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Let $G_1=\operatorname{GL}_2(\mathbb C)$ act on $V_1=\mathbb C^2$ via the standard multiplication. Denote this representation by $\operatorname{St}_2$. Let $G_2=\operatorname{SL}_3(\mathbb C^3)$ act on $V_2=\mathbb C^6$ via $\operatorname{Sym}^2(\mathbb C^3)$. Then we have the representation of $G_1\times G_2$ on $V_1\otimes V_2$. I am curious about the orbits of this action. Do we know there are a finite number of orbits? If so, how many orbits are there and how do we describe them?

I appreciate any comments/references.

(This seems to be an exercise from representation theory, and thus I am not sure if it is appropriate to ask this question here.)

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    $\begingroup$ First consider the actions of $G_i$ on their respective representations. That's pretty much basic linear algebra. Then think about how these orbits combine in the tensor product. $\endgroup$
    – Vít Tuček
    Commented Jan 21, 2020 at 21:51
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    $\begingroup$ @VítTuček Thanks for your comment. But that is basically the same as my question. It seems that one can not always build the orbits of the tensor product from each pieces in an easy way. One evidence of the last statement comes from the example of orbits of tensor products $\mathbb C^{n_1}\times \mathbb C^{n_2}\times \mathbb C^{n_3}$ of the group $GL_{n_1}\times GL_{n_2}\times GL_{n_3}$ mathoverflow.net/questions/67897/…. Even both $\mathbb C^{n_1}\times \mathbb C^{n_2} $ and $\mathbb C^{n_3}$ have finite number of orbits, the tensor does not ingeneral $\endgroup$
    – Q-Zh
    Commented Jan 21, 2020 at 21:59
  • $\begingroup$ You are right. I had a wrong mental picture of tensor product. $\endgroup$
    – Vít Tuček
    Commented Jan 21, 2020 at 22:10
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    $\begingroup$ Unless I missed something the question belongs to classical algebraic geometry (and also linear algebra) rather than representation theory. It pertains to classifying pencils of conics in the plane. $\endgroup$
    – Abdelmalek Abdesselam
    Commented Jan 21, 2020 at 23:03
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    $\begingroup$ @abdelmalekabdesselam Your comment made something click about both algebraic geometry and representation theory for me. $\endgroup$
    – Paul Siegel
    Commented Jan 22, 2020 at 11:25

1 Answer 1

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Yes, the number of orbits is finite. Indeed, as Abdelmalek mentioned, this is the question of classification of pencils of conics. The orbits are the following. First, assume that two conics in the pencil are non-proportional.

1) Assume that at least one of the conics in the pencil is nondegenerate. Then this conic is isomorphic to $\mathbb{P}^1$ and the intersection points of the pencil is a subscheme of length 4.

1a) If the 4 points are distinct, this is a quadruple of points in general position on $\mathbb{P}^2$, all such quadruples are conjugate, and in appropriate coordinates this pencil can be written as $$ \langle xy - xz, xz - yz \rangle. $$

1b) If two points collide, this is a triple a points and a tangent vector at one of them, so the pencil can be written in the form $$ \langle xy + xz, yz \rangle. $$

1c) If two pairs of points collide, the pencil can be written in the form $$ \langle x^2, yz \rangle. $$

1d) If three points collide, the pencil can be written in the form $$ \langle x^2 - yz, xy \rangle. $$

1e) If all four points collide, the pencil can be written in the form $$ \langle x^2 - yz, y^2 \rangle. $$

2) Assume now that all conics in the pencil are degenerate. This is possible in either of two cases.

2a) All conics contain a given line. Then each conic is the union of this line and an extra line. Extra lines also form a pencil, and its intersection point can lie on the fixed line or away from it. This gives two more orbits:

2a') $\langle xy, xz \rangle$.

2a'') $\langle xy, x^2 \rangle$.

2b) All conics have a fixed singular point, but no common lines: $\langle x^2, y^2 \rangle$.

3) Next, assume that all conics in the pencil are proportional (equivalently, one of the conics is zero). These orbits are parameterized by the rank of the conic.

3a) $\langle x^2 - yz, 0 \rangle$.

3b) $\langle xy, 0 \rangle$.

3c) $\langle x^2, 0 \rangle$.

3d) $\langle 0, 0 \rangle$.

So, altogether there are 12 orbits. However, I could forget something.

EDIT. The orbit closure order, I think, is the following.

Orbits of type 1 are ordered as $(1e) < (1d), (1c) < (1b) < (1a)$; orbits $(1d)$ and $(1e)$ are incomparable and both sit between $(1e)$ and $(1b)$.

$(2a') < (1d)$, and incomparable with $(1c)$ and $(1e)$.

$(2a'') < (2a')$, $(2a'') < (1e)$.

$(2b) < (1e)$, and incomparable with $(2a')$, $(2a'')$.

$(3a) < (1e)$, and incomparable with type 2 orbits.

$(3b) < (3a)$, $(3b) < (2b)$, $(3b) < (2a'')$.

$(3d) < (3c) < (3b)$.

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  • $\begingroup$ thank you very much for your excellent answer. For our purpose, we also need the closure relations of those orbits. Here is our guess: $(3d)\subset \overline{(3c)}\subset \overline{(3b)}\subset \overline{(3a)},$ $(3a)\subset \overline{(2a')}\subset \overline{(1c)}$, $(3a)\subset \overline{(2a'')}\subset \overline{(1c)}, (3a)\subset \overline{(2b)}\subset{\overline(1e)}\subset \overline{(1d)}\cap \overline{(1c)}$, $(1d)\subset \overline{(1b)}\subset \overline{(1a)}$, and $(1c)\subset \overline{(1b)}\subset \overline{1(a)}$. The notations are the same as in your answer. $\endgroup$
    – Q-Zh
    Commented Jan 28, 2020 at 21:56
  • $\begingroup$ Dear @Sasha, Could you confirm if the closure relations in the last comment. I am sorry for the messy notations. I know it is better to draw a tree about the closure relations. But I don't know how to do it in a comment. It will be great if you could provide some references. Thanks again. $\endgroup$
    – Q-Zh
    Commented Jan 28, 2020 at 21:59
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    $\begingroup$ I added a description of the order as I see it. It is not quite the same as you suggest: First $(3a)$ is not contained in the closures of type 2 orbits, because it contains a smooth conic, while those don't. Second, $(2a'') < (1e)$ because $x^2-yz$ in $(1e)$ can degenerate to $yz$. Third, $(2a') < (1d)$ because $x^2 - yz$ in $(1d)$ can degenerate to $yz$. Fourth, $(2a')$ is incomparable with $(1c)$, because $(1c)$ contains a double line, while $(2a')$ does not. $\endgroup$
    – Sasha
    Commented Jan 30, 2020 at 6:59
  • $\begingroup$ Thanks for your reply. I will think about it. $\endgroup$
    – Q-Zh
    Commented Jan 30, 2020 at 17:47

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