Why does the use of infinitesimals in deriving the Lie Algebra of some Lie group work? I'd like to know why the following "dirty" derivation of the lie algebra of a lie group works. Take for example the group $SU(2) = \{U \in \mathbb{C_{2\times2}} | UU^\dagger = 1, \text{det}(U) = 1\}$
The arguement goes something like the following, we consider $U = 1 + \epsilon H$ where $\epsilon$ is some "small" or infinitesimal number and $H$ is some 2x2 complex matrix. Then we simple plug in this expression for $U$ into the constraint for $SU(2)$ and we get
$$
(1 + i\epsilon H)(1 + i\epsilon H)^\dagger = 1 \\
$$
to first order this becomes $1 + i\epsilon H - i \epsilon H^\dagger = 1 \implies H = H^\dagger$
There are a world of problems with this derivation from a rigorous standpoint, for example addition is not defined within the group so its not even clear what $1 + i\epsilon H$ even means (or why it should preserve the group structure). Moreover the lie algebra is constructed in a very geometric manner which has to do with left invariant vector fields on the group manifold and so on... and yet I see this kind of derivation where we "differentiate" or "linear approximate" some expression like $U + \epsilon H$ work all the time.
We can think of the tangent space at a point intuitively as the space of linear approximations at that point, but this does not justify why this procedure works so well. What's going on here?
 A: An equivalent definition of the Lie algebra of a Lie group $G$ is the tangent space to the identity element of $G$. For matrix Lie groups, this would be the tangent space to the identity matrix $1$. Elements of $T_1G$ are tangent vectors $v$ to curves $\gamma(t)$ in $G$ such that $\gamma(0) = 1$ and $\dot\gamma(0) = v$. Matrix groups naturally embed into Euclidean space (some $\mathbb R^N$ or $\mathbb C^{N'}$), and it's a theorem of smooth manifold theory that any choice of embedding gives the same tangent space. By embedding $G$ into some Euclidean space, for any vector $v$ in that Euclidean space, there is a curve we can write of the form $\gamma(t) = 1 + tv$, using the vector space structure of the Euclidean space.
However, the curve as written may not lie tangent to the matrix group $G$—it's just jutting out into the large ambient Euclidean space we are embedded in. So to ensure that the curve we wrote will give us a tangent to $G$ at $1$, we ask that $\gamma(t)$ satisfy some constraint so that it lies in $G$ for all $t$ sufficiently small, at least to first-order, since the first-order information of $\gamma(t)$ at $t=0$ is all that we are trying to capture in the notion of "tangent vector." In the example you gave, we should ask that $\gamma(t)\gamma(t)^\dagger = 1$ to first-order at $t = 0$.
