# encoding/decoding in the CKKS are isometric ring isomorphisms?

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?

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 $$\Delta = 1$$.