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Please give some hints on the Complex projective group

  1. Oct 14, 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.

    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

    fzero

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    First of all, ##\mathbb{CP}^n## is called the complex projective space, in particular it is not a group, though it is a homogenous space as related to the unitary groups by the cosets that you mention.


    The last part should be ##SU(N)/(U(1)\times SU(N-1))##.

    Let's start with the basics, namely the definition of ##\mathbb{CP}^n##. We start with ##(z_1,\ldots z_{n+1})\in \mathbb{C}^{n+1}## and identify
    $$(z_1,\ldots z_{n+1})\sim (\lambda z_1,\ldots \lambda z_{n+1}),~~~\lambda \in \mathbb{C},~~~\lambda\neq 0.$$
    So first we see that the definition suggests that natural coordinates for ##\mathbb{CP}^n## can be defined in patches ##U_\nu## where at least one coordinate ##z_\nu \neq 0##. The ##n## ratios ##\zeta_\mu = z_\mu/z_\nu, ~~\mu\neq \nu## are invariant under the action above and form the local homogeneous coordinates on the patch ##U_\nu##.

    We can obtain some of the geometry of complex projective space by looking at the coordinates for a given patch, say the one where ##z_{n+1}\neq 0##. We choose the parameterization
    $$\begin{split} & z_i = \zeta_i z_{n+1}, ~~~i=1,\ldots, n, \\
    & z_{n+1} = \frac{r}{\sqrt{\sigma}} e^{i\psi}, ~~~\sigma = 1 + \sum_{i=1}^n |\zeta_i|^2. \end{split}$$
    So far this looks a bit funny, but let's look at
    $$ \sum_{\mu = 1}^{n+1} | z_\mu |^2 = (1+\sum_i |\zeta_i|^2) \frac{r^2}{\sigma} = r^2. $$


    For a given fixed ##r##, this is the equation of a sphere, ##S^{2n+1}##. Therefore we see that the set of coordinates ##(\psi, \zeta_i)## describe a unit sphere sitting in ##\mathbb{C}^{n+1}##. When we further fix ##\psi##, we obtain our ##\mathbb{CP}^n##. If we wanted to do a bit more work, we could show that the sphere actually has the structure of a fiber bundle over ##\mathbb{CP}^n##, with fiber the circle ##S^1=U(1)## represented by the coordinate ##\psi##. Since the fiber is a group, we can therefore say that the complex projective space can be obtained by quotienting a point in the total space by the group action:
    $$ \mathbb{CP}^n = S^{2n+1}/U(1).~~~(*)$$

    Now I claim that ##S^{2n+1} = SU(n+1)/SU(n)##. I won't completely prove this, but will outline it. As usual with spheres, we can consider a 1-1 correspondence between a point on the unit sphere and a unit vector extending from the origin of the total space. We can define a natural action of ##SU(n+1)## on vectors in ##\mathbb{C}^{n+1}## via the matrix action ## g v = v'##. The first part of the proof would then be to show that this action is transitive, namely that given any specified vector, say ## e_1 = (1, 0, \ldots , 0)##, we can obtain any point ##x## on the sphere via some group element ##g##, so that ## g e_1 = x##. The proof boils down to setting up some linear equations and showing the necessary matrix is actually unitary and has determinant 1. Next we would show that the subgroup of ##SU(n+1)## that leaves a specific vector invariant is ##SU(n)##. You could again show this by choosing say ##e_1## as before and looking at what matrices leave it invariant explicitly.

    So let's collect the results from the previous paragraphs. Given a specified unit vector, we can use ##SU(n+1)## transformations to map out the entire sphere. But there are an ##SU(n)## worth of transformations that actually don't do anything to the original unit vector, so we don't want to count them. Therefore we can take the group quotient and hence ##S^{2n+1} = SU(n+1)/SU(n)##, as claimed. Furthermore. if we wanted to include the explicit phase in ##g## so that we have ##U(n+1)## rather than ##SU(n+1)##, if we mod out by it at the end anyway, we see that we also have ##S^{2n+1} = U(n+1)/U(n)##.

    Putting these together with the result (*), we have
    $$ \mathbb{CP}^n = U(n+1)/(U(1)\times U(n)) = SU(n+1)/(U(1)\times SU(n)).$$


    So as you might have found, the homotopy theorem referred to is known as the long exact sequence for fiber bundles. In addition we have the formula for the homotopy groups of a product space. I don't know how much topology you've studied but certainly the first result probably wouldn't be proved until a couple of months into a graduate level course, so I don't know that I can even get started here on explaining more than the wikipedia page explains.

    Yes, that is true, which is why you should apply the theorem to the version where ##H= U(1)\times SU(n)##.
     
  4. Oct 16, 2015 #3
    Thank You for Your Great Answer! Thank you very much!!!
     
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