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I have stumbled on what to me is an interesting observation: over any commutative ring $F$, the general linear group is a subgroup of the split orthogonal group of twice the dimension: $$\mathrm{GL}(n,F) \lt \mathrm{O}(n,n,F)$$ This is in a sense the reverse of the more obvious subgroup relation $$\mathrm{O}(n,n,F) \lt \mathrm{GL}(2n,F)$$ in this case the dimension being the same.

My proof is simple enough – request this if needed.

My questions:

  1. Is my observation true?
  2. Is this well-known, and in what form would it normally be stated in texts (e.g. as a more general statement)?

EDIT: This is shown by considering the matrix embedding $$\mathrm{M}_n(F)\to\mathrm{M}_{2n}(F):T\mapsto\left[\begin{matrix}T & 0 \\ 0 & (T^{-1})^\mathrm{t}\end{matrix}\right]$$ restricted obviously to invertible $T$, with ${}^\mathrm{t}$ denoting the transpose.

qman
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    For the casual readers, here's a reference on the split orthogonal group. – Mark Schultz-Wu Mar 12 '17 at 05:20
  • Yes, this is true. I'm not sure what else you're looking for. – Qiaochu Yuan Mar 12 '17 at 05:52
  • I guess I was surprised that I could not google anything about it. Hence my second question: perhaps it was something more general. But mainly I just needed confirmation due to lack of finding it online. – qman Mar 12 '17 at 06:01
  • I'm curious what the embedding is. – anon Mar 12 '17 at 15:13
  • There are all sorts of facts a dedicated student can prove for themselves that aren't mentioned in textbooks for a variety of reasons, the basic ones being 1) it is not useful for anything and 2) it does not even make a good exercise. This is a sign of progress, really. – Qiaochu Yuan Mar 12 '17 at 23:28
  • @arctictern, the embedding is particularly straightforward. I'll edit it into the question. – qman Mar 13 '17 at 01:58
  • @QiaochuYuan: I'm not sure what you are referring to by "it is not useful for anything" – if you are referring to facts that have no identified use, that is the way to phase it. I expect that this would be a good example of something to pose as an exercise in a textbook, and indeed, I have a application in mind where it would most definitely be useful. – qman Mar 13 '17 at 01:58
  • For completeness, the case where 2 is not a unity in $F$ need clarification: Defining the orthogonal group via preservation of a bilinear form and a quadratic form do not seem to be equivalent. – qman Mar 13 '17 at 05:02
  • This is related to my answer to Geometric product between matrices. – mr_e_man Nov 26 '20 at 21:48

1 Answers1

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Here is some context in which to place these observations. To my mind, the most "conceptual" maps between groups are those which arise from restricting functors to automorphisms. A good example is the pair of embeddings

$$GL_n(\mathbb{C}) \to GL_{2n}(\mathbb{R})$$ and $$GL_n(\mathbb{R}) \to GL_n(\mathbb{C})$$

which are somewhat analogous to the pair of embeddings you describe. These correspond to the restriction of scalars / underlying real vector space functor from complex to real vector spaces, resp. the complexification functor $V \mapsto V \otimes_{\mathbb{R}} \mathbb{C}$ from real to complex vector spaces. The significance of looking specifically at the induced maps on automorphisms is that they are related to e.g. the corresponding underlying real vs. complexification functors for real and complex vector bundles.

The analogous statements for the split orthogonal group are the following. Let's define a quadratic space to be a vector space equipped with a quadratic form. There is a natural forgetful functor from quadratic spaces to vector spaces which is responsible for the "obvious" embeddings of various orthogonal groups into various general linear groups, including for the split orthogonal groups, which correspond to split quadratic spaces.

On the other hand, if $V$ is a vector space, then from it we can construct a quadratic space with underlying vector space $V \oplus V^{\ast}$ and quadratic form

$$q(v \oplus f) = f(v).$$

This quadratic space is split, and this construction is functorial with respect to automorphisms (but, unlike our previous example, not with respect to arbitrary maps). This functor corresponds to your embedding of the general linear group into a split orthogonal group.

Qiaochu Yuan
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  • It is interesting that you have been able to infer so much context from the question, suggesting a natural fit. I essentially was looking for the (ideally, smallest) orthogonal group containing $\mathrm{GL}(n,F)$. I had noticed $V \oplus V^{\ast}$ as a natural occurrence of the split quadratic form as opposed to imposing it on $V \oplus V$. Your confirmation is a boost to my amateur investigation. – qman Mar 13 '17 at 03:37