Automorphisms Group and Radical Polynomial of Standard Trilinear Forms

1. Introduction

Let V be a $K-$vector space of dimension n and let ${\wedge }^{3}V$ be the third-degree exterior power space of V over the field K. Any element $\omega$ of ${\wedge }^{3}V$ is named trivector on V. By virtue of the canomical identification ${\wedge }^{3}V\simeq {\left({\wedge }^{3}V\right)}^{*}$, there is no diffrence between trivectors and trilinear alternating form. The classification of trilinear alternating forms is the study of the action of $GL\left(V\right)$ on ${\wedge }^{3}V$: $f.\omega =\left({\wedge }^{3}f\right)\left(\omega \right)$. For $n\le 8$, this classification was completed for any fields and the number of orbits is finite. In dimension 9, the number of classes of trivectors over $\mathbb{C}$ is infinite (see [3]) and over ${\mathbb{F}}_2$ is finite (there are 317 classes, see [6]).

In this paper we examine the general cause for the trivector ${\omega }_{3k}$ in dimension $n=3k,k\ge 1$, called standard trilinear alternating form, the form of the transitive automorphism group ${\omega }_L$ over L in which L is the extension of the field K and the trivector in dimension $n=2k+1,k\ge 3$ with the isotropic hyperplane. We computed their groups of automorphisms and their radical polynomials. The main results are the Tables 1, 2, 3, 4 and 5 containing the trivectors ${\omega }_{3k}$ and ${\omega }_{2k+1}$ with the sizes of the group of automorphisms, the radical polynomials and the number of orbits of trivectors in dimension eight. The commutant of a trivector with a maximum rank of eight forms a Frobenius algebra with $dimV\ge 3dimC\left(\omega \right)$ (See [10] Theorem 2.8, p.49), we show that this result is not true for $n\ge 9$.

The motivation behind this research stems from the graph theory, complexity and cryptography, in which they are interested in the alternating trilenear form equivalence (ATFE) problem and code loops, see [12] and [15]. We note that Frobenius algebras are significant in the algebraic approach.

2. Preliminaries

Let $\omega :V^3\to K$ be a trilinear alternating form on a vector space V over a field $K,dimV=n$. The trivector $\omega$ satisfies the equality $\omega (x_{\sigma \left(1\right)},x_{\sigma \left(2\right)},x_{\sigma \left(3\right)})=sgn\left(\sigma \right)\omega (x_1,x_2,x_3)$ for every permutation $\sigma \in S_3$. Two forms ${\omega }_1$ and ${\omega }_2$ are equivalent, ${\omega }_1\sim {\omega }_2$, if there exists a homomorphism bijective $\varphi$ of V verifying:

$${\omega }_2(\varphi \left(x_1\right),\varphi \left(x_2\right),\varphi \left(x_3\right))={\omega }_1(x_1,x_2,x_3)\mathrm{for} \mathrm{every}{x}_1,x_2,x_3\in V$$

One of the possible approaches to the classification of trivectors is to use invariants. We recall three invariants of trivectors: the automorphism group $Aut\left(\omega \right)$, the radical polynomial $P\left(\omega \right)$ of $\omega$ introduced by J.Hora [6] and the commutant $C\left(\omega \right)$.

Definition 1.

The group of automorphisms of ω, $Aut\left(\omega \right)$, is defined by

$$Aut\left(\omega \right)=\{\varphi \in GL\left(V\right)/\varphi .\omega =\omega \}$$

Example 1.

Let ${\omega }_3$ be the trivector $\omega =e_1{e}_2{e}_3$, then $Aut\left({\omega }_3\right)=S{L}_3\left(K\right)$.

Definition 2.

The set $Rad\left(\omega \right)=\{v\in V,\omega (v,.,.)=0\}$ is called the radical of ω and denoted by $Rad\left(\omega \right)$. If $Rad\left(\omega \right)$ is trivial $\left(Rad\right(\omega )=\{0\left\}\right)$, then ω is called non degenerate.

Fix $v\in V$ and define radical $Ra{d}_{\omega }\left(v\right)$ of V as $Ra{d}_{\omega }\left(v\right)=\{u\in V,\omega (u,v,.)=0\}$. $Ra{d}_{\omega }\left(v\right)$ is a subspace of V. The rank of $v\in V$, $r_{\omega }\left(v\right)=n-dimRa{d}_{\omega }\left(v\right)$ is an even number.

Definition 3.

Let $K={\mathbb{F}}_q$ be a finite field. The polynomial $P\left(\omega \right)$ defined by:

$$P\left(\omega \right)=\sum _{v\in V}{x}^{r\left(v\right)}{y}^{n-r\left(v\right)}$$

or

$$P\left(\omega \right)=\sum _{i=0}^{n-1}{\alpha }_i{x}^i{y}^{n-i}$$

where ${\alpha }_i\in \mathbb{N}$ and ${\sum }_{i=0}^{n-1}{\alpha }_i=q^n$, is the radical polynomial of ω.

Definition 4.

Two vectors $u,v\in V^{*}=V-\left\{0\right\}$ are orthogonal $u{\perp }_{\omega }v$, if $u\in Rad\left(v\right)$. The subspaces $V_1$ and $V_2$ of V are orthogonal $V_1\perp V_2$, if $v_1{\perp }_{\omega }{v}_2$ for all $v_1\in V_1$ and $v_2\in V_2$.

Definition 5.

We say that a non degenerate trilinear alternating form ω on V is decomposable if $V=W_1\oplus W_2\oplus ...\oplus W_m,m\ge 2$, and $W_i\perp W_j$ whenever $i\ne j$. The restrictions of ω to $W_i$ are denoted by ${\omega }_i$. $P\left(\omega \right)$ is compatible with the orthogonal decomposition:

$$if \omega ={\sum }_i{\omega }_i, then P\left(\omega \right)={\prod }_{i}P\left({\omega }_i\right)$$

Example 2.

Let ${\omega }_6$ be the trivector ${\omega }_6=e_1{e}_2{e}_3+e_4{e}_5{e}_6$. The radical polynomial of ${\omega }_6$ is equal to

$$P\left({\omega }_6\right)=P\left(e_1{e}_2{e}_3\right)\times P\left(e_4{e}_5{e}_6\right)={(y^3+7x^{2}y)}^2=y^6+14x^2{y}^4+49x^4{y}^2$$

Definition 6.

The commutant of ω, $C\left(\omega \right)$ is

$$C\left(\omega \right)=\{f\in End(V):\omega (x,f\left(y\right),z)=\omega (x,y,f\left(z\right)),\mathrm{for}\mathrm{every}x,y,z\in V\}$$

Definition 7.

Let A be an algebra of finite dimension over a field K. We say that A is a F.algebra if $A\sim Hom(A,K)$ (isomorphic as A-modules).

Note that if A is finite-dimensional $K-$algebra over a field K. Then A is a Frobenius algebra if and only if there is a non-degenerate symmetric bilinear form $\varphi :A\times A\to K$ such that $\varphi (ab,c)=\varphi (a,bc)$ for all $a,b,c\in A$.

3. Invariants on Dimension $3k$

Standard Trilinear Form

Definition 8.

Let V be a $3k$- dimensional vector space over K and let $B=\{e_1,e_2,e_3,....,e_{3k-1},e_{3k}\}$ be a fixed basis of V. A standard trilinear alternating form ${\omega }_{3k}$ can be expressed as

$${\omega }_{3k}=e_1{e}_2{e}_3+e_4{e}_5{e}_6+e_7{e}_8{e}_9+....+e_{3k-2}{e}_{3k-1}{e}_{3k}={\sum }_{i=1}^k{e}_{3i-2}{e}_{3i-1}{e}_{3i}$$

Since ${\omega }_{3k}$ is a decomposable form, then $P\left({\omega }_{3k}\right)$ is compatible with the orthogonal decomposition and we have

$$P\left({\omega }_{3k}\right)=\prod _{i}P\left({\omega }_i\right)={(y^3+7x^{2}y)}^k$$

Proposition 1.

Let $Aut\left({\omega }_3\right)$ be the automorphism of ${\omega }_{3k}$, then it satisfies the exact sequence

$$1\to {\left[S{L}_3\left(K\right)\right]}^k\to Aut\left({\omega }_{3k}\right)\to S_k\to 1$$

i.e.

$$Aut\left({\omega }_{3k}\right)\simeq {\left[S{L}_3\left(K\right)\right]}^k⋊S_k$$

If $K={\mathbb{F}}_q,\left|Aut\left({\omega }_{3k}\right)\right|=k!q^{3k}{(q^3-1)}^k{(q^2-1)}^k$

Proof. 

The domain $R\left({\omega }_{3k}\right)=V_1\cup V_2\cup ...\cup V_k={\cup }_{i=1}^k{V}_i$ where $V_i=<e_{3i-2},e_{3i-1},e_{3i}>$, $i=1,...,k$ is invariant i.e. if $f\in Aut\left({\omega }_{3k}\right),f\left(R\left({\omega }_{3k}\right)\right)\subset R\left({\omega }_{3k}\right)$. So that $f\left(V_i\right)\subset V_{\sigma \left(i\right)}$ for permutation $\sigma \in S_k$, we can define a groups homomorphism.

$\phi :Aut\left({\omega }_{3k}\right)\to S_k,\phi \left(f\right)=\sigma$ where $\sigma =\left[\begin{array}{cccc}{V}_1& V_2& \cdots & V_k\\ V_{\sigma \left(1\right)}& V_{\sigma \left(2\right)}& \cdots & V_{\sigma \left(k\right)}\end{array}\right]$

$\phi$ is surjective, we deduce that the sequence

$$1\to ker\phi \to Aut\left({\omega }_{3k}\right)\to S_k\to 1$$

is exact.

Let $f\in ker\phi$, then $f\left(V_i\right)\subset V_i,i=1,..,k$ and from the equality ${\wedge }^{3}f\left({\omega }_{3k}\right)={\omega }_{3k}$, hence $f\left(e_{3i-2}{e}_{3i-1}{e}_{3i}\right)=e_{3i-2}{e}_{3i-1}{e}_{3i},i=1,...,k$

We can write the matrix of f as follows

$$M\left(f\right)=\left[\begin{array}{cccc}{A}_1& 0& \cdots & 0\\ 0& A_2& \cdots & 0\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \cdots & A_k\end{array}\right]$$

with $det{A}_i=1,i=1,...,k$ then $ker\phi \simeq {\left[S{L}_3\left(K\right)\right]}^k$. □

The first Galois cohomology $H^1(G,Aut\left(\omega \right))$ where $G=Gal(\overline{K}/K)$, $\overline{K}$ is the algebraic closure of K, distinguishes forms over K and ${\omega }_{\overline{K}}\simeq \omega$.

We consider $\omega ={\omega }_{3k}$. If L is the extension of K, there exists a trivector ${\omega }_L\in {\wedge }^{3}V$ such that ${\omega }_L\ncong {\omega }_{3k}$ and ${\omega }_L\otimes L\in {\wedge }^3\left(V{\otimes }_{K}L\right)$ is L-isomorphic to ${\omega }_{3k}$. Let C be the set of orbits of the forms of ${\omega }_{3k}$. Since $H^1(G,S{L}_3\left(K\right))=0$. The exact sequence of Galois cohomology sets gives us $C=H^1(G,S_k)$. Where $H^1(G,S_k)$$\simeq \{\mathrm{Etale}K-\mathrm{algebras}$ of degree $k\}$.

We obtain the following lemma

Lemma 1.

The trivector ${\omega }_{3k}$ has a K-form ${\omega }_L$ with the automorphisms group $Aut\left({\omega }_L\right)$ verify the following exact sequence:

$$1\to S{L}_3\left(L\right)\to Aut\left({\omega }_L\right)\to {\mathbb{Z}}_k\to 1$$

i.e.

$$Aut\left({\omega }_L\right)\simeq S{L}_3\left(L\right)⋊{\mathbb{Z}}_k$$

If $K={\mathbb{F}}_q$, $L={\mathbb{F}}_{q^k}$ and $|Aut\left({\omega }_L\right)|=k{q}^{3k}(q^{3k}-1)(q^{2k}-1)$.

Since $Aut\left({\omega }_L\right)$ contain $S{L}_3\left(L\right)$ and $Aut\left(S{L}_n\left({\mathbb{F}}_{q^k}\right)\right)$ contain the finite cyclic group $Gal({\mathbb{F}}_{q^k}/{\mathbb{F}}_q)\simeq {\mathbb{Z}}_k$, then $Aut\left({\omega }_L\right)$ is semi-direct product $S{L}_3\left(L\right)⋊{\mathbb{Z}}_k$.

The only non trivial trivector over ${\mathbb{F}}_q$ with a transitive automorphisms group is the trivector arising from the three dimensional determinant over ${\mathbb{F}}_{q^k}$ (see J. Hora[6], page 11). So ${\omega }_L$ is the only trivector in dimension $3k$ with the transitive automorphisms group.

The trivector ${\omega }_L$ is the only trivector with equal ranks of all non zero vectors v, $rg\left(v\right)=2k=3k-dimRad\left(v\right),v\ne 0$. Its radical polynomial is equal to

$$P\left({\omega }_L\right)=y^{3k}+(2^{2k}-1)x^{2k}{y}^k$$

We get the following Table 1.

Table 1. Automorphisms group and radical polynomial of ${\omega }_{3k}$.

Table 1. Automorphisms group and radical polynomial of ${\omega }_{3k}$.

Remark 1.

If $K={\mathbb{F}}_2$, $L={\mathbb{F}}_{2^3}$, we have

For $dimV=3$, ${\omega }_3=f_{3-1}$ and $\left|Aut\right({\omega }_3\left)\right|=168$

For $dimV=6$, ${\omega }_6=f_{6-1}$, $\left|Aut\right({\omega }_6\left)\right|=56448$ and ${\omega }_L=f_{6-3}$, $\left|Aut\right({\omega }_L\left)\right|=120960$

For $dimV=9$, ${\omega }_9=f_{9-007}$, $|Aut\left({\omega }_9\right)|=6.2^9{(2^3-1)}^3{(2^2-1)}^3=28449792$ and ${\omega }_L=f_{9-0004}$, $|Aut\left({\omega }_L\right)|=3.2^9(2^9-1)(2^6-1)=49448448$. (See J. Hora [6] Appendix A and B page 12-13).

Remark 2.

The trivector ${\omega }_9$ has three K-forms ${\omega }_9$, ${\omega }_L$ and ${\omega }_{L’}$ in which $L=K\left(t\right),t^3=a$ and $L’=K^{\prime }\times K,K^{\prime }=K\left(\alpha \right)$ where ${\alpha }^2=d(d\notin K^{{*}^2})$.

Proof. 

Let L be an extension of degree 3 of K and F is a space of dimension 3 over L. We consider a standard basis $\{e_1,e_2,e_3\}$ of F, the determinant form is defined by: $\phi :{\wedge }^{3}F⟶L$ where $\phi (e_1\wedge e_2\wedge e_3)=1$. The trace form is $Tr:L⟶K$, we put ${\omega }_L=T{r}_L\circ \phi :F^3⟶K$, ${\omega }_L$ is a trivector of rank nine on $V=F$.

We take $L=K\left(t\right)$ with $t^3=a$ in this case, the basis of V is

$\{e_1,e_2,e_3,e_4=t{e}_1,t{e}_5=t{e}_{2}t{e}_6=t{e}_{3}t{e}_7=t^2{e}_1,e_8=t^2{e}_2,e_9=t^3{e}_3\}$.

We can calculate: ${\omega }_L(e_1\wedge e_2\wedge e_3)=3,{\omega }_L(e_4\wedge e_5\wedge e_6)=3a,{\omega }_L(e_7\wedge e_8\wedge e_9)=3a^2,{\omega }_L(e_1\wedge e_5\wedge e_9)=3q,{\omega }_L(e_1\wedge e_6\wedge e_8)=-3a,{\omega }_L(e_2\wedge e_4\wedge e_9)=-3a,{\omega }_L(e_2\wedge e_6\wedge e_7)=3a,{\omega }_L(e_9\wedge e_4\wedge e_8)=3a,{\omega }_L(e_3\wedge e_5\wedge e_7)=-3a$ and ${\omega }_L(e_i\wedge e_j\wedge e_k)=0$ otherwise. We obtain

$${\displaystyle \frac{1}{3}}{\omega }_L=e_1{e}_2{e}_3+a{e}_4{e}_5{e}_6+a^2{e}_7{e}_8{e}_9+a{e}_1{e}_5{e}_9-a{e}_1{e}_6{e}_8+a{e}_2{e}_4{e}_9-a{e}_2{e}_6{e}_7+a{e}_3{e}_4{e}_8-a{e}_3{e}_5{e}_7$$

and it follows that

$${\omega }_L=e_1{e}_2{e}_3+a{e}_4{e}_5{e}_6+a^2{e}_7{e}_8{e}_9+a{e}_1{e}_5{e}_9-a{e}_1{e}_6{e}_8+a{e}_2{e}_4{e}_9-a{e}_2{e}_6{e}_7+a{e}_3{e}_4{e}_8-a{e}_3{e}_5{e}_7$$

If $L^{\prime }=K^{\prime }\times K$, $K^{\prime }=K\left(\alpha \right)$ where ${\alpha }^2=d(d\notin K^{{*}^2})$, is a quadratic extension of K, in this case, the basis of V is $\{e_1,e_2,e_3,e_4=\alpha e_1,e_5=\alpha e_2,e_6=\alpha e_3,e_7,e_8,e_9\}$. Then,

$${\omega }_L(e_1\wedge e_2\wedge e_3)=3,{\omega }_L(e_7\wedge e_8\wedge e_9)=3,{\omega }_L(e_1\wedge e_2\wedge e_5)=3d,{\omega }_L(e_2\wedge e_4\wedge e_6)=-3d,{\omega }_L(e_3\wedge e_4\wedge e_5)=3d,$$

and ${\omega }_L(e_i\wedge e_j\wedge e_k)$ otherwise.

We obtain

$${\displaystyle \frac{1}{3}}{\omega }_L=e_1{e}_2{e}_3+d{e}_1{e}_5{e}_6+d{e}_2{e}_4{e}_6+d{e}_3{e}_4{e}_5+e_7{e}_8{e}_9$$

and it follows that

$${\omega }_{L^{\prime }}=e_1{e}_2{e}_3+d{e}_1{e}_5{e}_6+d{e}_2{e}_4{e}_6+d{e}_3{e}_4{e}_5+e_7{e}_8{e}_9={\omega }_{6,1,d}+e_7{e}_8{e}_9$$

If $L=K\times K\times K=K^3$ then $F=L^3=K^9$, by the same method we obtain ${\omega }_9$. □

If $K={\mathbb{F}}_q$ finite field of order q, $L={\mathbb{F}}_{q^3}$ and $L’={\mathbb{F}}_{q^2}\times {\mathbb{F}}_q$.

We get the following Table 2.

Table 2. Automorphisms group and its size of ${\omega }_9$

Table 2. Automorphisms group and its size of ${\omega }_9$

Remark 3.

In dimension $3k$, the expression of the trivector ${\omega }_L$ seems to be a difficult problem (see Table 1.).

4. Invariants on Dimension $2k+1$

4.1. General Rule for the Non Trivial Forms with an Isotropic Hyperplane

Let V be a vector space of dimension n, $n=2k+1$ and let ${\omega }_{2k+1}$ be a trivector of rank $2k+1$. There exists a basis $B=\{e_1,e_2,e_3,...,e_{2k},e_{2k+1}\}$ such that

$${\omega }_{2k+1}=e_1(e_2{e}_3+e_4{e}_5+...+e_{2k}{e}_{2k+1})$$

It is the only trivector in dimension $2k+1,k\ge 2$ with an isotropic hyperplane (Hyperplane W such that the restriction on W is the zero form).

Proposition 2.

Let $Aut\left({\omega }_3\right)$ be the automorphism of ${\omega }_{3k}$, then it satisfies the exact sequence

$$1\to A’\to A\to K^{*}\to 1$$

$$1\to K^{2k}\to A’\to S{p}_{2k}\left(K\right)\to 1$$

i.e.

$$Aut\left({\omega }_{2k+1}\right)\simeq S{p}_{2k}\left(K\right)\times K^{*}\times K^{2k}$$

If $K={\mathbb{F}}_q$, then $|Aut\left({\omega }_{2k+1}\right)|=q^{k^2+2k}(q-1){\prod }_{i=1}^k(q^{2i}-1)$

Proof. 

We consider the set $V_1=\{x\in V/x.\omega =0\}$. If $x={\sum }_{i=1}^{2k+1}{\alpha }_i{e}_i$∈$V_1$, we have $x={\alpha }_1{e}_1$, then $V_1=<e_1>$ and ${\wedge }^{4}f\left(x\omega \right)=f\left(x\right){\wedge }^{3}f\left(\omega \right)=f\left(x\right)\omega =0$, for $f\in A=Aut\left({\omega }_{2k+1}\right)$. Then $f\left(V_1\right)\subset V_1$ and there exists $\lambda \in K^{*}$ / $f\left(e_1\right)=\lambda e_1$. The matrix of f with a basis $\{e_1,...,e_{3k}\}$ must have the form $\left[\begin{array}{cc}\lambda & X\\ 0& B\end{array}\right]$, this we define a group of homomorphism surjective $\phi :A\to K^{*}$, $\phi \left(f\right)=\lambda$.

Let $f\in A^{\prime }=Ker\phi$: $e_1{\wedge }^{2}f(e_2{e}_3+....+e_{2k}{e}_{2k+1})=e_1(e_2{e}_3+....+e_{2k}{e}_{2k+1})$, which means ${\wedge }^{2}f(e_2{e}_3+....+e_{2k}{e}_{2k+1})=e_2{e}_3+....+e_{2k}{e}_{2k+1}+x{e}_1$ with $x\in V_2=$

$<e_2,....,e_{2k+1}>$. We consider $g:V_2⟶V_2$ the linear application with the matrix $B:{\wedge }^{2}g(e_2{e}_3+....+e_{2k}{e}_{2k+1})=e_2{e}_3+....+e_{2k}{e}_{2k+1}$. The homomorphism

$\psi :A’\to S{p}_{2k}\left(K\right)$ defined by $\psi \left(f\right)=B$, is surjective and $ker\psi \simeq K^{2k}$. □

While observing the radical polynomial $P\left(\omega \right)$ of ${\omega }_5,{\omega }_7$ and ${\omega }_9$, we deduce

The radical polynomial of ${\omega }_{2k+1}$ is equal to

$$P\left({\omega }_{2k+1}\right)=y^{2k+1}+(2^{2k}-1)x^2{y}^{2k-1}+2^{2k}{x}^{2k}y$$

We get the following Table 3.

Table 3. Automorphisms group and radical polynomial of ${\omega }_{2k+1}$

Table 3. Automorphisms group and radical polynomial of ${\omega }_{2k+1}$

Remark 4.

If $K={\mathbb{F}}_2$ finite field of order 2, we have:

For $dimV=5$, ${\omega }_5=f_{5-2}$ and $\left|Aut\right({\omega }_5\left)\right|=11520$

For $dimV=7$, ${\omega }_7=f_{7-4}$ and $\left|Aut\right({\omega }_7\left)\right|=92897280$

For $dimV=9$, ${\omega }_9=f_{9-1}$ and $|Aut\left({\omega }_9\right)|=2^{24}(2-1){\prod }_{i=1}^4(2^{2i}-1)=12128668876800$, it is the class with the largest group of automorphisms. (See J. Hora [6] Appendix A and B page 12-13).

4.2. General Rule for the Form of the Type ${\omega }_{2k+1}=e_1{e}_2{e}_3+e_3{e}_4{e}_5+e_5{e}_6{e}_7+....+e_{2k-1}{e}_{2k}{e}_{2k+1}$

Let V be a vector space of dimension n, $n=2k+1$ and let ${\omega }_{2k+1}$ be a trivector of rank $2k+1$. There exists a basis $B=\{e_1,e_2,e_3,...,e_{2k},e_{2k+1}\}$ such that ${\omega }_{2k+1}=e_1{e}_2{e}_3+e_3{e}_4{e}_5+e_5{e}_6{e}_7+....+e_{2k-1}{e}_{2k}{e}_{2k+1}$

Proposition 3.

Let $Aut\left({\omega }_3\right)$ be the automorphism of ${\omega }_{3k}$, then it satisfies the exact sequence

$$1\to A’\to A\to {\mathbb{Z}}_2\to 1$$

$$1\to A^{\prime \prime }\to A’\to {\left(K^{*}\right)}^{k-1}\to 1$$

$$1\to K^{3k+1}\to A^{\prime \prime }\to S{L}_2\left(K\right)\times S{L}_2\left(K\right)\to 1$$

i.e.

$$A=Aut\left({\omega }_{2k+1}\right)={\left(S{L}_2\left(K\right)\right)}^2\times {\left(K^{*}\right)}^{k-1}\times K^{3k+1}\times {\mathbb{Z}}_2$$

If $K={\mathbb{F}}_q,\left|Aut\left({\omega }_{2k+1}\right)\right|=2q^{3k+3}{(q^2-1)}^2{(q-1)}^{k-1}$

Proof. 

We observe the domain (linear span) $R=<e_3,e_5,e_7,...,e_{2k+1}=K{e}_3\cup K{e}_5\cup ...\cup K{e}_{2k-1}$ is invariant. We obtain

$$1\to A’\to A\stackrel{\phi }{\to }{\mathbb{Z}}_2\to 1$$

$$\phi \left(f\right)=\left\{\begin{array}{cc}1\hfill & \mathrm{if}f\left(e_3\right)\subset e_3,f\left(e_5\right)\subset e_5,...,f\left(e_{2k+1}\right)\subset e_{2k+1}\hfill \\ 0\hfill & \mathrm{otherwise}\hfill \end{array}\right.$$

Then $A’=\{f/f\subset A,f\left(e_3\right)\subset e_3,f\left(e_5\right)\subset e_5,...,$$f\left(e_{2k+1}\right)\subset e_{2k+1}\}$ therefore, ${\wedge }^{k+1}f$ stabilizer each of the two subspaces $F_1=\{e_3,e_5,e_7,..,e_{2k-1},e_{2k},e_{2k+1}\}$ and

$F_2=\{e_1,e_2,e_3,e_5,e_7,...,e_{2k-1}\}$. We can define a group homomorphisms

$$\psi :A’⟶{\left(K^{*}\right)}^{k-1},\psi \left(f\right)=(f\left(e_3\right)/e_3,f\left(e_5\right)/e_5,...,f\left(e_{2k-1}\right)/e_{2k-1})$$

Then

$$1\to A^{\prime \prime }\to A’\to {\left(K^{*}\right)}^{k-1}\to 1$$

If $f\in A^{\prime \prime }=ker\psi$, f acts on $e_1{e}_2$ and $e_{2k}{e}_{2k+1}$ as $S{L}_2\left(K\right)\times S{L}_2\left(K\right)$.

We obtain the exact sequence

$$1\to K^{3k+1}\to A^{\prime \prime }\to S{L}_2\left(K\right)\times S{L}_2\left(K\right)\to 1$$

If f is the identity over $e_1,e_2,e_{2k}$ and $e_{2k+1}$ (modulo $P=\{e_3,e_5,...,e_{2k-1}\}$),

$f\left(e_i\right)=e_i+v_i,i\in \{1,2,2k,2k+1\}$ where $v_i\in P$, then

$f\left(e_4\right)=e_4+v_4$, $f\left(e_6\right)=e_6+v_6$,...,$f\left(e_{2k-2}\right)=e_{2k-2}+v_{2k-2}$

$A^{\prime \prime }$ is an additive group isomorphic to $K^{3k+1}$ □

We get the following Table 4.

Table 4. Automorphisms group and its size of ${\omega }_{2k+1}$

Table 4. Automorphisms group and its size of ${\omega }_{2k+1}$

Remark 5.

Remark 4.If $K={\mathbb{F}}_2$, we have

For $dimV=7$, ${\omega }_7=f_{7-1}$, $\left|Aut\right({\omega }_7\left)\right|=73728$

For $dimV=9$, ${\omega }_9=f_{9-23}$, $|Aut\left({\omega }_9\right)|=2.2^{15}{(2^2-1)}^2{(2-1)}^3=589824$ (See J. Hora [6] Appendix A and B page 12-13).

5. Commutant of a Trivector and Frobenius Algebra

Proposition 4.

There exists a vector space V of dimension nine and $\omega \in {\wedge }^{3}V$ in which $C\left(\omega \right)$ is not a F.algebra and $dimV<3dimC\left(\omega \right)$.

Proof. 

Let $\omega$ be a trivector of rank nine (See [3] Table 1):

$$\omega =e_1{e}_2{e}_9+e_1{e}_3{e}_5+e_1{e}_4{e}_6+e_2{e}_3{e}_7+e_2{e}_4{e}_8+e_3{e}_4{e}_9$$

We consider f, an element of the commutant and $M_B$, the matrix of f, where B is the standard basis, then by direct competitions we have:

$$\begin{array}{c}\hfill \omega \left(f\left(e_1\right),e_2,e_9\right)=\omega \left(e_1,f\left(e_2\right),e_9\right)=\omega \left(e_1,e_2,f\left(e_9\right)\right)=a_{11}=a_{22}=a_{99}\\ \hfill \omega \left(f\left(e_1\right),e_3,e_5\right)=\omega \left(e_1,f\left(e_3\right),e_5\right)=\omega \left(e_1,e_3,f\left(e_5\right)\right)=a_{11}=a_{33}=a_{55}\\ \hfill \omega \left(f\left(e_1\right),e_4,e_6\right)=\omega \left(e_1,f\left(e_4\right),e_6\right)=\omega \left(e_1,e_4,f\left(e_6\right)\right)=a_{11}=a_{44}=a_{66}\\ \hfill \omega \left(f\left(e_2\right),e_3,e_7\right)=\omega \left(e_2,f\left(e_3\right),e_7\right)=\omega \left(e_2,e_4,f\left(e_7\right)\right)=a_{22}=a_{33}=a_{77}\\ \hfill \omega \left(f\left(e_2\right),e_4,e_8\right)=\omega \left(e_2,f\left(e_4\right),e_8\right)=\omega \left(e_2,e_4,f\left(e_8\right)\right)=a_{22}=a_{44}=a_{88}\\ \hfill \omega \left(f\left(e_3\right),e_4,e_9\right)=\omega \left(e_3,f\left(e_4\right),e_9\right)=\omega \left(e_3,e_4,f\left(e_9\right)\right)=a_{33}=a_{44}=a_{99}\end{array}$$

Then,

$$a_{11}=a_{22}=a_{33}=a_{44}=a_{55}=a_{66}=a_{77}=a_{88}=a_{99}=\alpha$$

Also we have,

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \omega (f\left(e_1\right),e_2,e_3)=\omega (e_1,f\left(e_2\right),e_3)=\omega (e_1,e_2,f\left(e_3\right))\\ \hfill \Rightarrow a_{71}=-a_{52}=a_{93}=\beta \\ \hfill \omega (f\left(e_1\right),e_2,e_4)=\omega (e_1,f\left(e_2\right),e_4)=\omega (e_1,e_2,f\left(e_4\right))\\ \hfill \Rightarrow a_{81}=-a_{62}=a_{94}=\gamma \\ \hfill \omega (f\left(e_2\right),e_3,e_4)=\omega (e_2,f\left(e_3\right),e_4)=\omega (e_2,e_3,f\left(e_4\right))\\ \hfill \Rightarrow a_{92}=-a_{83}=a_{74}=\tau \\ \hfill \omega (f\left(e_1\right),e_3,e_4)=\omega (e_1,f\left(e_3\right),e_4)=\omega (e_1,e_3,f\left(e_4\right))\\ \hfill \Rightarrow a_{91}=-a_{63}=a_{54}=\mu \\ \hfill \omega \left(f\left(e_i\right),e_j,e_k\right)=\omega \left(e_i,f\left(e_j\right),e_k\right)=\omega \left(e_i,e_j,f\left(e_k\right)\right)=0& \Rightarrow a_{ij}=0,\mathrm{for}\mathrm{the}\mathrm{other}\mathrm{cases}.\hfill \end{array}\end{array}$$

$M_B\left(f\right)$ has the form

$$M_B\left(f\right)=\left(\begin{array}{ccccccccc}\alpha & 0& 0& 0& 0& 0& 0& 0& 0\\ 0& \alpha & 0& 0& 0& 0& 0& 0& 0\\ 0& 0& \alpha & 0& 0& 0& 0& 0& 0\\ 0& 0& 0& \alpha & 0& 0& 0& 0& 0\\ 0& -\beta & 0& \mu & \alpha & 0& 0& 0& 0\\ 0& -\gamma & -\mu & 0& 0& \alpha & 0& 0& 0\\ \beta & 0& 0& \tau & 0& 0& \alpha & 0& 0\\ \gamma & 0& -\tau & 0& 0& 0& 0& \alpha & 0\\ \mu & \tau & \beta & \gamma & 0& 0& 0& 0& \alpha \end{array}\right),$$

thus $C\left(\omega \right)=K\oplus V_1\oplus V_2\oplus V_3\oplus V_4$ with $V_i=K{\epsilon }_i,i=1,...,4$ and the matrices of ${\epsilon }_1$, ${\epsilon }_2$, ${\epsilon }_3$ and ${\epsilon }_4$ are presented by

$$M_B\left({\epsilon }_1\right)=\left(\begin{array}{ccccccccc}& & & & & & & & \\ & & & & & & & & \\ & & 0_{4\times 4}& & & & 0_{4\times 5}& & \\ & & & & & & & & \\ 0& -1& 0& 0& & & & & \\ 0& 0& 0& 0& & & & & \\ 1& 0& 0& 0& & & 0_{5\times 5}& & \\ 0& 0& 0& 0& & & & & \\ 0& 0& 1& 0& & & & \end{array}\right),$$

$$M_B\left({\epsilon }_2\right)=\left(\begin{array}{ccccccccc}& & & & & & & & \\ & & & & & & & & \\ & & 0_{4\times 4}& & & & 0_{4\times 5}& & \\ & & & & & & & & \\ 0& 0& 0& 0& & & & & \\ 0& -1& 0& 0& & & & & \\ 0& 0& 0& 0& & & 0_{5\times 5}& & \\ 1& 0& 0& 0& & & & & \\ 0& 0& 0& 1& & & & \end{array}\right),$$

$$M_B\left({\epsilon }_3\right)=\left(\begin{array}{ccccccccc}& & & & & & & & \\ & & & & & & & & \\ & & 0_{4times4}& & & & 0_{4\times 5}& & \\ & & & & & & & & \\ 0& 0& 0& 0& & & & & \\ 0& 0& 0& 0& & & & & \\ 0& 0& 0& 1& & & 0_{5\times 5}& & \\ 0& 0& -1& 0& & & & & \\ 0& 1& 0& 0& & & & \end{array}\right),$$

and

$$M_B\left({\epsilon }_4\right)=\left(\begin{array}{ccccccccc}& & & & & & & & \\ & & 0_{3\times 4}& & & & 0_{3\times 5}& & \\ & & & & & & & & \\ 0& 0& 0& 1& & & & & \\ 0& 0& -1& 0& & & & & \\ 0& 0& 0& 0& & & & & \\ 0& 0& 0& 0& & & 0_{6\times 5}& & \\ 0& 0& 0& 0& & & & & \\ 1& 0& 0& 0& & & & \end{array}\right).$$

By computation we prove that ${\epsilon }_1$, ${\epsilon }_2$, ${\epsilon }_3$ and ${\epsilon }_4$ satisfy:

${\epsilon }_1^2={\epsilon }_2^2={\epsilon }_3^2={\epsilon }_4^2={\epsilon }_1{\epsilon }_2={\epsilon }_1{\epsilon }_3={\epsilon }_1{\epsilon }_4={\epsilon }_2{\epsilon }_1={\epsilon }_3{\epsilon }_1={\epsilon }_4{\epsilon }_1={\epsilon }_2{\epsilon }_3={\epsilon }_2{\epsilon }_4={\epsilon }_3{\epsilon }_2={\epsilon }_4{\epsilon }_2={\epsilon }_3{\epsilon }_4={\epsilon }_4{\epsilon }_3=0.$

If $C\left(\omega \right)$ is a Frobenius algebra, there exists a non-degenerate symmetric bilinear form $\varphi :C\left(\omega \right)\times C\left(\omega \right)\to K$ in which $\varphi (fg,h)=\varphi (f,gh)$ where $f,g,h\in C\left(\omega \right).$ We put $C=M_B\left(\varphi \right)=\left(c_{ij}\right)$ the matrix of $\varphi$ in the basis $B=\left\{1,{\epsilon }_1,{\epsilon }_2,{\epsilon }_3,{\epsilon }_4\right\}$, then

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \varphi \left({\epsilon }_1,{\epsilon }_2\right)& =\varphi \left(1.{\epsilon }_1,{\epsilon }_2\right)=\varphi \left(1,{\epsilon }_1{\epsilon }_2\right)=\varphi \left({\epsilon }_3.{\epsilon }_1,{\epsilon }_2\right)=\varphi \left({\epsilon }_3,{\epsilon }_1{\epsilon }_2\right)\hfill \\ & =\varphi \left({\epsilon }_4.{\epsilon }_1,{\epsilon }_2\right)=\varphi \left({\epsilon }_4,{\epsilon }_1{\epsilon }_2\right)=0\mathrm{imply}\mathrm{that}{c}_{23}=0\hfill \\ \hfill \varphi \left({\epsilon }_2,{\epsilon }_1\right)& =\varphi \left(1.{\epsilon }_2,{\epsilon }_1\right)=\varphi \left(1,{\epsilon }_2{\epsilon }_1\right)=\varphi \left({\epsilon }_3.{\epsilon }_2,{\epsilon }_1\right)=\varphi \left({\epsilon }_3,{\epsilon }_2{\epsilon }_1\right)\hfill \\ & =\varphi \left({\epsilon }_4.{\epsilon }_2,{\epsilon }_1\right)=\varphi \left({\epsilon }_4,{\epsilon }_1{\epsilon }_2\right)=0\mathrm{imply}\mathrm{that}{c}_{32}=0\hfill \\ \hfill \varphi \left({\epsilon }_1,{\epsilon }_3\right)& =\varphi \left(1.{\epsilon }_1,{\epsilon }_3\right)=\varphi \left(1,{\epsilon }_1{\epsilon }_3\right)=\varphi \left({\epsilon }_2.{\epsilon }_1,{\epsilon }_3\right)=\varphi \left({\epsilon }_2,{\epsilon }_1{\epsilon }_3\right)\hfill \\ & =\varphi \left({\epsilon }_4.{\epsilon }_1,{\epsilon }_3\right)=\varphi \left({\epsilon }_4,{\epsilon }_1{\epsilon }_3\right)=0\mathrm{imply}\mathrm{that}{c}_{24}=0\hfill \\ \hfill \varphi \left({\epsilon }_3,{\epsilon }_1\right)& =\varphi \left(1.{\epsilon }_3,{\epsilon }_1\right)=\varphi \left(1,{\epsilon }_3{\epsilon }_1\right)=\varphi \left({\epsilon }_2.{\epsilon }_3,{\epsilon }_1\right)=\varphi \left({\epsilon }_2,{\epsilon }_3{\epsilon }_1\right)\hfill \\ & =\varphi \left({\epsilon }_4.{\epsilon }_3,{\epsilon }_1\right)=\varphi \left({\epsilon }_4,{\epsilon }_3{\epsilon }_1\right)=0\mathrm{imply}\mathrm{that}{c}_{42}=0\hfill \\ \hfill \varphi \left({\epsilon }_1,{\epsilon }_4\right)& =\varphi \left(1.{\epsilon }_1,{\epsilon }_4\right)=\varphi \left(1,{\epsilon }_1{\epsilon }_4\right)=\varphi \left({\epsilon }_2.{\epsilon }_1,{\epsilon }_4\right)=\varphi \left({\epsilon }_2,{\epsilon }_1{\epsilon }_4\right)\hfill \\ & =\varphi \left({\epsilon }_3.{\epsilon }_1,{\epsilon }_4\right)=\varphi \left({\epsilon }_3,{\epsilon }_1{\epsilon }_4\right)=0\mathrm{imply}\mathrm{that}{c}_{25}=0\hfill \\ \hfill \varphi \left({\epsilon }_4,{\epsilon }_1\right)& =\varphi \left(1.{\epsilon }_4,{\epsilon }_1\right)=\varphi \left(1,{\epsilon }_4{\epsilon }_1\right)=\varphi \left({\epsilon }_2.{\epsilon }_4,{\epsilon }_1\right)=\varphi \left({\epsilon }_2,{\epsilon }_4{\epsilon }_1\right)\hfill \\ & =\varphi \left({\epsilon }_3.{\epsilon }_4,{\epsilon }_1\right)=\varphi \left({\epsilon }_3,{\epsilon }_4{\epsilon }_1\right)=0\mathrm{imply}\mathrm{that}{c}_{52}=0\hfill \\ \hfill \varphi \left({\epsilon }_2,{\epsilon }_3\right)& =\varphi \left(1.{\epsilon }_2,{\epsilon }_3\right)=\varphi \left(1,{\epsilon }_2{\epsilon }_3\right)=\varphi \left({\epsilon }_1.{\epsilon }_2,{\epsilon }_3\right)=\varphi \left({\epsilon }_1,{\epsilon }_2{\epsilon }_3\right)\hfill \\ & =\varphi \left({\epsilon }_4.{\epsilon }_2,{\epsilon }_3\right)=\varphi \left({\epsilon }_4,{\epsilon }_2{\epsilon }_3\right)=0\mathrm{imply}\mathrm{that}{c}_{34}=0\hfill \\ \hfill \varphi \left({\epsilon }_3,{\epsilon }_2\right)& =\varphi \left(1.{\epsilon }_3,{\epsilon }_2\right)=\varphi \left(1,{\epsilon }_3{\epsilon }_2\right)=\varphi \left({\epsilon }_1.{\epsilon }_3,{\epsilon }_2\right)=\varphi \left({\epsilon }_1,{\epsilon }_3{\epsilon }_2\right)\hfill \\ & =\varphi \left({\epsilon }_4.{\epsilon }_3,{\epsilon }_2\right)=\varphi \left({\epsilon }_4,{\epsilon }_3{\epsilon }_2\right)=0\mathrm{imply}\mathrm{that}{c}_{43}=0\hfill \\ \hfill \varphi \left({\epsilon }_2,{\epsilon }_4\right)& =\varphi \left(1.{\epsilon }_2,{\epsilon }_4\right)=\varphi \left(1,{\epsilon }_2{\epsilon }_4\right)=\varphi \left({\epsilon }_1.{\epsilon }_2,{\epsilon }_4\right)=\varphi \left({\epsilon }_1,{\epsilon }_2{\epsilon }_4\right)\hfill \\ & =\varphi \left({\epsilon }_3.{\epsilon }_2,{\epsilon }_4\right)=\varphi \left({\epsilon }_3,{\epsilon }_2{\epsilon }_4\right)=0\mathrm{imply}\mathrm{that}{c}_{35}=0\hfill \\ \hfill \varphi \left({\epsilon }_4,{\epsilon }_2\right)& =\varphi \left(1.{\epsilon }_4,{\epsilon }_2\right)=\varphi \left(1,{\epsilon }_4{\epsilon }_2\right)=\varphi \left({\epsilon }_1.{\epsilon }_4,{\epsilon }_2\right)=\varphi \left({\epsilon }_1,{\epsilon }_4{\epsilon }_2\right)\hfill \\ & =\varphi \left({\epsilon }_3.{\epsilon }_4,{\epsilon }_2\right)=\varphi \left({\epsilon }_3,{\epsilon }_4{\epsilon }_2\right)=0\mathrm{imply}\mathrm{that}{c}_{53}=0\hfill \\ \hfill \varphi \left({\epsilon }_3,{\epsilon }_4\right)& =\varphi \left(1.{\epsilon }_3,{\epsilon }_4\right)=\varphi \left(1,{\epsilon }_3{\epsilon }_4\right)=\varphi \left({\epsilon }_1.{\epsilon }_3,{\epsilon }_4\right)=\varphi \left({\epsilon }_1,{\epsilon }_3{\epsilon }_4\right)\hfill \\ & =\varphi \left({\epsilon }_2.{\epsilon }_3,{\epsilon }_4\right)=\varphi \left({\epsilon }_2,{\epsilon }_3{\epsilon }_4\right)=0\mathrm{imply}\mathrm{that}{c}_{45}=0\hfill \\ \hfill \varphi \left({\epsilon }_4,{\epsilon }_3\right)& =\varphi \left(1.{\epsilon }_4,{\epsilon }_3\right)=\varphi \left(1,{\epsilon }_4{\epsilon }_3\right)=\varphi \left({\epsilon }_1.{\epsilon }_4,{\epsilon }_3\right)=\varphi \left({\epsilon }_1,{\epsilon }_4{\epsilon }_3\right)\hfill \\ & =\varphi \left({\epsilon }_2.{\epsilon }_4,{\epsilon }_3\right)=\varphi \left({\epsilon }_2,{\epsilon }_4{\epsilon }_3\right)=0\mathrm{imply}\mathrm{that}{c}_{54}=0\hfill \\ \hfill \varphi \left({\epsilon }_1,{\epsilon }_1\right)& =\varphi \left(1.{\epsilon }_1,{\epsilon }_1\right)=\varphi \left(1,{\epsilon }_1^2\right)=0\mathrm{imply}\mathrm{that}{c}_{22}=0\hfill \\ \hfill \varphi \left({\epsilon }_2,{\epsilon }_2\right)& =\varphi \left(1.{\epsilon }_2,{\epsilon }_2\right)=\varphi \left(1,{\epsilon }_2^2\right)=0\mathrm{imply}\mathrm{that}{c}_{33}=0\hfill \\ \hfill \varphi \left({\epsilon }_3,{\epsilon }_3\right)& =\varphi \left(1.{\epsilon }_1,{\epsilon }_3\right)=\varphi \left(1,{\epsilon }_3^2\right)=0\mathrm{imply}\mathrm{that}{c}_{44}=0\hfill \\ \hfill \varphi \left({\epsilon }_4,{\epsilon }_4\right)& =\varphi \left(1.{\epsilon }_4,{\epsilon }_4\right)=\varphi \left(1,{\epsilon }_4^2\right)=0\mathrm{imply}\mathrm{that}{c}_{55}=0\hfill \end{array}\end{array}$$

The matrix C is represented by

$$C=\left(\begin{array}{ccccc}{c}_{11}& c_{12}& c_{13}& c_{14}& c_{15}\\ c_{21}& 0& 0& 0& 0\\ c_{31}& 0& 0& 0& 0\\ c_{41}& 0& 0& 0& 0\\ c_{51}& 0& 0& 0& 0\end{array}\right).$$

The rank of $\varphi =2$, different from 5, hence this contradicts the non-degeneracy of $\varphi$. We conclude that $C\left(\omega \right)$ is not a F.algebra and $dimC\left(\omega \right)=5$. We deduce that $dimV<3dimC\left(\omega \right)$. □

6. Classification of Trivectors in Dimension 8 over a Finite Field of Characteristic 2

For $n=8$, the classification of trivectors over finite fields, except for characteristic 2 or 3, and over a finite field of two elements has been done in [7] and [5] respectively. More recently a classifications have appeared for a finite field of characteristic 3 [9]. We give the following classification over ${\mathbb{F}}_{2^m}$.

Theorem 1.

Let V be a vector space of dimension eight over a finite field ${\mathbb{F}}_q$ of characteristic 2, $q=2^m$. If m is odd, there are 20 inequivalent trivectors in ${\wedge }^{3}V$ which are of a full rank.

Table 5. The cardinality of the automorphisms groups $Aut\left({\omega }_i\right)$ for trivectors ${\omega }_i$ of rank 8 over ${\mathbb{F}}_{2^m}$. $d\in {\left({\mathbb{F}}_q^{*}\right)}^2$, $a\notin {\left({\mathbb{F}}_q^{*}\right)}^3$

Table 5. The cardinality of the automorphisms groups $Aut\left({\omega }_i\right)$ for trivectors ${\omega }_i$ of rank 8 over ${\mathbb{F}}_{2^m}$. $d\in {\left({\mathbb{F}}_q^{*}\right)}^2$, $a\notin {\left({\mathbb{F}}_q^{*}\right)}^3$

Remark 6.

This classification was done over ${\mathbb{F}}_2$ (see table 2 page 3468 in [5] or see Appendix A page 12 in [6]).

7. Weight Varieties of a Non-Degenerate Form

We can use the classification of trivectors in the theory of codes (See [13]).

Some undefined terms can be found in [13, page 426-429].

Similar arguments applied in [13] are used for determining the varieties $X_1\left({\omega }_{8,i}\right)$ and $X_2\left({\omega }_{8,i}\right)$ for some trivectors and we have:

Proposition 5.

The varieties $X_1\left({\omega }_{8,i}\right)$ and $X_2\left({\omega }_{8,i}\right)$ for $7\le i\le 11$ are given by:

8. Conclusions

In this paper, by using the invariants of the trivectors, we deduce the general rule for some trivectors. As a future work, one can calculate the cardinalities of $X_1\left({\omega }_{8,i}\right)$ and $X_2\left({\omega }_{8,i}\right)$ and use them to fully determine the spectrum of $C(3,8)$.