# Compact Lie groups with only 3 dimensional cohomology generators

Let $M$ be a compact connected semi-simple Lie group. Then by Hopf's Theorem $H^*(M;\mathbb Q)=\Lambda[\omega_1,...,\omega_s]$ where $\omega_i\in H^i(M;\mathbb Q)$ , $i\ge 3$ is odd.

1. For which $M$, ${\rm dim}(\omega_i)=3$ for all $i=1,...,s$? The examples are $SO(3)$ and quaternionic group $S^3$ (the only simply-connected Lie groups with 1-dimensional torus) as well as their products? Are there any other simply-connected compact Lie groups with this property?

2. Let the fundamental group $\pi_1 M= (\mathbb Z_2\oplus \cdots\oplus \mathbb Z_2)\oplus (\mathbb Z_{h_1}\oplus \cdots\oplus \mathbb Z_{h_s})$ be the sum of $s_2$ copies of $\mathbb Z_2$ and $s$ cyclic groups of rank $\ge 3$. Does it imply that ${\rm dim} M\ge 3s_2+5s$?

• Could you share a few words about your motivation for these questions, please? – Oldřich Spáčil Jun 23 '14 at 13:31
• $SO(3)$ is not simply-connected by doubly covered by $S^3$. I think the answer to 1. is only products of $S^3$'s. See "Foundations of Lie Theory and Lie Transformation Groups", edited by V.V. Gorbatsevich, A.L. Onishchik, E.B. Vinberg, p. 127, or Bourbaki. – Claudio Gorodski Jun 23 '14 at 14:17
• Claudio meant to say "$SO(3)$ is not simply-connected, but is doubly covered by $S^3$." – Ben Wieland Jun 24 '14 at 0:28
• The answer to the second part of the question 1 shoulc be no. You can use the Hurewicz theorem to see that all homology generators in dimension 3 are represented by a map from $S^3$, so you get a product from $S^3$ to $M$ which induces a rational homology equivalence. – user43326 Jun 24 '14 at 5:46
• The motivation of the question is to found out which self-maps of Lie groups $f:M\to M$ have the property: for each $k\in \mathbb N$ the least number of n-periodic points in the homotopy (continuous) class of $f$ ($\min_g\#{\rm Fix}(g^n); g \text{ is homotopic to } f\}$ ; can be realized by a smooth $g$. – user46227 Jun 24 '14 at 13:05

By the classification of simply connected compact Lie groups, the universal cover of such an M will be product of $SU(2)$s. So $M$ is a product of $SU(2)$, $SO(3)$, but also $SO(4)$ (which is $SU(2) \times SU(2)$ modulo the diagonal $\mathbb Z/2$) and possibly others.
As for question 2, note that you can assume without loss of generality that $M$ is center-free and so, again, that it splits as a product of simple center-free Lie groups. Then you just go through the list ($A_n$, ..., $E_8$). For instance, $PSU(n)$ has fundamental group $\mathbb Z/n$ and dimension $\geq 5$ if $n \geq 3$. Your bound should be ok although it's not exactly sharp.
• I don't see the point in "but also $SO(4)$". Either you say it's a quotient of a finite direct product of $SU(2)$'s by a central subgroup, or if you really want a full classification, there are infinitely many possibilities which are not nice to enumerate (for $n$ factors, they are indexed by the quotient of the Grassmannian of $(Z/2Z)_2^n$ by the symmetric group $S_n$)... If we want only those that are not direct products, we should remove those element in the Grassmannian that split as products according to the direct product decomposition of $(Z/2Z)^n$. – YCor Jun 30 '14 at 12:30
• True -- I just wanted to give one concrete example of a Lie group that is not a copy of $SU(2)$s and $SO(3)$s. – Tilman Jul 1 '14 at 14:21