How to show that the homotopy group $\pi_4(U(3))$ of a unitary group is finite

Question

I have dealt with the stable unitary groups and calculated the groups for $\pi_i(U(n))$ when $i=1,2,3$ and I know that $\pi_4(U(3))\cong \pi_4(U(4))$ and the fibration $U(n)\to U(n+1) \to S^{2n+1}$ which I've tried to use the LES but haven't gotten far. I'm aware of the isomorphism to $SU$, but I don't think that should be necessary here.

Answer 1

From $$ \begin{aligned} & \color{gray}{ \pi_5(U(2))\to \pi_5(U(3))\to} \pi_5(S^5) &\searrow \\ &&\swarrow \\ & \longleftarrow\longleftarrow\longleftarrow\longleftarrow\longleftarrow\longleftarrow\longleftarrow\longleftarrow\longleftarrow && \\ \swarrow\quad & \\ \searrow\quad & \\ &\pi_4(U(2))\to \pi_4(U(3))\to \underbrace{\pi_4(S^5)}_{=0} \end{aligned} $$ we get a surjection $\color{blue}{\pi_4(U(2)) \twoheadrightarrow\pi_4(U(3))}$. Using $$ \underbrace{U(1)}_{S^1}\to U(2)\to \underbrace{U(2)/U(1)}_{S^3}$$ we get the isomorphisms in higher degrees of the homotopy groups for $U(2)$ and $S^3$. In particular, $$ \pi_4(U(2))\cong \pi_4(S^3)\cong \Bbb Z/2\ , $$ the last isomorphism taken from the Wiki Table.

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To see that in the end $\pi_4(U(3))=0$, one has maybe to investigate the map $$ \Bbb Z\cong \pi_5(S^5)\to\pi_4(U(2))\to \pi_4(U(2)/U(1))=\pi_4(S^3)\overset{\text{Hopf}}\cong \pi_4(S^2) \cong \Bbb Z/2 $$ and show, it is not the zero map.

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The catlab-links are providing some more information. A good reference is the Mimura-Toda paper from 1963.

• ncatlab.org/nlab/show/unitary+group
• ncatlab.org/nlab/show/unitary+group#MimuraToda63

Answer 2

One can completely compute the rational homotopy groups of an H-space simply by looking at the cohomology. It is always a symmetric algebra (i.e. a tensor product of an exterior algebra on the odd degree generators and a polynomial algebra on the even degree generators), and we have the property that the rank of the nth rational homotopy group is equal to the number of generators of degree $n$.

In the case of $U(3)$ we have that its cohomology is $\Lambda[x_1,x_2,x_3]$ where $|x_i|=2i-1$. Since there is no generator in degree $4$, we see that $\pi_4(U(3)) \otimes \mathbb{Q}=0$ hence $\pi_4(U(3))$ is torsion.