What is the relation between unitary group, homotopy?

1. Oct 15, 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.

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:

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

andrewkirk

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.

3. Oct 15, 2015

PRB147

U(N),SU(N) are unitary and special unitary group respectively

4. Oct 15, 2015

andrewkirk

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.

5. Oct 16, 2015

PRB147

Thank you very much, I admire mathematician!

6. Oct 18, 2015

lavinia

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
7. Oct 19, 2015

PRB147

8. Oct 26, 2015

zinq

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.
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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)​