Think about the dihedral group $D_4$ acting on the polynomial algebra $\mathbb C[x_1, \cdots, x_4]$ via generating permutations $(x_1\ x_2)$, $(x_3\ x_4)$, and $(x_1\ x_3)(x_2\ x_4)$. I'd like to look at the polynomials that are invariant under the first two permutations and change sign under the last one; the two obvious examples are $$P_n = x_1^n + x_2^n - x_3^n - x_4^n$$ and $$Q_n = x_1^nx_2^n - x_3^nx_4^n.$$

Normally I'd think about the ring of invariants, but I don't even have a ring here - $P_1^2$ doesn't have the right invariance properties!

My main question is: How can I effectively describe the polynomials that obey these (anti)symmetries?

It seems like I'm going to end up having to consider the $D_4$-invariant polynomials, since if I have a polynomial $s$ that's $D_4$-partially-anti-invariant (as above), and a polynomial $t$ that's $D_4$-invariant, then clearly $st$ will be $D_4$-partially-anti-invariant. But I've got no clue how to describe the (vector space of) polynomials that are $D_4$-partially-anti-invariant in the first place.

EDIT: After reading a bit about this and doing some thinking, I can make the question more explicit. Write $R = \mathbb{C}[x_1, x_2, x_3, x_4]$, let $R^{D_4}$ be the ring of $D_4$-invariant polynomials (under the obvious permutation action), and let $\chi: D_4 \to \mathbb{C}^{\times}$ be the character above. Then I'd like a description (or a good reference for how to construct such a description) of the semi-invariants $R^{D_4}_{\chi}$ - those polynomials $p$ such that $\sigma(p) = \chi(\sigma)\cdot p$ - as a module over the ring of invariants $R^{D_4}$.

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    $\begingroup$ You have an action of group $G$ and a character $\chi:G\to\mathbb C^\times$, and what you are looking for is the so called semi-invariants. Indeed, this set is a module over the invariants, and in many cases we know how to describe them. Google a bit for examples. $\endgroup$ May 5, 2016 at 1:18
  • $\begingroup$ If this is for an assignment then it's useless and there's probably a much simpler answer. $\endgroup$ May 5, 2016 at 2:55
  • $\begingroup$ @MarianoSuárez-Alvarez: thanks for the tip - I'll take a look! $\endgroup$
    – Thurmond
    May 5, 2016 at 3:32
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    $\begingroup$ @MattSamuel: A description in terms of Schubert polynomials could be useful, and would certainly be appreciated. And no, this is not for an assignment. I'm trying to generalize work of Khovanov-Rozansky - see page 6 of arxiv.org/pdf/math/0401268v2.pdf for the relevant construction. $\endgroup$
    – Thurmond
    May 5, 2016 at 3:34

1 Answer 1


What you want are the $D_4$-invariant elements of $\mathbb{C}[x_1, ...,x_4]$, but with a weird group action, which isn't as simple as permuting the variables. To describe the group action you want, I'll need to describe some other homomorphisms first.

There is a group homomorphism $h: D_4 \mapsto \{1, -1\}$ that can be thought of as follows: an element of $D_4$ can be thought of as a symmetry of a square, and we can ask whether the symmetry swaps the diagonals or keeps them the same. If it swaps them, map to $-1$, otherwise, map to $1$. There is also a group homomorphism $i: D_4 \mapsto S_4$ based on the permutation of the corners, which we'll label, in clockwise order, 1, 3, 2, 4 (this labelling is chosen based on the specific generators you wanted). We can now describe the desired action of $D_4$ on $\mathbb{C}[x_1, ...,x_4]$: an element $u$ of $D_4$ acts by performing the permutation $i(u)$ on the variables, and then multiplying by $h(u)$. Now the polynomials you are interested are exactly the $D_4$-invariant ones, with this group action.

  • $\begingroup$ This is at odds with the claim made by the OP that his almost invariants do not form a ring. $\endgroup$ May 5, 2016 at 1:15
  • $\begingroup$ No, it doesn't, because the group isn't acting by ring automorphisms, they are just linear maps. $\endgroup$
    – Tom Price
    May 5, 2016 at 1:28
  • $\begingroup$ In fact, since being a semi-invariant of a group action is equivalent to being an invariant of another related group action, my answer is actually equivalent to your comment. $\endgroup$
    – Tom Price
    May 5, 2016 at 1:49
  • $\begingroup$ This answer makes sense. But after having thought about what @MarianoSuárez-Alvarez said in the comments to the question, it's clear to me that I'd like a nicer description of the semi-invariants $\mathbb{C}[x_1,\cdots,x_4]^{D_4}_h$ as a module over the invariants $\mathbb{C}[x_1,\cdots,x_4]^{D_4}$. Your answer tells me how to do this on each homogeneous piece of $\mathbb{C}[x_1,\cdots,x_4]$, which is nice, but doesn't tell me anything about the multiplication. If I just wanted homogeneous pieces I could sum over the group acting on a monomial and twist and see what I get. $\endgroup$
    – Thurmond
    May 5, 2016 at 14:14
  • $\begingroup$ Ok, I think I misunderstood your question. By "effectively describe the polynomials" I thought you meant that you just wanted a nice definition of these polynomials, but I gather that you are actually looking for some sort of parametrization (which also interacts nicely with multiplication). $\endgroup$
    – Tom Price
    May 5, 2016 at 15:45

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