# Abstract Nonsense

## Alternating Forms of Maximal Degree

Point of post: In this post I discuss the topic of alternating forms of maximal degree as is discussed in Halmos’s section 31. I also give proof of the last theorem discussed in section 30.

Motivation

In this post we give two quick observations about alternating forms $m$-forms on an $m$-dimensional space. This will show, in essence that the space of alternating $m$-linear forms has dimension $1$. So, the alternating forms make up a relatively small slice, loosely speaking, of the entirety of $\text{Mult}_m\left(\mathscr{V}\right)$. The title is named as it is because we call a $m$-linear alternating form on an $m$-dimensional space an alternating form of maximal degree. The reason is that, as we shall see, if $k>m$ the only $k$-linear alternating form is the zero form.

Theorem: Let $m>0$ and $\mathscr{V}$ be a $m$-dimensional $F$-space. Then, if $\text{Alt}_k\left(\mathscr{V}\right)$ is the set of all $k$-linear, skew-symmetric forms on $\mathscr{V}$ then $\text{Alt}_k\left(\mathscr{V}\right)$ is a subspace of $\text{Mult}_k\left(\mathscr{V}\right)$.

Proof: Evidently $\text{Alt}_k\left(\mathscr{V}\right)$ is non-empty since $\bold{0}\in\text{Alt}_k\left(\mathscr{V}\right)$. So, now suppose that $\alpha,\beta\in F$ and $K,K'\in\text{Alt}_k\left(\mathscr{V}\right)$. Note then that if $x\in\mathscr{V}$ and $r,s\in[k]$ then

\displaystyle \begin{aligned}\left(\alpha K+\beta K'\right)(x_1,\cdots,\underbrace{x}_{r^{\text{th}}},\cdots,\underbrace{x}_{s^{\text{th}}},\cdots,x_k) &= \alpha K\left(x_1,\cdots,x,\cdots,x,\cdots,x_n\right)+\beta K'\left(x_1,\cdots,x,\cdots,x,\cdots,x_n\right)\\ &= \alpha 0+\beta 0\\ &= 0\end{aligned}

and since $x,r$ and $s$ were arbitrary it follows that $\alpha K+\beta K'$ is alternating and thus in$\text{Alt}_k\left(\mathscr{V}\right)$. $\blacksquare$

We now claim that $\displaystyle \dim_F\text{Alt}_m\left(\mathscr{V}\right)\leqslant 1$. To do this, we’ll show that any two alternating forms of maximal degree are linearly dependent.

Theorem: Let $\mathscr{V}$ be a $m$-dimensional $F$-space with basis $\{x_1,\cdots,x_m\}$ and $K,K'\in\text{Alt}_m\left(\mathscr{V}\right)$. Then, $K$ and $K'$ are linearly dependent.

Proof: Note that since any field is a $1$-dimensional vector space over itself it follows that the two scalars $K\left(x_1,\cdots,x_m\right)$ and $K'\left(x_1,\cdots,x_m\right)$ are linearly dependent and so there exists a non-zero scalar $\alpha$ such that $K(x_1,\cdots,x_m)=\alpha K'(x_1,\cdots,x_m)$. Note then by an earlier theorem we have that for any $\displaystyle \sum_{j_1=1}^{m}\alpha_{1,j_1}x_{j_1},\cdots,\sum_{j_m=1}^{m}\alpha_{m,j_m}x_{j_m}\in\mathscr{V}$ it’s true that

\displaystyle \begin{aligned}K\left(\sum_{j_1=1}^{m}\alpha_{1,j_1}x_{j_1},\cdots,\sum_{j_m=1}^{m}\alpha_{m,j_m}x_{j_m}\right) &= \sum_{\pi\in S_m}\text{sgn}(\pi)\prod_{k=1}^{m}\alpha_{k,j_k}K\left(x_1,\cdots,x_m\right)\\ &= \sum_{\pi\in S_m}\text{sgn}(\pi)\prod_{k=1}^{m}\alpha_{k,\pi(k)}\alpha K'\left(x_1,\cdots,x_m\right)\\ &= \alpha\sum_{\pi\in S_m}\text{sgn}(\pi)\prod_{k=1}^{n}\alpha_{k,\pi(k)} K'\left(x_1,\cdots,x_m\right)\\ &= \alpha K'\left(\sum_{j_1=1}^{m}\alpha_{1,j_1}x_{j_1},\cdots,\sum_{j_m=1}^{m}\alpha_{m,j_m}x_{j_m}\right)\end{aligned}

and since $\displaystyle \sum_{j_1=1}^{m}\alpha_{1,j_1}x_{j_1},\cdots,\sum_{j_m=1}^{m}\alpha_{m,j_m}x_{j_m}\in\mathscr{V}$ were arbitrary it follows that $K=\alpha K'$ as desired. $\blacksquare$

We now prove that in fact, if $\dim_F\mathscr{V}=m$ then $\dim_F\text{Alt}_m\left(\mathscr{V}\right)=1$ by showing the existence of a non-zero alternating $m$-form. We follow the proof given in Halmos closely:

Theorem: Let $\mathscr{V}$ be a $m$-dimensional $F$-space and $k\in[m]$. Then, $\text{Alt}_k\left(\mathscr{V}\right)$ is non-trivial.

Proof: We proceed by induction. For $k=1$ this is true since evidently $\text{Hom}\left(\mathscr{V},F\right)=\text{Alt}_1\left(\mathscr{V}\right)\cong \mathscr{V}$. Now, suppose that $\text{Alt}_k\left(\mathscr{V}\right)$ for some $1\leqslant k  and thus has some non-zero $K\in\text{Alt}_k\left(\mathscr{V}\right)$. Now since $K\ne\bold{0}$ we may find $x^0_1,\cdots,x^0_k\in\mathscr{V}$ such that $K\left(x^0_1,\cdots,x^0_k\right)\ne 0$ . So, now since $k we may find some $x^0_{k+1}\in\mathscr{V}-\text{span}\{x^0_1,\cdots,x^0_k\}$ and so evidently we may find some $\varphi\in\text{Hom}\left(\mathscr{V},F\right)$ such that $\varphi(x^0_1)=\cdots=\varphi(x^0_k)=0$ and $\varphi(x^0_{k+1})\ne 0$ (for example, we may extend $\{x^0_1,\cdots,x^0_{k+1}\}$ to some basis (since they must be l.i. since the value of an alternating multilinear form on a dependent set is zero) $\mathscr{B}$ and so we may define $\varphi$ to be the unique linear functional such that $\varphi\left(\mathscr{B}-\{x^0_{k+1}\}\right)=\{0\}$ and $\varphi(x^0_{k+1})=1$).

We then define $\tilde{K}\in\text{Mult}_{k+1}\left(\mathscr{V}\right)$ by

$\displaystyle \tilde{K}(x_1,\cdots,x_{k+1})=\sum_{j=1}^{k}\left((j,k+1)K\right)(x_1,\cdots,x_k)\varphi(x_{k+1})-K(x_1,\cdots,x_k)\varphi(x_{k+1})\quad\mathbf{(1)}$

We now claim that $\tilde{K}$ is non-zero and alternating. The first fact is clear since

$\tilde{K}\left(x^0_1,\cdots,x^0_k,x^0_{k+1}\right)=-K(x^0_1,\cdots,x^0_k)\varphi(x^0_{k+1})\ne 0$

We suppose next that $x_1,\cdots,x_k,x_{k+1}$ are vectors and $r,s\in[k+1]$ (distinct) and $x_r=x_s$. Note that if $x_r,x_s$ occur simultaneously in the slots of $K$ in the expression for $\mathbf{(1)}$ then that term is zero.

From here we break into two cases, namely $s=k+1$ and $s\leqslant k$. If the first is true then all the terms of $\mathbf{(1)}$ vanish except

$(r,k+1)K(x_1,\cdots,x_k)\varphi(x_{k+1})-K(x_1,\cdots,x_k)\varphi(x_{k+1})$

and since $x_r=x_{k+1}$ the transposition $(r,k+1)$ does not change the value of $K(x_1,\cdots,x_k)$ and thus the above vanishes. If $j\leqslant k$ then the number of terms in $\mathbf{(1)}$ which are non-vanishing is two, and since one of these two may be obtained by applying $(r,s)$ to the other it follows that it vanishes as well. Thus, $\tilde{K}$ is alternating and thus the induction is complete. $\blacksquare$

Corollary: If $\mathcal{V}$ is a $m$-dimensional $F$-space, then $\dim_F\text{Alt}_m\left(\mathscr{V}\right)=1$

Proof: Since we proved any two $m$-forms must be linearly dependent it follows that $\dim_F\text{Alt}_m\left(\mathscr{V}\right)\leqslant 1$ and since, as was just proven, $\text{Alt}_m\left(\mathscr{V}\right)$ is not trivial it follows that $\dim_F\text{Alt}_m\left(\mathscr{V}\right)\geqslant 1$. $\blacksquare$

References: Halmos, Paul R.  Finite-dimensional Vector Spaces,. New York: Springer-Verlag, 1974. Print