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I want to find the underlying manifolds of Lie Groups $\mbox{SU}(n)$ for general $n$.

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My lecturer told us last year that the only n-spheres that admit a Lie group structure are $\mathbb{S}^1$ and $\mathbb{S}^3$. Moreover I see several question describing why this is so, such as here and here, and understand the arguments therein.

However each Lie group must have some underlying manifold by definition. However I do not know how to find what the manifold is for any given Lie group.

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My motivation for this question is to understand the Lie algebra relation $$ \mathfrak{su}(4) \simeq \mathfrak{so}(6) $$ and as such there is some relation between the Lie groups $\mbox{SU}(4)$ and $\mbox{SO}(6)$, with my neive guess being

$$ \mbox{SU}(4) / \mathbb{Z}_2 \simeq \mbox{SO}(6) $$

I don't have any very good reason for this however, but it feels right. I note that

$$ \mbox{dim}[\mbox{SU}(4)] = 15 \quad \mbox{ and } \quad \mbox{dim}[\mbox{SO}(6)] = 15$$

so we don't immediately think that no such relation should exist. Moreover, I'm encouraged by the relation $\mbox{SU}(2) / \mathbb{Z}_2 \simeq \mbox{SO}(3) $, which I understand and can calculate.

Explicitly, one first notes that both $\mathfrak{su}(2)$ and $\mathfrak{so}(3)$ have the same Lie algebra, namely

$$ [T_i, T_j ] = \varepsilon^{ijk} T_k $$

Then one shows that $\mbox{SU}(2)$ has the 3-sphere $\mathbb{S}^3$ as underlying manifold, since any $U \in \mbox{SU}(2)$ can be written like

$$ U = a_0 I_2 + i \vec{a} \cdot \vec{\sigma} $$

with

$$ a_0^2 + \vec{a} \cdot \vec{a} = 1 $$

and this is the equation for the unit 3-sphere $\mathbb{S}^3$, while it can be shown that $\mbox{SO}(3)$ has underlying manifold $\mathbb{RP}^3$, and it remains to note that $\mathbb{S}^3$ is a double cover of $\mathbb{RP}^3$, which motivates the quotient by $\mathbb{Z}_2$, leading to the result.

Now, I have tried thinking about using this approach in my case of $\mbox{SU}(4)$ and $\mbox{SO}(6)$, but I make no progress.

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Perhaps a related point, we can show that the action of $\mbox{SU}(2)$ on $\mathbb{C}^2 \simeq \mathbb{R}^4$ has as orbits 3-spheres $\mathbb{S}^3$ by considering the fact that that the action of $\mbox{SU}(2)$ on $\mathbb{C}^2$ preserves the Hilbert norm, giving a constraint

$$ ||\vec{z}||^2 = |z_1|^2 + |z_2|^2 = x_1^2 +y_1^2 + x_2^2 +y_2^2 $$

for $\vec{z} = (z_1,z_2) = (x_1 + i y_1, x_2 + i y_2) \in \mathbb{C}^2$.

Now this I have had success (I hope!) in generalising, and I find that the action of $\mbox{SU}(n)$ on $\mathbb{C}^n \simeq \mathbb{R}^{2n}$ should have as orbits (2n-1)-spheres $\mathbb{S}^{2n-1} \subset \mathbb{R}^{2n}$. I am unsure if this actually helps me with my original question however.

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Finally, I'll note that my motivation for wanting to find a relation between $\mbox{SU}(4)$ and $\mbox{SO}(6)$ comes from studying the AdS/CFT correspondence (I'm a TP). In $\mathcal{N}=4$ Super-Yang-Mills theory we get one vector supermultiplet of states, which can be written as one $\mathcal{N}=1$ vector supermultiplet and three $\mathcal{N}=1$ chiral supermultiplet, consisting of a total of four gauge vector fields $A_{\mu}^i(x)$ and six scalar fields $\phi^{[ij]}(x)$ for $i = 1,2,3,4$ which then have an $\mbox{SU}(4)$ R-symmetry (an $\mathcal{N}=N$ supersymmetric field theory has an $\mbox{SU}(N)$ R-symmetry). However, in order to compare this to the type IIB superstring theory on $AdS_5 \times \mathbb{S}^5$ we need to match this $\mbox{SU}(4)$ R-symmetry with the $\mbox{SO}(6)$ rotation symmetry of the 5-sphere \mathbb{S}^5$.

In the AdS/ CFT papers that I have read the TPs (who are not known for making accurate group theoretic statements! (See, for example, the top of page 30 and the bottom of page 72 of what is otherwise a brilliantly clear paper 0712.0689 ) ) simply write $\mbox{SU}(4) \simeq\mbox{SO}(6)$, which I know is deffinitely not correct! For starters, $\mbox{SU}(4)$ is simply connected, while $\mbox{SO}(6)$ is not (my thanks to QMechanic for pointing this nice quick check out to me many questions ago!).

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To conlude, I want to find the relation between $\mbox{SU}(4)$ and $\mbox{SO}(6)$, and I think that not being able to find the underlying manifold of either $\mbox{SU}(4)$ or $\mbox{SO}(6)$ is what is stopping me (of course, perhaps I'm taking the complete incorrect approach...).

Thank you to anyone who reads this and can help me. I'm sorry if it's a little long, but I wanted to let you know what I've tried thinking about, and what my motivation for the question is.

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  • $\begingroup$ When you write such a long text, it is good to graphically emphasize your actual question so that it is easily found. $\endgroup$ – Mariano Suárez-Álvarez Jan 12 '15 at 23:24
  • $\begingroup$ @MarianoSuárez-Alvarez : what do you mean by 'graphically emphasize' ? I was aware of this danger, so I tried to put the question at the top, and then again at the end. $\endgroup$ – Flint72 Jan 12 '15 at 23:36
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    $\begingroup$ @KCd I think OP is looking for a geometric description, or at least topological. Not the way we ( mathematicians ) would have phrased the question, but I think that's what is being asked. More importantly I think OP is asking if $SO(6)$ can be seen as a quotient of $SU(4)$, which would "explain" the isomorphism of Lie Algebras. Presumably the most satisfying answer would be similar to the Dirac belt trick for visualizing $SU(2)/\mathbb{Z}_2 \simeq SO(3)$. OP, although I know your conjecture is true, I don't have a good geometric description. $\endgroup$ – Callus Jan 13 '15 at 0:29
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    $\begingroup$ You might try reading the answers to this post from a year ago; it seems like the easiest way to think about the isomorphism $SU(4) \cong SO(6)/\mathbb{Z}^2$ is to make $SU(4)$ act on $\mathbb{C}^6$ in a nice way. It's not exactly a direct picture of the manifolds -- but I'm not sure we have a name or a picture for the underlying manifold of $SU(4)$ that's fundamentally different from "the underlying manifold of $SU(4)$". $\endgroup$ – mollyerin Jan 13 '15 at 0:37
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    $\begingroup$ It is an old question, but future readers might be helped by reading Jordan Watts answer to this Mathoverflow question: mathoverflow.net/questions/69352/topology-of-su3 $\endgroup$ – CvZ Sep 9 '15 at 2:21
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[NB: This was originally posted as a comment above, but I've been asked to repost as an answer so it can be accepted.]

You might try reading the answers to this post from a year ago; it seems like the easiest way to think about the isomorphism $SU(4) \cong SO(6)/\mathbb{Z}^2$ is to make $SU(4)$ act on $\mathbb{C}^6$ in a nice way.

It's not exactly a direct picture of the manifolds -- but I'm not sure we have a name or a picture for the underlying manifold of $SU(4)$ that's fundamentally different from "the underlying manifold of $SU(4)$".

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