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

I am attempting to get an understanding on what a depiction is, and also a teacher offered me a straightforward instance of standing for the team $Z_{12}$ as the twelve origins of unity, or equivalent $2\times 2$ matrices. Currently I am asking yourself just how $\operatorname{GL}(1,\mathbb{C})$ and also $\operatorname{GL}(2,\mathbb{R})$ relate, given that the components of both teams are automorphisms of the intricate numbers. $\operatorname{GL}(\mathbb{C})$, the team of automorphisms of C, is (to my understanding) isomorphic to both $\operatorname{GL}(1,\mathbb{C})$ and also $\operatorname{GL}(2,\mathbb{R})$ given that the intricate numbers are a 2 - dimensional vector room over $\mathbb{R}$. Yet it does not feel like these 2 teams are isomorphic per various other.

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2019-12-02 02:50:11
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Answers: 3

I would certainly such as to supplement the above solutions with 2 vital instances. Take into consideration a hyperbolic makeover $H(x, y) = (2x, y/2)$ and also a shear $S(x,y) = (x+y, y)$, both components of $GL(2,R)$. If you attract just how these act upon factors of $R^2$ you will certainly see that these misshape angles. On the various other hand, components of $GL(1,C)$ constantly maintain angles (and also hence so do holomorphic maps, far from crucial points).

That is, the added algebraic framework maintained by $GL(1,C)$ converts to the geometric building of maintaining angles and also "handedness".

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2019-12-03 05:33:36
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$GL(1,\mathbb{C})$ can be taken the team of automorphisms of $\mathbb{C}$ as a facility vector room. Each such automorphism is offered by reproduction by a nonzero facility number. On the various other hand, $GL(2,\mathbb{R})$ can be taken the team of automorphisms of $\mathbb{C}$ as a actual vector room, and also therefore it is bigger. In regards to matrices, $GL(1,\mathbb{C})$ represents all $2$ - by - $2$ actual matrices of the kind $\begin{bmatrix} a & -b \\ b & a \end{bmatrix}$ with $a^2+b^2\gt0$, which is a correct subgroup of the team $GL(2,\mathbb{R})$ of all $2$ - by - $2$ actual invertible matrices.

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2019-12-03 04:23:47
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No, they are not isomorphic. $GL(1,\mathbb{C})$ infuses right into $GL(2, \mathbb{R})$ yet the last contains the set of actual automorphisms of $\mathbb{R}^2$, while the previous contains the set of all intricate automorphisms of $\mathbb{C}$. As an example, a component of $GL(1, \mathbb{C})$ is the compound of a scaling (extension) by a favorable actual number and also a turning, so is $\mathbb{R}_{>0} \times SO(2) = \mathbb{R}_{>0} \times S^1$. Nonetheless, $GL(2, \mathbb{R})$ consists of extra difficult points (as an example representations, points which also if scaled do not come to be orthogonal makeovers).

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2019-12-03 04:20:29
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