Duality

Dual Space & Covector

The dual space of a vector space $V$, denoted as $V^{\ast }$, is the set of all linear maps from $V$ to $\mathbb{R}$. i.e. $V^{\ast }=\mathrm{Hom}(V,\mathbb{R})$. And elements of $V^{\ast }$ are called covectors or dual vectors.

Proposition

The dual space is a vector space.

Proof It is clear enough to show that both associativity, commutativity hold. We identify the constant function $v\mapsto 0$ as the additive identity. $\square$

Proposition

Suppose $v_1,v_2,\dots ,v_n$ is a basis for $V$. Then ${\phi }_1,{\phi }_2,\dots ,{\phi }_n$ defined by ${\phi }_i(v_j)={\delta }_{ij}$forms a basis for $V^{\ast }$.

Proof

Corollary

For any vector space $V$, $\dim V=\dim V^{\ast }$, and further $V^{\ast }\cong V$.

Proof This a consequence of the above proposition, because $v_i\mapsto {\phi }_i$ is a linear isomorphism. $\square$

Riesz representation

In a inner product space $(V,⟨\cdot ,\cdot ⟩)$, for any $\phi \in V^{\ast }$, there is a unique $w\in V$ such that $\phi =⟨w,\cdot ⟩$. Proof We identify any $\phi \in V^{\ast }$ by identifying its values on the basis vectors. Suppose $v_1,v_2,\dots v_n$ form a basis for $V$. Then we have

Proposition

There is a natural isomorphism between $V$ and $(V^{\ast }{)}^{\ast }$. That is, the double dual functor is naturally isomorphic to the identity functor on the category of vector spaces.

Proof Define the natural transformation $\eta :\mathrm{id}\Rightarrow (-{)}^{\ast \ast }$ by ${\eta }_V(v)(\alpha ):=\alpha (v),\text{for all }\alpha \in V^{\ast },v\in V,V\in {\text{Vect}}_0$We shall check the following diagram commutes for any $V,W\in {\text{Vect}}_0$: For any $v\in V$ and $\alpha \in W^{\ast }$, there holds $\begin{array}{r}(f^{\ast \ast }\circ {\eta }_V)(v)(\alpha )=f^{\ast \ast }({\eta }_{V}v)(\alpha )=({\eta }_{V}v\circ f^{\ast })(\alpha )={\eta }_{V}v(f^{\ast }\alpha )={\eta }_{V}v(\alpha \circ f)=\alpha (fv),\\ ({\eta }_W\circ f)(v)(\alpha )={\eta }_W(fv)(\alpha )=\alpha (fv).\end{array}$Thus the diagram commutes, and $\eta$ is a natural transformation. Moreover, ${\eta }_V$ is an isomorphism for any $V\in {\text{Vect}}_0$, thus $\eta$ is a natural isomorphism. $\square$

Geometric Interpretation

Geometrically, this definition uses no bases or coordinates—it is intrinsically defined by the structure of vector spaces. That’s why “natural” is often equated with basis-independence or coordinate-free definition in geometry. Concretely, if one identify $f$ as the coordinate, then the above commutative diagram basically means coordinate free.