Good questions — there are two separate group/algebra pairs in play, and the confusion comes from mixing them. Let me lay out the full cast of characters precisely, then answer the “why doesn’t exponentiation just hand me the determinant” question (spoiler: it does, almost).

The cast: two groups, two Lie algebras, two maps connecting them¶

	source side	target side
group	$G = GL(n,\mathbb C)$ : invertible $n\times n$ matrices, under matrix multiplication	$\mathbb C^* = GL(1,\mathbb C)$ : nonzero numbers, under multiplication
Lie algebra	$\mathfrak{gl}(n,\mathbb C) =$ all $n\times n$ matrices (not just invertible!), bracket $[X,Y]=XY-YX$	$\mathbb C$ : all complex numbers, bracket $\equiv 0$ (abelian)
exp map	matrix exponential $X\mapsto e^X$	ordinary $z\mapsto e^z$
basis of the algebra	the $n^2$ matrix units $E_{ij}$	the number 1

Two clarifications of phrases in your question:

“The determinants form a group $GL(n,\mathbb C)\to\mathbb C^*$ ” — not quite: $\det$ is not a group, it is a map between two groups (a homomorphism). The values of $\det$ live in the group $\mathbb C^*$ . “One-dimensional representation” means exactly this: each group element $A$ is represented by a $1\times 1$ invertible matrix, i.e. a nonzero number $f(A)$ , such that composition goes to multiplication, $f(AB)=f(A)f(B)$ .
“The $E_{ij}$ have a one-dimensional representation $a\operatorname{tr}X + b\,\overline{\operatorname{tr}X}$ ” — the $E_{ij}$ don’t have a representation; they are elements (a basis) of the algebra $\mathfrak{gl}(n,\mathbb C)$ . The expression $\phi(X) = a\operatorname{tr}X+b\,\overline{\operatorname{tr}X}$ is the representation map of the whole algebra, the algebra-level shadow of $f$ . Evaluated on the basis it gives numbers: $\phi(E_{ij})=0$ for $i\neq j$ (forced: those are commutators), and the diagonal direction carries the one free “charge” $a$ (plus $b$ for the anti-holomorphic part). That’s the entire representation theory at the algebra level: one number.

So the complete picture is a commuting square — the same map at two levels, with the exponentials as the vertical legs:

\begin{array}{ccc} \mathfrak{gl}(n,\mathbb C) & \xrightarrow{\ \ \phi\,=\,\operatorname{tr}\ \ } & \mathbb C \\[2pt] \big\downarrow{\scriptstyle \exp} & & \big\downarrow{\scriptstyle e^{(\cdot)}} \\[2pt] GL(n,\mathbb C) & \xrightarrow{\ \ f\,=\,\det\ \ } & \mathbb C^* \end{array} \qquad\qquad f(e^X) = e^{\phi(X)}.

(1)

Generators map to generators ( $\operatorname{tr}$ , the top arrow), group elements map to group elements ( $\det$ , the bottom arrow), and exponentiation commutes with everything. The exact analogue you already know is $U(1)$ : group $\{e^{i\theta}\}$ , algebra $i\mathbb R$ , representations $e^{i\theta}\mapsto e^{ik\theta}$ with integer charge $k$ — the algebra-level rep is “multiply by $k$ ”, the group-level rep is its exponentiation. Here $G = GL(n,\mathbb C)$ , the algebra-level rep is “take $a\cdot\operatorname{tr}$ ”, and $\det$ is the charge-1 representation.

“Can’t I just exponentiate and get the determinant directly?”¶

Yes — that is literally what the determinant is. What you exponentiate is not a single generator, but the algebra representation $\phi=\operatorname{tr}$ , carried through both vertical legs of the square:

\boxed{\ \det A = e^{\operatorname{tr}\log A}\ }

(2)

i.e.: take any $X$ with $e^X = A$ (a logarithm of $A$ ), trace it, exponentiate the number. Check it on examples you know:

$A = \operatorname{diag}(d_1,d_2)$ : $X = \operatorname{diag}(\log d_1,\log d_2)$ , $\operatorname{tr}X = \log d_1+\log d_2$ , $e^{\operatorname{tr}X} = d_1 d_2$ . ✓
A rotation $R(\theta)$ : $X = \theta(E_{21}-E_{12})$ , $\operatorname{tr}X = 0$ , so $\det R = e^0 = 1$ . Rotations preserve volume because their generator is traceless. Same for boosts/transvections — that’s Liouville again.

So why did the notes need any further work? Exactly two fine points separate the boxed formula from a finished theory, and they are where all the actual content lives:

Fine point 1: $\log A$ is multivalued — well-definedness is a real condition. The vertical legs are not injective: many $X$ give the same $A$ (already in the target: $e^z = e^{z+2\pi i}$ ). The recipe $e^{\operatorname{tr}X}$ defines a function of $A$ only if changing the choice of $X$ doesn’t change the answer. The ambiguity in $X$ shifts $\operatorname{tr}X$ by $2\pi i\,\mathbb Z$ , and $e^{2\pi i m}=1$ — so for $\phi = \operatorname{tr}$ the ambiguity cancels and $\det$ is honestly well-defined. But for the general charge $\phi = a\operatorname{tr} + b\,\overline{\operatorname{tr}}$ , the same check forces $a-b\in\mathbb Z$ : charge quantization, identical in mechanism to why $U(1)$ representations need integer $k$ . Topologically: $GL(n,\mathbb C)$ contains a noncontractible loop (wind the phase of one diagonal entry by $2\pi$ ), $\mathbb C^*\cong\mathbb R_{>0}\times U(1)$ contains a circle, and a representation must map winding numbers to winding numbers integrally. This is the global information that the algebra alone — which only sees infinitesimal neighborhoods — cannot know.

Fine point 2: $e^{\operatorname{tr}\log A}$ is correct but not closed-form. To use it you must produce a logarithm (or the eigenvalues) of $A$ — an analytic task, and the statement “every $A\in GL(n,\mathbb C)$ has a log” is itself a theorem. The Leibniz polynomial is the same function written without logarithms: an explicit polynomial in the entries. That’s the only role of the singlet/ $\varepsilon$ -tensor construction (section 4.4): it manufactures, in closed form, the function whose generator is $\operatorname{tr}$ , so you never have to invert $\exp$ . And the Gauss factorization merely substitutes for “every $A$ is reachable by exponentials” with finite algebra instead of analysis.

So the one-sentence answer to your last question: exponentiating the trace does give the determinant directly — $\det = \exp\circ\operatorname{tr}\circ\log$ — and the rest of the derivation is just (1) the quantization check that this is single-valued, and (2) rewriting it as a polynomial (Leibniz) so that no logarithm is needed.

If you want, I can restructure 10-lie.md around this picture — the commuting square first, the two fine points after — and cut the parts that duplicate determinant_homomorphism.md (the Gauss factorization and assembly steps), keeping only references to it.