Please give some hints on the Complex projective group

[PRB147]

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 possesses topological solitons for arbitrary N in the planar geometry. It is
the CP^{N-1} sigma model,†whose group manifold is
CP^{N-1}=ƒ U(N)/†[U(1)\bigotimes U(N-1)]‡ =SU(N)/[†U(1)\bigotimes SU(N)\bigotimes SU(N-1)‡]
The homotopy theorem tells
\pi_2(CP^{N-1})=Z
since \pi_2(G/H)…=\pi_1(„H)… (when G is simply connected) and \pi_n(„G\bigotimes G')=\pi_n… („G)\bigoplus \pi_n(„G')…. It is also called the SU(N) sigma model.

Would anyone gives a more detailed hints to the following sentences:

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

U(N) seems to be not simply connected.

 

[fzero]

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.

Would anyone gives a more detailed hints to the following sentences:

CP^{N-1}=ƒ U(N)/†[U(1)\bigotimes U(N-1)]‡ =SU(N)/[†U(1)\bigotimes SU(N)\bigotimes SU(N-1)‡]

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
$$ \mathbb{CP}^n = U(n+1)/(U(1)\times U(n)) = SU(n+1)/(U(1)\times SU(n)).$$

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

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.

U(N) seems to be not simply connected.

Yes, that is true, which is why you should apply the theorem to the version where $H= U(1)\times SU(n)$.

 

[PRB147]

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 $H= U(1)\times SU(n)$.

Thank You for Your Great Answer! Thank you very much!