Consider the unitary group $\mathcal U(n)$ of $n \times n$ unitary matrices, generated by the Lie algebra $\mathfrak u(n)=span(iH_1, … iH_{n^2-1})$, acting on the space $\mathbb C(n)$.
By considering only the smaller group $A = e^{\mathfrak{a}}, \mathfrak{a} = span(iH_1, …, iH_{m^2-1}), m<n$, how can I find a mapping of group element from $\mathcal U(n)$ to $A$ and mapping of states from $\mathbb C(n)$ to its corresponding subspace $\mathbb C(m)$?
There seems to be a bit of confusion here. Firstly, $\mathfrak{u}_n$ is not the span of $n$ matrices. It is $n^2 - 1$ dimensional (as a real vector space). Presumably you are then asking about taking a subalgebra isomorphic to $\mathfrak{u}_m$, $m<n$. Especially the way you are saying that, it seems like you are implicitly choosing a subspace $\mathbb{C}^m \leq \mathbb{C}^n$ for this to act on. Since we have an inner product kicking around we can identify an ortho-complement $\mathbb{C}^{n-m}\leq \mathbb{C}^n$ while we are at it so that $\mathbb{C}^m \oplus \mathbb{C}^{n-m} = \mathbb{C}^n$. But then we have a natural map from $\mathbb{C}^n$ to $\mathbb{C}^m$ given by (ortho-)projection.
Trying to map from the big Lie group onto this smaller subgroup however is not so friendly. Thinking just on the Lie algebra level, we could take $\mathfrak{u}_n = \mathfrak{u}_m \oplus \mathfrak{u}_m^\perp$ (using the Killing form) and define a projection that way, but this is only a direct sum of vector spaces and so does not define a splitting of $\operatorname{U}(n)$ as a product of groups.
Note this last is a general rule, a subalgebra $\mathfrak{h}$ of a Lie algebra $\mathfrak{g}$ may or may not have a natural complement $\mathfrak{m}$ in the Lie algebra (i.e. $\mathfrak{g} = \mathfrak{h} \oplus \mathfrak{m}$ as vector spaces) but $\mathfrak{m}$ will, in general, fail to be a subalgebra (i.e. $[\mathfrak{m},\mathfrak{m}] \not\subset \mathfrak{m}$) and is even more rarely a direct sum decomposition (which requires $[\mathfrak{h},\mathfrak{m}] = 0$) so there isn't in general a friendly decomposition in the group that corresponds to this.