Let $V = (\mathbb{R}^n, g)$, where $g$ is the Euclidean inner product on $V$. Denote by $G$ the orthogonal group $O(V) = O(n)$ and by $\mathfrak{g}$ the Lie algebra of $G$.

Let $W \subset \Lambda^2V^* \odot \Lambda^2V^*$ be the subset satisfying the algebraic Bianchi identity. More precisely, let $R(v_1,v_2,v_3,v_4)$ denote an element of $\Lambda^2V^* \odot \Lambda^2V^*$. Thus $R$ is skew-symmetric in $v_1$ and $v_2$ and it is also skew-symmetric in $v_3$ and $v_4$. Moreover

$$ R(v_3,v_4,v_1,v_2) = R(v_1,v_2,v_3,v_4). $$

Then $R \in W$ if and only if, in addition to the conditions above, $R$ also satisfies the following identity (known as the algebraic Bianchi identity):

$$ R(v_1,v_2,v_3,v_4) + R(v_2,v_3,v_1,v_4) + R(v_3,v_1,v_2,v_4) = 0. $$

Now my question can be formulated. What is an explicit description of the ring $\mathbb{C}^G[W]$ of $G$-invariant polynomials on $W$ (with $W$ being the space of algebraic curvature tensors, if I may call it so) and $G$ acting on $W$ by restricting its natural action on $\Lambda^2V^* \odot \Lambda^2V^*$.

Also, if one fixes a degree $d > 0$, what is an explicit description of the space of $G$-invariant homogeneous polynomials in $W$ of degree $d$?

I was thinking at first about the Chern-Weil homomorphism, but I think this only gives a proper subspace of $G$-invariant polynomials on $W$, and not all of them (I am not a 100% sure). This has probably been studied in the literature. I don't have access to MathSciNet anymore though (due to some budget cuts by my University).

Edit 1: I notice some overlap with the post Invariant polynomials in curvature tensor vs. characteristic classes, but the posts are sufficiently different.


2 Answers 2


I am not sure that this has a "nice" answer. Your question can be reformulated as follows. Let $\mathcal{A}_n$ be the space of algebraic curvature tensors on $\mathbb{R}^n$. A homogenous polynomial $P$ on $\mathcal{A}_n$ is the same as an element of $S^k\mathcal{A}_n$, the $k$-th symmetric tensor power of $\mathcal{A}_n$. Now if $H_k$ is the space of homogeneous polynomials of degree $k$ on $\mathcal{A}_n$, then $H_k \subset S^k \mathcal{A}_n$ is a subrepresentation of $G$.

In other words, a recipe to obtain an answer to your question for specific $k$, $n$, is the following. Decompose the $G$-representation $S^k \mathcal{A}_n$ into irreducible $G$-representations and count the number of trivial representations among those. This can be done for low $k$, $n$ using software such as LiE.

Note that as a $G$-representation, the space $\mathcal{A}_n$ splits into the direct sum of three irreducible representations: $$ \mathcal{A}_n = \mathbb{R} \oplus S^2_0(\mathbb{R}^n) \oplus \mathcal{W},$$ where $\mathcal{W}$ is the space of Weyl curvature tensors (i.e. those curvature tensors that are additionally entirely trace-free).

A quick check on LiE shows that there are plenty such polynomials: For example, looking for polynomials that depend on the Weyl part alone and $n$ large, there is one of degree 2 and four of degree 3. I doubt there is a good general answer.

  • $\begingroup$ I should check out the LiE software. I had heard of it before, but never got to install it and try it out. Your answer and @RobertBryant's answer are close in spirit. I am not sure which one to choose. They are both good answers, which are both telling me that I am not asking a very good question, in a way :) (or at least, I am not asking a question with a very nice answer). $\endgroup$
    – Malkoun
    Jul 25, 2020 at 19:52
  • $\begingroup$ @Malkoun - On the contrary, I think you're being told this is a very good question - just that the history of these sorts of questions is that they don't have very nice answers, until they do (but the nice answer might not be explicit in the sense of 'explicit' that you were looking for). $\endgroup$ Jul 26, 2020 at 5:15
  • $\begingroup$ Since this was kind of a "tie" between your answer and @RobertBryant's answer, which are close in spirit, I just went for the oldest one. The reader should read both answers though, as they are really both "accepted answers", but the website does not allow me to choose more than one "accepted answer". $\endgroup$
    – Malkoun
    Jul 26, 2020 at 11:24

I think this is unlikely to have a very nice answer. When $n=2$ and $n=3$, the answer is simple, but, already for $n=4$, it's not likely to be easy to give a set of generators and relations for the $\mathrm{O}(n)$-invariant polynomials on the vector space $\mathcal{R}_n$ of algebraic curvature tensors in dimension $n$. (I'm avoiding the OP's notation of $W$ for this space because it doesn't explicitly reference the dimension $n$ and I don't want to confuse it with the space of Weyl curvature tensors.)

Since $\mathcal{R}_n$ has dimension $\tfrac1{12}n^2(n^2-1)$ and since, for $n>2$ the generic element of $\mathcal{R}_n$ has only a finite stabilizer in $\mathrm{O}(n)$, the dimension of the ring of $\mathrm{O}(n)$-invariant polynomials on $\mathcal{R}_n$ will be $$ \frac1{12}n^2(n^2-1) - \frac12n(n-1) = \frac1{12}(n+3)n(n-1)(n-2), $$ so there will always be at least that many independent generators and, when $n>3$, many more, plus a bunch of relations, since the quotient space will not be 'smooth' near the origin.

Once one gets above the low degrees when $n>3$, to compute the dimensions of the graded pieces of this ring will be complicated (essentially, one is asking for the Hilbert series of the ring of invariants). (However, the dimension of the grade 1 piece is 1, and the dimension of the grade $2$ piece is $2$ for $n=3$ and $3$ for $n>3$. If one were using $\mathrm{SO}(4)$ for $n=4$, the dimension of the grade $2$ piece would be $4$.)

I imagine that the answers for $n=4$ are known (though I don't know them) since it is, in principle, just a representation-theoretic computation.

  • $\begingroup$ I see. Thank you for your answer Prof. Bryant. This was actually a question on the side that I was hoping would help me figure out a pattern between some formulas. As far as this post is concerned, your answer and @MatthiasLudewig's answer are close in spirit. $\endgroup$
    – Malkoun
    Jul 25, 2020 at 19:49
  • $\begingroup$ Hi Robert, you calculate the dimension of the ring (algebra?) of $O(n)$-invariant polynomials on $\mathcal{R}_n$, but shouldn't this be an infinite-dimensional space? What do you mean exactly here? $\endgroup$ Jul 25, 2020 at 20:34
  • 2
    $\begingroup$ @MatthiasLudewig: I don't mean the dimension as a vector space over $\mathbb{R}$, which would be infinite, of course. I mean the Krull dimension of the ring of $\mathrm{O}(n)$-invariant polynomials as an $\mathbb{R}$-algebra. For example, the Krull dimension of the ring $\mathbb{R}[x_1,x_2,\ldots,x_n]$ as an $\mathbb{R}$-algebra is $n$. The Krull dimension of the quotient ring $\mathbb{R}[x,y]/(y^3-x^2)$ is $1$. In the case at hand, it's also equial to the maximum number of algebraically independent elements of the ring of invariants. $\endgroup$ Jul 25, 2020 at 22:05
  • 2
    $\begingroup$ Just to note that as a representation of $\mathrm{GL}(V)$, $\mathcal{R}_n$ is the irreducible highest weight representation $\Delta^{(2,2)}(V)$ labelled by the partition $(2,2)$. One class of $\mathrm{O}(V)$ invariants come from invariants for $\mathrm{SL}(V)$, which correspond, in degree $d$, to Schur functions labelled by rectangular partitions in the plethysm $s_{(d)} \circ s_{(2,2)}$. Decomposing such plethysms is known to be a tough open problem in algebraic combinatorics. $\endgroup$ Jul 26, 2020 at 11:24
  • 1
    $\begingroup$ @MarkWildon: At least in the n=4 case, it's easy to see what the $\mathrm{SL}(4,\mathbb{R})$-invariant polynomials are on $\mathcal{R}_4$: In this case, this representation is just the symmetric traceless endomorphisms of $\Lambda^2(\mathbb{R}^4)\simeq \mathbb{R}^{(3,3)}$, via the double cover $\mathrm{SL}(4,\mathbb{R})\to\mathrm{SO}^\circ(3,3)$. Thus, there are such 5 invariants, one each of degree $2$, $3$, $4$, $5$, and $6$, that are independent and generate the ring of $\mathrm{SL}(4,\mathbb{R})$-invariants on $\mathcal{R}_4\simeq\mathbb{R}^{20}$. $\endgroup$ Jul 26, 2020 at 11:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.