This is an edited version of the original question taking into account the comments below by Bruce. The original formulation was imprecise.

Let $\mathfrak{g}$ denote a complex simple Lie algebra of type $F_4$. Its smallest nontrivial irreducible representation is 26-dimensional. Let's call it $V$. This question is about the invariants of $\mathfrak{g}$ in this representation.

It is well-known that $\mathfrak{g}$ leaves invariant a quadratic form $Q \in \operatorname{Sym}^2 V$ and a cubic form $C \in \operatorname{Sym}^3V$ on $V$. Indeed, $\mathfrak{g}$ can be characterised as the Lie subalgebra of $\mathfrak{sl}(V)$ which leaves invariant $Q$ and $C$. Since $V$ is irreducible, $Q$ is non-degenerate and we may use it to identify $V$ with $V^*$ as $\mathfrak{g}$ modules.

It seems to be part of the group theoretical folklore in the Physics literature (starting possibly with this paper) that any $\mathfrak{g}$-invariant tensor on $V$ --- that is, any $\mathfrak{g}$-invariant element of in $\bigoplus_{n\geq 0} V^{\otimes n}$ --- can be constructed out of $Q$ (and its inverse), $C$ and a nonzero "volume element" $\nu \in \Lambda^{26}V$ via products in the tensor algebra and contractions.

For example, we can construct six invariant tensors out of $Q$ and $C$ in degree $4$ $$ Q_{ab}Q_{cd} \qquad Q_{ac}Q_{bd} \qquad Q_{ad}Q_{bc} \qquad C_{abe}C_{cdf} Q^{ef} \qquad C_{ace}C_{bdf} Q^{ef} \qquad C_{ade}C_{bcf} Q^{ef} $$ which satisfy a linear relation, since there is only a 5-dimensional space of such tensors.

Now, a quick calculation in LiE reveals that there is a $\mathfrak{g}$-invariant tensor $\Phi \in \Lambda^9 V$

 > alt_tensor(9,[0,0,0,1],F4)|[0,0,0,0]

which cannot be constructed out of $Q$, $C$ and $\nu$ in the aforementioned way.

One possible way to understand $\Phi$ is to think in terms of the $\mathfrak{so}(9)$ subalgebra of $\mathfrak{g}$. Under $\mathfrak{so}(9)$, $V$ breaks up as a direct sum of the trivial ($\Lambda^0$), vector ($\Lambda^1$) and spinor ($\Delta$) irreducible representations: $$ V = \Lambda^0 \oplus \Lambda^1 \oplus \Delta $$

There are precisely two $\mathfrak{so}(9)$-invariants in $\Lambda^9 V$: one is the volume form on $\Lambda^1$ and the other is the "volume" form on $\Lambda^0$ wedged with the $\mathfrak{so}(9)$-invariant 8-form on $\Delta$. Notice that $(\mathfrak{so}(9),\Delta)$ is the holonomy representation for the Cayley plane $F_4/\operatorname{Spin}(9)$, which is well-known to have a parallel self-dual $8$-form. Then $\Phi$ is some linear combination of these two $\mathfrak{so}(9)$-invariants, which I have yet to work out.


I have two questions and, as usual, I would be very grateful for any pointers to the relevant literature:

  1. Can every invariant tensor be constructed out of $Q$ (and its inverse) $C$, $\nu$ and $\Phi$ by products and contractions?

  2. Is there a more convenient (for calculations) description of $\Phi$? In particular, I would like to know about the relation of the form $\Phi \otimes \Phi = \cdots$.

Thank you in advance.

  • 2
    $\begingroup$ Modern (but to some extent pre-computer) mathematics textbooks follow Weyl's tradition, with little said about tensor invariants for exceptional Lie algebras or groups. I'm not sure how much physics motivation there is, but the only book I've noticed has (probably not by coincidence) the same author as the paper you link to: Cvitanovi´c, Predrag, Group theory. Birdtracks, Lie’s, and exceptional groups. Princeton University Press, Princeton, NJ, 2008. I wonder if it makes any sense to approach $F_4$ as a "folding" of the more symmetric $E_6$ for your purposes? $\endgroup$ May 6, 2012 at 23:31
  • 1
    $\begingroup$ If you just want the dimension of the space of invariant tensors you can use alt_tensor(9,[0,0,0,1],F4)|[0,0,0,0] $\endgroup$ May 7, 2012 at 6:27
  • 2
    $\begingroup$ @Jose, Abdelmalek: My expectation is that all invariant tensors can be constructed (in the agreed sense) by the tensors $Q_{ab}$, $Q^{ab}$ and $C_{abc}$ (including $\nu$ and $\Phi$). However I do not have a proof so this should be taken as an opinion. I would be very interested in settling this problem. $\endgroup$ May 9, 2012 at 21:03
  • 1
    $\begingroup$ For what it's worth, and to partially answer @RobertBryant's comment, the Hilbert-Poincaré series $\sum_{n=0}^{+\infty} c_n t^n$ where $c_n = \dim((\mathcal{S}^n V)^{F_4})$ is the dimension of the symmetric $F_4$-invariants experimentally appears to be $\frac{1}{(1-t^2)(1-t^3)}$ in the sense that it coincides with it to $O(t^{31})$. Concerning $\dim((\bigwedge^n V)^{F_4})$, it is $1$ iff $n \in \{0,9,17,26\}$ and $0$ otherwise. $\endgroup$
    – Gro-Tsen
    Apr 28 at 15:59
  • 1
    $\begingroup$ @Gro-Tsen: Your formula for the Hilbert-Poincaré series of the polynomial invariants of $\mathrm{F}_4$ acting on its 26-dimensional representaion is correct because it is the polar representation associated to the rank 2 symmetric space $\mathrm{E_6}/\mathrm{F}_4$. As usual in such a case, every invariant polynomial is determined by its restriction to the abelian section $\mathbb{R}^2\simeq{\frak{a}}\subset V$, and these must be invariant under the Weyl group of the symmetric space, and hence this algebra is freely generated by $x^2+y^2$ and $x^3-3xy^2$. $\endgroup$ Apr 28 at 19:56

3 Answers 3


This is more of an extended comment/enlargement of the original question/wild speculation than an answer. Due to my poor recollection of reading, a while back, Klein's Erlangen program and Franz Meyer's review on classical invariant theory, I thought Jose's Q 1 was an easy consequence of a 19th's century theorem which would say (disclaimer: what follows is meta-math):

`Let $G$ be a group acting on some vector space $V$. Let $A$ denote a collection of tensors in some tensor powers of $V$ and its dual (such as $Q$ and $C$ above). Let $H$ be the subgroup of $G$ defined as the stabilizer of $A$. Suppose we know the first fundamental theorem for $G$: namely a description of invariants of $G$ using contractions of a finite set of elementary pieces. Then one can get the FFT of $H$ simply by adjoining the pieces in the collection $A$.'

Suppose for instance one looks at a polynomial $P({\mathbf{F}})$ where $\mathbf{F}$ is some generic $V$-tensor. Suppose $P(g{\mathbf{F}})=P(\mathbf{F})$ for all $g$ in $H$. Then the previous meta-theorem is the statement that there exists a polynomial $\Phi(\mathbf{F},\mathbf{A})$ involving this time a generic tensor or collection of tensors $\mathbf{A}$ of the same format as $A$, such that:

1) $\Phi$ is invariant by the full group $G$, i.e., $\Phi(g\mathbf{F},g\mathbf{A})=\Phi(\mathbf{F},\mathbf{A})$ for all $g$ in $G$.

2) the specialization of $\Phi(\mathbf{F},\mathbf{A})$ to $\mathbf{A}:=A$ is the original subgroup invariant $P(\mathbf{F})$.

If this were true then the FFT for $G=GL(V)$ would be the mother of all FFT's. The statement works for going from $GL(n)$ to $SL(n)$ and taking $A$ to be the epsilon Levi-civita tensor. For orthogonal and symplectic groups one takes for $A$ a symmetric or antisymmetric form. But it works also if $G=SL(2)$ and $A=e_1=(1, 0)^T$ so that $H$ is the group of upper triangular matrices (I guess for $n>2$ one would need $A=e_1, e_1\wedge e_2,\ldots$). I thought the meta-theorem was mentioned in Franz Meyer's review, but in fact the only case he explicitly talks about is the last one. He calls that the study of "peninvariants" which is the same as seminvariants sources of covariants of binary forms etc.

Now here is a meta-meta-proof of the meta-theorem:

Let $\Phi(\mathbf{F},\mathbf{A}):=P(g\mathbf{F})$ where $g$ is some element of $G$ such that $\mathbf{A}=g^{-1}A$.

This is well defined since $g_1^{-1}A=g_2^{-1}A$ implies by definition of $H$ that $g_2 g_1^{-1}\in H$. Therefore $P(g_2\mathbf{F})=P((g_2 g_1^{-1})g_1\mathbf{F})= P(g_1\mathbf{F})$ by $H$-invariance of $P$.

This is a $G$-invariant because by construction, if $\mathbf{A}=g^{-1}A$ then $g'\mathbf{A}=(g g'^{-1})^{-1}A$ and so $\Phi(g'\mathbf{F},g'\mathbf{A})=P((gg'^{-1})g'\mathbf{F}) =P(g\mathbf{F})=\Phi(\mathbf{F},\mathbf{A})$.

Now of course there is a catch: one seems to need the $G$-orbit of $A$ to essentially (Zariski dense?) be all of the space where $\mathbf{A}$ belongs. This holds in the previous instances of the meta-theorem. Now when $A=C,Q$ as in Jose's problem, I am less sure. So I am not as convinced now by what I said in my comment above that Jose's $\Phi$ should be expressible in terms of $Q,C,\nu$ alone. Although, Bruce seems to be able to show this, so stay tuned.

In fact one does not seem to need the orbit of $A$ to fill everything, but one needs the existence of a $G$-invariant extension of the $\Phi$ I defined to the full space of $\mathbf{A}$. An idea to do this is to find some extension of $\Phi$ not necessarily invariant and then average it over $G$. For instance if $G=SL(n)$, one needs to take the Haar measure average over $SU(n)$. Usually if one does this the risk is to get zero, but this cannot happen here since the restriction of the average to $A$ should be the original $P$ which must be nonzero, otherwise the problem of expressing $P$ is moot. This sounds too good to be true, in particular in view of Nagata's counterexample for finite generation of rings of invariants. A specialist of algebraic groups would be better able to delineate the boundaries of what is solid math and what is pure fancy in the above arguments.

Since Jose's question 1 really is: is there a FFT for $F_4$? I briefly searched the literature and found this recent paper by Bruno Blind. It has some good references, in particular by Gerald Schwarz who solved this problem for $G_2$, and several papers by Iltyakov (who proves things about $F_4$ but uses notations and definitions I do not understand).

By the way, another paper by Schwarz in AIF is about binary cubics. This seems to be related to the instance of the meta-theorem where $G=SL(4)$ and $H$ is an imbedded $SL(2)$ using the 3rd symmetric power map. The $A$ in this case, I guess would be the twisted cubic curve. In general, one would need a hypersurface given by the Veronese imbedding (Hesse's transfer principle referred to in Klein's program). Classics studied the kind of things addressed in Schwarz's AIF article under the heading of "invariant types" see the book by Grace and Young Ch. XV and XVI.

The above setup also makes sense if $H$ is finite. I wonder if one can prove the fundamental theorem of symmetric polynomials (in a very complicated way) along these lines. I was toying with $A=x_1\cdots x_n$ or $x_1^p+\cdots+x_n^p$ for some well chosen power $p$ but one needs to get rid of a torus or roots of unity as well as solve the extension problem.

Update: I recently came across this article by Alexander Schrijver "Tensor subalgebras and first fundamental theorems in invariant theory". J. Algebra 319 (2008), no. 3, 1305–1319. It is related to what I said above since it deduces the FFT for classical groups from that of $GL(n)$.

Update 2: A better "classical" reference (than Klein's of Meyer's) on the above method is the book by Turnbull "The Theory of Determinants, Matrices and Invariants". In Chapter XXI (in the 1960 Edition) he calls such FFT's "Adjunction Theorems", see in particular Sections 10 and 11.

Update 3: There is another interesting reference with relevance to the main theme of this post, namely, relations between FFTs for invariant theories of different groups which is how I understand the Erlangen Program. It is the recent preprint by Deligne, Lehrer and Zhang inspired by the classic article by Atiyah, Patodi and Singer where the FFT for the orthogonal group is deduced from of that of the general linear group.


First you use two phrases "can be constructed out of" and "complete generating set". I think I understand what you mean here but it is possible there is a misunderstanding.

I was hoping you might get a better answer as the following is really a strategy and not a complete proof.

Take the category generated by your tensors (I hope we are in agreement as to what that means). Then this is a rigid symmetric category. In fact it is what Deligne calls a tannakian category together with a fibre functor. Then Tannakian reconstruction reconstructs an affine group scheme which has this category as its representation category (I think in the physics literature they would refer to Doplicher-Roberts).

There is a discussion of reconstruction at

Tannakian Formalism

Then if this is not $F_4$, what is it?

P.S. Do you need the volume form? I have not seen this in this context.

  • $\begingroup$ It is entirely possible that I may be going about this is a very naive way, but let me try to explain what I meant by the two offending phrases. The tensor algebra $\bigotimes V$ is a noncommutative algebra on which $F_4$ acts by automorphisms. The invariants therefore form a subalgebra and one can ask whether this algebra is finitely generated -- which I believe it is -- and, if so, one can ask about possible generating sets. Actually, since $Q$ is nondegenerate, I should add also its inverse to the set in the question. Does this make more sense now? $\endgroup$ May 8, 2012 at 12:46
  • $\begingroup$ That does make sense but it is not what I had in mind. Since you gave a reference to Cvitanic I understood his meaning. Namely, in index notation, you can take all the tensors you consider but then you can also raise and lower indices as well as contract indices. I am not offended; this makes a significant difference to the question. $\endgroup$ May 8, 2012 at 13:18
  • $\begingroup$ Indeed, in Cvitanović's notation, I have $Q_{ab}$, $Q^{ab}$, $C_{abc}$ and the $\varepsilon_{a...z}$. And any other invariant tensor $T_{ab...}$ should be made out of the generating set by products and contractions. $\endgroup$ May 8, 2012 at 13:23
  • $\begingroup$ I have just checked that the space of invariant tensors in degree 5 is 15. In your algebra I only see 2! $\endgroup$ May 8, 2012 at 13:24
  • $\begingroup$ Oh! so you are allowing contractions? $\endgroup$ May 8, 2012 at 13:26

This is more of a comment but the margins are too small to contain it.

The $F_4$ is a subgroup of $O(27)$. Thus it is easier to construct invariants in $\mathbb{R}^{27}$. See wikipedia. In this case there are at least 4 invariants:

$$A_n = \delta_{0n}+\delta_{1n}+\delta_{2n}$$ $$Q_{nm} = \delta_{nm}$$ $$\varepsilon^{n_1n_2...n_{27}}$$ $$C_{nmo} = [\text{something complicated}]$$

Note $Q$ and $\varepsilon$ are the symmetric and totally anti-symmetric invariants of $O(27)$. And $A$ and $C$ reduce the symmetry group to $F_4$.

Thus if we set $A_n x^n=C_1$, $Q_{nm}x^nx^m=C_2$, $C_{nmo}x^n x^m x^o=C_3$ the space has symmetry $F_4$ assuming $x$ are commuting variables. (In fact $\varepsilon_{n_1n_2...n_{27}}x^{n_1}x^{n_2}...x^{n_{27}}=0$ might be thought of as the definition that the $x$'s commute as this certainly wouldn't be true for Grassmann variables, for example. That is to say we are talking about commutative geometry here.)

Here the indices label dimensions (they are not powers).

We could use the first invariant to eliminate one of the dimensions which would alter the other invariants.

Also we note that $$Q^{aa'} \propto \varepsilon^{abcd...}\varepsilon^{a'b'c'd'...}Q_{bb'}Q_{cc'}Q_{dd'}...$$

So I don't think, the inverse of $Q$ is not a separate invariant tensor. (And anyway, in this case where $Q_{ab}=\delta_{ab}$ the inverse is obvious!)

Since $F_4$ is a subgroup of $O(27)$ it means that every transform $M$ satisfies $M^T=M^{-1}$ which means we can contract the tensors together. The only interesting ones are $C$, $\varepsilon$ and $A$. For example we might find invariants such as:

$$T_{ab} = A_n C_{nab}$$ And perhaps: $$U_{ABCD...} = C_{Aaa'}C_{Bbb'}....\varepsilon^{abcdefghijklmnopqrstuvwxyz\omega}\varepsilon^{a'b'c'd'e'f'g'h'i'j'k'l'm'n'o'p'q'r's't'u'v'w'x'y'z'\omega'}$$

Perhaps this is a better way at counting the invariants of $F_4$.

As for your original question of whether there are any more invariants? Other than made out of these? I don't know but I have a feeling not as this new invariant should imply the simpler invariants.


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