It is often said that vector spaces are not naturally isomorphic to dual spaces, because the dual functor is not naturally isomorphic to the identity functor. But the latter is a rather trivial statement, because you can’t have a natural transformation between a contravariant functor and a covariant functor. So I’d like to see if it it can be modified into something less trivial.
Let $VectIso$ be the category with vector spaces as objects and bijective linear transformations as morphisms. Let $D:VectIso\rightarrow VectIso$ be a covariant functor defined by $D(V)=V^*$ for any vector space $V$ and $D(T)(f)=f\circ T^{-1}$ for any bijective linear transformation $T:V\rightarrow W$ and any linear functional $f\in V^*$.
Then my question is, is $D$ naturally isomorphic to the identity functor? I assume the answer is no, but how would you prove it?
Best Answer
Suppose we had a natural transformation $\eta: 1 \Rightarrow D$. Let $V$ be a vector space and $T \in GL(V)$. The naturality condition asserts that for all $v \in V$, we have $$\eta_V(Tv) = \eta_V(v) \circ T^{-1}.$$ Fix $v \in V \setminus \{0\}$, and let $f = \eta_V(v)$. Since $GL(V)$ acts transitively on $V \setminus \{0\}$, we see that $\eta_V: V \to V^*$ is completely determined by $f$ according to the equation above.
We need to check that $\eta_V$ is thus well-defined. In particular, we need to have $$f = f \circ T$$ for all $T$ fixing $v$.
Let us be more concrete. Let $V = \mathbb{R} \langle e_1, e_2, e_3 \rangle$ and $v = e_1$. Consider transformations $T \in GL(V)$ as follows: $$\begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \qquad \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}, \qquad \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}.$$ These matrices all fix $e_1$. Plugging them into the equation $f = f \circ T$, we find \begin{align} f(e_2) &= f(e_1) + f(e_2) \\ f(e_3) &= f(e_2) + f(e_3) \\ f(e_2) &= f(e_3) \end{align}
Consequently, we have $f(e_1) = f(e_2) = f(e_3) = 0$, so that $f$ is the zero linear functional. But then $\eta_V(e_1) = 0 = \eta_V(0)$, contradicting the fact that $\eta_V$ is a linear isomorphism. So no natural transformation $\eta: 1 \Rightarrow D$ can exist.