# How is $\operatorname{GL}(1,\mathbb{C})$ related to $\operatorname{GL}(2,\mathbb{R})$?

I am trying to get a grasp on what a representation is, and a professor gave me a simple example of representing the group $Z_{12}$ as the twelve roots of unity, or corresponding $2\times 2$ matrices. Now I am wondering how $\operatorname{GL}(1,\mathbb{C})$ and $\operatorname{GL}(2,\mathbb{R})$ are related, since the elements of both groups are automorphisms of the complex numbers. $\operatorname{GL}(\mathbb{C})$, the group of automorphisms of C, is (to my understanding) isomorphic to both $\operatorname{GL}(1,\mathbb{C})$ and $\operatorname{GL}(2,\mathbb{R})$ since the complex numbers are a two-dimensional vector space over $\mathbb{R}$. But it doesn't seem like these two groups are isomorphic to each other.

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An easy way to see that these are distinct is to note that they have different dimensions: $\dim_\mathbb{C}(\mathfrak{gl}_1(\mathbb{C}))=1$, while $\dim_\mathbb{R}(\mathfrak{gl}_2(\mathbb{R}))=4$. –  Aaron Mazel-Gee Nov 30 '10 at 6:44
@Aaron: if the OP knew how to compute dimensions of Lie algebras, presumably he/she wouldn't be asking this question. –  Qiaochu Yuan Nov 30 '10 at 22:48
Okay. Sorry if that was rude or anything. I know very little about representation theory, and I definitely didn't know that people study representations of Lie groups totally independently of Lie algebras. –  Aaron Mazel-Gee Dec 1 '10 at 7:26

No, they're not isomorphic. $GL(1,\mathbb{C})$ injects into $GL(2, \mathbb{R})$ but the latter consists of the set of real automorphisms of $\mathbb{R}^2$, while the former consists of the set of all complex automorphisms of $\mathbb{C}$. For instance, an element of $GL(1, \mathbb{C})$ is the composite of a scaling (dilation) by a positive real number and a rotation, so is $\mathbb{R}_{>0} \times SO(2) = \mathbb{R}_{>0} \times S^1$. However, $GL(2, \mathbb{R})$ includes more complicated things (e.g. reflections, things which even if scaled do not become orthogonal transformations).

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That makes sense, and that's why I'm having trouble. They are not isomorphic, but consider the following Wikipedia entry, If V is a vector space over the field F, then we write GL(V) or Aut(V) for the group of all automorphisms of V, i.e. the set of all bijective linear transformations V → V, together with functional composition as group operation. If the dimension of V is n, then GL(V) and GL(n, F) are isomorphic. We can look at C as a 1-dimensional vector space over itself, or as a 2-dimensional vector space over the reals. Shouldn't this imply the two are isomorphic? What am I missing? –  Sam Nov 30 '10 at 4:34
@Sam: Implicit in the definition of a vector space is a fixed base field. When you say "the group of all automorphisms of $V$", the base field is already chosen. The vector space $\mathbb{C}$ over $\mathbb{C}$ and the vector space $\mathbb{C}$ over $\mathbb{R}$ are different vector spaces, even though they have the same underlying set and additive structure, so you should not expect their automorphism groups to be isomorphic. (And the answer above gives you detail on why they are not.) –  Jonas Meyer Nov 30 '10 at 4:42
To elaborate, the point is that being a $\mathbb{C}$-automorphism is a stronger condition than being an $\mathbb{R}$-automorphism, because the $\mathbb{C}$-structure is richer than the $\mathbb{R}$-structure, even though the underlying sets are the same. –  Akhil Mathew Nov 30 '10 at 4:45
Thank you. It's much more clear now, although some of these concepts are still hard to grasp. That answers my question. –  Sam Nov 30 '10 at 4:48
...where by $\mathbb{R}^*$ you mean $\mathbb{R}^{> 0}$, of course. –  Pete L. Clark Nov 30 '10 at 6:48

The one group is commutative, the other is not. So they can not be isomorphic.

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$GL(1,\mathbb{C})$ can be thought of as the group of automorphisms of $\mathbb{C}$ as a complex vector space. Each such automorphism is given by multiplication by a nonzero complex number. On the other hand, $GL(2,\mathbb{R})$ can be thought of as the group of automorphisms of $\mathbb{C}$ as a real vector space, and hence it is larger. In terms of matrices, $GL(1,\mathbb{C})$ corresponds to all $2$-by-$2$ real matrices of the form $\begin{bmatrix} a & -b \\ b & a \end{bmatrix}$ with $a^2+b^2\gt0$, which is a proper subgroup of the group $GL(2,\mathbb{R})$ of all $2$-by-$2$ real invertible matrices.

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That makes sense. I guess I'm still having trouble understanding how the group of automorphisms is different based on how you view the vector space. –  Sam Nov 30 '10 at 4:41
@Sam: I gave one reply on Akhil's answer. A rough way to think of it is that being able to pull out complex scalars is stronger than being able to pull out real scalars, since the latter can be thought of as a special case of the former. –  Jonas Meyer Nov 30 '10 at 4:46
@Sam: By the way, Akhil's answer gives a good conceptual way to see that the groups are not isomorphic, but you could also see this with some examples. E.g., complex conjugation gives a real automorphism that is not a complex automorphism, and shows that there are more order 2 real automorphisms than complex automorphisms. –  Jonas Meyer Nov 30 '10 at 6:03
@Sam: the same mathematical object often has several different automorphism groups, depending upon what kind of automorphisms are being considered. For instance the automorphism group of $\mathbb{R}$ as a metric space (i.e., its isometry group) is much smaller than its automorphism group as a topological space, which is (very!) much smaller than its automorphism group as a set (i.e., the bijections). Something similar is happening here: a complex vector space has "more structure" than its underlying real vector space, hence fewer automorphisms. –  Pete L. Clark Jun 7 '11 at 16:55

I'd like to supplement the above answers with two important examples. Consider a hyperbolic transformation $H(x, y) = (2x, y/2)$ and a shear $S(x,y) = (x+y, y)$, both elements of $GL(2,R)$. If you draw how these act on points of $R^2$ you will see that these distort angles. On the other hand, elements of $GL(1,C)$ always preserve angles (and thus so do holomorphic maps, away from critical points).

That is, the extra algebraic structure preserved by $GL(1,C)$ translates to the geometric property of preserving angles and "handedness".

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