(t)$ is a nonzero linear combination of field automorphisms. But field automorphisms are known to be linearly independent (see for example Marcus Number Fields Ch. 4 Ex. 15.) so we are done. $\qquad \square$

This group, $\hat{H}^{-1}(L/K,A)$ is part of a series of infinite groups, $\hat{H}^i(L/K,A)$ for $i\in \mathbb{Z}$, called the Tate cohomology groups. There’s also a similar series of groups $H^i(L/K,A)$, which are the regular Galois cohomology groups. Together, these form extremely refined invariants of a $\mathbb{Z}[G]$-module $A$ that can enable a ton of algebraic number theory and arithmetic geometry. Hopefully, the concrete problem of determining the structure of $\text{SO}_2(K)$ has helped motivate why some of these constructions are interesting.

  1. In case 1, $K(i)$ makes no sense since $i$ is already in $K$. Instead, you want to work with $K[x]/(x^2-1)$. What’s the structure of this ring? ↩