The Double Dual and Natural Isomorphism

The canonical double-dual embedding is far more than a curiosity of pure linear algebra; it is a structural template reused throughout mathematics. In functional analysis, a normed space $V$ is called reflexive when the analogous evaluation map $V \to V^{**}$ (built using continuous, rather than all, linear functionals) is a bijection; Hilbert spaces and $L^p$ spaces for $1<p<\infty$ are reflexive, while $L^1$ and $C[0,1]$ are not, and the failure of reflexivity governs deep results about weak compactness and the existence of minimizers in optimization problems. In differential geometry, the double dual identification underlies the standard convention that a tangent vector at a point can be regarded as a linear functional on covectors — indeed as a "directional derivative" operator — with no arbitrary choice of coordinates, which is essential to defining tensors invariantly. In category theory and theoretical computer science, the double-dual map is the motivating example used to teach the definition of a natural transformation, a concept now central to the semantics of functional programming languages (via the Curry–Howard correspondence and parametric polymorphism, where "naturality" corresponds to functions that behave uniformly across all types). In representation theory, natural maps like $\Lambda$ are used to identify a representation with its double dual representation coherently across an entire category of modules, which matters when constructing invariant pairings and traces. Even in numerical linear algebra, the conceptual habit of tracking which isomorphisms are basis-dependent (like coordinate maps to $\mathbb{F}^n$) versus basis-independent (like $\Lambda$) helps explain why certain algorithms are coordinate-free while others are not, guiding when a change of basis silently changes an answer.

Ask the tutor: Can you show me why no natural (basis-free) isomorphism $V \to V^*$ can exist, using the naturality square the way it was used for $\Lambda$?

Think of a vector $v \in V$ not as a static arrow but as a machine for testing linear functionals: feed it any $\varphi \in V^*$, and it outputs the number $\varphi(v)$. This "testing" operation is itself linear in $\varphi$ — scale $\varphi$ and the output scales, add two functionals and the outputs add — so the assignment $\varphi \mapsto \varphi(v)$ is a genuine element of $V^{**}$. The map $\Lambda$ just packages "$v$ as a tester" into a single object. The reason this differs fundamentally from the isomorphism $V \cong V^*$ is that constructing $V^*$'s elements from $V$'s elements ordinarily requires a basis: given a basis $(e_1, \dots, e_n)$ of $V$, one builds the dual basis $(e_1^*, \dots, e_n^*)$ and declares $e_i \mapsto e_i^*$. Change the basis of $V$ to $(2e_1, e_2, \dots)$, and the corresponding "isomorphism" sends $e_1$ to a completely different functional. There is no way to single out one particular isomorphism $V \to V^*$ using only the vector space structure of $V$ — any such choice secretly encodes a basis (or, in the presence of extra structure, an inner product). By contrast, $\Lambda(v)(\varphi) = \varphi(v)$ requires zero choices. It is the same formula whether you use the standard basis, a rotated basis, or no basis at all. This is what "natural" means: a single rule, uniform across all vector spaces and all linear maps between them, that produces the isomorphism automatically.

Example 1: Explicit computation of $\Lambda$ in $\mathbb{R}^3$, with an asymmetric basis of $V^*$. Let $V = \mathbb{R}^3$ with standard basis $e_1, e_2, e_3$. Consider the (deliberately non-standard, asymmetric) basis of $V^*$ given by

One checks these are linearly independent (their coefficient matrix $\begin{pmatrix}1&1&0\\0&1&-1\\0&0&2\end{pmatrix}$ has determinant $2 \neq 0$), so $(\varphi_1, \varphi_2, \varphi_3)$ is a basis of $V^* \cong \mathbb{R}^3$. Let $\varphi_1^{**}, \varphi_2^{**}, \varphi_3^{**}$ denote the dual basis of $V^{**}$ associated to $(\varphi_1,\varphi_2,\varphi_3)$, i.e. $\varphi_i^{**}(\varphi_j) = \delta_{ij}$. Take the vector $v = (3, -1, 2) \in V$. We compute $\Lambda(v)$ two ways. Direct evaluation. By definition $\Lambda(v)(\varphi) = \varphi(v)$ for any $\varphi \in V^*$. In particular:

Expressing $\Lambda(v)$ in the dual basis of $V^{**}$. Since $\Lambda(v) \in V^{**}$ and $(\varphi_1^{**}, \varphi_2^{**}, \varphi_3^{**})$ is a basis of $V^{**}$, we can write $\Lambda(v) = a_1 \varphi_1^{**} + a_2 \varphi_2^{**} + a_3 \varphi_3^{**}$ for unique scalars $a_i$. Evaluating both sides at $\varphi_j$ and using $\varphi_i^{**}(\varphi_j) = \delta_{ij}$ gives $a_j = \Lambda(v)(\varphi_j)$. So

This confirms concretely that $\Lambda(v)$ is a genuine, computable element of $V^{**}$ — not an abstract placeholder — and that its coordinates in the dual-of-dual basis are just the values $\varphi_i(v)$, exactly as the definition promises. Note also that the coordinate vector $(2,-3,4)$ of $\Lambda(v)$ in the basis $(\varphi_i^{**})$ is different from the coordinate vector $(3,-1,2)$ of $v$ in the basis $(e_i)$ — the numbers change because the bases differ, but this has no bearing on the fact that $\Lambda$ itself was defined without reference to any basis at all. The formula $\Lambda(v)(\varphi) = \varphi(v)$ never mentioned $(\varphi_i)$ or $(e_i)$; we only introduced them to display coordinates.

Example 2: Why $V \cong V^*$ is not natural — a basis-change computation that breaks any fixed formula. Let $V = \mathbb{R}^2$. Fix the standard basis $\mathcal{E} = (e_1, e_2)$ with dual basis $\mathcal{E}^* = (e_1^*, e_2^*)$, and define the isomorphism $S: V \to V^*$ by $S(e_1) = e_1^*$, $S(e_2) = e_2^*$ (extended linearly), i.e. $S(ae_1+be_2) = a e_1^* + b e_2^*$. Now change basis: let $\mathcal{F} = (f_1, f_2)$ with $f_1 = e_1 + e_2$, $f_2 = e_2$ (an asymmetric, non-orthogonal choice). This is indeed a basis since the change-of-basis matrix $\begin{pmatrix}1&0\\1&1\end{pmatrix}$ is invertible. Compute the dual basis $\mathcal{F}^* = (f_1^*, f_2^*)$ directly from $f_1^*(f_i) = \delta_{1i}$, $f_2^*(f_i) = \delta_{2i}$. Writing $f_1^* = \alpha e_1^* + \beta e_2^*$: we need $f_1^*(f_1) = \alpha + \beta = 1$ and $f_1^*(f_2) = \beta = 0$, giving $f_1^* = e_1^*$. Writing $f_2^* = \gamma e_1^* + \delta e_2^*$: we need $f_2^*(f_1) = \gamma + \delta = 0$ and $f_2^*(f_2) = \delta = 1$, giving $\gamma = -1$, so $f_2^* = -e_1^* + e_2^*$. Define the "basis isomorphism" $S': V \to V^*$ using $\mathcal{F}$ instead: $S'(f_1) = f_1^*$, $S'(f_2) = f_2^*$. Let's evaluate both $S$ and $S'$ on the same vector $v = f_1 = e_1 + e_2$:

These are different functionals: $S(v)(e_1) = 1$ but $S'(v)(e_1) = 1$; check instead at $e_2$: $S(v)(e_2) = 1$ while $S'(v)(e_2) = f_1^*(e_2)$. Since $e_2 = f_2$, $f_1^*(e_2) = f_1^*(f_2) = 0$. So $S(v)(e_2) = 1 \neq 0 = S'(v)(e_2)$. Thus $S(v) \neq S'(v)$ for the same vector $v$: the "isomorphism to the dual" genuinely depends on which basis is used to construct it, confirming there is no way to write a single basis-free formula for an isomorphism $V \to V^*$ analogous to $\Lambda(v)(\varphi) = \varphi(v)$. Contrast this with $\Lambda: V \to V^{**}$, where the formula $\Lambda(v)(\varphi) = \varphi(v)$ gave the same answer regardless of which auxiliary basis we used to display coordinates in Example 1 — that is precisely the content of naturality.