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SL(3,R) Generators and their Interpretation

  1. May 24, 2012 #1
    Hi,

    I would like to study [itex]SL(3,\mathbb{R})[/itex] a little. I was motivated to look into them a bit because of their volume-preserving nature. Now this group has [itex]n^2-1=9-1=8[/itex] generators and I found out through some reading that three of them are the generators for rotation (duh Judah, volume preserving), and the other five have to do with stretching. Both of these can be represented as spherical tensors, one as [itex]T^{(1)}_{m}[/itex] and the other as [itex]T^{(2)}_{m}[/itex] for [itex]-l\leq m\leq l[/itex]. So these other five generators are a 2x2 symmetric traceless tensor. Now for the life of me, I can't find out what these other five generators do in real life. What's their physical interpretation explicitly?

    My thinking was that the xy, yz, xz components of the symmetric traceless tensor are the generators for shearing, and xx, yy, zz, are something goofy having to do with another type of stretching, with the additional constraint that [itex]T_{xx}+T_{yy}+T_{zz}=0[/itex] . But, I'm not sure.

    On top of that, I would like to find out how to write out the infinitesimal generators, much like [itex]L_{z}=\partial_{\phi}[/itex], but for the other five generators.

    Does anyone know a good book, that would cover how to develop such generators, and can anyone lend me some physics insight into what these generators each do?

    Thanks,
     
  2. jcsd
  3. May 24, 2012 #2

    vanhees71

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    Perhaps, it's most intuitive to think of symmetric 2nd-rank tensors in terms of the stress tensor. For a homogeneous medium like a fluid the traceless part describes shear stress and the trace part bulk stress. While shear stress doesn't lead to changes in the volume of the fluid elements the bulk stress does, and that's why you don't have this part in the game. So, if you take a cube, the shear stress just deforms this to a more general parallelipiped and bulk stress changes it's volume without deforming its shape.

    The antisymmetric part of a general 2nd-rank tensor describes rigid rotations of the fluid element.
     
  4. May 24, 2012 #3
  5. May 25, 2012 #4

    vanhees71

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  6. May 25, 2012 #5
    As far as I know, SL(n, R) is the set of all nxn real matrices with determinant 1. They "preserve volume", because if you consider them as linear homogeneous transformations, then the Jacobian of the transformation is 1, hence, volume elements are preserved.

    The distance of a point from the origin is, in general, given by:
    [tex]
    s'^2 = \langle \mathbf{x}' \vert \mathbf{x}' \rangle = \langle \mathbf{x} \vert \, \left( \hat{A}^{\top} \cdot \hat{A} \right) \, \vert \mathbf{x} \rangle
    [/tex]
    Then, the real symmetric matrix [itex]\hat{g} = \hat{A}^{\top} \cdot \hat{A}[/itex] ([itex]\hat{g}^{\top} = \hat{g}[/itex]) plays the role of a metric tensor. Notice that [itex]\det \hat{g} = 1[/itex].

    Now, every real symmetric matrix may be diagonalized by the use of an orthogonal matrix [itex]\hat{O}[/itex]:
    [tex]
    \hat{g} \cdot \hat{O} = \hat{O} \cdot \hat{\Lambda}, \ \hat{O}^{\top} \cdot \hat{O} = \hat{O} \cdot \hat{O}^{\top} = \hat{1}
    [/tex]
    where [itex]\hat{\Lambda}[/itex] is a real diagonal matrix. Notice that the eigenvalues of this matrix, being the distance of the corresponding eigenvector [itex]\vert \mathbf{x} \rangle[/itex] are non-negative (they are zero only for the zero eigenvector, which is a trivial eignevector, therefore, they are actually positive). Notice that [itex]\mathrm{\Pi}_{\alpha = 1}^{n}{\lambda_{\alpha}} = 1[/itex], which again corroborates the strict positive definiteness of the eigenvalues. We may define "the square root of the metric tensor" very easily:
    [tex]
    \sqrt{\hat{g}} = \hat{O} \cdot \sqrt{\hat{\Lambda}} \cdot \hat{O}^{\top}
    [/tex]

    Suppose we write:
    [tex]
    \hat{A} = \hat{X} \cdot \sqrt{\hat{g}} \Rightarrow \hat{A}^{\top} = \sqrt{\hat{g}} \cdot \hat{X}^{\top}
    [/tex]
    Then:
    [tex]
    \hat{g} = \hat{A}^{\top} \cdot \hat{A} = \sqrt{g} \cdot \hat{X}^{\top} \cdot \hat{X} \cdot \sqrt{\hat{g}} \Rightarrow \hat{X}^{\top} \cdot \hat{X} = \hat{1}
    [/tex]
    This means that [itex]\hat{X}[/itex] is another orthogonal matrix. Thus, we are led to the following decomposition of the transformation matrix:
    [tex]
    \hat{A} = \hat{P} \cdot \sqrt{\hat{\Lambda}} \cdot \hat{O}^{\top}, \ \hat{P} \equiv \hat{X} \cdot \hat{O}
    [/tex]

    This is called "polar decomposition of a matrix". Let us do a counting of the elements. A general n×n matrix has n2 elements. But, because we require that the determinant is 1, we actually have n2-1 elements.

    The group of all orthogonal n×n matrices is denoted by O(n). If we also require that the matrix has determinant one, then these are special orthogonal matrices SO(n). Now, we may exponentiate a real special orthogonal matrix as:
    [tex]
    \hat{O} = \exp \left( \hat{M} \right), \ \hat{O}^{\ast} = \hat{O} \Rightarrow \hat{M}^{\ast} = \hat{M}, \ \hat{O}^{\top} = \hat{O}^{-1} \Rightarrow \hat{M}^{\top} = -\hat{M}
    [/tex]
    Thus, the matrix [itex]\hat{M}[/itex] is real, and antisymmetric. It means it has zeros along the main diagonal, and that the (real) entries in the upper triangle are the negative of the entries on the lower triangle. Notice that [itex]\det \hat{O} = \exp \left(\mathrm{Tr} \hat{M} \right) = e^{0} = 1[/itex], i.e. the matrix is guaranteed to have determinant 1, i.e. it is special. There are n(n - 1)/2 entries on the main diagonal. This is the number of independent elements of [itex]\hat{M}[/itex], and, consequently, of [itex]\hat{O}[/itex] as well. What is the significance of an orthogonal matrix. If you look above, it leaves the distance of each point from the origin fixed. This is a property of rotations and reflections, thus, orthogonal matrices correspond to rotations and reflections. One may show that proper rotations are represented by a special orthogonal matrix. Then, the number of parameters is equal to the number of pairs of coordinate axes in which planes we may rotate. A reflection may always be represented as [itex]\hat{R} = \hat{O}^{\top} \cdot \hat{P} \cdot \hat{O}[/itex], where the special orthogonal matrix [itex]\hat{O}[/itex] brings the reflection plane to, say, the [itex]x^{1}x^{2}[/itex], and [itex]\hat{P}[/itex] is a reflection w.r.t. this plane. Notice that [itex]\hat{P}^2 = \hat{1}[/itex], and [itex]\det \hat{P} = -1[/itex].

    The number of independent entries for the positive diagonal matrix [itex]\Lambda[/itex] is n - 1, because of the requirement [itex]\mathrm{\Pi}_{\alpha = 1}^{n}{\lambda_{\alpha}} = 1[/itex]. Thus, the number of parameter for the general SL(n, R) is
    [tex]
    2 \times \frac{n (n - 1)}{2} + (n - 1) = (n - 1) (n + 1) = n^2 -1
    [/tex]
    as expected!

    This is how we may interpret a general SL(n, R) transformation then. A distance preserving coordinate tranformation (rotation+reflection represented by [itex]\hat{O}^{\top}[/itex]). This has n(n - 1)/2 generators. A volume-preserving rescaling of the axes, represented by [itex]\sqrt{\hat{\Lambda}}[/itex]. This has n-1 generators. You may represent the matrix in exponential form [itex]\sqrt{\Lambda} = \exp (\hat{S}), \ \mathrm{Tr} \hat{S} = 0[/itex], where [itex]\hat{S}[/itex] is a real, traceless, diagonal matrix. Finally, we perform another distance-preserving transformation, represented by the orthogonal matrix [itex]\hat{P}[/itex].
     
  7. May 25, 2012 #6
    wow, thanks.
     
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