Matrix doodles

Discussion in 'Physics & Math' started by arfa brane, Sep 18, 2016.

  1. arfa brane call me arf Valued Senior Member

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    5,373
    Suppose you'd like to map the rotation group of a square to a square matrix.

    First of all a 2 x 2 matrix should suffice to describe the four vertices, i.e. something like:

    \( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \)
    or if you want to represent a 4-cycle: (clockwise labeling):

    \( \begin{pmatrix} a & b \\ d & c \end{pmatrix} \)
    The permutation matrix \( \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \), acting on the left of any 2 x 2 matrix will swap the rows. Acting on the right it will swap the columns.

    \( \begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} a & b \\ d & c \end{pmatrix} = \begin{pmatrix} d & c \\ a & b \end{pmatrix} \)

    \(\begin{pmatrix} a & b \\ d & c \end{pmatrix} \begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix}b & a \\ c & d \end{pmatrix}\)
    So with the matrix transpose, a rotation or cyclic permutation of the elements of a 2 x 2 matrix can be defined:

    \( \begin{pmatrix} a & b \\ d & c \end{pmatrix}^{\intercal}\begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix}d & a \\ c & b \end{pmatrix} = \biggl {(} \begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} a & b \\ d & c \end{pmatrix} \biggr {)}^{\intercal} \)
    But you can also map the four vertices {a,b,c,d} to a column vector:

    \( \begin{pmatrix} a \\ b \\ c \\ d \end{pmatrix} \)​

    Then rotations are represented by 4 x 4 matrices, a subset of the 24 possible permutations of four symmetric objects. Can the above representation of rotating a square be extended to the whole of \( S_4 \), which is the symmetry group of a solid cube ? A cubic matrix will be needed, and a way to define rotations of the entire matrix (or equivalently, rotations of an abstract system of coordinates, given by a set of indices).

    Start with a second 2 x 2 matrix like:

    \( \begin{pmatrix} e & f \\ h & g \end{pmatrix} \)
    With connections between the rows/columns of this matrix to the first one. Index-wise, this is just a pair of 2 x 2 matrices connected, or 'sliced' as \( M_{1ij} + M_{2ij} \). Obviously the indices will allow three distinct slices of the 2 x 2 x 2 matrix.
     
    Last edited: Sep 18, 2016
    magnetoschild likes this.
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  3. Confused2 Registered Senior Member

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  5. arfa brane call me arf Valued Senior Member

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    Not sure what you mean by "encapsulated" so much, but to be a tensor, a multidimensional array has to have certain transformation properties. You need, I think, to start with vector spaces and these have to have co- and contravariant properties.

    A bare 2 x 2 x 2 cubic array of "numbers" or just symbols like a, b, c . . . with the transformation I've defined so far, might not qualify as a tensor. I need to define a direction vector somewhere along with what happens to it when it gets transported around under the rotation I've defined.
     
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  7. rpenner Fully Wired Staff Member

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    What do you mean?

    The cyclic permutation group of 4 vertices? The rotational group of a square with distinct vertices in a 2-D embedding space? The rotational group of a square with indistinguishable vertices in a 2-D embedding space? The rotational group of a square with distinct vertices in a 2-D embedding space? The rotational group of an oriented square with indistinguishable vertices in a 3-D embedding space? The rotational group of an unoriented square with indistinguishable vertices in a 3-D embedding space?
     
  8. arfa brane call me arf Valued Senior Member

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    Hmm. The rotation group or subgroup of symmetries of a square, or any polygon in a Euclidean domain, is closed under composition, any composition of rotations is a rotation.
    This isn't true for reflections in the plane because compositions of reflections are rotations, and transposing a 2 x 2 matrix is analogous to reflecting two vertices of a square about a diagonal line.

    In a matrix, all the elements are distinguished with indices. The transpose operation with another 'reflection' through a horizontal or vertical line of symmetry gives a 'rotation'.

    So if say,

    \( \sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \)
    Then if A is any 2 x 2 matrix, we have:

    \( (A^{\intercal}\sigma_x)^2 = \sigma_x A \sigma_x \)
    and \( \sigma_x = \sigma_x^{-1} \)​
     
    Last edited: Sep 21, 2016
  9. arfa brane call me arf Valued Senior Member

    Messages:
    5,373
    \( (A^{\intercal}\sigma_x)^2 = \sigma_x A \sigma_x \)
    and \( \sigma_x = \sigma_x^{-1} \)

    Screwed up the notation. It should be:
    \( (A^{\intercal}\sigma_x)^{\intercal}\sigma_x = \sigma_x A \sigma_x \)​


    Still thinking about how to "repeat" either expression so you get the identity permutation on both sides.
     
  10. arfa brane call me arf Valued Senior Member

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    5,373
    Well, there are the equalities since, from the above:

    \( (A^{\intercal} \sigma_x)^{\intercal} = \sigma_x A \), so
    \( (A^{\intercal} \sigma_x) = (\sigma_x A) ^{\intercal}\),

    by a symmetry, we have:

    \( (\sigma_x A^{\intercal}) = (A \sigma_x )^{\intercal} \)

    But \( \sigma_x A \sigma_x \) "twice", is
    \( \sigma_x (\sigma_x A \sigma_x ) \sigma_x = \sigma_x^2 A \sigma_x^2 = A \)​
     
  11. arfa brane call me arf Valued Senior Member

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    5,373
    Just a thought about what \( \sigma_x A \sigma_x \) means.

    Mathematically the rows are exchanged (transposed) and the columns are exchanged. It doesn't matter which is "done first".
    However, it should be clear that

    \( \sigma_x (A \sigma_x) = ( \sigma_x A) \sigma_x \)
    Which although the parentheses are redundant, shows there is a rule of composition.
     
  12. arfa brane call me arf Valued Senior Member

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    5,373
    With the matrix transpose, we have a symmetric kind of operation:

    \( (\sigma_x A \sigma_x)^{\intercal} = \sigma_x A^{\intercal} \sigma_x \)
    But when \( \sigma_x \) appears just once, either on the left or the right of A, the matrix transpose also transposes the position of \( \sigma_x \):

    \( (\sigma_x A)^{\intercal} = A^{\intercal}\sigma_x,\; (A\sigma_x)^{\intercal} = \sigma_x A^{\intercal} \)​
     

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