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It is well known that the fundamental class of a compact Lie group $G$ is stably spherical (see "H-Spaces and Duality" by Browder and Spanier, or "Thom Complexes" by Atiyah), and there is a stable equivalence $G\simeq A\vee S^n$ for some subcomplex $A$, with $n$ being the dimension of $G$.

My question relates to exactly when the attaching map of the top cell becomes trivial? How many suspensions are required for the top cell to split off? Is there an exact answer, or bounds for some special cases perhaps?

As an example, Mimura shows in "On the Number of Multiplications on $SU(3)$ and $Sp(2)$" that this occurs for $Sp(2)$ after exactly two suspensions, $S^2\wedge Sp(2)\simeq S^5\cup e_9\vee S^{12}$, and for $SU(3)$ after exactly three suspensions, $S^3\wedge SU(3)\simeq S^6\cup e_8\vee S^{11}$. These are very special cases. Are more general results or estimates available?

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    $\begingroup$ For abelian groups you need at most one suspension. I imagine there are several more simple cases like this. $\endgroup$ – Ryan Budney Sep 10 '15 at 16:12
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I don't know the minimal number of suspensions required, but for the classical groups $O(n)$, $U(n)$ and $Sp(n)$ the existence of a bound quadratic in $n$ follows from Miller's stable splittings:

  • H. Miller. Stable splittings of Stiefel manifolds. Topology 24 (1985) 411-419.

The result is that there is a splitting of $O(n)$, $U(n)$ and $Sp(n)$ as $\bigvee_{k=1}^n\operatorname{Gr}(k,n)^{\operatorname{ad}_k}$, i.e., as a wedge of Thom spaces over the respective Grassmannians. The number of suspensions required to be able to define the splitting map (up to the level $k$ of the decomposition) is the maximum of the embedding dimensions of the Grassmannians involved, cf. Remark 3.9 of Miller's paper. The full splitting exists after $\max\{2\dim_\mathbb{R}\mathbb{K}\cdot k(n-k)\mid 1\leq k\leq n\}$ suspensions where $\mathbb{K}=\mathbb{R},\mathbb{C},\mathbb{H}$ for the three classical series.

As another reference, the following paper gives a discussion of Miller's splitting for $U(n)$ and shows that the top cell splits off after $n^2$ suspensions (see page 462):

  • N. Kitchloo. Cohomology splittings of Stiefel manifolds. J. London Math. Soc. 64 (2001), 457-471.

Both papers give explicit constructions of splitting maps and relate the top cell to the adjoint representation of the respective Lie group. Probably this would be true generally, but I don't know of splitting results for the exceptional groups.

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Perhaps worth noting here:

There is an old open conjecture, the Hirsch conjecture, which says that a stably parallelizable (closed) smooth $n$-manifold $M$ always embeds in $\Bbb R^{[(3/2)n]}$ (approximately). The conjecture has been verified when $M$ is $[n/4]$-connected.

A compact Lie group is parallelizable so, if the conjecture is true, then the top cell will split off after approximately $[n/2]$-suspensions if the Lie group has dimension $n$.

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Let me just give the naive answer. The homotopy category of $n$-connected $(n+k)$-dimensional CW-complexes is stable for $k<n-1$, by the Freudenthal suspension theorem etc. So the $(\dim G+3)$-fold suspension splits in general, if $G$ is connected the $(\dim G+2)$-fold suspension suffices, and if $G$ is simply connected suspending $\dim G$ times is OK ($\pi_2 G=0$).

This naive bound gives $n^2+2$ for $U(n)$. As Matthias says $n^2$ is sufficient, which is pretty close. For $O(n)$ the bound is $\frac{n^2-n}{2}+3$, which for $n\geq 6$ improves John's estimate. For $SU(n)$ it is $n^2-1$, which is far from optimal as you indicate. The same happens for $Sp(n)$.

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