Assume that $V=V(m,\mathbb{R})$, $W=W(n,\mathbb{R})$, and $g:V\to V^* $ and $G:W\to W^*$ are two isomorphisms. Given a map $f:V\to W$, define the adjoint of $f$, denoted by $\tilde{f}$, by $G(w,fv)=g(v,\tilde{f}w)$ (by Nakahar's book, page 98), where $v\in V$ and $w\in W$. Why does $\tilde{f}w$ exist so that $G(w,fv)=g(\tilde{f}w,v)$ and what is it exactly?
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6$\begingroup$ It seems a question for Mathematics $\endgroup$– GiorgioP-DoomsdayClockIsAt-90Commented Aug 14, 2023 at 5:33
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$\begingroup$ Does this help? It discusses existence, well-definedness and so on for (finite-dimensional, complex) Hilbert spaces... So it may be also helpful for your question. $\endgroup$– JakobCommented Aug 14, 2023 at 15:50
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$\begingroup$ Essentially, $G$ and $g$ are dot products, $f$ is an $n\times m$ matrix, and $\tilde f$ is its transpose, an $m\times n$ matrix. $\endgroup$– Akiva WeinbergerCommented Aug 14, 2023 at 16:00
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1$\begingroup$ I think there's some confusion with notation. Either $G$ is a function from $W$ to $W^*$, in which case you should write $G(w)(fw)$, or $G$ is a bilinear function from $W\times W$ to $\Bbb R$ in which case you'd write $G(w,fw)$ as you've done. $\endgroup$– Akiva WeinbergerCommented Aug 14, 2023 at 16:02
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$\begingroup$ Dear @Jakob , thanks for the comment. It helped so much. But I have some problems: here $V$ and $W$ are not Hilber spaces, $f$ is a linear map (necessarily not bounded) from $V$ to $W$ (so not a operator). So does your answer still hold here? I mean Riesz Theorem or te definition of adjoint operator. $\endgroup$– MahtabCommented Aug 15, 2023 at 4:17
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By the nice @Jakob comments and links, I answer my question in this way:
We define $\tilde{f}:W\to V$ by $\tilde{f}=g^{-1}\circ f^* \circ G$, where $f^*:W^* \to V^*$ is the pullback of $f$ (i.e. $f^* (g)=g\circ f$). Also we have $$G(w,fv)=G(w)(f(v))=f^*\circ G(w)(v)=g\circ \tilde{f}(w)(v)=g(\tilde{f}(w))(v)=g(\tilde{f}w,v).$$