The tesseract flattened

A problem-of-the-week in Todd and Vishal’s blog led me to the following two-dimensional view of the tesseract (four-dimensional cube), on the originality of which I can only make a weak claim:

tesseract

One could comment quite a bit on this graph, remembering the partly shown edges: (1, 1, 1, 1) is adjacent to (0, 1, 1, 1), less visibly (1, 1, 1, 0) is adjacent to (1, 0, 1, 0), and so on. Some reflections may be represented by a single axis (like the ones running through the middle row or middle column, affecting the first and second coordinate, respectively); other reflections are represented by two axes (like the pair running through the top and bottom rows and affecting the third coordinate or the pair running through the right and left columns and affecting the fourth coordinate). More tesseract isometries may be detected depending on the reader’s imagination!

Back to the original problem, that is the coloring of the vertices of the octeract (eight-dimensional cube) in 8 colors in such a way that every vertex is adjacent to vertices of all 8 colors, let me first provide two solutions for the corresponding problem for the tesseract (and 4 colors):

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Visually inspired by the rather obvious step from the square to the tesseract, let me now provide two x two = four octeract colorings as follows:

Omitting coordinate details, let me just clarify that each vertex’s neighbors are located one and three vertices away from it in the four directions; neighbors of vertices near the edges of the 16 x 16 square are determined as in the case of the tesseract. Inside each octeract you can see 16 tesseracts of two kinds, each of which corresponds to the two tesseract colorings described above.

The same way the turning by 90 degrees of the yellow and green edges leads from the tesseract coloring on the left to the tesseract coloring on the right, the turning by 90 degrees of the lighter-colored (‘new’) tesseracts inside the four octeracts above leads to four more octeract colorings:

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How many non-isomorphic colorings do we get from the eight colorings above? Replacing every pair of vertices of the same color by an edge we obtain the following eight ‘frames’:

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It is clear now that we have constructed four non-isomorphic colorings. Are there any more?

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5 Responses to “The tesseract flattened”

  1. Μαρία Says:

    Ευχές για την αυριανή διάλεξη.
    Αν μπορέσω, θα την παρακολουθήσω.

  2. gbaloglou Says:

    Δεν πηγα και ασχημα 🙂 Ελπιζω να βγει η παρουσιαση συντομα στο διαδικτυο για οσους δεν ηταν εκει…

  3. gbaloglou Says:

    Escher εδώ 🙂

  4. Panos I. Papadopoulos Says:

    Να μου εξηγήσεις τι ακριβώς γίνεται εδώ. Πάντως σαν άρθρο είναι εκπληκτικό, τουλάχιστον όσον αφορά τα μορφολογικά του στοιχεία.

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