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Let $G=\mbox{GL}(\mathbb{Z}/p\times\mathbb{Z}/p )\times \mbox{GL}(\mathbb{Z}/p\times\mathbb{Z}/p ) \times \mbox{GL}(\mathbb{Z}/p\times\mathbb{Z}/p )$, $p$ prime, act on $(\mathbb{Z}/p\times \mathbb{Z}/p)\otimes (\mathbb{Z}/p\times \mathbb{Z}/p) \otimes (\mathbb{Z}/p\times \mathbb{Z}/p)$ by the following mapping $$ (\alpha,\beta,\gamma)(w_1\otimes w_2\otimes w_3)=\alpha w_1\otimes \beta w_2 \otimes \gamma w_3. $$

Let $\Sigma_1=e_1 \otimes e_1 \otimes e_1 + e_1 \otimes e_2 \otimes e_2 + e_2 \otimes e_1 \otimes e_ 2$ and $\Sigma_2=e_1\otimes e_1 \otimes e_1 + e_2 \otimes e_2 \otimes e_2$.

First, I want to determine the rank of $\Sigma_1$ and $\Sigma_2$, then I want to compute the cardinality of their orbits.

I think that I know how to tackle the first part of this problem: I can write $\Sigma_2$ as $e_1\otimes M_1 + e_2 \otimes M_2$ and see that the rank of the coeff. matrices $M_1$ and $M_2$ is equal to 1 and since $M_1$ and $M_2$ are linealy independent, I can conclude that $\mbox{rank}(\Sigma_2)=2$. (Here, I also use the fact that $\mbox{span}\{M_1,M_2\}$ has a basis of $e_1\otimes e_1$ and $e_2\otimes e_2$ tensors). By doing something similar I can conclude that $\mbox{rank}(\Sigma_1)=3$.

  • My question is: How can I determine the cardinalities of their orbits?
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1 Answer 1

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Orbit-Stabilizer Thm: If $G$ acts on a set $X$, then $[G:\mathrm{Stab}(x)]=|\mathrm{Orb}(x)|$ for all $x\in X$.

(The fibers of the map $G\to\mathrm{Orb}(x)$ defined by $g\mapsto gx$ are the cosets of $\mathrm{Stab}(x)$.)

Thus to compute orbit sizes, we should compute indices of stabilizers, which for finite groups only requires us to compute the sizes of stabilizers and the original group.

Set $G=\mathrm{GL}_2(\mathbb{F}_p)$. It acts on $\mathbb{F}_p^2$ via $(\begin{smallmatrix}a&b\\c&d\end{smallmatrix})e_1=ae_1+ce_2$ and $(\begin{smallmatrix}a&b\\c&d\end{smallmatrix})e_2=be_1+de_2$ where $e_1,e_2$ are the standard coordinate vectors. There is an induced action on $\mathbb{F}_p^2\otimes\mathbb{F}_p^2\otimes\mathbb{F}_p^2$ determined by the "diagonal" rule $g(u\otimes v\otimes w)=(gu)\otimes(gv)\otimes(gw)$.

Let's compute the stabilizer of $\Sigma_2$. We get

$$(\begin{smallmatrix}a&b\\c&d\end{smallmatrix})(e_1\otimes e_1\otimes e_1+e_2\otimes e_2\otimes e_2)$$

$$= (ae_1+ce_2)\otimes(ae_1+ce_2)\otimes(ae_1+ce_2)+(be_1+de_2)\otimes(be_1+de_2)\otimes(be_1+de_2)$$

$$\begin{array}{l}=(a^3+b^3)e_{111}+(a^2c+b^2d)(e_{112}+e_{121}+e_{211}) \\ \phantom{=} +(ac^2+bd^2)(e_{122}+e_{212}+e_{221})+(c^3+d^3)e_{222} \end{array} $$

where I am abbreviating $e_i\otimes e_j\otimes e_k$ as $e_{ijk}$ - note these form a basis. Setting equal to $e_{111}+e_{222}$ and equating coefficients yields the system of equations

$$\begin{array}{ccc} a^3+b^3 & = & c^3+d^3 & = & 1 \\ a^2c+b^2d & = & ac^2+bd^2 & = & 0 \end{array}$$

If we multiply the last two nontrivial expressions by $u$ and $v$ respectively and add we get

$$ac(au+cv)+bd(bu+dv)=0 $$

Since $(\begin{smallmatrix}a&b\\c&d\end{smallmatrix})\in\mathrm{GL}_2(\mathbb{F}_p)$ is invertible, so is its transpose, so $(\begin{smallmatrix}a&c\\b&d\end{smallmatrix})(\begin{smallmatrix}u\\v\end{smallmatrix})=(\begin{smallmatrix}au+cv \\ bu+dv\end{smallmatrix})$ can be any vector, which implies $ac=bd=0$. The solutions $a=b=0$ and $c=d=0$ are inconsistent with the first equations in the system, so we get $a=d=0$ or $c=b=0$, which tells us

$$\mathrm{Stab}_{\mathrm{GL}_2(\mathbb{F}_p)}(\Sigma_2)=\{(\begin{smallmatrix}1&0\\0&1\end{smallmatrix}),(\begin{smallmatrix}0&1\\1&0\end{smallmatrix})\}.$$

Since $|\mathrm{GL}_2(\mathbb{F}_p)|=(p^2-1)(p^2-p)$ (to form an invertible $2\times 2$ matrix, there are $p^2-1$ choices for the first column (any but the $0$ vector), and subsequently $p^2-p$ choices for the second (any but the multiples of the first column)), we can finally say that

$$|\mathrm{Orb}(\Sigma_2)|=(p^2-1)(p^2-p)/2. $$

I haven't done $\Sigma_1$, but I assume it proceeds the same way.

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