Adjoint of the Dual Map

Question

Given a linear map $f:X\longrightarrow Y$ between finite dimensional Hilbert spaces, we can define the dual map $f^*:Y^*\longrightarrow X^*$ by $f^*(\phi) = \phi\circ f$. In a paper on quantum computation that I'm reading the adjoint of the dual map is mentioned, it is given the name $f_*$. I'm trying to understand what $f_*$ looks like.

Here is my progress so far:
Clearly $f_*:X^*\longrightarrow Y^*$. Now let $(x_i)_i,(y_i)_i$ be bases of $X$ and $Y$ respectively and $(x_i^*)_i,(y_i^*)_i$ the dual bases, then \begin{align*} \langle y_j^*,f_*(x_i^*)\rangle_{Y^*} = \sum_k \langle y_j^*(y_k),f_*(x_i^*)(y_k)\rangle_{\mathbb{C}} = f_*(x_i^*)(y_j) \end{align*} where I used, that $y_j^*(y_k) = \delta_{jk}$. But since $f_*$ is the adjoint of $f^*$ we can also write $$\langle y_j^*,f_*(x_i^*)\rangle_{Y^*} = \langle f^*(y_j^*),x_i^*\rangle_{X^*}$$ Thus \begin{align*} f_*(x_i^*)(y_j) &= \langle f^*(y_j^*),x_i^*\rangle_{X^*} = \langle y_j^*\circ f,x_i^*\rangle_{X^*} = \sum_l \langle y_j^*(f(x_l)),x_i^*(x_l)\rangle_\mathbb{C}\\ &= \overline{y_j^*(f(x_i))} \end{align*} where the overline denotes complex conjugation. This determines the evaluation at basis vectors. To find an expression for general $\phi\in X^*$,$y\in Y$ one could use linear combinations of the above. Now my question is: Is there a way of expressing $f_*$ that does not explicitely employ the use of a basis? Like for $f^*$ where we can write $f^*(\phi) = \phi\circ f$ without the need to define a basis

Answer 1

When I first read your question my intuition was that—because the dual and the adjoint of a Hilbert space operator are "equivalent" & because the adjoint is an involution (i.e. $(A^\dagger)^\dagger=A$)—the adjoint of the dual has to lead back to the original operator "somehow". In order to turn this into a precise statement we certainly need the

Riesz representation theorem: Every Hilbert space and its dual are related via the real-linear bijective map \begin{align*} \Phi_X:X&\to X^*\\ x&\mapsto \Phi(x) \end{align*} which acts via $(\Phi(x))(x')=\langle x,x'\rangle$.

In quantum info notation one would express this via $\Phi(x)=\langle x|$. However, for this question it will be beneficial to assign an explicit symbol to this map.

With this we can—as a first step—relate the dual $f^*:Y^*\to X^*$ of a bounded linear map $f:X\to Y$ with its adjoint operator $f^\dagger:Y\to X$ (i.e. $\langle f^\dagger y,x\rangle=\langle y,fx\rangle$ for all $x\in X$, $y\in Y$). This yields the precise formulation of the initial claim that "the dual and the adjoint are equivalent".

Lemma. $f^*=\Phi_X\circ f^\dagger\circ \Phi_Y^{-1}$

Proof. Equivalently, we will show that $ f^\dagger\circ \Phi_Y^{-1}=\Phi_X^{-1}\circ f^*$. Given arbitrary $x\in X$, $y\in Y^*$ we compute \begin{align*} \langle (f^\dagger\circ \Phi_Y^{-1})(y),x\rangle&=\langle\Phi_Y^{-1}(y),fx\rangle=\big(\Phi_Y(\Phi_Y^{-1}y)\big)(fx)\\ &=y(fx)=(f^*\circ y)(x)\\ &=\big(\Phi_X(\Phi_X^{-1}(f^*\circ y))\big)(x)\\ &=\langle \Phi_X^{-1}(f^*\circ y),x\rangle=\langle (\Phi_X^{-1}\circ f^*)(y),x\rangle\,.\tag*{$\square$} \end{align*}

Now if we could just "take the adjoint" of the above lemma we'd obtain a relation between $(f^*)^\dagger=:f_*$ and $(f^\dagger)^\dagger=f$. However, this adjoint would be on the level of dual spaces; hence this only makes sense once we know that the dual of a Hilbert space is a Hilbert space, as well. Indeed this is true when equipping the dual with the inner product \begin{align*} \langle\cdot,\cdot\rangle_{X^*}:X^*\times X^*&\to\mathbb C\\ (x_1,x_2)&\mapsto\langle\Phi_X^{-1}(x_1),\Phi_X^{-1}(x_2)\rangle_X\,. \end{align*}

Again in physicists notation this would read $\langle \ \langle x_1|\,,\,\langle x_2|\ \rangle_{X^*}=\langle x_1,x_2\rangle$ for all $x_1,x_2\in X$. Altogether we can now relate $f_*$ to the original $f$:

Proposition. It holds that $f_*=\Phi_Y\circ f\circ \Phi_X^{-1}$.

Proof. By definition, $f_*=(f^*)^\dagger$ where $(\cdot)^\dagger$ is now the adjoint on the level of dual (Hilbert!) spaces. Using the above lemma we obtain $ f_*=(f^*)^\dagger=(\Phi_X\circ f^\dagger\circ \Phi_Y^{-1})^\dagger $. Now the adjoint acts on compositions via $(A\circ B)^\dagger=B^\dagger \circ A^\dagger$ which implies $$ f_*=(f^*)^\dagger=(\Phi_X\circ f^\dagger\circ \Phi_Y^{-1})^\dagger=(\Phi_Y^{-1})^\dagger\circ (f^\dagger)^\dagger\circ \Phi_X^\dagger=\Phi_Y\circ f\circ\Phi_X^{-1}\,. $$ In the last step we used that the adjoint is an involution as well as the redaily verified fact the Riesz isomorphism is a unitary transformation, that is, $\Phi^\dagger=\Phi^{-1}$.$\quad\square$

This is the precise formulation of our initial intuition that "the adjoint of the dual has to lead back to the original operator". Applied to any $\langle x|\in X^*$ (where $x\in X$) this proposition yields: $$ \boxed{f_*(\langle x|)=\Phi_Y(fx)=\langle fx|} $$ P.S. Be aware that at no point did we assume that our Hilbert spaces are finite-dimensional. Indeed the relation between $f_*$ and $f$ holds for arbitrary Hilbert spaces.