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As far as I understand, one has (at least?) two choices to introduce infinite matrix groups:

Either, one can say they are all subgroups of the general linear group over the complex numbers numbers $\mathbf{GL}(n,\mathbb{C})$. Then, $\mathbf{GL}(n,\mathbb{R})$ is a subgroup of $\mathbf{GL}(n,\mathbb{C})$.

Or one can say that a infinite matrix group is a subgroup of the general linear group $\mathbf{GL}(n,\mathbb{R})$. In this case we can say that $\mathbf{GL}(m,\mathbb{C})\subset \mathbf{GL}(2m,\mathbb{R})$, while a single complex number is represented using a $2\times 2$ matrix.

Is there any particular (e.g. technical) reason to prefer one over the other?


edit:

These following comments from Gallier ("Geometric Methods and Applications", second edition, 2011, p.475) made me wonder:

"As in the real case, the groups $\mathbf{GL}(n, \mathbb{C})$, $SL(n, \mathbb{C})$, $\mathbf{U}(n)$, and $\mathbf{SU}(n)$ are also topological groups (viewed as subspaces of $\mathbb{R}^{2n^2}$ ), and in fact, smooth real manifolds. Such objects are called (real) Lie groups. [...]

It is also possible to define complex Lie groups, which means that they are topological groups and smooth complex manifolds. It turns out that $\mathbf{GL}(n, \mathbb{C})$ and $\mathbf{SL}(n, \mathbb{C})$ are complex manifolds, but not $\mathbf{U}(n)$ and $\mathbf{SU}(n)$. One should be very careful to observe that even though the Lie algebras ${\frak sl}(n, \mathbb{C})$, ${\frak u}(n)$, and ${\frak su}(n)$ consist of matrices with complex coefficients, we view them as real vector spaces. The Lie algebra ${\frak sl}(n, \mathbb{C})$ is also a complex vector space, but ${\frak u}(n)$ and ${\frak su}(n)$ are not! Indeed, if A is a skew-Hermitian matrix, iA is not skew-Hermitian, but Hermitian!"

So, all infinite matrix groups are real manifolds, but only some of them are complex manifolds. So, isn't it conceptually "nicer" to say that $\mathbf{GL}(m,\mathbb{C})\subset \mathbf{GL}(2m,\mathbb{R})$?

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    You showed the definitions are equivalent. What reasons do you imagine would then lead to prefer one over the other?2012-02-14
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    Didactic reasons for instance. But after your comment, I guess it is best to mention that both are equivalent. Thanks!2012-02-14
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    Perhaps not technical, but I much prefer $GL(n,\mathbb{R})\leq GL(n,\mathbb{C})$ simply because $\mathbb{C}$ is an extension of $\mathbb{R}$2012-02-14
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    An $n \times n$ matrix over $\mathbb C$ can be represented by $2n$ real numbers, whereas a $2n \times 2n$ matrix over $\mathbb R$ requires $4n$ so, as a computationally inclined person, I prefer ${\rm GL}(n,{\mathbb C})$.2012-02-14
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    Am I being dense or are we ignoring the set of all matrices over $\mathbb{R}$ as being a group under addition? This to me should be included in the definition of "infinite matrix group", after all it IS one.2012-08-26
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    @fretty: that's included in the present definition: you can embed $\mathbb{R}^n$ as an additive group into $GL(n+1,\mathbb{R})$ [see here](http://math.stackexchange.com/questions/178703/can-bbbr2-be-given-the-structure-of-a-matrix-lie-group/178708#178708); and $M_n(\mathbb{R})$ is just $\mathbb{R}^{n^2}$.2012-08-26
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    Ah I didnt think that through lol.2012-08-26
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    Since $U(n)$ and $SU(n)$ are *defined* using complex matrices (as group of unitary $n\times n$ matrices), it's IMHO the natural thing to consider it as subgroup of $GL(n,\mathbb C)$. Also, if you want to ever introduce tensor products of them (such as $SU(2)\otimes SU(2)$), which are important e.g. in quantum information, you should be damned sure that you get the correct group when working in a $GL(2n,\mathbb R)$ representation (the tensor products are defined for the standard complex matrix representation, and trying to use other representations might give different groups).2012-08-26
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    Note that a very trivial example where you get the wrong tensor product is $U(1)\otimes U(1)$. Since unitary $1\times 1$ matrices are just complex numbers of the form $\mathrm e^{\mathrm i\phi}$ and the tensor product for them reduces to the normal product on complex numbers, it is obvious that $U(1)\otimes U(1)=U(1)$. On the other hand, the $\mathbb R^{2\times 2}$ representation of $U(1)$ is the standard representation of $SO(2)$, and it is easy to see with this representation that $SO(2)\otimes SO(2)\ne SO(2)$. Especially $I\otimes A\ne A\otimes I$ because $\dim A>1$.2012-08-26

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