An unexpected(?) way to identify a 2d vector space with its dual Let $X$ be a finite-dimensional vector space over the field $\mathbb{R}$. Denote the dual of $X$ by $X^*$. 

Definition: Let us say that a (not necessarily linear) mapping $x \mapsto x^* : X \to X^*$ is natural with respect to a linear transformation $T:X \to X$ if it makes the diagram
  $$\begin{array}{ccc}
X & \overset{T}\longrightarrow & X \\
\downarrow &  & \downarrow \\
X^* & \overset{T^*}\longleftarrow & X^* 
\end{array}$$
  commutative. Here, $T^*: X^* \to X^*$ sends $\varphi \mapsto \varphi T$. 

Obviously $x^* \equiv 0$ is natural with respect to $T$ for every linear transformation $T : X \to X$. However, one basically expects no other maps $X \to X^*$ to have this property because of this sort of thing. A cute thing does happen in 1-dimension. 

Example 1: Let $X$ be a 1-dimensional vector space. Define $0^*=0$ and, for each nonzero $x \in X$, define $x^* \in X^*$ to be the unique functional such that $x^*(x)=1$. Then, the (admittedly nonlinear) mapping $x \mapsto x^* : X \to X^*$ is natural with respect to every linear transformation $T: X \to X^*$.

Of course, the case $\dim(X)=1$ is quite special (the only linear transformations are scalar multiplications). In higher dimensions, I figured that naturality with respect $GL(X)$, or maybe even just $SL(X)$, should be enough to force $x^*\equiv 0$. I approached this question using the following easy observation as a starting point.

Observation: Let $X$ be a finite-dimensional vector space. Suppose $x \in X$ and $y \in X \setminus \mathrm{span}(x)$.
  
  
*
  
*If $\dim(X) \geq 2$, then there exists $T \in GL(X)$ such that $Tx = x$, $Ty = 2y$. For example, if $X=\mathbb{R}^2$, $x=(1,0)$, $y=(0,1)$, then $T = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}$ works. 
  
*If $\dim(X) \geq 3$, then there exists $T \in SL(X)$ such that $Tx = x$, $Ty = 2y$. For example, if $X=\mathbb{R}^3$, $x=(1,0,0)$, $y=(0,1,0)$, then $T = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & \frac{1}{2} \end{bmatrix}$ works.
  

This observation leads to a proof of the following Proposition

Proposition: Let $X$ be a finite dimensional vector space and consider a mapping $x \mapsto x^* : X \to X^*$. If either
  
  
*
  
*$\dim(X) \geq 2$ and $x \mapsto x^*$ is natural with respect to every $T \in GL(X)$ or
  
*$\dim(X) \geq 3$ and $x \mapsto x^*$ is natural with respect to every $T \in SL(X)$,
  
  
  then $x^* \equiv 0$. 
Proof: Assume the first statement, and fix $x \in X$. Since $\dim(X) \geq 2$, in order to show $x^*=0$, it suffices to show $x^*(y)=0$ for every vector $y \in X \setminus \mathrm{span}(x)$. For such $y$, there exists $T \in GL(X)$ such that $Tx = x$ and $Ty=2 y$. Then, $x^*(y) = (Tx)^*(Ty)=x^*(2y) = 2 x^*(y)$, so  $x^*(y)=0$, as desired. The proof of the second part is basically the same

The hypothesis $\dim(X) \geq 2$ cannot be dropped from (1), as the Example 1 above showed. I was very surprised to discover, however, that the hypothesis $\dim(X) \geq 3$ cannot be dropped from (2)!

Example 2: Consider $X=\mathbb{R}^2$, for notational convenience. So, we can represent elements of $X$ by column vectors and elements of $X^*$ by row vectors. Consider the mapping
  $$ \begin{bmatrix}
x \\
y \\
\end{bmatrix} \mapsto 
\begin{bmatrix}
-y & x \\
\end{bmatrix}
: X \to X^* $$
  This mapping is natural with respect to every $T \in SL(2,\mathbb{R})$. 

I find the preceding example very surprising. The take away is that, given a 2-dimensional vector space $X$, it is possible to define a linear bijection $X \to X^*$ which gets along with every volume preserving coordinate change of $X$. My somewhat vague question is:

Question: Are there some connections between Example 2 and other "singular examples" in mathematics? Is there some reason I should not be surprised that there exists an especially natural way to identify a 2-dimensional vector space with its dual?

 A: The data of an identification of a finite-dimensional vector space $V$ with its dual $V^{\ast}$ is precisely the data of a nondegenerate bilinear form $B : V \otimes V \to k$ ($k$ the underlying field). The identification will be natural with respect to the group $\text{Aut}(B)$ of invertible linear transformations $V \to V$ which respect this bilinear form. There are two major special cases of this:


*

*If $B$ is symmetric, then $\text{Aut}(B)$ is a version of the orthogonal group.

*If $B$ is skew-symmetric, then $\text{Aut}(B)$ is a version of the symplectic group. 


A skew-symmetric bilinear form can be represented by a map $\Lambda^2(V) \to k$. When $\dim V = 2$, $\Lambda^2(V)$ is the top exterior power, and you can check that the data of a nondegenerate skew-symmetric bilinear form in this case is equivalent to the data of an isomorphism $\Lambda^2(V) \to k$, or equivalently the data of a volume form. And so you can check that when $\dim V = 2$ the symplectic group is the special linear group. 
The global version of this statement is that picking a symplectic form on a $2$-manifold is equivalent to picking a volume form. In general, a symplectic form determines a volume form but is not determined by it; locally this says that in general the symplectic group is strictly contained in the special linear group. 
