Consider the representation of $\textrm{SO}(4)$ on $\textrm{Sym}^2(\wedge^2\mathbb{R}^4)$ induced by the standard representation of $\textrm{SO}(4)$ on $\mathbb{R}^4$. I am interested in the ring of invariants of this representation, i.e. the ring of all polynomial functions on $\textrm{Sym}^2(\wedge^2\mathbb{R}^4)$ that are invariant.


Here is a way of constructing some of these invariants. The space of $\textrm{SO}(4)$-linear maps $\wedge^2\mathbb{R}^4\to\wedge^2\mathbb{R}^4$ is $2$-dimensional: We have the identity $I$ and the Hodge star operator $\star$. Thus the coefficients of the polynomial $$\det(x\cdot I+y\cdot \star+A)\in\mathbb{R}[x,y]$$are polynomials in the entries of $A\in\textrm{Sym}^2(\wedge^2\mathbb{R}^4)$ which are $\textrm{SO}(4)$-invariant. My question is: Do these polynomials already generate the ring of invariants?



Invariant ring of $\textrm{Sym}^2(\wedge^2\mathbb{R}^4)$ under $\textrm{SO}(4)$
The answer is 'no', though I don't know an easy way to see this without doing an explicit calculation. Here is where to look though, if you want to do the calculation yourself:


Things work out a bit better if one uses indeterminates $z = x+y$ and $w = x-y$. Then one has an expansion
$$
\det\bigl(x{\cdot}I + y{\cdot}\star + A\bigr) = \sum\_{0\le i,j\le 3} P\_{ij}(A)\,z^{3-i}w^{3-j}
$$
where $P\_{ij}(A)$ is a polynomial in $A$ of degree $i{+}j$.


One finds that $P\_{00}=1$. Meanwhile, the polynomials $P\_{10}$, $P\_{01}$, $P\_{20}$, $P\_{11}$, $P\_{02}$, $P\_{30}$, $P\_{21}$, $P\_{12}$, and $P\_{03}$ are independent and generate all of the polynomial invariants of degree $3$ or less. Moreover, the three quartic polynomials $P\_{3,1}$, $P\_{2,2}$, and $P\_{1,3}$ are algebraically independent over the ring generated by the lower degree $P\_{ij}$.


However, the $P\_{ij}$ with $i+j\le 4$ do not generate all of the invariant polynomials of degree $4$. There is one further quartic polynomial invariant, say $Q$, that is not a polynomial in the $P\_{ij}$ with $i+j\le 4$. Since the remaining three coefficients $P\_{3,2}$, $P\_{2,3}$ and $P\_{3,3}$ are all of degree 5 or higher, $Q$ does not lie in the ring generated by the $P\_{ij}$.


**Added detail:** $\mathrm{SO}(4)$ preserves the splitting $\Lambda^2(\mathbb{R}^4)= \Lambda^2\_+(\mathbb{R}^4)\oplus\Lambda^2\_-(\mathbb{R}^4)$ into self-dual and anti-selfdual forms, i.e., the split into the eigenspaces of $\star$, which has trace-zero and square $\star^2 = I$. It follows that we can write block decompositions
$$
\star = \begin{pmatrix} I\_3&0\_3\\0\_3&-I\_3\end{pmatrix}
\quad\text{and}\quad 
A = \begin{pmatrix} a&b\\b^T&c\end{pmatrix},
$$
where $a=a^T$, $b$ and $c=c^T$ are $3$-by-$3$ matrices. The action of $\mathrm{SO}(4)$ on $A$ can be described as
$$
g\cdot A = \begin{pmatrix} g\_+ag\_+^T&g\_+bg\_-^T\\g\_-b^Tg\_+^T&g\_-cg\_-^T\end{pmatrix}
$$
where $g\_\pm:\mathrm{SO}(4)\to\mathrm{SO}(3)$ are homomorphisms such that $(g\_+,g\_-):\mathrm{SO}(4)\to\mathrm{SO}(3)\times\mathrm{SO}(3)$ is surjective, with kernel $\{\pm I\_4\}$. We can write
$$
\det(x{\cdot}I\_6 + y{\cdot}\star + A)
= \det\begin{pmatrix} zI\_3+a & b\\ b^T& wI\_3 + c\end{pmatrix}
= \sum\_{0\le i,j\le 3} P\_{ij}(A)\,z^{3-i}w^{3-j},
$$
and we can refine our notation by writing $P\_{ij}(A) = P\_{ij}(a,b,c)$.


Now, because of the way $a$, $b$, and $c$ transform under the action of $\mathrm{SO}(4)$, we see, for example, that $\mathrm{tr}(a^k)$ and $\mathrm{tr}(c^l)$ are invariant polynomials, as is $\mathrm{tr}(bb^T)$ (though $\mathrm{tr}(b)$ is not), and, indeed, we see, by inspection, that, for example,
$$
P\_{10} = \mathrm{tr}(a)\quad\text{and}\quad P\_{01} = \mathrm{tr}(c),
$$
and a little reflection shows that
$$
P\_{20} = \tfrac12\bigl(\mathrm{tr}(a)^2-\mathrm{tr}(a^2)\bigr)
\quad\text{and}\quad
P\_{02} = \tfrac12\bigl(\mathrm{tr}(c)^2-\mathrm{tr}(c^2)\bigr)
$$
while it's not hard to see that
$$
P\_{11} = \mathrm{tr}(a)\mathrm{tr}(c) - \mathrm{tr}(bb^T).
$$
To go further, imagine that we have an alphabet of 4 letters, $a$, $b$, $b^T$, and $c$, and we declare a valid `word' to be string of these letters subject to the following rules: $a$ can be followed only by $a$ or $b$, $b$ can only be followed by $c$ or $b^T$, $c$ can only be followed by $c$ or $b^T$, $b^T$ can only be followed by $a$ or $b$, and finally, that the number of $b$s in the word is the same as the number of $b^T$s. Thus, for example, $a^4$, $abb^T$, and $abcb^T$ are valid words, but $abc$ is not valid. What one sees that is a vaild word $w$, when interpreted as a product of matrices, transforms under $\mathrm{SO}(4)$ into $g\_\pm w g\_\pm^T$, and hence $\mathrm{tr}(w)$ is an invariant polynomial for any valid word.


Now we can go on: For example
$P\_{30} = \det(a)$ which is a weighted homogeneous polynomial in $\mathrm{tr}(a)$, $\mathrm{tr}(a^2)$, and $\mathrm{tr}(a^3)$, and similarly with $P\_{03}= \det(c)$. Moreover, one finds that
$$
P\_{21} = P\_{20}P\_{01}-P\_{10}(P\_{10}P\_{01}-P\_{11})
+\mathrm{tr}(abb^T)
$$
with a similar formula for $P\_{12}$. Thus, all of the polynomials $P\_{ij}$ for $0<i+j\le 3$ are expressed as polynomials in the 9 (independent) invariants
$$
\mathrm{tr}(a),\mathrm{tr}(c),\ \mathrm{tr}(a^2),\mathrm{tr}(c^2),\mathrm{tr}(bb^T),\ \mathrm{tr}(a^3),\mathrm{tr}(abb^T),\mathrm{tr}(cb^Tb), \mathrm{tr}(c^3),
$$
and conversely. These are all of the invariants of degree $3$ or less. (Of course, $\mathrm{tr}(a^4)$ is a polynomial in these. Not all valid words yield an invariant that cannot be expressed as a polynomial in invariants of lower degree.)


But now, one sees where the discrepancy comes in. The quartic polynomials $P\_{31}$, $P\_{22}$, $P\_{13}$ can be expressed in terms of the 'word-trace' invariants, but there are ony three of them, whereas there are *four* new quartic word-trace invariants:
$$
\mathrm{tr}(aabb^T),\ \mathrm{tr}(bb^Tbb^T),\ \mathrm{tr}(ccb^Tb),\ \mathrm{tr}(abcb^T)
$$
and it is not hard to see that no nontrivial linear combination of them can be written as a polynomial in the word-trace invariants of lower degree.


In fact, the situation gets worse: There are only two $P\_{ij}$ of degree 5, but there are *four* new word-trace invariants of degree 5, and $P\_{33}$ is the only one of degree 6 while there are *two* new word-trace invariants of degree 6.


