Adjoint of projection onto direct sum of Hilbert spaces Let $K_n$ be Hilbert spaces and define \begin{equation*} K := \bigoplus_{\ell_2} K_n = \left\{ (x_1,x_2,\ldots) \in \bigoplus_{n=1}^\infty K_n : \sum_{n=1}^\infty \|x_n\|^2 < \infty \right\} \end{equation*}
It is easy to see that $K$ is a Hilbert space with the inner product $(x,y) = \displaystyle{\sum_{n=1}^\infty (x_n,y_n) } $. I'm interested in the adjoint of the operator projection \begin{equation*} \begin{split} & \pi_n :\bigoplus_{\ell_2}K_n \rightarrow K_n \\ & (x_1, \ldots,x_n, \ldots) \mapsto x_n. \end{split} \end{equation*} My idea is that ${\pi_n}^*$ is the natural inclusion from $K_n$ to $K$, but I can't find the correct way to prove it. Is my idea right? How can I prove it?
Than you for your help.
 A: You are right, the adjoint of $\pi_n : K \to K_n$ is the canonical inclusion $\iota_n : K_n \to K$.
Let $(x_1, x_2, \ldots) \in K$ and $y_n \in K_n$. We have:
\begin{align}
\Big\langle \pi_n(x_1, x_2, \ldots), y_n\Big\rangle_K &= \langle x_n, y_n\rangle_{K_n} \\
&= \Big\langle (x_1, x_2, \ldots), (\underbrace{0, \ldots, 0}_{n-1}, y_n, 0, \ldots)\Big\rangle_{K}\\
&= \Big\langle (x_1, x_2, \ldots), \iota_n(y_n)\Big\rangle_{K}\\
\end{align}
Therefore $\pi_n^* = \iota_n$.
A: The adjoint $\pi_n^*:K_n\to K$ of $\pi_n$ is uniquely defined via the conditon
$$
\langle x,y_n\rangle_{K_n}= \boxed{\langle x,\pi_n y\rangle_{K_n}=\langle \pi_n^* x,y\rangle_K}=\sum_{j=1}^\infty \langle (\pi_n^*x)_j,y_j\rangle_{K_j}\\=\langle (\pi_n^*x)_n,y_n\rangle_{K_n}+\sum_{\substack{j=1\\j\neq n}}^\infty \langle (\pi_n^*x)_j,y_j\rangle_{K_j}
$$
for all $x\in K_n$, $y\in K$. So as we can freely chose $x,y_1,y_2,\ldots$ this implies $(\pi_n^*x)_n=x$ and $(\pi_n^*x)_j=0$ for $j\neq n$ so as conjectured, $\pi_n^*$ is the natural embedding of $K_n$ into $K$.
Edit: it is probably easier to consider the approach from mechanodroid's answer so
$$
\langle (0,\ldots,0,x,0,\ldots),y\rangle_K=\langle x,y_n\rangle_{K_n}=\ldots=\langle \pi_n^*x,y\rangle_K,
$$
but in the end, both ways obviously yield the same result.
