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Suppose that group $G_i$ acts on finite dimensional complex vector space $V_i$ with finite number of orbits for $i=1,2$. Let $G=G_1\times G_2, V=V_1\otimes_{\mathbb C} V_2$ and let $G$ act on $V$ by $(g_1,g_2).v_1\otimes v_2=(g_1v_1)\otimes (g_2v_2)$. [Edit: Here a $G$ action on $V$ means a linear representation.] Is there a general way to describe the orbits of the group action of $G$ on $V$ in terms of the orbits of the $G_i$ action on $V_i$? (Edit: if it is easier, one can assume the representation $V_i$ of $G_i$ is irreducible.)

If the above question has no general answer, here is a more concrete one. Suppose that $G_1=GL_2(\mathbb C)$ and $V_1=\mathbb C^2$. Suppose that $G_2$ acts on a complex vector space $V_2$ with 4 orbits $0, C_1,C_2,C_3$, then is there a way to describe the orbits of the natural $G_1\times G_2$ action on $V_1\otimes_{\mathbb C}V_2 $ in terms of $C_1,C_2,C_3$?

The following are some efforts I tried for the concrete example. The space $V_1\otimes V_2$ can be interpreted as $Hom_{\mathbb C}(V_1^*,V_2)\cong M_{2\times n}(\mathbb C)$, where $n=dim V_2$. The action of $G_1\times G_2$ on $Hom(V_1^*,V_2)$ is given by $(g_1,g_2).f=g_2fg_1^{-1}.$ We can consider the ranks of $f$. If $rk(f)=0,$ there is only one orbit. If $rk(f)=1$, there are 3 orbits depending on the image of $f$. But I don't know what happened if $rk(f)=2$. I am very confused about the last case. I am wondering if there is a general method to solve this kind problem.

Thanks in advance.

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    $\begingroup$ Do you know that the action of $G_1\times G_2$ on $V_1\otimes V_2$ is well defined? $\endgroup$ Dec 7, 2019 at 7:07
  • $\begingroup$ @OliverJones Yes. It is assumed to be well-defined. $\endgroup$
    – Q. Zhang
    Dec 7, 2019 at 18:34
  • $\begingroup$ What do you mean "assumed"? It either is well-defined or there are conditions on the action that make it well-defined. Which is it? $\endgroup$ Dec 7, 2019 at 22:52
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    $\begingroup$ @OliverJones The tensor product representation of $G_1 \times G_2$ from representations of $G_1$ and $G_2$ is a standard construction in representation theory. Why are querying it? But I am not sure whether you can enumerate the orbits of its action just from a knowledge of the numbers of the orbits of $G_1$ and $G_2$. $\endgroup$
    – Derek Holt
    Dec 8, 2019 at 11:56
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    $\begingroup$ @QingZhang The number of orbits is irrelevant. What complicates the description of an orbit is the action of $G_1\times G_2$ on the indecomposable vectors in $V_1\otimes V_2$. $\endgroup$ Dec 10, 2019 at 0:05

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