Inclusion operator on half-integer weight modular forms and its adjoint We have an inclusion $\iota: S_{k+1/2}(8N) \hookrightarrow S_{k+1/2}(16N)$, whose adjoint with respect to the Petersson scalar product is apparently given by $$\iota^{*} = Tr: \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + \begin{pmatrix} 1 & 0 \\ -4N & 1 \end{pmatrix}.$$ For example, see p. 131 of this text (Theory of Newforms of Half-integral Weight). I am confused how this computation arises. First of all, the inclusion would just correspond to the identity matrix $I$ right? In this case, I would think the adjoint operator would also correspond to identity. My second question is why this supposed adjoint looks like a trace.
One problem is that the paper I've referenced doesn't give a definition of the inclusion they mention. So I'm most likely missing something... it could be the $V$ operator, yes?
 A: Let $\langle \cdot,\cdot \rangle_{\Gamma_0(8N)}$ and $\langle \cdot,\cdot \rangle_{\Gamma_0(16N)}$ denote the Peterrson scaler products on the spaces $S_{k+1/2}(8N)$ and $S_{k+1/2}(16N)$ respectively. Let $\iota: S_{k+1/2}(8N) \hookrightarrow S_{k+1/2}(16N)$ denote the inclusion operator; and let $\iota^*$ denote its adjoint operator. Now, this means that $\iota^*$ is the $\it unique$ operator which satisfies
$$ \langle f|\iota,g \rangle_{\Gamma_0(16N)} = \langle f,g|\iota^* \rangle_{\Gamma_0(8N)}$$
for any $f\in S_{k+1/2}(8N)$ and $g\in S_{k+1/2}(16N)$. Hence (since adjoint of an operator in unique), it is enough to prove that the operator
$$Tr: \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + \begin{pmatrix} 1 & 0 \\ -4N & 1 \end{pmatrix}
$$ satisfies
$$ \langle f|\iota,g \rangle_{\Gamma_0(16N)} = \langle f,g|Tr\rangle_{\Gamma_0(8N)}.$$ The above computation has been done for the case of integral weight modular forms in Cohen and Stromberg, Modular forms: A Classical Approach, Lemma 13.3.21 (page 535).
At the moment I do not have the calculations for the half-integral weight case, but I will try to put them. I hope this helps.
Some thoughts: I had the exact same doubts when I came upon these operators. Like, adjoint of inclusion (which is kinda identity)should be identity. But if one observes that by inclusion we take the form $f$ to a higher level, so its adjoint should be a level lowering operator. (Trace operator is a level lowering one).
By the operator $V$ in the question, if it is the operator $f \mid V(m) := \sum_{n} a(n)q^{mn},$ then its trace is something else. I do not know(at the moment) in this case, but for the integral weight case, it is given in the second part of the above-mentioned lemma in Cohen-Stromberg book. (Note that the notation for this operator is changed there, it is $B(d)$.)
Please add a comment if any clarification is required.
