Reasoning about Lie theory and the Exponential Map

Question

I'm having a little difficulty wrapping my head around Lie theory (I'm a computer scientist, so perhaps that's to be expected).

Specifically, considering the following definition from Wikipedia for the exponential map. What is the significance of the identity element? I think that this could be to do with the derivative of exp being equal to the function itself, but I'm not quite sure if I'm on the right track here. Any nudges in the right direction would be highly appreciated.

"Let G be a Lie group and $\mathfrak{g}$ be its Lie algebra (thought of as the tangent space to the identity element of G). The exponential map is a map $\exp\colon \mathfrak g \to G$ which can be defined in several different ways."

Answer 1

I am novice in Lie theory myself, but I'll try to answer your question about the significance of identity element. There is, in fact, no significance. The identity element is as good as any other element in the connected component of Lie group, and the tangent to the identity element can be moved anywhere, as long as we stay in the connected component. However, it is easier to perform calculations involving the identity element than those that involve other elements. This is why the identity element is used.

Answer 2

The elements of the Lie algebra can be thought of as "infinitesimal elements," so it's natural to think of them as being infinitesimally close to the identity; if we denote by $\epsilon$ an infinitesimal that squares to zero, then you can informally think of the Lie algebra as the set of elements of the form $I + \epsilon X$ where $I$ is the identity. For example, the elements of the orthogonal group satisfy $A \cdot A^T = I$, so the infinitesimal elements of the orthogonal group satisfy $(I + \epsilon X)(I + \epsilon X^T) = I + \epsilon (X + X^T) = I$, or $X + X^T = 0$. Hence the Lie algebra of the orthogonal group is precisely the space of skew-symmetric matrices. (This argument can be made completely formal if we work with algebraic groups instead of Lie groups.)

There is a precise sense in which the Lie algebra consists of "infinitesimal elements": whenever a Lie group $G$ acts on a manifold $M$ by symmetries, the Lie algebra $\mathfrak{g}$ acts by differential operators on the smooth functions $M \to \mathbb{R}$. So they "infinitesimally generate" symmetries of $M$ (and the precise sense in which this is true is the exponential map).

If you prefer, there is an equivalent definition which does not privilege the identity element: the elements of the Lie algebra are the left-invariant vector fields on the Lie group. (Since a Lie group acts transitively on itself, a left-invariant vector field is determined by its value at any group element, and again, it's natural to look at the identity element.)

Answer 3

The answers given are both really good; I just want to add an additional reason why the identity element is preferred in consideration of the exponential map. The reason is because the exponential map gives a one to one correspondence between tangent vectors at the identity and one parameter subgroups of $G$. A one parameter subgroup of $G$ is a homomorphism $\mathbb R \to G$, or equivalently a curve $\gamma(t)\in G$ such that $\gamma(t+s) = \gamma(t)\gamma(s)$. Since $\gamma$ is a group homomorphism, we need $\gamma(0)$ to be the identity $e\in G$ (so here we are using the special property of $e$) and any such one parameter subgroup is determined by $\gamma'(0) \in T_e G$. Conversely, given $X \in T_e G$, the curve $t\mapsto \exp(tX)$ is a one-parameter subgroup with derivative $X$ at time 0.