Let $SU$ denote the infinite unitary group. Does the quotient space $SU/SU(n)$ admit a delooping $X$? One could also ask that this space $X$ sit in a fiber sequence $BSU(n)\to BSU\to X$, but this is not strictly part of the question. Note that $SU/SU(n)$ is not a topological group, because $SU(n)$ is not normal in $SU$ --- but this doesn't prohibit $SU/SU(n)$ from admitting a delooping. Perhaps a geometric construction of a H-space structure can be given by viewing the space as a Stiefel manifold. Note that $SU$ is an infinite loop space by Bott periodicity.


1 Answer 1


I'll work with mod $2$ cohomology. Note that $H^*(BSU(2))$ is polynomial on $c_2$ (in degree $4$) and $H^*(BSU)$ is polynomial on $c_k$ for $k\geq 2$. Here $c_k$ has degree $2k$ and so $H^6(BSU)=\{0,c_3\}$. If $X$ exists then it seems we should have $H^*(X)$ polynomial on generators in degrees $6,8,10,\dotsc$. In particular, $H^6(X)$ should be generated by $c_3$ and $H^{10}(X)$ should be generated by a single element that is $c_5$ modulo decomposables. However, $H^*(X)$ should also be closed under the action of the Steenrod algebra so it should contain $\text{Sq}^4(c_3)$, which is $c_2c_3$ by a calculation with symmetric polynomials. This is inconsistent, so $X$ cannot exist. I would guess that this line of argument can be improved to show that $SU/SU(2)$ does not deloop, but I have not tried to work out the details.

  • $\begingroup$ Thanks. I guess the equation $\mathrm{Sq}^4(c_m) = c_2 c_m$ shows more generally that there's no delooping of $SU/SU(n)$ with a map from $BSU$ whose homotopy fiber is $BSU(n)$. $\endgroup$
    – skd
    Nov 5, 2019 at 2:25

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