Let $X$ be a Hilbert space with associated canonical isomorphism $I:X\rightarrow X^\ast$ (by the Riesz representation theorem). If $A:X\rightarrow X$ is a linear operator on $X$, then its dual $A^\ast:X^\ast\rightarrow X^\ast$ and its adjoint (or "transpose") $A^\dagger:X\rightarrow X$ are related to one another by $I\circ A^\dagger=A^\ast\circ I$. Is there an enlightening categorical viewpoint on the connection between $A^\ast$ and $A^\dagger$?
[Math] Dual and adjoint operator
adjoint-functorscategory-theoryhilbert-spaceslinear algebrareference-request
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First, there's something called the Riesz Representation Theorem. To understand it, start by fixing a vector $v \in H$. We can now use the dot product to define a continuous linear functional $L_v: H \rightarrow F$ by $$L_v(w) = \langle w,v \rangle.$$ What the Riesz Representation Theorem says is that every continuous linear functional $\phi:H \rightarrow F$ arises in this way! That is, given any element $\phi$ of the dual space $H^\ast$, there is some $v \in H$ so that $\phi(w) = \langle w, v \rangle$. So this is why we can sort of think as linear functionals as elements of the Hilbert space, and vice versa. In more mathematical terms, we say that the dual of $H$ is isomorphic to $H$.
Second, a note on dual spaces: I believe that typically the space $H^\ast$ is defined to be the set of continuous linear functionals. This distinction is important, as there are different types of duals. So far, I've been talking about the topological dual. However, there is an algebraic dual, which I have seen denoted $H^\star$, which is just all linear functionals $H \rightarrow F$, no continuity assumed. The Reisz Representation Theorem concerns only the topological dual. (The duals are actually the same for finite dimensional Hilbert spaces, but I don't believe the Hilbert spaces encountered in QM are.)
Third, adjoints: Given any Hilbert spaces $H, K$, and a continuous linear functional $A: H \rightarrow K$, there is a continuous linear map called the adjoint $A^\ast:K \rightarrow H$ (note that it goes the other way) that is defined by the equation $\langle Av, w \rangle = \langle v, A^\ast w \rangle$. You typically only see the case $K = H$. So no, the adjoint of an operator $A: H \rightarrow H$ is not an element of $H^\ast$, since the members of $H^\ast$ are continuous linear functionals from $H$ into $F$, and $H^\ast$ goes from $H$ into $H$.
Unfortunately, I don't know much about QM, so this last bit is just speculating on how I think the notation works. If you consider kets $|\phi \rangle \in H$ to be an element of the Hilbert space, then there is a continuous linear functional that I suppose you could call $\langle \phi |$ defined by $\langle \phi | v \rangle = \langle v, \phi \rangle.$ And conversely, given a bra $\langle \phi |$, by the Reisz Representation Theorem, there is a bra $| \phi \rangle$ so that $\langle \phi | v \rangle = \langle \phi , v \rangle.$
A simple way to think about this is to identify $V$ and $V^*$. Then the proof is slick, but it conceals the difference between $V$ and $V^*$, and the dependence of their identification on the dot product: $$(S^Tx,x)=(x,Sx)=(Sx,x)>0.$$
In some cases, e.g. in tensor calculus and differential geometry, it is sometimes important to make this identification explicit (it corresponds to raising and lowering indices in coordinate notation). To make it explicit, it is convenient to think of $S^*:V^*\to V^*$ instead of $S^T$, and write $\langle\phi,x\rangle$ instead of $\phi(x)$ for the value of $\phi\in V^*$ on $x\in V$. The $\langle\cdot,\cdot\rangle$ is called the canonical pairing, and it saves us from writing a lot of nested parentheses. Then $S^*$ is defined by $\langle S^*\phi,x\rangle:=\langle\phi,S x\rangle$, and (a unique) functional $\phi$ is corresponded to $x$ via $\langle\phi,y\rangle=(x,y)$. There is also the induced inner product on $V^*$ with similar definition and properties. Now the calculation goes like this: $$(S^*\phi,\phi)=\langle S^*\phi,x\rangle=\langle\phi,S x\rangle=(x,Sx)=(Sx,x)>0$$
There is a way to make this independent of an inner product. We should consider not operators on the same space, but operators from a space to its dual $S:V\to V^*$. They are used in tensor calculus (to contract indices), and in functional analysis, e.g. by Lions et al.
It is natural to call such operators positive definite when $\langle Sx,x\rangle>0$ for $x\neq0$. Defining $S^*$ by $\langle S^*x,y\rangle=\langle Sy,x\rangle$ we get $S^*:V^{**}\to V^*$. Note how neither inner product, nor basis, nor any other extra structure is needed so far. Moreover, $V$ can be identified with a subspace of its double dual $V^{**}$ also canonically, see injection into the double-dual. Namely, $x\mapsto\langle\cdot,x\rangle$ defines a functional on $V^*$ canonically corresponded to $x$. For finite-dimensional spaces $V$ maps onto the entire $V^{**}$. In general, when this happens the space is called reflexive. Now we recover the simplicity: $$\langle S^*x,x\rangle=\langle Sx,x\rangle>0.$$
Best Answer
More generally, if $A : X \to Y$ is a linear operator, then we have $I_X \circ A^\dagger=A^\ast\circ I_Y$, as in the following commutative diagram:
$$\begin{array}{c} Y & \xrightarrow{A^\dagger} & X \\ I_Y \downarrow ~~~~~&& ~~~~\downarrow I_X \\ Y^* & \xrightarrow{A^*} & X^* \end{array}$$
This is just a reformulation of the definition of $A^\dagger$: $$A^*(I_Y(y))(x)=I_Y(y)(A(x))=\langle A(x),y \rangle = \langle x,A^\dagger(y)\rangle=I_X(A^\dagger(y))(x).$$
And the diagram tells us that $I$ is a natural isomorphism from the functor $\mathrm{Hilb}^{\mathrm{op}} \to \mathrm{Hilb}$ mapping $X \mapsto X$ and $A \mapsto A^\dagger$ to the functor $\mathrm{Hilb}^{\mathrm{op}} \to \mathrm{Hilb}$ mapping $X \mapsto X^*$ and $A \mapsto A^*$.
Strictly speaking, if we work over $\mathbb{C}$, we have to take the complex-transpose $\overline{X}$ to define an isomorphism $I_X : \overline{X} \cong X^*$, and the first functor has to be modified accordingly.