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How can we find $\pi_{d}(U(N+M)/U(N) \times U(M))$ ? Is there any way to visualize the target space ? I am specifically interested in the $d=1,2$ & $(N,M)=(2,2),(2,4)$ cases. Thanks.

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Have you tried looking at the long exact sequence for the fibration $U(N)\times U(M)\hookrightarrow U(N+M)\to U(N+M)/U(N)\times U(M)$? That is usually good enough to get the low degree groups. –  Mariano Suárez-Alvarez Sep 8 '13 at 4:57
    
Please include the definitions of what you are asking when you ask a question. –  Vishal Sep 8 '13 at 5:16
    
What are $M,N, M +N, U$? How does $d$ get into the act? –  Ronnie Brown Sep 8 '13 at 10:18

1 Answer 1

Let us write $X=U(N+M)/U(N)\times U(M)$. The fibration $$U(N)\times U(M)\hookrightarrow U(N+M)\to U(N+M)/U(N)\times U(M)$$ gives as a long exact sequence (because all spaces are connected) \begin{align} \tag{$\star$} 0\leftarrow\pi_1(X)\leftarrow\pi_1(U(N+M))&\leftarrow\pi_1(U(N)\times U(M))\leftarrow \\ &\leftarrow\pi_2(X)\leftarrow\pi_2(U(N+M))\leftarrow\cdots \end{align} If $K\geq1$, there is a fibration $$U(K-1)\hookrightarrow U(K)\to U(K)/U(K-1)=S^{2K-1}$$ from which we get again a long exact sequence \begin{align} 0\leftarrow\pi_1(S^{2K-1})&\leftarrow\pi_1(U(K))\leftarrow\pi_1(U(K-1))\leftarrow \\ &\leftarrow\pi_2(S^{2K-1})\leftarrow\pi_2(U(K))\leftarrow\pi_2(U(K-1))\leftarrow\pi_3(S^{2K-1})\cdots \end{align} Now:

  • If $K\geq2$, then $2K-1\geq3$ and therefore $\pi_1(S^{2K-1})=\pi_2(S^{2K-1})=0$, and so we see that the inclusion $U(K-1)\to U(K)$ induces an isomorphism in $\pi_1$. It follows by induction that $\pi_1(U(K))=\pi_1(U(1))\cong\mathbb Z$ for $K\geq1$
  • If $K\geq3$, then $2K-1\geq5$ and $\pi_2(S^{2K-1})=\pi_3(S^{2K-1})=0$, and the exact sequence shows that the inclusion $U(K-1)\to U(K)$ induces an isomorphism in $\pi_2$. It follows by induction that $\pi_2(U(K))\cong\pi_2(U(2))$ for $K\geq2$.
  • When $K=2$, the long exact sequence tells us that $$\pi_2(S^3)\leftarrow\pi_2(U(2))\leftarrow\pi_2(U(1))$$ is exact, and we know that the first and third groups are zero, so $\pi_2(U(2))=0$. The previous point tells us then that $\pi_2(U(K))=0$ for all $K\geq1$.

We now go back to the sequence $(\star)$ and let us suppose that $N,M\geq1$. Using the computations for $\pi_1$ and $\pi_2$ of $U(K)$ that we have now, and the fact that homotopy groups play nicely with products. We are left with the sequence $$ 0\leftarrow\pi_1(X)\leftarrow\pi_1(U(N+M))\leftarrow\pi_1(U(N))\times\pi_1(U(M))\xleftarrow{\partial}\pi_2(X)\leftarrow0 $$ We know that the arrow $\pi_1(U(N+M))\leftarrow\pi_1(U(N))\times\pi_1(U(M))$ is sujective (because already the restriction to th efirst factor, $\pi_1(U(N+M))\leftarrow\pi_1(U(N))$ is surjective according to the first point above!) It follows that $\pi_1(X)=0$ and $\pi_2(X)\cong\mathbb Z$.

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I come from a Physics background. The above Homotopy group is relevant in the context of 'Topological defects' in Condensed matter physics. I am only comfortable with thinking about mappings from circle / sphere to sphere / torus. I am finding your answer highly technical. Can you please suggest a textbook where I find such stuff (basically the idea of fibration) ? Thanks a lot for your answer. –  Akshay Kumar Sep 8 '13 at 16:44
    
Well, computing homottopy groups is a rather technical enterprise! You can learn all this in books like Switzer's or Whitehead's on homotopy theory. –  Mariano Suárez-Alvarez Sep 8 '13 at 19:16
    
For the case of Z classification, can we write down a formula that gives me the particular integer class (topological invariant) to which a given mapping belongs to ? –  Akshay Kumar Sep 9 '13 at 5:58
    
The map $\partial$ in the last exact sequence is quite explicit (because our fibration is very explicit!) and can in principle be computed. You should read a bit about the long exact sequence for homotopy groups. –  Mariano Suárez-Alvarez Sep 9 '13 at 6:14

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