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U(n) is the group of n by n unitary complex matrices and SO(2n) is the group of 2n by 2n real orthogonal matrices with determinant 1.So far I can show that how to get an injective group homomorphism from U(n) to SO(2n). This shows that U(n) is isomorphic to a subgroup of SO(2n).But I have no idea whether U(n) is normal in SO(2n)? If someone could explain this to me or just point out some reference for me,I will appreciate your help. Thanks.

edited Jul 6 2010 at 12:10

asked Jul 6 2010 at 4:10

Huichi Huang
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Have you checked with the references at en.wikipedia.org/wiki/Classical_group ? By the way, your tagging is misleading: group-theory and lie-groups are definitely more efficient. – Wadim Zudilin Jul 6 2010 at 4:30

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One of the reasons this is so difficult may be that $U(n)$ is not a normal subgroup of $SO(n)$ at all (unless $n=1$): when $n>2$ the Lie algebra $so(2n)$ is simple and does not have any ideals and when $n=2$ the only ideals are copies of $su(2)$, not $u(2)$. – algori Jul 6 2010 at 4:57

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..that is $U(n)$ is not a normal subgroup of $SO(2n)$.. – algori Jul 6 2010 at 5:03

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Finding an injection from a group $A$ to a group $B$ doesn't show that $A$ is a subgroup of $B$; it shows that $A$ is isomorphic to a subgroup of $B$. The distinction is important (in general, though I don't know about the particular case you ask about), e.g., the group $D_8$ of symmetries of a square has several subgroups isomorphic to the cyclic group of order 2, but one of those subgroups is normal in $D_8$ and the others aren't. – Gerry Myerson Jul 6 2010 at 5:40

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Following algori's observations, shouldn't the title of the question perhaps be changed to remove the word "why"? – Yemon Choi Jul 6 2010 at 5:59

Huichi Huang · Answer 1 · 2010-07-06 12:33:08Z UTC

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Thanks. The answers clarify many things for me. The reason that I ask this question is that I am confused about what's the meaning of the quotient $SO(2n)/U(n)$ if $U(n)$ is not isomorphic to a normal subgroup of $SO(2n)$?

answered Jul 6 2010 at 12:33

Huichi Huang
187●5

The quotient SO(2n)/U(n) simply means to take the (right) coset space, which is NOT a group in this case. However, using the natural map $\pi:SO(2n)\rightarrow SO(2n)/U(n)$, we put a topology on $SO(2n)/U(n)$ which turns $SO(2n)/U(n)$ into a (topological) manifold. Further, we can declare $\pi$ to be a submersion which puts a differentiable structure on $SO(2n)/U(n)$, making it a smooth manifold. (All of this applies equally well to any $G/H$ with $G$ and $H$ compact. If things are not compact, one must be a little careful.) – Jason DeVito Jul 6 2010 at 12:53

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Huichi -- if $G$ is a group and $H$ is a subgroup of $G$, normal or not, one can consider the set $G/H$ formed by the left cosets $gH,g\in G$. If $G$ is a Lie group and $H$ is a closed subgroup (hence, automatically a Lie subgroup), then $G/H$ is a smooth manifold of dimension $\dim G-\dim H$. – algori Jul 6 2010 at 13:04

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Jason -- indeed one must be careful when considering actions of non-compact groups but in this case there aren't any nasty surprises: the quotient of a any Lie group by any closed subgroup is a manifold. – algori Jul 6 2010 at 13:16

@algori - Thank you! Unfortunately (fortunately?), I really only work with compact spaces so I often forget when theorems can be extended from "compact" to, say, "closed subgroup", or "proper map", etc. – Jason DeVito Jul 6 2010 at 13:36

@Jason: I know that's equivalent, but isn't $G/H$ technically the left coset space $\{aH\}$? The left cosets are the ones that have a left action of $G$. – Greg Kuperberg Jul 6 2010 at 17:46

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Is U(n) a normal subgroup of SO(2n)?

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