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What is the relation between unitary group, homotopy?

  1. Oct 15, 2015 #1
    The O(N) nonlinear sigma model has topological solitons only when N=3 in the
    planar geometry. There exists a generalization of the O(3) sigma model so that the
    new model possess topological solitons for arbitrary N in the planar geometry. It is
    the CP^{N-1} sigma model,†whose group manifold is
    [tex]CP^{N-1}=ƒ U(N)/†[U(1)\bigotimes U(N-1)]‡ =SU(N)/[†U(1)\bigotimes SU(N)\bigotimes SU(N-1)‡][/tex]
    The homotopy theorem tells
    [tex] \pi_2(CP^{N-1})=Z [/tex]
    since [tex]\pi_2(G/H)…=\pi_1(„H)…[/tex] (when G is simply connected) and [tex]\pi_n(„G\bigotimes G')=\pi_n… („G)\bigoplus \pi_n(„G') [/tex]…. It is also called the SU(N) sigma model.

    I don't understand the following sentences, what is the meaning of the following math expression?Would anyone gives a more detailed hints to the following sentences:

    [tex]CP^{N-1}=ƒ U(N)/†[U(1)\bigotimes U(N-1)]‡ =SU(N)/[†U(1)\bigotimes SU(N)\bigotimes SU(N-1)‡][/tex]
    The homotopy theorem tells
    [tex] \pi_2(CP^{N-1})=Z [/tex]
    since [tex]\pi_2(G/H)…=\pi_1(„H)…[/tex] (when G is simply connected) and [tex]\pi_n(„G\bigotimes G')=\pi_n… („G)\bigoplus \pi_n(„G') [/tex]

    U(N) seems to be not simply connected.
     
  2. jcsd
  3. Oct 15, 2015 #2

    andrewkirk

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    What sort of mathematical objects are the ##U(N)## and ##SU(N)## items? If they are vector or tensor spaces, the expression with the ##\otimes## symbols can be interpreted as a tensor product.
     
  4. Oct 15, 2015 #3
    U(N),SU(N) are unitary and special unitary group respectively
     
  5. Oct 15, 2015 #4

    andrewkirk

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    Based on this, I would guess that the ##\otimes## symbol is supposed to indicate semidirect product, which is more usually denoted by ##\rtimes##. The link gives an identification of ##U(1)## with a subgroup of ##U(N)##, and we could use the same principle to identify ##U(N-1)## with a subgroup, viz the group of ##n\times n## matrices formed by adding a row of zeros above and column of zeros to the left of a matrix in ##U(N-1)## and then putting a 1 in the top-left cell. Then we can take the semidirect product of those two subgroups.
     
  6. Oct 16, 2015 #5
    Thank you very much, I admire mathematician!
     
  7. Oct 18, 2015 #6

    lavinia

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    The correct expressions are

    ##CP^{n-1} = U(n)/(U(n-1)##x##U(1)) = SU(n)/S(U(n-1)##x##U(1))## where x means direct product(not semi direct product) of groups algebraically and Cartesian product topologically.

    ##\bigotimes## usually means tensor product which in this case is meaningless.

    From the long homotopy sequence of the fibration, ##U(1)\rightarrow S^{2n-1}\rightarrow CP^{n-1}## one has

    ##π_2(S^{2n-1})\rightarrow π_2(CP^{n-1})\rightarrow π_1(U(1)) \rightarrow π_1(S^{2n-1})## it follows that ##π_2(CP^{n-1}) = Z## since the homotopy groups of the 2n-1 sphere are zero below dimension 2n-1 and the fundamental group of the circle is Z.

    ##SU(n)## is simply connected which can be shown by induction starting with ##SU(1)## which is the trivial group. From the long exact sequence of the fibration,

    ## S(U(n-1)##x##U(1))\rightarrow SU(n)\rightarrow CP^{n-1}## one has

    ## π_2(SU(n))\rightarrow π_2(CP^{n-1})\rightarrow Z \rightarrow 0##

    You need to show that ## π_2(SU(n))## is zero.

    BTW: ##U(n)## is not simply connected. The complex determinant defines a continuous homomorphism from ##U(n)## onto ##U(1)## with kernel ##SU(n)##. Since ##SU(n)## is simply connected, the exact sequence of the fibration shows that ##π_1(U(n)) = Z##.
     
    Last edited: Oct 18, 2015
  8. Oct 19, 2015 #7
    Thank You very much lavinia for your great answer!!!
     
  9. Oct 26, 2015 #8
    Lavinia wrote "You need to show that π2(SU(n)) is zero."

    It is an interesting fact that for any Lie group G at all, π2(G) = 0. This is a rather deep fact using algebraic topology.
    ------------------------------------------------------------------------------------------------------------------------------------------------------

    A somewhat simpler fact is that, loosely speaking, a Lie group resembles the cartesian product of odd-dimensional spheres. The precise statement is somewhat technical, but it is an amazing theorem:

    The rational cohomology ring H*(G; ℚ) of a Lie group G is the same as the rational cohomology ring of some product of odd-dimensional spheres.
    (This is "simpler" in the sense that its proof follows from the easy-to-prove observation that the cohomology ring of a Lie group is a Hopf algebra.)

    For some examples:

    H*(U(n); ℚ) = H*(S1 × S3 × ... × S2n-1; ℚ)​

    H*(SU(n); ℚ) = H*(S3 × S5 × ... × S2n-1; ℚ)​

    H*(Sp(n); ℚ) = H*(S3 × S7 × ... × S4n-1; ℚ)​
     
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