# Homotopy classes of maps between special unitary Lie groups

I am sorry to mislead the notations: $$SU(n)$$ should be replaced by $$PSU(n)$$. I will reformulate it now.

We consider the special unitary Lie group $$SU(n)$$. Then its center is $$\mathbb{Z}_n$$ and we define $$PSU(n)=SU(n)/\mathbb{Z}_n$$. Then $$\pi_1(PSU(n))=\mathbb{Z}_n$$ and $$H^3(PSU(n))=\mathbb{Z}$$.

Does there exist a map $$f:PSU(4)\to PSU(2)$$ inducing : $$f^*:H^3(PSU(2);\mathbb{Z})\to H^3(PSU(4);\mathbb{Z})$$, $$f^*(x)=dx$$ for an odd $$d$$

and the non-zero homomorphism $$f_{\#}:\pi_1(PSU(4))=\mathbb{Z}_4\to \mathbb{Z}_2=\pi_1(PSU(2))$$ ?

• Aren’t the special unitary groups simply connected? For example, SU(2) is the 3-sphere, so it shouldn’t have a fundamental group... – Dylan Wilson Mar 14 at 12:32
• The special unitary group has a center of dimension $n-1$ no ? Not finite in any case but $n=1$ – Bleuderk Mar 14 at 15:35
• @bleuderk I think you’re thinking of the maximal torus. – Dylan Wilson Mar 14 at 23:35
• oops, indeed, thanks for pointing this out – Bleuderk Mar 15 at 11:31

No. Such a map would give rise to a map $$PSU(4)/(\mathbb Z/2) \to SU(2)$$ inducing multiplication by $$d$$ on $$H^3$$ and hence to a map $$SU(4) \to SU(2)$$ inducing multiplication by $$2d$$ on $$H^3$$ and such a map can not exist.
Indeed, let $$\Sigma \mathbb{C}P^3 \to SU(4)$$ be the axial map (see for instance I.M.James, "The topology of Stiefel manifolds", LMS, p. 22, where the map is called $$\phi$$) inducing an isomorphism on homology in degree $$\leq 7$$. Localizing at $$2$$ for convenience, if the map $$SU(4) \to SU(2)$$ as above were to exist then we would be able (2-locally) to extend a degree 2 map $$S^3 \to S^3$$ to $$\Sigma \mathbb{C}P^3$$. This is not possible even stably.
$$2$$-locally the stable homotopy groups of spheres are $$\pi_1 S^0 \cong \mathbb{Z}/2 \eta, \ \pi_2 S^0 \cong \mathbb{Z}/2 \eta^2, \ \pi^3 S^0 \cong \mathbb{Z}/8 \nu$$ with $$\eta^3=4\nu$$. Looking at the stable homotopy long exact sequence of the cofiber sequence $$S^1 \xrightarrow{\eta} S^0 \to \Sigma^{-2}\mathbb{C}P^2$$ we see that $$\pi_3 \Sigma^{-2}\mathbb{C}P^2 \cong \mathbb{Z}/4$$, generated by $$S^3 \xrightarrow{\nu} S^0 \to \Sigma^{-2}\mathbb{C}P^2$$ The attaching map $$S^3 \to \Sigma^{-2}\mathbb{C}P^2$$ of the top cell in $$\Sigma^{-2}\mathbb{C}P^3$$ must be $$2\nu$$: indeed it can't be an odd multiple of $$\nu$$ because the action of the Steenrod square $$Sq^4$$ on $$H^0(\Sigma^{-2}\mathbb{C}P^3;\mathbb{Z}/2)$$ is trivial and it can't be $$0$$ for then the top cell would split off which is contradicted by the behaviour of Adams operations on complex $$K$$-theory - see for instance Adams, "Vector fields on spheres", Annals of Math (1962) Theorem 7.2.
Writing $$t\colon \Sigma^{-2}\mathbb{C}P^2 \to S^0$$ for the map which has degree $$2$$ on the bottom cell we see that the composite $$S^3 \xrightarrow{2\nu} \Sigma^{-2}\mathbb{C}P^2 \xrightarrow{t} S^0$$ will therefore equal $$0\neq 4\nu \in \pi_{3}S^0$$ and therefore the map $$\Sigma^{-2}\mathbb{C}P^2 \xrightarrow{t} S^0$$ can not possibly extend to $$\Sigma^{-2}\mathbb{C}P^3$$.