# Which mathematical theory investigates distortions?

For transformations like rotations, translations or boosts Lie theory is the appropriate theory. Which theory talks in similar, systematic terms about distortions

$$M \vec{v} = m \vec{v} \quad \forall \ \vec{v} \quad ?$$

In Lie theory we can learn a lot about certain transformations/symmetries by looking at the corresponding Lie algebra and deriving representations of the transformations in question.

I'm searching for a similar treatment of distortions. Concretely this means looking at infinitesimal elements, deriving the corresponding algebra that encodes the information about the transformation and deriving systematically representations.

I wasn't able to find some "distortion group" and therefore I concluded there must be a good reason for this. Maybe we need here another framework. If it can be done using Lie theory, the correct name of the "distortion group" would help me immensely

• Doesn't Lie theory still encompass that? – Robin Goodfellow Apr 3 '15 at 9:08
• Is $m$ just a scalar? Then you're just doing linear algebra. – Qiaochu Yuan Apr 3 '15 at 9:09
• @QiaochuYuan yes $m$ is just a scalar. Of course things boil down to linear algebra, just as in Lie theory. Nevertheless, Lie theory isn't just linear algebra – Tim Apr 3 '15 at 9:26
• I mean, what do you want to know about scalar multiplication? It would help if your question were more specific. – Qiaochu Yuan Apr 3 '15 at 9:27
• @RobinGoodfellow I searched for quite a while for some treatment of distortions using Lie theory, but couldn't find something. If it can be done, the name of the group would be immensly useful – Tim Apr 3 '15 at 9:28

If I understand your question correctly, you want to understand the structure, Lie algebra, and representation theory of the Lie group $\mathbb{R}^{\times}$ of invertible scalars. Another name for this Lie group is the general linear group $\text{GL}_1(\mathbb{R})$. This is a very easy Lie group to understand, since it's $1$-dimensional and therefore abelian. It has two connected components, the positive reals and the negative reals, and the positive reals are isomorphic to $\mathbb{R}$ (which is its Lie algebra) via the exponential map
$$\mathbb{R} \ni t \mapsto e^t \in \mathbb{R}^{\times}_{>0}.$$
Because this group is $1$-dimensional and abelian, its Lie algebra is also $1$-dimensional and abelian, and so there are no interesting Lie brackets.
A finite-dimensional complex representation of the Lie algebra $\mathbb{R}$ is given by a choice of square complex matrix. The classification of representations is the classification of complex matrices up to conjugation, which is described by Jordan normal form. The classification of real representations is a bit more complicated but can be deduced from the structure theorem for finitely generated modules over a PID applied to $\mathbb{R}[x]$ (which should be interpreted in this context as the universal enveloping algebra of the Lie algebra $\mathbb{R}$).
There is absolutely no need to go beyond standard linear algebra and Lie theory here. Replacing $\mathbb{R}$ with $\mathbb{C}$ is also not an issue.