# Abstract Nonsense

## Halmos Section 29,30 and 31: Multilinear Forms, Alternating Multilinear Forms, Alternating Multilinear Forms of Maximal Degree

Point of post: In this post we solve the problems given at the end of the sections 29,30 and 31 in Halmos’s book

1.

Problem: Give an example of a vector space $\mathscr{V}$ and a multilinear form on $\mathscr{V}$ which is skew-symmetric but not alternating.

Proof: By virtue of a previous theorem it’s clear that that the ground field for $\mathscr{V}$ have characteristic $2$. So, let’s just consider $\mathscr{V}$ to be $\mathbb{Z}_2$ over itself. Then, we can consider

$K:\mathbb{Z}_2\boxplus\mathbb{Z}_2\to\mathbb{Z}_2:(x,y)\mapsto xy$

This is evidently bilinear since

$K(\alpha x+\beta x',y)=(\alpha x+\beta x')y=\alpha xy+\beta xy'=\alpha K(x,y)+\beta K(x',y)$

since it’s symmetric it follows that $K$ is bilinear. Also, it’s clear that it’s skew-symmetric since (recalling that $-1=1$ in $\mathbb{Z}_2$ $\pi K=\pm K=K$. It’s not alternating however since $K(1,1)=1\ne 0$.

2.

Problem: Give an example of a non-zero alternating $k$-linear form $K$ on an $n$-dimensional space ($k) such that $K(x_1,\cdots,x_n)=0$ for some linearly independent set of vectors $x_1,\cdots,x_n$.

Proof:

Plan of attack: What if we considered a non-zero alternating $k+1$-form but fixed the last entry. So, to be explicit what if we looked at some non-zero $K\in\text{Alt}_{k+1}\left(\mathscr{V}\right)$, which we know exists by prior theorem since $k+1\leqslant n$. Then, we could find some $x_1^0,\cdots,x_k^0,x_0$ (the superscripts are just meant so you don’t confused these fixed vectors with the variable vectors $x_1,\cdots,x_k$ which are soon to be used) such that $K\left(x_1^0,\cdots,x_k^0,x_0\right)\ne 0$. So, what if we considered $\tilde{K}\in\text{Alt}_{k}\left(\mathscr{V}\right)$ given by $\tilde{K}\left(x_1,\cdots,x_k\right)=K\left(x_1,\cdots,x_k,x_0\right)$. Indeed this function is multilinear since $K$ is multilinear and alternating since $K$ is alternating. Moreover, $K$ is non-zero since $\tilde{K}\left(x_1^0,\cdots,x_k^0\right)=K\left(x_1^0,\cdots,x_k^0,x_0\right)\ne 0$. But, what if we picked a set of $k$ vectors $\{x_1,\cdots,x_k\}$ which are linearly independent but for which $\{x_1,\cdots,x_k,x_0\}$ isnt? Then $\tilde{K}\left(x_1,\cdots,x_k\right)=K\left(x_1,\cdots,x_k,x_0\right)=0$ since $K$ is alternating and $\{x_1,\cdots,x_k,x_0\}$ is linearly dependent. So with this in mind let’s consider an actual:

Anti-climactic Example

Consider

$K:\mathbb{R}^3\boxplus\mathbb{R}^3\to\mathbb{R}:\left(\begin{bmatrix}x_1\\ x_2\\ x_3\end{bmatrix},\begin{bmatrix}y_1\\ y_2\\ y_3\end{bmatrix}\right)\mapsto \det\begin{bmatrix}x_1 & y_1 & 1\\ x_2 & y_2 & 1\\ x_3 & y_3 &0\end{bmatrix}$

clearly then $K\in\text{Alt}_2\left(\mathbb{R}^3\right)$ and $K$ is non-zero but

$\displaystyle K\left(\begin{bmatrix}1\\ 0\\ 0\end{bmatrix},\begin{bmatrix}0\\ 1\\ 0\end{bmatrix}\right)=\det\begin{bmatrix}1 & 0 &1\\ 0 & 1& 1\\ 0 & 0 & 0\end{bmatrix}=0$

even though $\left\{\begin{bmatrix}1\\ 0\\ 0\end{bmatrix},\begin{bmatrix}0\\ 1\\ 0\end{bmatrix}\right\}$ is a linearly independent set of vectors.

3.

Problem: What is the dimension of all symmetric $k$-linear forms ($k)? What about the skew-symmetric? What about the alternating ones?

Proof: Personally, and maybe this problem just psyched me out, but I thought this was a tough one to prove. We begin by classifying the skew-symmetric forms.

Claim:Let $\mathscr{V}$ be an $n$-dimensional space, then the dimension of $\displaystyle \text{SkewSym}_k\left(\mathscr{V}\right)={n\choose k}$.

Proof: Fix a basis $\{x_1,\cdots,x_n\}=\mathcal{B}$ for $\mathscr{V}$. Then, for each $\{\ell_1,\cdots,\ell_k\}$ with $\ell_1<\ell_2<\cdots<\ell_k$ let $K$ be the unique $k$-linear form on $\mathscr{V}$ such that

$\displaystyle K_{\ell_1,\cdots,\ell_k}\left(x_{r_1},\cdots,x_{r_k}\right)=\prod_{j=1}^{k}\delta_{\ell_j,r_j}$

Then, let $J_{\ell_1,\cdots,\ell_k}=\text{Asym}\left(K_{\ell_1,\cdots,\ell_k}\right)$ where $\text{Asym}$ is the “antisymmetrizer function” described in the post on skew-symmetric forms. Note then that if $\{\ell_1,\cdots,\ell_k\}\ne\{r_1,\cdots,r_k\}$ then $J_{\ell_1,\cdots,\ell_k}\left(x_{r_1},\cdots,x_{r_k}\right)=0$ (note that this no longer evaluates to zero on tuples who have the same elements as $\{x_{\ell_1},\cdots,x_{\ell_k}\}$ but a possibly different order of arrangement) and that $K(x_{\ell_{\pi(1)}},\cdots,x_{\pi(\ell_k)})=\text{sgn}(\pi)$.  We claim that

$\mathscr{B}\overset{\text{def.}}{=}\text{span}\left\{J_{\ell_1,\cdots,\ell_k}:\{\ell_1,\cdots,\ell_k\}\subseteq[n]\right\}=\text{SkewSym}_k\left(\mathscr{V}\right)$

To see that suppose that

$\displaystyle \sum_{\{x_{\ell_1},\cdots,x_{\ell_k}\}\subseteq\mathcal{B}:\ell_1<\cdots<\ell_k}\alpha_{\ell_1,\cdots,\ell_1}J_{\ell_1,\cdots,\ell_k}=\bold{0}$

Note then that for each $\{r_1,\cdots,r_k\}\subseteq[n]$ with $r_1<\cdots we see that

$\displaystyle \alpha_{r_1,\cdots,r_k}=\sum_{\{x_{\ell_1},\cdots,x_{\ell_k}\}\subseteq\mathcal{B}:\ell_1<\cdots<\ell_k}\alpha_{\ell_1,\cdots,\ell_k}K_{\ell_1,\cdots,\ell_k}\left(x_{r_1},\cdots,x_{r_k}\right)=0$

from where linear independence follows. Now, to see that $\text{span }\mathscr{B}=\text{Alt}_k\left(\mathscr{V}\right)$ we prove that for any $K\in\text{Alt}_k\left(\mathscr{V}\right)$ that

$\displaystyle K=\sum_{\{x_{\ell_1},\cdots,x_{\ell_k}\}\subseteq\mathcal{B}:\ell_1<\cdot<\ell_k}K\left(x_{\ell_1},\cdots,x_{\ell_k}\right) J_{\ell_1,\cdots,\ell_k}\quad\mathbf{(1)}$

to do this it suffices to show that the right hand side $\mathbf{(1)}$ equals the left hand side for all elements of $\mathcal{B}^k$. To do this we let $(x_{r_1},\cdots,x_{r_k})\in\mathcal{B}^k$ be arbitrary. If any of the coordinates of the tuples are equal then both sides of $\mathbf{(1)}$ are zero since both sides are skew-symmetric and thus alternating, so assume not. Then, there exists some $\pi\in S_k$ such that $r_{\pi(1)}<\cdots we note then that

$\displaystyle K(x_{r_1},\cdots,x_{r_1})=\text{sgn}(\pi)K(x_{r_{\pi(1)}},\cdots,x_{r_{\pi(k)}})=\sum_{\{x_{\ell_1},\cdots,x_{\ell_k}\}\subseteq\mathcal{B}:\ell_1<\cdots<\ell_k}K\left(x_{\ell_1},\cdots,x_{\ell_k}\right) J_{\ell_1,\cdots,\ell_k}\left(x_{r_1},\cdots,x_{r_k}\right)$

Thus, noting that the number of $\{\ell_1,\cdots,\ell_k\}\subseteq[n]$ for which $\ell_1<\cdots<\ell_k$ is $\displaystyle {n\choose k}$ finishes the argument. $\blacksquare$
Corollary: If the characteristic of the ground field for $\mathscr{V}$ has characteristic greater than two then $\displaystyle \dim_F\text{Alt}_k\left(\mathscr{V}\right)={n\choose k}$

Next we prove the result for symmetric forms, which for some reason I found considerably easier

Claim If $\mathscr{V}$ is an $n$-dimensional space then $\displaystyle \dim_F\text{Sym}_k\left(\mathscr{V}\right)={n+k-1\choose k}$

Proof: Fix a basis $\{x_1,\cdots,x_n\}=\mathcal{B}$. It’s clear upon expansion of an arbitrary $k$-linear form though that each such form is really just the linear combination of monomials of the form $\alpha_{1,f(1)}\cdots\alpha_{k,f(k)}$ and that these monomials are linearly independent symmetric forms. The claim follows by noticing that the number of such monomials is $\displaystyle {n+k-1\choose k}$. The conclusion follows. $\blacksquare$

References:

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