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Decomposition of Lie Algebras

  1. Nov 13, 2005 #1
    Hi guys - long time reader first time poster!
    I'm currently getting to grips with the topic of Lie Algebras, and I've come across something that's baffled me somewhat. I've been asked to show:

    [tex]so(4) = su(2) \oplus su(2)[/tex]

    Where the lower so(n) denotes the Lie Algebra of SO(n) etc. Now, in a previous question, I was asked to show:

    [tex]u(2) = su(2) \oplus \mathbb{R} [/tex]

    Where [tex]\mathbb{R}[/tex] denotes the set of constant (real) multiples of the matrices of the form [tex]i\mathbb{I}_2[/tex]. This was easy enough; I showed that for each [tex]v\in u(2)[/tex] there exists a [tex]x \in su(2)[/tex] and a [tex]y \in \mathbb{R}[/tex] such that v=x+y, and also that if [tex]A \in su(2) \cap \mathbb{R} [/tex] then A=0.

    However, in this new case, I assume the [tex]\oplus[/tex] means the the matrix direct sum, but surely if this is the case, then it's false - since the matrices in so(4) take the form:

    [tex]\left(\begin{array}{cccc}0&a&b&c\\-a&0&d&e\\-b&-d&0&f\\-c&-e&-f&0\end{array}\right)[/tex]

    Which is not of the form of a matrix direct sum. If anyone could give me any hints as to where my confusion lies, I'd be very greatful - although if you could keep the hints sufficiently vague, as to not to do all the work for me!
     
  2. jcsd
  3. Nov 13, 2005 #2
    Actually, I should add. My only (perhaps?) contructive thought was the existance of a certain isomorphism which takes su(2) to a subset of so(4). For a general [tex]v \in su(2) [/tex] we have:

    [tex] v = \left(\begin{array}{cc}ia&ic+b\\ic-b&-ia\end{array}\right) = \left(\begin{array}{cc}0&b\\-b&0\end{array}\right) + i\left(\begin{array}{cc}a&c\\c&-a\end{array}\right) = A + iB[/tex]

    Then there is an isomorphism G such that:

    [tex] G(v) = \left(\begin{array}{cc}A&-B\\B&A\end{array}\right) [/tex]

    Since we have:

    [tex] G(v_1 v_2) = \left(\begin{array}{cc}A_1 A_2 - B_1 B_2&-(A_1 B_2 + B_1 A_2 )\\A_1 B_2 + B_1 A_2&A_1 A_2 - B_1 B_2\end{array}\right) = \left(\begin{array}{cc}A_1&-B_1\\B_1&A_1\end{array}\right)\left(\begin{array}{cc}A_2&-B_2\\B_2&A_2\end{array}\right) = G(v_1)G(v_2) [/tex]

    And so:

    [tex]G([v_1,v_2]) = [G(v_1),G(v_2)] [/tex]

    But this doesn't give me all of so(4).
     
    Last edited: Nov 13, 2005
  4. Nov 14, 2005 #3

    George Jones

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    Are you working with real or complex Lie algebras?

    It is true that for complex Lie algebras,

    [tex]
    so_{4} \mathbb{C} \cong su_{2} \mathbb{C} \oplus su_{2} \mathbb{C},
    [/tex]

    but I don't think that a similar relationship holds for the real Lie algebras so(4) and su(2).

    The set of matrices

    [tex]
    \left\{\left(\begin{array}{cc}i&0\\0&-i\end{array}\right), \left(\begin{array}{cc}0&i\\i&0\end{array}\right), \left(\begin{array}{cc}0&1\\-1&0\end{array}\right)\right\}
    [/tex]

    is a basis for both the real Lie algebra su(2) and the complex Lie algebra [tex]su_{2} \mathbb{C}[/tex]. In the former, all real linear combinations of basis elements are allowed; in the latter, all complex linear combinations of basis elements are allowed.

    If you would like, I will give some broad hints in a future post.

    Regards,
    George
     
    Last edited: Nov 14, 2005
  5. Nov 14, 2005 #4

    matt grime

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    Have you considered that there might be more than one basis you can think of?
     
  6. Nov 14, 2005 #5

    matt grime

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    the latex for setting lie algebras is \mathfrak{text}, though i don't know if we support it here:
    [tex]\mathfrak{sl}_2[/tex]

    incidentally, sl_2 and so_3 are the same lie algebra, but if you pick the standard basis they dont' look it.
     
  7. Nov 14, 2005 #6
    Apologies George - I should have said, the Lie Algebras are complex. Any hints (however vague) would be much appreciated. :)
    Hmmm, ok - so at the moment I'm considering a basis of [tex]\mathfrak{so}(4)[/tex] of the form:

    [tex]\left{\left(\begin{array}{cccc}0&1&0&0\\-1&0&0&0\\0&0&0&0\\0&0&0&0\end{array}\right), \left(\begin{array}{cccc}0&0&1&0\\0&0&0&0\\-1&0&0&0\\0&0&0&0\end{array}\right), \quad \textrm{etc.}\right}[/tex]

    But perhaps if I think about suitable combinations of these, I'll get something more like the form I'm looking for?

    The help is much appreciated guys. :)
     
    Last edited: Nov 14, 2005
  8. Nov 14, 2005 #7

    George Jones

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    In more detail: the real Lie sl(2,C) has a complex structure. When this is taken into account, sl(2,C) with complex structure is isomorphic to the complexification of the real lie algebra so_3.

    This is why angular momentum theory in physics quantum theory texts looks like the representation theory of sl_2 given in math texts.

    Regards,
    George
     
  9. Nov 16, 2005 #8
    Well, I thought I'd get back to you guys and let you know it's all sorted. Here's how I went about it:

    1) Contruct a basis for [tex]\mathfrak{so}_4[/tex] with the following matrices:


    [tex]\left\{ \left{\left(\begin{array}{cccc}
    0&1&0&0\\
    -1&0&0&0\\
    0&0&0&1\\
    0&0&-1&0\end{array}\right),
    \left(\begin{array}{cccc}
    0&0&0&1\\
    0&0&-1&0\\
    0&1&0&0\\
    -1&0&0&0\end{array}\right),
    \left{\left(\begin{array}{cccc}
    0&0&1&0\\
    0&0&0&-1\\
    -1&0&0&0\\
    0&1&0&0\end{array}\right),
    \left(\begin{array}{cccc}
    0&0&-1&0\\
    0&0&0&-1\\
    1&0&0&0\\
    0&1&0&0\end{array}\right),
    \left{\left(\begin{array}{cccc}
    0&0&0&1\\
    0&0&-1&0\\
    0&1&0&0\\
    -1&0&0&0\end{array}\right),
    \left(\begin{array}{cccc}
    0&-1&0&0\\
    1&0&0&0\\
    0&0&0&1\\
    0&0&-1&0\end{array}\right) \right\} [/tex]

    Which I'll refer to as [tex]\left\{ a_1, a_2, a_3, b_1, b_2, b_3 \right\}[/tex]. Then upon computing the Lie brackets, we see:

    [tex] [a_i, a_j] = \epsilon_{ijk} a_k \qquad [b_i, b_j] = \epsilon_{ijk} b_k [/tex]

    And so [tex]\left\{a_i\right\} \cong \mathfrak{su}_2[/tex] and also [tex]\left\{b_i\right\} \cong \mathfrak{su}_2[/tex]. Since [tex]\mathfrak{so}_4={a_i} \oplus {b_i}[/tex] we have:

    [tex]\mathfrak{so}_4 \cong \mathfrak{su}_2 \oplus \mathfrak{su}_2 [/tex]

    Which I think is ok.
     
    Last edited: Nov 16, 2005
  10. Nov 16, 2005 #9

    George Jones

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    a_2 = b_2. Is this just a typo?

    Regards,
    George
     
  11. Nov 16, 2005 #10
    Sure is (too many matrix elements!). It ([tex]a_2[/tex]) should be:

    [tex]\left(\begin{array}{cccc}
    0&0&0&1\\
    0&0&1&0\\
    0&-1&0&0\\
    -1&0&0&0\end{array}\right)[/tex]
     
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