If $V=W$ (vector spaces), then $\det A^{\dagger}=\det A$, where $A$ is a map and $A^{\dagger}$ is its adjoint

Question

Let $(V,g)$ and $(W,h)$ be scalar product spaces, and let $A:V\to W$ be linear. We define $A^{\dagger}:W\to V$ by $$A^{\dagger}=S\circ A^*\circ\mathcal{F}$$ where $S$ is the sharp map corresponding to $g$ and $\mathcal{F}$ is the flat map corresponding to $h$.

Side note for more clarity: sharp map is the inverse of the flat map, and the flat map corresponding to a scalar product is defined as $F(v)(w)=g(v,w)$.

There's a theorem I'm looking at in the book: With the above setup, if $V=W$, then $\det A^{\dagger}=\det A $. The proof goes like this:

With $\mathcal{F}=F$, we have $S=\mathcal{F}^{-1}$, so the result follows from (theorems referenced from earlier chapters).

I'm confused about whether this should necessarily be true. If a single vector space can admit more than one scalar product, won't that cause a problem? Because then the flat maps won't necessarily match.

Answer 1

You may define the adjoint $A^\dagger$ directly on the level of scalar products via $$g(v, A^\dagger w)=\langle v| A^\dagger w\rangle_V \:\stackrel{!}{=}\: \langle Av | w\rangle_W = h(Av, w) \quad\forall v\in V, w\in W$$ which makes obvious the dependence of $A^\dagger$ on the given scalar products.
According to the equivalent definition you wrote above $$\begin{array} {}V^* & \stackrel{A^*}{\longleftarrow} & W^* \\ \downarrow{S} & & \uparrow{F} \\ V & \stackrel{A^{\dagger}}{\longleftarrow} & W \end{array}$$ we have the refined view on where the dependency on the scalar products sits, namely in the vertical isomorphisms $F$ and $S$ (being conjugate-linear if the ground field is $\mathbb C$). They are obtained from the Riesz representation theorem which identifies a Hilbert space with its dual, in a canonical way as no choice of a basis is required.
The upper horizontal arrow $A^*$ denotes the dual map.

If $V=W$, then only one scalar product is around, and $A^{\dagger}=F^{-1}\circ A^*\circ F$.

Hope this helps to dissolve your confusion.

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Note that you introduced a slightly unusual notation. Usually, or at last very often, $A^*$ denotes the adjoint (which for matrices is given by transposition and conjugation), whereas the dual map is denoted by $A'$.

Finally two related links, the first one refers to $\,\det(A^{\dagger})=\det(A)\,$
Geometric interpretation of $\det(A^T) = \det(A)$
the second is an answer of mine regarding Index raising and lowering aka flat and sharp maps
https://math.stackexchange.com/a/2980908/316749