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From the book "Einstein manifolds" by Arthur L. Besse (at section 7.B), Lie groups $Sp(n)$, $Sp(n)\cdot U(1)$, $SU(2n)$ and $U(2n)$ constitute the complete list of Lie subgroups of $U(2n)$ acting transitively and effectively on the sphere $S^{4n-1}$.

Consider natural inclusions of $Sp(n)$, $Sp(n)\cdot U(1)$, $SU(2n)$ into $U(2n)$ and consider their actions on $S^{4n-1}\subset\mathbb{C}^{2n}$ as linear transformations (matrix action). Suppose that there is a matrix subgroup $G$ of $U(2n)$ acting transitively on $S^{4n-1}$ (thus isomorphic to above models). Is this $G$ conjugate (by an element of $U(2n)$) isomorphic to models?

For instance, if $G\subset U(2n)$ acting transitively on $S^{4n-1}$ and $G$ is isomorphic to $Sp(n)$, then is there $M\in U(2n)$ such that $MGM^{-1}=Sp(n)$?

Thanks.

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Yes, there is always such an $M$.

To see this, first note that saying two representations of $G$ on a vector space $V$ are equivalent is the same as saying the two images of $G$ in $Gl(V)$ are conjugate by an element of $Gl(V)$. A theorem due to Mal'cev which can be found in

Mal'cev. On semisimple subgroups of lie groups. Amer. Math. Soc. Trans., 1:172-273, 1950.

proves that if $V$ is given a Hermitian inner product and both images of $G$ lie in $U(V)\subseteq Gl(V)$, then they are equivalent representations iff the images are conjugate by an element of $U(V)$. (This is also true replacing $U(n)$ by any odd dimensional orthogonal group, $SU(n)$, or $Sp(n)$, but needs to be modified slightly in the case of even dimensional orthogonal groups.)

So, if each of the groups in your list has a unique nontrivial $2n$-dimensional representation, we'll be done. Unfortunately, this is true of $Sp(n)$ only. The group $Sp(n)\times S^1$ has infinitely many (given as the tensor product of the standard representation with any nontrival 1-d rep of $S^1$). On the other hand, $SU(2n)$ has precisely two given by the standard representation or the complex conjugate of the standard representation. Finally, $U(2n)$ has two infinite familes given on the cover $SU(2n)\times S^1$ as a tensor product of either of $SU(2n)$s reps with any nontrivial $1$-dim rep of $S^1$.

Let me just focus on the case of $Sp(n)\times S^1$, the other cases being similar. Consider the tensor product of the standard $2n$-dimensional rep of $Sp(n)$ together with any nontrivial $1$-dimensional rep of $S^1$, say, of weight $k$. By precomposing with the cover $\pi:Sp(n)\times S^1\rightarrow Sp(n)\times S^1$ where $\pi(A,z) = (A,z^k)$, we see that the image of $Sp(n)\times S^1$ in $U(2n)$ under this representation is the same as if we use the weight $1$ (standard) rep of $S^1$. Now, Mal'cev's theorem implies that the image of $Sp(n)\times S^1$ is conjugate to the standard $Sp(n)\times S^1$.

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Thanks for your answer. But I can't find any literature dealing with representation of $Sp(n)$, $Sp(n)\times S^1$ and so on (I am not actually working on Lie group theory). I will appreciate it very much if you tell me the literature to show that the $2n$-dimensional representation of $Sp(n)$ is unique and related results. –  user33387 Jun 24 '13 at 5:45
    
I use Fulton and Harris's book for almost all of my representation theory needs. In particular, page 406 gives a formula for the dimension of all irreducible reps of Sp(n) in terms of natural numbers written over nodes of its Dynkin diagram. One simply needs to check the cases where all nodes are $0$ except one which is a $1$. The other result I used is that an irreducible representation of a product is a tensor product of irreducible representations, which is not too hard to prove (but I don't know of a reference off hand.) –  Jason DeVito Jun 25 '13 at 14:49
    
Looking back on this, I wrote something false: If two representations of $G$ are equivalent, then the images are conjugate. The converse, however, is very false. For example, consider $G\times G \rightarrow G\rightarrow Gl(V)$ where the first map is either of the two projections. Then the images are conjugate trivially, but so long as the representation of $G$ is nontrivial, the two reps are not equivalent. –  Jason DeVito Feb 25 at 19:54
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