encoding/decoding in the CKKS are isometric ring isomorphisms?

Question

I am working on my master thesis which has as a main subject the CKKS algorithm. I am following the paper https://eprint.iacr.org/2016/421.pdf in which on page 8 it is mentioned that the encoding/decoding in the CKKS are isometric ring isomorphisms between $(S, \|\cdot \|_{\infty}^{can})$ and $(\mathbb{C}^{N/2}, \|\cdot \|_{\infty})$ where $S = \mathbb{R}[X]/(\Phi_M(X))$, $\Phi_M(X)$ a cyclotomic polynomial of degree $N = \phi(M), \|\cdot \|_{\infty}$ is the infinity norm and $\| m\|_{\infty}^{can} = \|\sigma (m)\|_{\infty}$ with $\sigma$ to be the canonical embedding between $S$ and $\mathbb{C}^N.$ I am trying to prove that the norms are preserved but all my attempts didn't work. What I am trying to prove is $\| Dec (m) \|_{\infty}=\|m\|_{\infty}^{can}$

Attempt 1 : $\begin{align*} \| Dec (m) \|_{\infty} = \| \pi \circ \sigma (\Delta^{-1}m)\|_{\infty} =\max\limits_i |\pi \circ \sigma (\Delta^{-1}m_i)| = \max\limits_i |\sigma (\Delta^{-1}m_i)|_{\infty} = \|\Delta^{-1}m\|_{\infty}^{can}= \Delta^{-1} \|m\|_{\infty}^{can} \end{align*} $
So I am ending up with a $\Delta^{-1}$ factor the I don't want

Attempt 2: On this one, I only focus on the form of the polynomials that I will use in the CKKS algorithm, or in other words, I assumed that my polynomial $m$ already carries a $\Delta$ factor, i.e, $m = \Delta m_1$ some polynomial $m_1$, but still:
\begin{align*} \|Dec(m)\|_{\infty} = \Delta^{-1} \|m\|_{\infty}^{can} = \Delta^{-1}\|\Delta m_1\|_{\infty}^{can} = \|m_1\|_{\infty}^{can} \end{align*}

getting a different polynomial from the one I started with, which doesn't seem to be what I need.

Can anyone help me out?

Thanks in advance

Answer 1

The statement "encoding and decoding are ismetric ... " appears to occur before any discussion of $\Delta$ occurs, so my reading is that they are discussing in the case of $\Delta = 1$ with that statement.

Separately, when they later discuss $\mathsf{Ecd}$ and $\mathsf{Dcd}$, they mention

$\mathsf{Dcd}(m; \Delta)$. For an input polynomial $m \in R$, output the vector $z = \pi \circ \sigma(\Delta^{-1} \cdot m)$, i.e., the entry of $z$ of index $j \in T$ is $z_j = \Delta^{-1} \cdot m(\zeta_M^j)$.

This is to say that (in the case $\Delta\neq 1$) they appear to additionally get this $\Delta^{-1}$ factor, similarly to how you do. This further suggests to me they only meant the spaces are isometric for $\Delta = 1$.