Integrating Lie group infinitesimal generators

Question

My professor used this definition for the Lie group generators:

$$X_1=\lim_{\varepsilon\to 0}\frac{g(\varepsilon,0,0,0,...,0)-g(0,0,0,0,...,0)}{\varepsilon}\space;\space g\in G$$ (The first one, just for the example)

Which looks like the derivative of the group around the Identity element in a specific "direction". [And really, when representing group elements using matrices, we take their derivative to find the generators].

My question: Why do we use generators using the exponent ($e^{\alpha_1X_1}$) and not using integrals, which make more sense to me because they add up all the infinitesimal contributions (the generators)? [Like we do for differential forms for example]

This can also be asked the other way around - can we use exponents to add up differential forms instead of integrating them?

(edit) My guess After thinking about this a bit more I think, like in a Taylor series, we try to span the whole group only using knowledge about the derivative at a specific local point of the group (because the generator is actually the derivative at the Identity element "point"). This raises two more questions for me:

1. Can "ordinary" function $f(x)$ also be recovered using only the first approximation $f'(0)$ and exponentiation? [i.e $f(x)=e^{xf'(0)}$]
2. Is there a more "general" derivative of the group that we can also integrate, and is not "local" [in analog to ordinary functions - is there $f'(x)$ for any x?]

Answer 1

Part of the problem here is what does integrating mean in this sense? There are various notions floating around here but we are no longer integrating a simple real valued function or even integrating a path in a vector space so we have to be a bit more careful. Not all of our ideas for what integration means will be the same thing anymore.

In this case you are integrating a vector $X$ in the tangent space $T_{e}G$ at the identity into an element of the manifold $g\in G$ (in some neighbourhood of the identity). How does that make sense? Well the natural thing to try is to find a curve through the identity whose derivative is $X$. In this way we can find some sort of inverse to the "differentiation" we have just done which took a curve in the group and gave us a vector.

So how to find a curve $\gamma:\mathbb{R}\to G$ which goes through the identity ($\gamma(0) = e$) and whose derivative there is $X$ ($\gamma'(0) = X$). Well it turns out that $\gamma(t) := \exp(tX)$ is exactly the right choice of curve. Note of course that there are many other possibilities but this gives us a natural and canonical choice. These curves are also known as $1$-parameter subgroups. If we want a specific point rather than a curve we need only ask for a specific value of $t$, say $\gamma(1)= \exp(X)$.

More generally, given a vector field on a manifold it is possible to integrate this into a curve on the manifold. Indeed a family of curves since we must pick a starting point (initial condition). These are called integral curves. Note in the Lie group case the $1$-parameter subgroup is exactly the integral curve (passing through $e$) of the left-invariant vector field generated by $X$.

Differential forms are another beast again. We have to decide what we want integration to be. Does it mean the opposite of the exterior derivative (in which case there will not be an answer unless the form is exact)? Or does it mean taking a $k$-form over a $k$-dimensional manifold as an infinitesimal "$k$-density" and integrating that over the manifold.

As to differentiating elsewhere on $G$ that is also allowed. However if we differentiate a curve $\gamma$ on $G$ the result will be a vector field which is not the same thing as a curve in $\mathfrak{g}$ so we must be careful. We can try to rectify this by considering $\gamma(t)^{-1}\gamma'(t)$ which is indeed a curve in $\mathfrak{g}$ but we need to then understand what this object actually is.