# What are the orbits of tensor product under the action of a product of groups?

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.

• Do you know that the action of $G_1\times G_2$ on $V_1\otimes V_2$ is well defined? Dec 7, 2019 at 7:07
• @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 $G_2$. Dec 8, 2019 at 11:56
• @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 $V_1\otimes V_2$. Dec 10, 2019 at 0:05