7
$\begingroup$

The first few unstable homotopy groups of the unitary groups $U(n)$ were calculated by Borel-Hirzebruch, Toda, and Kervaire, and they are all torsion. There is a paper by Matsunaga (details below) in which the p-primary parts of the next few homotopy groups are calculated. This paper claims to be giving a complete description of these homotopy groups, but I don't see any explanation of why there are no free abelian factors.

My question is: Why $\pi_{2n+i} (U(n))$ is a torsion group for $i=3,4,5$?

More generally, is it know whether or not $\pi_* (U(n))$ is torsion for all $*>2n-1$?

Matsunaga's paper is:

MR182010 Matsunaga, Hiromichi Unstable homotopy groups of unitary groups (odd primary components). Osaka J. Math. 1 1964 no. 1, 15–24.

It's freely available here: https://www.jstage.jst.go.jp/article/kyushumfs/15/1/15_1_72/_article

There's actually an errata list for the article that's not contained in the freely available .pdf. It just corrects various notational errors (though not the obvious notational error in the main theorem, in which part b should refer to the 3-primary summand).

Improve this question
$\endgroup$
1
  • 2
    $\begingroup$ It's interesting to note that the analogous statement for the special orthogonal group does not hold. For example, $\pi_7(SO(6)) = \mathbb{Z}$. This follows from the fact that $\operatorname{Spin}(6) = SU(4)$, the first $23$ homotopy groups of which were calculated by Mimura and Toda in this paper. The point is that using complex Bott periodicity to calculate homotopy groups of $SU(n)$ applies for longer than using real Bott periodicity to calculate the homotopy groups of $SO(n)$. $\endgroup$
    – Michael Albanese
    Commented Oct 3, 2017 at 19:19

2 Answers 2

Reset to default
16
$\begingroup$

Being an $H$-space, $U(n)$ has the rational homotopy type of a product of odd-dimensional spheres. As we know its cohomology, these are $S^1 \times S^3 \times S^5 \times \cdots \times S^{2n-1}$. In particular, its rational homotopy groups vanish above degree $2n-1$.

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ Pedantic nitpick: "being an $H$-space with finitely generated homology..." $\endgroup$
    – Qiaochu Yuan
    Commented Nov 12, 2014 at 8:23
  • $\begingroup$ Does one not even need finite dimensional cohomology and not just finitely generated? The result is usually proved by the classification of finite-dimensional, graded commutative, graded cocommutative Hopf algebra over a field of characteristic 0. A theorem of Hopf says that it is a free exterior algebra with generators of odd degree. Does it really suffice to have finitely generated cohomology? $\endgroup$
    – archipelago
    Commented Nov 12, 2014 at 20:58
  • 1
    $\begingroup$ @archipelago: by "finitely generated homology" I mean that the homology (not the cohomology) is finitely generated as a graded vector space (not a ring). Sorry for the confusion. I wanted to use a term that generalizes smoothly to the case where we are not necessarily working rationally or with an $H$-space. Of course it does not suffice for either the homology or the cohomology to be finitely generated as a ring as the example of $K(\mathbb{Z}, 2n)$ shows. $\endgroup$
    – Qiaochu Yuan
    Commented Nov 12, 2014 at 21:18
12
$\begingroup$

A Sullivan minimal model for $U(n)$ is given by $\Lambda(x_1,x_3,\ldots,x_{2n-1})$ with zero differential. In fact for a Lie group, a minimal model is always a free algebra on odd generators and in this case, you can compute the degree of those generators by computing the cohomology of $U(n)$.

The graded group $\pi_*(U(n))\otimes\mathbb{Q}$ is the vector space of indecomposable elements in the minimal model. Thus you get one non-torsion factors in the $i$-th homotopy of $U(n)$ for $i=1,3,\ldots,2n-1$ and all the other homotopy groups must be torsion.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .