# Dyslexia in group actions of $SL_2$ on binary cubic forms

The group $SL_2$ (say, $SL_2(\mathbb{C})$, but we could take $SL_2$ of anything else, or probably regard $SL_2$ as a group scheme) acts on binary cubic forms. (Or binary $n$-ic forms in general.) What is the correct form of the action?

Let me give two different answers to this, both of which I have seen in the literature.

(1) Regard binary cubic forms as homomorphisms $\mathbb{A}^2 \rightarrow \mathbb{A}$, where $\mathbb{A}$ is affine space (i.e. just $\mathbb{C}$). We can define an action of $SL_2$ on $\mathbb{A}^2$ as follows: Write $g \in SL_2$ as a matrix, $v \in \mathbb{A}^2$ as a column vector, and then the action is just given by $g v$ (usual matrix multiplication). The action on $\mathbb{A}^1$ is trivial. With these conventions, for a binary cubic form $f$, write $(gf)(v) = f(g^{-1} v)$. This seems to be the representation-theoretic point of view; at any rate it is consistent with p. 4 of Fulton-Harris and this definition is given on p. 14 of Olver's Classical Invariant Theory.

(2) Use the definition $(gf)(v) = f(v g)$, where this time we write $v$ as a row vector, so that $v g$ is well-defined. This definition is quite common, appearing for example in Bhargava, Shankar, and Tsimerman's paper here among many other places (including papers I've co-authored!)

These definitions are equivalent (neither is "wrong") but they're not the same. Definition (1) feels "right" to me, where it seems that the point of (2) is to identify binary cubic forms with $\mathbb{A}^4$ instead of its dual. We would like to write $(gf)(v) = f(g(v))$, but this doesn't work for reasons that are more or less explained in (1).

A couple questions: First of all, is my explanation above accurate?

If it is, is there a good highbrow explanation of (2)? Somehow it feels like a hacky workaround to me. I think the notation (2) is much more natural than (1) in the paper I linked to (I am not trying to argue with their choice of notation) but it feels like cheating a bit, and it is confusing that the same group action is defined in different ways in the literature.

Is there a better perspective than the one I have offered, or do I just need to grin and bear it?

Thank you!

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Perhaps some general remarks are in order. First, let $M$ be a monoid. Recall that a left action of $M$ is a set $S$ together with a map $M \times S \to S$ satisfying the usual associativity conditions. Equivalently, it is a set $S$ together with a homomorphism $M \to \text{End}(S)$ of monoids. A right action of $M$ is a set $S$ together with a map $S \times M \to S$ satisfying the usual associativity conditions. Equivalently, it is a set $S$ together with a homomorphism $M^{op} \to \text{End}(S)$, where $M^{op}$ is the opposite monoid (the monoid with the same underlying set as $M$ but whose multiplication is in the other direction) of $M$.

In general $M$ and $M^{op}$ are non-isomorphic, so left and right actions are genuinely different and must be carefully distinguished. Moreover, if $S$ is a left action and $T$ is another set, then the set of functions $\text{Hom}(S, T)$ inherits a right action via

$$m : f(s) \mapsto f(ms)$$

and there is no way to turn this into a left action. (Category theorists should feel free to replace "monoid," "left action," and "right action" above with "category," "functor," and "contravariant functor.")

If $M$ is a group $G$ then it comes equipped with a canonical isomorphism to its opposite, namely $g \mapsto g^{-1}$. This gives a canonical way to turn all right actions into left actions, which is what we do when we pass from a representation $V$ of a group $G$ to the dual representation $V^{\ast}$ (which for a monoid is a right action if it acted from the left on $V$).

Some specifics. Let $V$ be a vector space ("column vectors") and $G$ a group on which $V$ acts. Then $G$ acts on the space of polynomial functions on $V$, which can be identified with the symmetric algebra $S(V^{\ast})$ (since $V^{\ast}$ corresponds to linear polynomials on $V$). This gives the first action.

$G$ also acts on $V^{\ast}$ ("row vectors"), which induces an action on the space of polynomial functions on $V^{\ast}$, which can be identified with the symmetric algebra $S(V)$. This gives the second action. (You should write it as $v(g^{-1})^{-1}$ because you are implicitly taking the dual twice.)

The two actions are isomorphic in this case because $V$ is self-dual. Indeed, if $V$ is a $2$-dimensional vector space over a field $k$ then $G$ lies in $\text{SL}_2(k)$ if and only if it preserves the exterior product $V \times V \to \Lambda^2(V)$, and choosing an identification of $\Lambda^2(V)$ with the underlying field yields a $G$-equivariant dual pairing $V \times V \to k$.

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This is a fantastic answer. Thank you very much. – Frank Thorne Aug 4 '12 at 13:33
I think one point I hadn't fully appreciated until I read this is that our choice of notation for function composition is arbitrary, but once we make this choice, this dictates that left actions and column vectors should naturally correspond to covariant functors. – Frank Thorne Aug 4 '12 at 13:37