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The complex spin groups $Spin^C(n)$ appear in the fibration

$Spin(n)\rightarrow Spin^C(n)\rightarrow\ S^1$

which must split since $BSpin(n)$ is 3-connected to give a homotopy equivalence

$Spin^C(n)\simeq Spin(n)\times S^1$

This is, however, only an equivalence of spaces, and not of topological groups.

In low dimensions there are the 'accidental' equivalences of lie groups $Spin(3)\cong SU(2)$, $Spin(4)\cong SU(2)\times SU(2)$, $Spin(5)\cong Sp(2)$ and $Spin(6)\cong SU(4)$.

My question is if these lead to 'accidental' isomorphisms of groups $Spin^C(3)\cong SU(2)$, $Spin^C(4)\cong U(2)\times SU(2)$ and $Spin(6)\cong SU(4)$?

The first of these isomorphism appears in the literature but I cannot find references for the other two above conjectures. They are certainly true homotopically, but are they true as topological groups?

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We have $Spin^C(n) = (Spin(n)\times S^1)/(\mathbb{Z}/2)$, where $\mathbb{Z}/2$ is generated by $(-1,-1)$ and the first $-1$ is the nontrivial element of the kernel of $Spin(n)\to SO(n)$. Since the accidental isomorphisms come from the spin representation of $Spin(n)$, this element corresponds to $-id$ in $SU(2),Sp(2),SU(4)$ and to the element $(-id,-id)$ of $SU(2)\times SU(2)$. The map $Spin^C(n)\to S^1$ is then given by $(g,z)\mapsto z^2$. On the other hand, we have $(SU(n)\times S^1)/(\mathbb{Z}/n)\cong U(n)$, where $\mathbb{Z}/n$ is generated by $\left(e^{\frac{2\pi i}{n}} id,e^{-\frac{2\pi i}{n}}\right)$ and the isomorphism is given by $(A,z)\mapsto Az$.

This shows that $Spin^C(6)$ is a two-fold cover of $U(4)$, since we can get the quotient by $\mathbb{Z}/4$ by quotienting by $\mathbb{Z}/2$ two times.

$Spin^C(4)\cong (SU(2)\times SU(2)\times S^1)/(-id,-id,-1)$ is not isomorphic to $U(2)\times SU(2)$: The first group has two commuting automorphisms of order $2$, where the first exchanges the two factors of $SU(2)$ and the second is complex conjugation on the factor $S^1$. One easily sees that these give a subgroup $\mathbb{Z}/2\times \mathbb{Z}/2$ of the outer automorphism group of $Spin^C(n)$, for instance by noting that every inner automorphism acts trivially on the center of the connected two-fold cover $SU(2)\times SU(2)\times S^1$. On the other hand, every automorphism $\alpha$ of the group $U(2)\times SU(2)$ must preserve the center $S^1\times\mathbb{Z}/2$ which has a unique element $(-1,1)$ of order $2$ in the connected component of the identity, so it must fix this element. It also has to preserve the commutator subgroup $SU(2)\times SU(2)$. But it is easy to see (for instance by looking at its semisimple Lie algebra) that the outer automorphism group of $SU(2)\times SU(2)$ is $\mathbb{Z}/2$, and an automorphism is not inner iff it sends $(-1,1)$ to $(1,-1)$. In particular the restriction of $\alpha$ to $SU(2)\times SU(2)$ must be given by conjugation by some element $(g,h)\in SU(2)\times SU(2)$. It follows that every class of the outer automorphism group of $U(2)\times SU(2)$ has a representative which restricts to the identity on $SU(2)\times SU(2)$. This representative induces an automorphism of the connected component of the center $S^1\times\{1\}$, and since this group and $SU(2)\times SU(2)$ jointly generate $U(2)\times SU(2)$, it is uniquely determined by this automorphism. This gives a monomorphism $Out(U(2)\times SU(2)) \to Aut(S^1)\cong \mathbb{Z}/2$, and this is indeed an isomorphism (there is outer automorphism of order $2$ given by complex conjugation on the first factor).

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