Invariants of $O(2) \times O(2)$ under simultaneous conjugation Let $G= \textrm{O}(2)$ be the group of orthogonal $2 \times 2$ matrices over $\mathbb{C}$. $G$ acts on $G \times G$ by conjugation: $g \cdot (a,b) :=(g  a g^{T}, g b g^T)$. This induces an action on the coordinate ring $A$ of the affine variety $G \times G$. Since $G$ is reductive, the invariant ring $A^G$ is finitely generated.
Question: What are the generators of $A^G$? Is there a generalization to orthogonal groups of size $>2$?
 A: It might be a good idea to study $S=\operatorname{SO}_2(\mathbb{C})$ first. This is a very special case, because $S$ is abelian and a curve. Because of these special properties, I currently don't know how to generalize the following to higher dimension.
Short proof that $S$ is abelian: $S$ acts by conjugation on itself. Each orbit under this action is connected because $S$ is connected. Since $S$ has dimension one, if any orbit is not a point, then it must be all of $S$. Hence, if $S$ were not abelian, there would be no fixed points under conjugation. However, $1\in S$ is fixed.
Now since $S$ is abelian, the action by conjugation of $S$ on $S\times S$ is trivial, the coordinate ring $\mathbb C[S\times S]^{S}=\mathbb C[S\times S]$. The group $G=\operatorname{O}(2)$ is disconnected and has two components that are conjugates of $S=\operatorname{SO}(2)$. We can choose 
$$t:=\begin{pmatrix}0&1\\1&0\end{pmatrix}$$
to get $G=S\sqcup tS$. In fact, $\langle t\rangle\cong\mathbb Z_2$ and $G=S\rtimes\mathbb Z_2$ in this way. Then,
$$G\times G=(S\times S)\sqcup(tS\times S)\sqcup(S\times tS)\sqcup(tS\times tS).$$
and each of these components is invariant under the conjugacy action of $S$. Hence, the coordinate ring $A$ of $G\times G$ is a direct product of four rings $A=A_{00}\times A_{10}\times A_{01}\times A_{11}$ where $A_{ij}=\mathbb C[t^iS\times t^jS]$. Let us study the orbit of $t$ under $S$:
$$\begin{pmatrix} x_1 & x_2 \\ y_1 & y_2 \end{pmatrix}
\begin{pmatrix}0&1\\1&0\end{pmatrix}
\begin{pmatrix} x_1 & y_1 \\ x_2 & y_2 \end{pmatrix}
=
\begin{pmatrix} x_2 & x_1 \\ y_2 & y_1 \end{pmatrix}
\begin{pmatrix} x_1 & y_1 \\ x_2 & y_2 \end{pmatrix}
=
\begin{pmatrix} 2x_1x_2 & y_1x_2+y_2x_1 \\ y_1x_2+y_2x_1 & 2y_1y_2 \end{pmatrix}
=t\begin{pmatrix} y_1x_2+y_2x_1 & 2y_1y_2 \\ 2x_1x_2 & y_1x_2+y_2x_1 \end{pmatrix}
$$
Now this is a nontrivial orbit! And by the same argument as above, we conclude that the orbit of $t$ must be all of $tS$. In fact, we can show that $(S\sqcup tS)/S$ and $(tS\sqcup S)/S$ and $(tS\sqcup tS)/S$ are all isomorphic to $S$ via


*

*projection to the first coordinate

*projection to the second coordinate

*multiplication $(g,h)\mapsto g^{-1}h$


Note that all these morphisms are equivariant with respect to the conjugacy action.
I will give an example for the last one because it is probably the least obvious
Consider the multiplication morphism
\begin{align*}
\mu: tS\times tS &\longrightarrow S \\
 (g,h) & \longmapsto g^{-1}h
\end{align*}
This is well defined because the $t$'s cancel. It is also surjective because $\mu(t,ts)=s$ for any $s\in S$. Furthermore, for any $u\in G$, we have
$$\mu(ugu^T,uhu^T)=(ugu^T)^Tuhu^T=ug^Tu^Tuhu^T=ug^Thu^T=u\cdot\mu(g,h)\cdot u^T$$
which means that $\mu$ is equivariant with respect to the action of $G$ by conjugation. In particular, it is equivariant with respect to the action of $S$ by conjugation. By the universal property of the quotient, $\mu$ therefore factors as a morphism $\bar\mu:(tS\times tS)/S\twoheadrightarrow S$ which is surjective. I claim that it is also injective. Essentially, I am proving that $S$ is also a quotient of $tS\times tS$ by $S$. Indeed, we have $\mu^{-1}(s)=\{ (t\tilde s,t\tilde s s) \mid \tilde s\in S \}$ because $\mu(t\tilde s,th)=s$ means $\tilde s^{-1}h=s$, so $h=\tilde ss$. Such a fiber is exactly the orbit of $(t,ts)$ under the conjugacy action of $S$. Hence, the fibers of $\mu$ are the orbits of $S$ and consequently, the induced map $\bar\mu$ is injective. Since this is a bijective morphism that maps to a smooth variety, it is an isomorphism by some corollary to Zariski's Main theorem.
Hence, $A^S = A_{00}^S\times A_{01}^S \times A_{10}^S\times A_{11}^S=\mathbb C[S\times S]\times \mathbb C[S]^3$, where $G$ acts on $\mathbb C[S]$ by conjugation. We are left to pass from conjugation by $S$ to conjugation by $G$. We have
$$A^G = A^{S\rtimes\langle t\rangle}= (A^S)^{\langle t\rangle}= \mathbb C[S\times S]^{\langle t\rangle} \times (\mathbb C[S]^3)^{\langle t\rangle},$$
where the second equality follows because for any $S$-invariant $f\in A$, the function $t.f$ is still $S$-invariant: Indeed, for $s\in S$ we have $s.t.f=st.f=t.(t^{-1}st).f=t.f$ because $t^{-1}st\in S$ leaves $f$ invariant.
Let $B:=\mathbb C[S]$. We study the invariants of this ring under a very finite group, namely $\langle t\rangle\cong\mathbb Z_2$. By general lore, $B^{\langle t\rangle}\hookrightarrow B$ is an integral extension of degree two. We have 
$$B=\mathbb C[x_1,x_2,y_1,y_2]$$
where we use the coordinates
$$\begin{pmatrix} x_1 & x_2 \\ y_1 & y_2 \end{pmatrix},\quad\text{such that}\quad
t\begin{pmatrix} x_1 & x_2 \\ y_1 & y_2 \end{pmatrix}t=
\begin{pmatrix} y_2 & y_1 \\ x_2 & x_1 \end{pmatrix}$$
so the action of $t$ permutes these generators by switching $x_1$ and $y_2$ as well as $x_2$ and $y_1$.
Now, the ring $B':=\mathbb C[x_1+y_2, x_2+y_1, x_1y_2, x_2y_1]$ is contained in $B^{\langle t\rangle}$. Furthermore, $B'[x_1,x_2]=B$ and this is a degree two extension. Indeed, $x_1$ is a zero of 
$$T^2 - (x_1+y_2) T + x_1y_2$$ 
and we then have the equality
$$x_2(x_1-y_2)=x_1x_2-x_2y_2=x_1x_2+x_1y_1-1=x_1(x_2+y_1)-1.$$
Therefore, when we work in the quotient field, we have
$$x_2 = \frac{x_1(x_2+y_1)-1}{(x_1-y_2)}$$
which means that $x_2$ is contained in the quotient field of $B'[x_1]$. Therefore, the quotient field of $B$, which is the quotient field of $B'[x_1,x_2]$, is the same as the quotient field of $B'[x_1]$. Thus, $B$ is a degree two extension of $B'$ and $B'$ contains only invariants, thus we must have $B'=B^{\langle t\rangle}$.
We are left to determine $\mathbb C[S\times S]^{\langle t\rangle}$, but that should work in the exact same way.
I admit that this is a little rough around the edges, but it already turned out to be quite long. Maybe someone can give a shorter, better and more precise description or you can fill the gaps. I am happy to edit this answer if you have specific requests.
