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I am trying to characterize the group $U(2)$.

I am trying to find an explicit homeomorphism for the known fact $U(2)\cong SU(2)\times S^1$.

Since every element $A$ of $U(2)$ has $|\det(A)|=1$, a natural option would be the map $A\mapsto (e^{-i\theta}A, e^{i\theta})$ where $\det(A)=e^{i\theta}$.

The only problem is I'm not sure how to show that this map is surjective or that the inverse is continuous (as I can't find an explicit formula).

Perhaps this map won't work after all, but if it does, how could I prove these two properties?

CuriousKid7
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    Not really the map you wrote, because $\det\left(\frac{1}{\det A}A\right)=\frac1{\det A}$, since $\det(\alpha A)=\alpha^n\det A$. So, to put it in your notation $e^{-i\theta}A\notin SU(2)$. –  Aug 10 '17 at 08:03
  • @JohnMa I've read that $U(2)$ is homeomorphic to $S^3 \times S^1$. Is this not true? – CuriousKid7 Aug 10 '17 at 08:05
  • Probably I will retract my comment, until I have a proof (of either statements). The direct product sounds too good to be true. @CuriousKid7 –  Aug 10 '17 at 08:07
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    The answer the question linked below says that $U(n)$ is the semi-direct product of $SU(N)$ with $U(1)$.

    https://math.stackexchange.com/questions/843874/what-is-the-manifold-structure-of-un

     – Peter Aug 10 '17 at 08:14
  • @JohnMa FYI a google search shows this paper https://arxiv.org/pdf/1505.02365.pdf ( on bottom of page 11) claiming that the homeomorpism is "well-known." – CuriousKid7 Aug 10 '17 at 08:31

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The group homomorphism $\mathrm{SU}(2)\times\mathrm{U}(1)\to\mathrm{U}(2)$ given by $(A,\lambda)\mapsto \lambda A$ is not one-to-one, its kernel is $\mathbb{Z}_2$ with nontrivial element $(-I,-1)$.

However, the map $\mathrm{SU}(2)\times\mathrm{U}(1)\to\mathrm{U}(2)$ given by $(A,\lambda)\mapsto[\begin{smallmatrix}\lambda&0\\0&1\end{smallmatrix}]A$ is a diffeomorphism, it's just not a group homomorphism. Indeed, using this copy of $\mathrm{U}(1)$ inside $\mathrm{U}(2)$, we see that $\mathrm{U}(2)$ is the internal semidirect product of $\mathrm{SU}(2)$ and $\mathrm{U}(1)$ (because every element is uniquely a product of two such elements, and $\mathrm{SU}(2)$ is normal).

anon
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  • Could you please explain why the second map is a homeomorphism? – CuriousKid7 Aug 10 '17 at 08:44
  • Just check off all the items in the checklist that is the definition of "diffeomorphism." Is smooth, has inverse, inverse is smooth. Or just continuous if all you want is "homeomorphism." If you think of ${\rm U}(1)$ as a Lie subgroup of ${\rm U}(1)$ in the upper left corner, then the map ${\rm SU}(2)\times{\rm U}(1)\to{\rm U}(2)$ is just a restriction of the map ${\rm U}(2)\times{\rm U}(2)\to{\rm U}(2)$, which is smooth because ${\rm U}(2)$ is a Lie group. – anon Aug 10 '17 at 08:45
  • Right, but I don't see surjectivity or that inverse is continuous. – CuriousKid7 Aug 10 '17 at 08:47
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    The inverse is $([\begin{smallmatrix}\det g & 0 \ 0 & 1\end{smallmatrix}]^{-1}g,,,\det g)$, and $\det$ is smooth. The fact it even has an inverse implies surjectivity. – anon Aug 10 '17 at 08:49
  • Last question, why is the homomorphism you described surjective? – CuriousKid7 Aug 10 '17 at 11:07
  • @CuriousKid7 Do you mean the non-injective homomorphism you described and is mentioned in the first sentence of my answer? Or the homeomorphism I describe in my second paragraph? If the latter, I already told you - the fact it even has an inverse implies surjectivity. If the former, you should be able to handle this problem: you need to write an arbitrary $g\in{\rm U}(2)$ as a scalar $\lambda$ times an $A\in{\rm SU}(2)$. What choices for $\lambda$ do you have? What does that force $A$ to be? – anon Aug 10 '17 at 11:30
  • $\lambda$ must be of modulus $1$, but I really don't see what $A$ would be. Is there a general form of a unitary matrix? – CuriousKid7 Aug 10 '17 at 11:42
  • @CuriousKid7 You do not need the general form of a unitary matrix. I ask you again: given that $g=\lambda A$, where $g\in{\rm U}(2)$, $\lambda\in S^1$, $A\in{\rm SU}(2)$, what choices for $\lambda$ are there? It is not the case that $\lambda$ can be just any number of modulus $1$. Actually figure out what $\lambda$ is from $g$! (There are two possible choices for $\lambda$.) I assume now you are talking about the group homomorphism in my first sentence of my answer. – anon Aug 10 '17 at 11:45