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Suppose I wanted to describe the subgroup of $GL_n(\mathbb{R})$ of matrices of the form

$$ \left [ \begin{array}{cc} A & B \\ 0 & C \\ \end{array} \right ] $$

where $A \in GL_k(\mathbb{R}),\; C \in GL_{n-k}(\mathbb{R})$ and say for argument $B \in M_{k, n-k}(\mathbb{R})$. What is the proper way to express what this subgroup is using the the other groups $GL_k(\mathbb{R}), GL_{n-k}(\mathbb{R})$, and $M_{k,n-k}(\mathbb{R})$? For instance, I know that the special Euclidean group has the form $SE(n) = SO(n) \ltimes \mathbb{R}^n$, but I'm not sure where this derivation came from. Ultimately I'd like to be able to recreate simple descriptions for matrix subgroups of this type. Ultimately my goal is to take quotients of certain matrix Lie groups, so I'd like to know when subgroups of this type can be represented as direct or semi-direct products of certain subgroups.

Reference requests and full explanations are welcome.

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  • $\begingroup$ I am deleting my answer in view of your comment. $\endgroup$ – P Vanchinathan Apr 15 '15 at 3:37
  • $\begingroup$ The example you have given is a group acting on Grassmannian transitively. To know when a subgroup is a direct, or semi-direct product, some standard examples are useful: the group of upper triangular marices is semi-direct product of the subgroup with 1's in the diagonal and its diagonal subgroup. And the n there is Levi decomposition for parabolic subgroups (those containing triangular group) at least in the case of matrix group over complex numbers. $\endgroup$ – P Vanchinathan Apr 15 '15 at 3:42
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This corresponds to a part of the Grassmanian Manifold corrosponding to $GL(n)$. In particular, it is the part corrosponding to $(0, k, n)$ sequeunce in the Flag decomposition, often called a Flag variety (or generalized flag variety). It acts fleely and transitivly, and infact, it is not hard to convince yourself that, if you call your group $G_{k, n}$, $BG_{k, n}$ is precisely this part of the manifold (If your new that the weird habit of putting $B$ infront of groups, it's called a classifying space). This is just nonsense though, but is kinda what I think Vanchinathan was getting at in the comments. Now for your question.

We have a projection mapping from $G_{k, n}\to GL(k)\times GL(n-k)$ given by forgetting the upper triangular part. Checking that this is a group homomorphism is an easy exercise. Now the kernel of this map consists of matrices whose $A=I_k$ and $C=I_{n-k}$. Now the multiplication on this normal subgroup can be computed easily, and infact is given by adding the $C$-components. Thus the subgroup can be made isomorphic to $M_{k, n-k}$ under addition. Thus we have an exact sequence $0\to \mathbb{R}^{k(n-k)}\to G_{k,n}\to GL(k)\times GL(n-k)\to 0$. Now this splits (by the clear inclusion on either side), so by the splitting lemma, we can write it as a semi-direct product: $G_{k, n}=GL(k)\times GL(n-k)\ltimes \mathbb{R}^{k(n-k)}$.

This gives the isomorphism for $SE(n)$ by changing $GL=SO$ and $n=1$ noting that $SO(1)=*$.

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