Differentiation in group space In a few physics papers (lattice gauge theory papers, to be more specific) I've seen the following definition for differentiation on group space
$$
\frac{\partial}{\partial U} f(U) = \frac{\partial}{\partial \alpha_a} f(e^{i \alpha_a T^a} U)\,,
$$
where $U \in SU(N)$ and the $T^a$ are the generators of the group. My questions are: why is it defined in such a way? Mathematically speaking, how can it be proved?
Thanks in advance.
 A: This has to do with the fact that Lie groups are always parallelizable (they have trivial tangent bundle). Paraphrasing Wikipedia, the differential of the left translation $h\mapsto gh$ can "move all tangent vectors around in a nice way".
What does it means in terms of derivatives?
Let $t\mapsto g(t)$ be a 1-parameter subgroup of $G$. So, $g(0)=I$ (identity).
Now, let $u\in G$. All points in a neighborhood of $u$ can be written as $g(t)u$, for a suitable 1-parameter subgroup $g$, and for $t$ "near enough" to $0$. And obviously, $g(0)u=u$. So:
$$
u+\Delta u = g(\Delta t)u.
$$
Since $g(\Delta t)$ is in a neighborhood of the identity, for $\Delta t$ small enough (but still finite) it can be expressed as the exponential of a Lie algebra element (element of the tangent space at the identity): 
$$
g(\Delta t) = \exp(i\Delta t\,\beta_kT_k).
$$
$i\beta_kT_k$ generates a 1-dimensional subspace of the Lie algebra. $i\Delta t \beta_kT_k$ identifies a point on that line which is near the origin.
$i\Delta t \beta_kT_k$ corresponds to the quantity $i\alpha_aT_a$ in your question.
Now if $f$ is a function on $G$:
$$
\frac{df}{du} = \lim_{\Delta u\to 0} \frac{f(u+\Delta u)-f(u)}{\Delta u}.
$$
By the chain rule:
$$
\frac{df(u(t))}{dt} = \lim_{\Delta t\to 0} \frac{f(g(\Delta t)u)-f(u)}{\Delta t} = \frac{d}{dt} f(g(t)u).
$$
At $u$, so in $t=0$:
$$
\frac{d}{dt} f(g(t)u)\bigg|_{t=0} = \frac{d}{dt} f(e^{i\Delta t\,\beta_kT_k} u)\bigg|_{t=0} .
$$
This is a directional derivative along the vector $\beta_k T_k$. The partial derivatives are then:
$$
\frac{\partial}{\partial \alpha_k} f(g(t)u)\bigg|_{t=0} = \frac{\partial}{\partial \alpha_k} f(e^{i\alpha_kT_k} u)\bigg|_{t=0} .
$$
You are not strictly deriving by the elements of the group, but rather by the parameters that generate it in the Lie algebra. This nevertheless motivates the notation. Intuitively, "varying $u$" is equivalent to "multiplying $u$ by an element near the identity", which in turn means "multiplying $u$ by the exponential of a Lie algebra element".
I think on Y. Makeenko, "Methods on Contemporary Gauge Theory" this is explained. But I am not 100% sure.
