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\begin{document}% \tiny \scriptsize \footnotesize \small \normalsize \large \Large \LARGE \huge \Huge 
\begin{center}{\Huge Self-dual vector spaces}\bigskip 

Pierre-Yves Gaillard\footnote{ORCID https://orcid.org/0000-0002-7960-1698} 
\end{center}%\centerline{Self-dual vector spaces}\bigskip

\noi Let $K$ be a field, $V$ a $K$-vector space, $\GL(V)$ the automorphism group of $V$, and $V^*:=\Hom_K(V,K)$ the dual of $V$, equipped with its natural $\GL(V)$-module structure. Say that $V$ is \textbf{self-dual} if there is a $\GL(V)$-module isomorphism $V\simeq V^*$. 

The purpose of this text is to answer the question: which vector spaces are self-dual? 

A necessary condition is that $V$ be finite dimensional (this follows from the Erdős–Kaplansky Theorem), and a sufficient condition is that $V$ be the zero vector space. So, from now on we assume that $V$ is nonzero and finite dimensional. We denote the dimension of $V$ by $n$ (note $n\ge1$), and the cardinality of $K$ by $\kappa$. 

\begin{thm*} 
The vector space $V$ is self-dual if and only if $\kappa+n\le4$. Moreover $\dim\Hom_{\GL(V)}(V,V^*)$ depends only on $\kappa$ and $n$. Denoting this integer by $d(\kappa,n)$, we have $d(\kappa,n)=1$ if $\kappa+n\le4$, and $d(\kappa,n)=0$ otherwise. (Recall that we assume $V\ne0$.)
\end{thm*}

Loosely speaking: the only nonzero self-dual vector spaces are $\F_2,\F_3$ and $(\F_2)^2$. The unique $\GL((\F_2)^2)$-isomorphism $\phi:(\F_2)^2\to((\F_2)^2)^*$ is given by 
$$
\left(\phi\begin{pmatrix}a\\ c\end{pmatrix}\right)\begin{pmatrix}b\\ d\end{pmatrix}=\det\begin{pmatrix}a&b\\ c&d\end{pmatrix}=ad+bc.
$$ 
Let us add for completeness sake that, for $\kappa$ equal to 2 or 3, a $\GL(\F_\kappa)$-isomorphism $\psi:\F_\kappa\to(\F_\kappa)^*$ is given by $(\psi(x))(y)=xy$. 

Note that $\Hom_{\GL(V)}(V,V^*)$ is isomorphic to the space $B(V)$ of $\GL(V)$-invariant bilinear forms on $V$. 

To prove the theorem, it suffices to verify the statement about $d(\kappa,n)$. The case $n=1$ is left to the reader as an easy exercise. So we suppose from now on $n\ge2$. Suppose there is a nonzero $f$ in $B(V)$. We must show $\kappa=n=2$ and $f=\det$. 

Let $e_1,\ldots,e_n$ be a basis of $V$, and set $a:=f(e_1,e_1),b:=f(e_1,e_2)$. By invariance we have $f(v,v)=a$ for all $v\ne0$, and $f(v,w)=b$ for all linearly independent $v$ and $w$; in particular $(a,b)\ne(0,0)$. We have 
$$
b=f(e_1,e_1+e_2)=f(e_1,e_1)+f(e_1,e_2)=a+b\implies a=0\ne b.
$$ 
For $0\ne\lambda\in K$ we have $b=f(e_1,\lambda e_2)=\lambda b$, hence $\lambda=1$, hence $\kappa=2$, hence $b=1$, hence $f=\det$ if $n=2$. If $n\ge3$ we have 
$$
1=f(e_1,e_2+e_3)=f(e_1,e_2)+f(e_1,e_3)=1+1=0,
$$ 
contradiction. This completes the proof.  

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