# Noether's bound for anticommutative invariant theory (diff. forms instead of polynomials)?

EDIT: Now with a concrete request to CAS experts (see the end of the post).

Let $G$ be a finite group, and $V$ a finite-dimensional representation of $G$. The classical invariant theory of $G$ and $V$ is a study of the $G$-invariants in the $K$-algebra $K\left[V\right]=\mathrm{S}\left(V^{\ast}\right)$, where $\mathrm{S}$ means the symmetric algebra. One has results like the Noether theorem, which says that over a field of characteristic $0$, the $G$-invariants in this $K$-algebra $\mathrm{S}\left(V^{\ast}\right)$ are generated (as a $K$-algebra) by the $G$-invariants in its submodule $\mathrm{S}^1\left(V^{\ast}\right)\oplus \mathrm{S}^2\left(V^{\ast}\right)\oplus ...\oplus \mathrm{S}^{\left|G\right|}\left(V^{\ast}\right)$.

Is there a similar theory, with similar bounds, for the $K$-algebra $\wedge \left(V^{\ast}\right)$ ? Actually this looks a bit simpler, because $\wedge^k \left(V^{\ast}\right)$ is nonzero only for $k\leq\dim V$; but $\dim V$ might still be greater than $\left|G\right|$. If there is an analogue of Weyl's theorem (Theorem A in 8.3 of Kraft-Procesi - see the Example in §8.5 for how I want to apply it), then we could limit ourselves to the case of $V$ being the regular representation, and then Noether's bound would follow. But there is no direct analogue of Weyl's theorem (in the sense of replacing $K\left[V^p\right]$ by $\wedge \left(V^{\ast p}\right)$ and $K\left[V^n\right]$ by $\wedge \left(V^{\ast n}\right)$).

The only thing from classical invariant theory which I can extend to this "anticommutative" (because $\wedge^k \left(V^{\ast}\right)$ can be seen as "anticommutative" polynomials on $V$) invariant theory is the Molien series, which takes the form

$\sum\limits_{d=0}^{\infty} \dim \left( \left(\wedge^d V\right)^G\right) T^d = \frac{1}{\left|G\right|}\sum\limits_{g\in G} \det\left(1+Tg\right)$

in our anticommutative setting, again only in the characteristic $0$ case. (Note that I have switched from $V$ to $V^{\ast}$, but this should not matter in the end since the the dimension of the invariants in some given representation always equals the dimension of the invariants in its dual - when we are in the characteristic $0$ case at least.)

Concrete request to anyone who has some experience with computer algebra systems:

Consider the exterior algebra freely generated by the differential forms $dx_1$, $dx_2$, ..., $dx_n$, $dy_1$, $dy_2$, ..., $dy_n$, $dz_1$, $dz_2$, ..., $dz_n$ over $\mathbb Q$. (This is the exterior algebra of a $3n$-dimensional vector space.)

Let the symmetric group $S_3$ act on this algebra by permuting $x$, $y$, $z$ (leaving the numeric indices unchanged).

For every $k\in\left\lbrace 1,2,...,n\right\rbrace$, the sum $dx_1\wedge dx_2\wedge ...\wedge dx_k + dy_1\wedge dy_2\wedge ...\wedge dy_k + dz_1\wedge dz_2\wedge ...\wedge dz_k$ is an invariant under this action. Does it lie in the subalgebra generated by invariants of smaller degree? (Note that it is enough to consider homogeneous invariants, so that there cannot be degree-reducing cancellations. Also note that it is easy to find a generating set of all invariants of given degree just by writing out all the monomials and averaging them over the action of $S_3$.) A negative answer for $n=k=7$ (or higher) would destroy the Noether bound in the anticommutative case.

I played with the idea of coding the exterior algebra as an algebraic type in Haskell, but finding out whether an element is generated by elements of smaller degree means solving a system of linear equations, and I don't have enough experience to code Gaussian elimination in Haskell. Other than that, there must surely be a CAS that knows how to work with anticommuting variables?

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Where is the dependency on $d$ in the coefficients of the series on the LHS of the equality? –  Qfwfq Sep 13 '10 at 12:07
Thanks, seems I'm not able to write a single page without typos. –  darij grinberg Sep 13 '10 at 12:21
This is more in the direction of you question title than your actual question (hint!), but have you looked at Roger Howe's paper "Remarks on classical invariant theory" and his Schur's lectures? Some of this theory is also in Goodman and Wallach, if I remember correctly. In particular, one of the techniques of generating invariants in the Grassman algebra is taking the top exterior power of a span of a $G$-orbit (which is a decomposable $k$-vector). By the way, what is Weyl's theorem that you referred to? –  Victor Protsak Sep 13 '10 at 17:21
Oh, thanks. Do you mean I should search for "super invariant theory"? I have already, to no avail. No, I could not locate the Schur notes (when I see them in a library, I will scan them), but I have his paper. I don't see where this is considered in Goodman-Wallach, but thanks for the reference (it's new to me). The Weyl theorem I referred to is Theorem A on page 73 of math.unibas.ch/~kraft/Papers/KP-Primer.pdf (there may be a minor typo here: instead of $K\left[V^p\right]^G=\left< K\left[V^n\right]^G \right>_{\mathrm{GL}_p}$, the proof only shows that $K\left[V^p\right]^G$ ... –  darij grinberg Sep 13 '10 at 17:43
There is a discussion of exterior invariant theory in Chapter 9 of L. Smith, Polynomial Invariants of Finite Groups, though he does not discuss an analogue of Noether's theorem. –  Richard Stanley Sep 14 '10 at 2:03

There is a recent work of Weiqiang Wang and his student Jinkui Wan on spin-invariant theory. It appears to be the first of its kind.

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Aha, thanks - this seems to handle the case of the group $S_n$ with its standard representation. No hints on the general group case, however... –  darij grinberg Oct 25 '10 at 23:59

PARTIAL answer: By a theorem of Solomon (Louis Solomon, Invariants of Finite Reflection Groups, Nagoya Math. J. Volume 22 (1963), pp. 57-64, also Julia Hartmann and Anne V. Shepler, Reflection Groups and Differential Forms), the Noether bound holds when the action of $G$ on $V$ is a reflection group. (Moreover, in this case, even the tensor product $S\left(V\right)\otimes \wedge\left(V\right)$ is generated in degree $\leq \left| G\right|$. I am wondering what can be said about $S\left(V\right)\otimes \wedge\left(W\right)$ for two different reflection representations $V$ and $W$ of $G$.)

This doesn't say anything about the $S_3$-module I am considering to be a potential counterexample, though ($S_3$ doesn't act as a reflection group on it). I would still be very indebted for some Sage code to work out this case.

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The Maple CAS works with anticommuting variables - see its Physics package - and also with differential geometry in general - see its DifferentialGeometry package.

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