What is the relation between unitary group, homotopy?

[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.

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:

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

 

[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.

 

[PRB147]

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.

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

 

[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.

 

[PRB147]

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.

Thank you very much, I admire mathematician!

 

[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 $π_1(U(n)) = Z$.

 

[PRB147]

Thank You very much lavinia for your great answer!

 

[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 × ... × S^2n-1; ℚ)​

H^*(SU(n); ℚ) = H^*(S³ × S⁵ × ... × S^2n-1; ℚ)​

H^*(Sp(n); ℚ) = H^*(S³ × S⁷ × ... × S^4n-1; ℚ)​