Selfadjoint Endomorphism

Question

Question:

Let $p >1$ be an integer, let $G = \mathbb{Z}/(p)$, and let $V = \mathbb{C}^G$, which is an inner product space over $\mathbb{C}$ with inner product defined by $\langle f, g\rangle =\sum_{n\in G}f(n)\overline{g(n)}$. Let $\alpha$ be the endomorphism of $V$ defined by $\alpha(f): n \mapsto f(n+1) + f (n - 1)$. Is $\alpha$ selfadjoint?

My work:

Selfadjoint: $\langle \alpha(f),g \rangle = \langle f,\alpha(g)\rangle$

Additionally, we have Proposition 17.2: Let $V$ be an inner product space over $\mathbb{C}$ and let $\alpha \in End(V)$ have an adjoint. If $\langle \alpha(v), v\rangle \in \mathbb{R}$ for all $v \in V$ , then $\alpha$ is selfadjoint.

Also, we take note of the fact that $G=\mathbb{Z} / (p)= \{0,1,...,p-1\}$ and thus, $n\in G \subset \mathbb{Z}$.

\begin{align*} \langle \alpha(f), f \rangle &= \langle f(n+1) + f(n-1), f(n) \rangle = \langle f(n+1),f(n) \rangle + \langle f(n-1),f(n)\rangle \\[1ex] &= \sum_{n\in G}f(n+1)\overline{f(n)}+\sum_{n\in G}f(n-1)\overline{f(n)} \end{align*}

And then I'm stuck. Any help is greatly appreciated.

Answer 1

Following your calculation, we have

$$ \overline{\left< \alpha(f), f \right>} = \sum_{n \in G} f(n) \overline{f(n+1)} + \sum_{n \in G} f(n) \overline{f(n-1)} = \sum_{m \in G} f(m-1) \overline{f(m)} + \sum_{m' \in G} f(m' + 1) \overline{f(m')} = \left< \alpha(f), f \right> $$

where we used the change of variables $m = n + 1$ and $m' = n - 1$ for the sums. This shows that $\left< \alpha(f), f \right>$ is real for all $f \in V$.

Answer 2

If you set $n-1=m$, then the second summation becomes $$ \sum_{m\in G}f(m)\overline{f(m+1)}= \sum_{n\in G}f(n)\overline{f(n+1)} $$ So \begin{align} \langle \alpha(f),f\rangle &=\sum_{n\in G}(\alpha(f)(n),f(n))\\ &=\sum_{n\in G}(f(n+1)+f(n-1))\overline{f(n)}\\ &=\sum_{n\in G}f(n+1)\overline{f(n)}+ \sum_{n\in G}f(n-1)\overline{f(n)}\\ &=\sum_{n\in G}f(n+1)\overline{f(n)}+ \sum_{n\in G}f(n)\overline{f(n+1)}\\ \end{align} which is real.