According to the book "Cylindric Algebras" by Henkin, Monk and Tarski ([1], p162), a cylindric algebra satisfies the following axioms:
$c_{\mathcal{\kappa}}0=0$
$x\leq c_{\mathcal{\kappa}}x$
$c_{\mathcal{\kappa}}(x\cdot c_{\mathcal{\kappa}}y)= c_{\mathcal{\kappa}}x\cdot c_{\mathcal{\kappa}}y$
$c_{\mathcal{\kappa}}c_{\lambda}x=c_{\lambda}c_{\mathcal{\kappa}}x$
$d_{\mathcal{\kappa,\kappa}}=1$
If $\mathcal{\kappa}\notin\{\lambda,\mu\}$, then $d_{\lambda\mu}=c_{\mathcal{\kappa}}(d_{\lambda\mathcal{\kappa}}\cdot d_{\mathcal{\kappa}\mu})$
If $\mathcal{\kappa}\ne\lambda$, then $c_{\mathcal{\kappa}}(d_{\mathcal{\kappa}\lambda}\cdot x)\cdot c_{\mathcal{\kappa}}(d_{\mathcal{\kappa}\lambda}\cdot -x)=0$
Where $c_{\mathcal{\kappa}}$ is called a cylindrification and $d_{\mathcal{\kappa}}$ a diagonal element.
The meaning of cylindrification is clear to me as it is intuitively like a rectangle in the Cartesian space. However, I am not sure about the meaning of $d_{\mathcal{\kappa}}$. By a comment in p164 [1], a Boolean algebra can be transformed into a Cylindric algebra by letting $c_{\mathcal{\kappa}}x=x$ and $d_{\mathcal{\kappa}\lambda}=1$. So can $d_{\mathcal{\kappa}}$ be intuitively thought as a diagonal (plane) in the Cartesian space?
It sounds like you're thinking of cylindrical algebras geometrically, but the prototypical example of a cylindrical algebra is the set of formulas$^1$ in some fixed first-order language with $c_\kappa$ interpreted as the operator $\varphi\mapsto \exists x_\kappa \varphi$ and $d_{\kappa\lambda}$ interpreted as the formula $x_\kappa=x_\lambda$. Classically of course we have that each of the following is a tautology:
$x_\kappa=x_\kappa$.
(Assuming $\kappa\not\in\{\lambda,\mu\})\quad$ $x_\lambda=x_\mu\leftrightarrow\exists x_\kappa(x_\mu=x_\kappa\wedge x_\kappa=x_\lambda)$.
(Assuming $\kappa\not=\lambda$, and $\chi$ is some fixed formula$)\quad$ $\neg[\exists x_\kappa(x_\kappa=x_\lambda\wedge \chi)\wedge\exists x_\kappa(x_\kappa=x_\lambda\wedge\neg\chi)].$
These correspond to the last three rules you mention.
$^1$Actually, that's not quite true. The right picture is to fix a structure $\mathfrak{M}$ in the relevant language, and then look at equivalence classes of formulas under the relation $\varphi\approx\psi$ iff $\varphi$ and $\psi$ have the same sets of satisfying variable assignments over $\mathfrak{M}$ (with variables restricted to some predetermined set of $x_\kappa$s). Alternatively you can think of the elements of the cylindrical algebra as particular sets of variable assignments themselves, but I think that's less intuitive.