Origin of notation of a dual space

Question

Classically, there’s discordance in the notation of the dual of a vector space over a field $K$. Of course we can all agree that the algebraic dual is the vector space of linear maps from $V$ to its base field. As such it can be represented as $$\operatorname{Hom}_K(V, K).$$

But notations are plenty as referring to this space such as $V’$, $V^*$, $V^\vee$ and even $V^\#$.

I’d like to know the origin of these notations and recommendations in which to use.

In the past I went for $V^*$, but many people I discussed with found confusing the fact than any linear map $T : V \to W$ induced another map $T^* : W^* \to V^*$ that is called the transpose of $T$. And NOT the conjugate transpose.

Soon I shifted towards $V^\vee$ but this hides the fact that if $V$ is a finite dimensional Hilbert space, then the linear map corresponding to a vector is just matrix multiplication by its conjugate transpose. This is given by the Ríesz representation theorem.

So I’d like to know which notation is best and where it comes from.

Answer 1

It seems to me that your question is really about the conjugate transpose, not about dual spaces. Let's be very clear about how this works.

If $V, W$ are two finite-dimensional vector spaces, over any field, and $F : V \to W$ is a linear map, then as you say it induces a dual map $F^{\ast} : W^{\ast} \to V^{\ast}$. I would call this map the dual, not the transpose. The relationship to taking the transpose of a matrix is that if $B = \{ v_i \}, C = \{ w_j \}$ are bases of $V, W$, then they induce dual bases $B^{\ast} = \{ v_i^{\ast} \}, C^{\ast} = \{ w_j^{\ast} \}$ of the dual spaces $V^{\ast}$ and $W^{\ast}$, defined by the condition $v_i^{\ast} v_j = \delta_{ij}$ and similarly for $w$. If we write $_C F_B$ for the matrix of a linear map with respect to bases then we have the relationship

$$_{B^{\ast}} F^{\ast}_{C^{\ast}} = \, \left( _C F_B \right)^T$$

In words, the matrix of the dual with respect to the dual bases is the transpose of the original matrix with respect to the original bases, or said another way, the transpose is the coordinate-dependent version of the coordinate-invariant concept of the dual. This is the relationship between duals and transposes over any field. Note that over an arbitrary field it doesn't even make sense to talk about the conjugate transpose.

Now let $V, W$ be complex Hilbert spaces and $F : V \to W$ be a bounded linear map. Then $F$ again induces a dual map $F^{\ast} : W^{\ast} \to V^{\ast}$, where now $(-)^{\ast}$ denotes the continuous dual. The special feature of the Hilbert space case is that the inner product provides a canonical identification

$$V \ni v \mapsto \langle v, - \rangle \in V^{\ast}$$

between $V$ and its continuous dual (here we adopt the convention that the inner product is linear in the second argument), which is antilinear (because the inner product is antilinear in the first argument). Applying these identifications to both $W^{\ast}$ and $V^{\ast}$ produces a third map, the (Hermitian) adjoint $F^{\dagger} : W \to V$, which is now a linear map from $W$ to $V$ (with no duals), and which can equivalently be defined as the unique map satisfying $\langle Fv, w \rangle = \langle v, F^{\dagger} w \rangle$ as usual.

Now suppose that $B = \{ v_i \}, C = \{ w_j \}$ are two orthonormal bases of $V, W$. Then we can ask for both the matrix $_C F_B$ of $F$ with respect to these bases, as well as the matrix $_B F^{\dagger}_C$ of the adjoint with respect to these same bases. And the relationship between these matrices is that the latter is the conjugate transpose of the former:

$$_B F^{\dagger}_C = \overline{ _C F_B}^T.$$

You will also see the conjugate transpose written $(-)^{\dagger}$, and it is for this reason. Said another way, the conjugate transpose is the coordinate-dependent version of the coordinate-invariant concept of the adjoint. The conjugation here exactly comes from the fact that the identification between a Hilbert space and its dual given by the inner product is antilinear. So what is being hidden here isn't about dual spaces, it's about the Riesz representation theorem and its antilinearity, and about the difference between the dual and the adjoint.

(Note also that orthonormal bases are, more or less by definition, bases which are their own dual bases with respect to the inner product. This is why the above result looks so similar but not quite identical to the result about transposes and dual bases.)

It is possible to make the antilinearity of the Riesz map more explicit notationally by writing its domain as $\overline{V}$, the complex conjugate, which is the same as $V$ except that scalar multiplication is twisted by the complex conjugate. Then we can say that the Riesz map is really a linear isomorphism $\overline{V} \cong V^{\ast}$, and applying it gives us a slightly different version of the adjoint which goes from $\overline{W}$ to $\overline{V}$. I'm not aware of any notation for this map and you will almost never see anyone talk about this, but this distinction is relevant in some other contexts.

---

Which notation you want to use for the dual space is a matter of taste. Personally I like $V^{\ast}$ and find the others a little ugly and hard to read. I also recommend using different notation for dual maps, transpose matrices, and adjoints, as above.