# Please give some hints on the Complex projective group

1. Oct 14, 2015

### 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 possess topological solitons for arbitrary N in the planar geometry. It is
the CP^{N-1} sigma model,whose group manifold is
$$CP^{N-1}= U(N)/[U(1)\bigotimes U(N-1)] =SU(N)/[U(1)\bigotimes SU(N)\bigotimes SU(N-1)]$$
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:

$$CP^{N-1}= U(N)/[U(1)\bigotimes U(N-1)] =SU(N)/[U(1)\bigotimes SU(N)\bigotimes SU(N-1)]$$
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.

2. Oct 15, 2015

### 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.

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

3. Oct 16, 2015