There is a canonical inclusion $i$ of $U(n-1)$ into $U(n)$. Using the fibration $U(n-1)\rightarrow U(n)\rightarrow S^{2n-1}$ and the associated long exact sequence of homotopy one gets that $i_*$ induces an isomorphism $\pi_k(U(n-1))\cong\pi_k(U(n))$ for sufficiently large n. This allows one to form "stable" homotopy groups $\pi_k^s(U)$ of the unitary groups. (Note: "Stable" has a different meaning in comparison to "stable" homotopy groups of the spheres.)

Now the article on unitary groups on ncatlab claims that we can form the limit $U$ of the unitary groups to obtain the isomorphism $\pi_k^s(U)\cong \pi_k(U)$. My question is:

Is this construction valid? And, if yes, how do I see this?

Assume we are given a $f:S^k\rightarrow U$ representing $[f]\in \pi_k(U)$. I can't find a reason why $f$ should be homotopy-equivalent to a map $g$ with image contained in a $U(n)\subset U$. If the unitary groups $U(n)$ were open, this would follow from compactness of $S^k$.

  • $\begingroup$ Can you give $U$ the structure of a CW complex where each $U(n)$ is a subcomplex? If so, then you can use cellular approximation. $\endgroup$ – Jason DeVito Mar 31 at 13:55
  • $\begingroup$ I know that there are more sophisticated techniques to obtain a CW structure, however I do not know any elementary way of seeing this: In comparison to the way of obtaining the CW decomposition of the complex projective plane the sphere appears to be on the "wrong" side of the fibration. $\endgroup$ – bright vader Mar 31 at 14:26
  • $\begingroup$ I think you can use Matt E's answer here math.stackexchange.com/questions/285351/… to construct a CW structure on each $U(n)$ so that the natural inclusions are inclusions of a subcomplex. $\endgroup$ – Jason DeVito Mar 31 at 14:47

Homotopy groups commute with "nice" filtered colimits. For example, if you have a direct limit of cofibrations, then the homotopy groups commute with the direct limit. This follows from the fact that the spheres are compact.

For your specific question, the image of $S^k$ is compact and so under some coherent CW decomposition (which these $U(n)$ have) it lies in finitely many cells. Necessarily, there must be some $U(n)$ that contains the entire image.

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  • $\begingroup$ Thank you for your answer! However, I am not able to see how it follows from the compactness of the spheres that the homotopy groups commute with the direct limit. If we were considering a filtered colimit of open inclusions, I would get the argument. And for the CW-decomposition: As far as I know, there is no elementary way to see this. $\endgroup$ – bright vader Mar 31 at 14:23
  • $\begingroup$ It definitely is not elementary that it has a CW decomposition, see Milnor and Stashev. Suppose you have a sequence of cofibrations, then any map from a compact space into the colimit factors through one of the spaces in the diagram: suppose it did not, then we could find a sequence of points in the image of the space such that each successive point is in strictly higher filtration. This sequence must have a limit point which lies inside the image. This point lies in some space in the diagram, and using some alternative characterizations of cofibration we can find an open set around this space $\endgroup$ – Connor Malin Mar 31 at 14:40
  • $\begingroup$ that none of the points in the sequence of filtration higher than this space lie in which gives a contradiction to these points converging. It is also possible to prove this by taking successive CW approximations and taking a limit of these and showing this approximates the colimit. In nicer cases you might also try to take small open neighborhoods of the spaces and use the open inclusion version. $\endgroup$ – Connor Malin Mar 31 at 14:41
  • $\begingroup$ I am sorry that I need to ask this, but about what property of cofibrations are you talking? $\endgroup$ – bright vader Apr 1 at 9:28
  • $\begingroup$ In the details of showing it is a contradiction, you are probably going to use the fact that a cofibration $X_n \rightarrow X$ is a neighborhood deformation retraction pair $\endgroup$ – Connor Malin Apr 1 at 16:28

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