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Comments on Does the $3+4$ coloring $\chi$ determine the homomorphism $\rho_{\chi}$?

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Does the $3+4$ coloring $\chi$ determine the homomorphism $\rho_{\chi}$?

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Consider the seed:

$$ \mathcal S = (\mathcal I, \Gamma, \Pi) $$

where $\mathcal I$ is a geometric realization, $\Gamma \subset \mathcal I$ is the distinguished $1$-skeleton, and $\Pi$ is the piecewise isometry group generated by "Rubik style moves" on the instrument $\mathscr I= (\mathcal I, \Gamma).$

In the undecorated setting I learned that there is no natural nontrivial homomorphism from the abstract $2 \times 2$ Rubik's cube group to $\mathrm{PGL}(3,2)$. I realized that the correct source of projective symmetry is not the bare group, rather a decorated realization of $\mathcal S$, which is what I was implicitly assuming, and should have made clear in that post.

A Fano decorated seed is a tuple:

$$ \mathcal S_\chi=(\mathcal I,\Gamma,\Pi,\chi), $$

where \(\chi\) is a \(3+4\) partition of seven distinguished cells of \((\mathcal I,\Gamma)\), written $$ \mathcal C=\lbrace m_x,m_y,m_z,s_{xy},s_{xz},s_{yz},s_{xyz} \rbrace. $$

Here $m_x,m_y,m_z$ denote the three meridian cells, and $s_{xy},s_{xz},s_{yz},s_{xyz}$ denote the remaining four cells.

We identify these seven cells with the seven nonzero vectors of $\mathbf F_2^3$ by the rule $$ m_x=e_1,\qquad m_y=e_2,\qquad m_z=e_3, $$ $$ s_{xy}=e_1+e_2,\qquad s_{xz}=e_1+e_3,\qquad s_{yz}=e_2+e_3,\qquad s_{xyz}=e_1+e_2+e_3. $$

This determines a canonical Fano plane incidence structure on \(\mathcal C\).

The line set \(\mathcal L\) associated to \(\mathcal C\) is the collection of seven triples $$ \mathcal L= \Big\lbrace \lbrace m_x,m_y,s_{xy}\rbrace, \lbrace m_x,m_z,s_{xz} \rbrace, \lbrace m_y,m_z,s_{yz} \rbrace, \lbrace s_{xy},s_{xz},s_{yz} \rbrace, $$ $$ \lbrace m_x,s_{yz},s_{xyz} \rbrace, \lbrace m_y,s_{xz},s_{xyz} \rbrace, \lbrace m_z,s_{xy},s_{xyz} \rbrace \Big \rbrace. $$ Equivalently, under the identification \(\mathcal C\cong \mathbf F_2^3\setminus\{0\}\), the lines are exactly the zero-sum triples $$ {a,b,a+b}, $$ or equivalently the triples \(\{a,b,c\}\) satisfying $$ a+b+c=0. $$

Thus $(\mathcal C,\mathcal L)$ is canonically isomorphic to the Fano plane. There is a canonical homomorphism $$ \rho_\chi:\Pi\longrightarrow \mathrm{PGL}(3,2). $$

It is important to state that the projective symmetry is a property of the decorated realization, not of the undecorated group in isolation. One could easily define a different color partition $\chi$ allowing for the manifestation of a different symmetry group than the Fano plane symmetry.

I created this depiction of the Fano decorated seed $ \mathcal S_\chi=(\mathcal I,\Gamma,\Pi,\chi) $:

Fano decorated seed

Question. Does the $3+4$ coloring $\chi$ determine the homomorphism $$ \rho_\chi:\Pi\longrightarrow \mathrm{PGL}(3,2) $$ intrinsically, uniquely up to conjugacy, as the induced incidence-preserving action of $\Pi$ on the associated Fano plane?

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Sorry, but to be very candid this makes no sense. What are the "three meridian cells"? What is the "d... (3 comments)
Sorry, but to be very candid this makes no sense. What are the "three meridian cells"? What is the "d...
clemens‭ wrote 5 months ago

Sorry, but to be very candid this makes no sense. What are the "three meridian cells"? What is the "distinguished 1-skeleton"? What is the difference between $\Gamma$ and $\mathcal{I}$? If anyone can make head or tail of this, I'd invite them to respond.

zetaspace‭ wrote 5 months ago

I will do my best to clarify. The picture above is very helpful here. $\Gamma$ is the 1-skeleton. It is a subset of the 2-complex, $\mathcal I$. In the diagram above, you can see that different cells are marked (colored). See those 1d arcs that wrap around the meridians of the object? Those comprise $\Gamma$. While we could have an unmarked, or, "bare" instrument $\mathscr I = (\mathcal I, \Gamma)$, the markings distinguish important symmetries which are washed out by the large unmarked complex. So, in this question, I focus on a specific 7-marked system. Then I identify those markings with the 7 points of the fano plane. Then I ask whether the action of $\Pi$ preserves incidence structure of the Fano plane.

clemens‭ wrote 5 months ago · edited 5 months ago

That makes more sense. So $\mathcal{I}$ is a 2-complex, and $\Gamma$ is its 1-skeleton. Does $\mathcal{I}$ then have 24 faces, 26 vertices, and 48 edges?

That being said you still didn't answer what the "three meridian cells" are, nor why there would be "four remaining cells". If you pick out cells that are identified with the Fano plane then yes, you will get a Fano-plane structure. Are you asking us to check whether the cells you describe do in fact follow that structure?