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 $f_{\#}:\pi_1(PSU(4))=\mathbb{Z}_4\to \mathbb{Z}_2=\pi_1(PSU(2))$ ?

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


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