The determinant as the exponentiated trace

This note explains, in physics language, how the determinant arises from Lie theory: the trace is the generator, the determinant is the finite transformation built from it, and the Leibniz formula is derived at the end as a matrix element — never assumed. It is a companion to the elementary derivation in Determinant From Homomorphism (same result, no calculus, no continuity) and to Part III of Deriving the determinant from homogeneity and multiplicativity (same argument in mathematical language); proofs that appear verbatim there are referenced, not repeated.

1. The cast of characters: two groups, two algebras, two maps¶

There are two group/algebra pairs in play, and keeping them apart is most of the battle:

Three clarifications, because the words are easy to mix up:

• $\det$ is not a group; it is a map between groups. Its values fill out the group $\mathbb C^*$, but $\det$ itself is a homomorphism $G\to\mathbb C^*$: it converts matrix multiplication into number multiplication. “One-dimensional representation” means exactly this: each group element $A$ is represented by an invertible $1\times1$ matrix — a nonzero number $f(A)$ — such that composition becomes multiplication, $f(AB)=f(A)f(B)$.

• The $E_{ij}$ do not have a representation; they are elements. They form a basis of the source algebra $\mathfrak{gl}(n,\mathbb C)$, the same way $J_i,K_i$ form a basis of the Lorentz algebra. A representation is a map defined on the whole algebra (or group).

• There are two levels of representation, one per row of the square below. Group level: $f:G\to\mathbb C^*$ with $f(AB)=f(A)f(B)$. Algebra level: its derivative at the identity, $\phi:=df_I:\mathfrak{gl}(n,\mathbb C)\to\mathbb C$, a linear map satisfying $\phi([X,Y])=[\phi(X),\phi(Y)]=0$ — the bracket on the right lives in the abelian target, hence vanishes. For $f=\det$ the algebra-level map will turn out to be $\phi=\operatorname{tr}$; on the basis: $\phi(E_{ij})=0$ for $i\neq j$, and $\phi(E_{ii})=1$ — each diagonal generator carries one unit of the single available “charge.”

Everything in this note is the study of one commuting square:

Generators map to generators (top arrow), finite transformations map to finite transformations (bottom arrow), and exponentiation commutes with everything.

Because the same letters tend to blur together, here is every symbol of this note with its type. In particular, $\phi$ is not a Lie algebra — it is a linear map between the two Lie algebras (the top arrow of the square), exactly as $\det$ is a map between the two groups (the bottom arrow):

2. The bottom-left corner: $E_{ij}$ exponentiates to a transvection (a Galilean boost)¶

The pattern is the one from the Lorentz group: a finite transformation is the exponential of a generator, $\Lambda=e^{\zeta K}$. Take the off-diagonal generator $E_{ij}$ ($i\neq j$). It is nilpotent, $E_{ij}^2=0$, so the exponential series terminates after two terms:

A transvection is literally the exponential of the generator $E_{ij}$ — nothing fancier. And you already know a transvection from physics: a Galilean boost. Acting on $(x,t)$,

which is $T_{12}(v)=\exp(vE_{12})$. The composition law $T_{ij}(c)T_{ij}(c')=T_{ij}(c+c')$, used in Step 2 of determinant_homomorphism.md, is just “boost velocities add” — the statement that $c\mapsto e^{cE_{ij}}$ is a one-parameter subgroup. The conjugation formula $D\,T_{ij}(c)\,D^{-1}=T_{ij}(c\,d_i/d_j)$ is the adjoint action you know from “a boost conjugated by a rotation is a boost in the rotated direction”: here a boost conjugated by a scaling is a boost with rescaled velocity.

The diagonal generators exponentiate just as explicitly: $e^{t\,E_{ii}}=\operatorname{diag}(1,\dots,\underset{i}{e^{t}},\dots,1)$, a scaling of one axis. So there are two families of finite transformations whose generators we fully control — transvections (nilpotent generators, boost-like) and diagonal matrices (commuting generators, scaling-like) — and they will carry the entire derivation.

3. Differentiating the bottom arrow: the generator of $f$, and the flow formula¶

Given a smooth one-dimensional representation $f$, its generator-level map is defined by expanding near the identity:

with $\phi:\mathfrak{gl}(n,\mathbb C)\to\mathbb C$ linear. For $f=\det$ you know this expansion by heart:

i.e. the generator of $\det$ is $\operatorname{tr}$ — that is the top arrow of the square evaluated on the known example.

The compatibility $f(e^X)=e^{\phi(X)}$ (the statement that the square commutes) is not an extra assumption; it follows from multiplicativity, and the mechanism is worth seeing twice. First as time slicing: the finite transformation built by composing the same infinitesimal step $I+\frac{X}{N}$ over and over is, by definition, the exponential,

(matrix compound interest; a finite rotation from $N$ tiny rotations). Apply $f$ and let each hypothesis do its one job:

The first equality is multiplicativity ($f$ of a product of $N$ slices is the product of $N$ numbers), the second is the defining expansion of $\phi$ applied to each slice with $\varepsilon=1/N$; the $O(1/N^2)$ error per slice totals $O(1/N)$ and vanishes. Same mechanism as time-slicing a path integral.

Second, as a differential equation: let $u(t):=f(e^{tX})$. Since $tX$ and $\varepsilon X$ commute (multiples of the same $X$ — no Baker–Campbell–Hausdorff needed), $e^{(t+\varepsilon)X}=e^{\varepsilon X}e^{tX}$, so

whose solution is $u(t)=e^{t\phi(X)}$. We call this the flow formula:

For $f=\det$, $\phi=\operatorname{tr}$, it is the identity used in one-loop effective actions and fermion determinants, $\det(e^X)=e^{\operatorname{tr}X}$, “$\log\det=\operatorname{Tr}\log$”. Physically it is Liouville’s formula: the linear flow $\dot x=Xx$ has flow map $e^{tX}$, the determinant of the flow map is the volume-expansion factor of a comoving blob, and $\operatorname{tr}X=\partial_i(Xx)_i=\nabla\cdot v$ is the divergence of the velocity field — “volume grows at rate $\nabla\cdot v$” integrates to $e^{t\operatorname{tr}X}$. The determinant is the finite-time volume factor; the trace is its instantaneous rate.

4. The top arrow is forced: $\phi = a\operatorname{tr} + b\,\overline{\operatorname{tr}}$¶

Here is the structural heart. The group-level statement is a one-liner: the target $\mathbb C^*$ is abelian, and an abelian charge cannot see group commutators,

because the four values are numbers and numbers commute. The algebra-level consequence is $\phi([X,Y])=0$ — but getting there requires a bridge between the group commutator $ABA^{-1}B^{-1}$ and the Lie bracket $[X,Y]=XY-YX$. We build the bridge first (4.1), cross it carefully (4.2), and then ask which matrices are brackets (4.3).

4.1 The Lie bracket is the infinitesimal group commutator¶

Run the flow of $X$ for a small time $s$, the flow of $Y$ for a small time $t$, then undo both, in order:

If $X$ and $Y$ commuted as matrices, the four factors would reorder and cancel, $C=I$. In general, expand every exponential, keeping all terms up to total degree 2 in $(s,t)$:

Multiply the left pair and the right pair:

(note $v_2=u_2$: both signs flip, and the cross term $(-sX)(-tY)=st\,XY$ comes out the same). Now multiply the two halves and collect by degree:

Degree 1: $u_1+v_1=0$ — to first order the flows do commute; whatever survives is second order. Degree 2:

The pure $s^2$ and $t^2$ terms cancel identically — only the bilinear term survives, and it is exactly the bracket:

This is what “the Lie bracket is the infinitesimal group commutator” means: the bracket measures the leading failure of two flows to close. You know it from rotations — a small rotation about $x$, a small one about $y$, then undoing both leaves a small rotation about $z$ of angle $st$: that is $[J_x,J_y]=J_z$ made tangible. (Same phenomenon as parallel parking: “steer” and “drive” don’t commute, and their commutator is the sideways translation you actually want.)

4.2 Crossing the bridge: $\phi([X,Y])=0$, every step¶

The cleanest complete derivation goes through conjugation; the direct second-order route is at the end.

Step 1 (group level: the charge is conjugation-blind). For any $g,h\in GL(n,\mathbb C)$:

since the values are commuting numbers. (This is Step 1 of determinant_homomorphism.md, reappearing.)

Step 2 (conjugation of an exponential is the exponential of the conjugated generator). For any invertible $g$ and any matrix $M$, $(gMg^{-1})^k=gM^kg^{-1}$ — the interior $g^{-1}g$ pairs cancel. Apply this to every term of the exponential series:

Step 3 (transfer to the algebra: $\phi$ is conjugation-invariant). Apply $f$ to Step 2 and use Step 1 on the left, the flow formula of section 3 on both sides:

Two exponentials of linear functions of $t$ that agree for all $t$ have equal slopes at $t=0$ (differentiate), so

Step 4 (Hadamard’s lemma — the physicist’s $e^{A}Be^{-A}$ formula). Take $g=e^{sX}$ and expand the conjugated generator in $s$. Let $G(s):=e^{sX}\,Y\,e^{-sX}$, a curve of matrices in $\mathfrak{gl}(n,\mathbb C)$. Differentiate:

In particular $G'(0)=[X,Y]$, so the Taylor expansion is

(iterating gives the full series $Y+s[X,Y]+\tfrac{s^2}{2}[X,[X,Y]]+\cdots=e^{s\,\mathrm{ad}_X}Y$, the adjoint action — “$\mathrm{Ad}=\exp(\mathrm{ad})$” — but the first-order term is all we need).

Step 5 (conclude). Insert Step 4 into Step 3 and use linearity of $\phi$:

The coefficient of $s$ must vanish:

$\square$

The second-order route, for comparison. One can instead apply $f$ directly to the group commutator of 4.1. Define $F(s,t):=f\big(C(s,t)\big)$. By multiplicativity and commuting values (the one-liner at the top of this section), $F(s,t)=1$ exactly, for all $s,t$. On the other hand, by 4.1 and the defining expansion of $\phi$,

and a function identically equal to 1 has all Taylor coefficients zero — in particular the $st$ coefficient, $\phi([X,Y])=0$. (Pedantic check that no second-derivative-of-$f$ terms pollute the $st$ coefficient: along the axes the commutator is trivial, $C(s,0)=C(0,t)=I$ identically, so both first partials of $C$ vanish at the origin and the mixed partial of $F$ is $\phi$ applied to the mixed partial of $C$, which is $[X,Y]$.) The conjugation route of Steps 1–5 is the same argument reorganized so that only first-order expansions are ever needed.

4.3 Which matrices are brackets: $[\mathfrak{gl}_n,\mathfrak{gl}_n]=\mathfrak{sl}_n$¶

Cyclicity of the trace gives $\operatorname{tr}[X,Y]=0$, so brackets are traceless. The converse also holds, and the $E_{ij}$ are exactly the witnesses:

and these span all traceless matrices. So $[\mathfrak{gl}_n,\mathfrak{gl}_n]=\mathfrak{sl}_n$ exactly, and $\phi$ must vanish on every traceless matrix — on the basis, $\phi(E_{ij})=0$ for $i\neq j$ and all $\phi(E_{ii})$ equal (their differences are commutators). A functional that kills the traceless part can only depend on the trace. If $\phi$ were complex-linear that would already give $\phi(X)=a\operatorname{tr}X$; smoothness of $f$ only guarantees real-linearity, which admits one more term:

— the $b$ term is the anti-holomorphic branch, whose integrated form will be $\overline{\det}$.

Compare with the groups you know: for $SU(2)$ or the Lorentz group, every generator is itself a commutator ($[J_i,J_j]=\epsilon_{ijk}J_k$, …), the algebra is “all commutator,” and so there is no nontrivial det-like function on them — their only one-dimensional representation is trivial. $GL(n)$ is special precisely because $\mathfrak{gl}_n=\mathfrak{sl}_n\oplus\mathbb C\cdot I$ has exactly one abelian direction — overall scaling, measured by the trace. The determinant exists because of, and only because of, that one direction.

5. Down the right leg: $\det=\exp\circ\operatorname{tr}\circ\log$, and the quantization of the charge¶

Take stock of which arrows of the square are now known. The top arrow is known: section 4 forced $\phi=a\operatorname{tr}+b\,\overline{\operatorname{tr}}$. The two vertical legs are known: they are the exponentials. The bottom arrow $f$ is the unknown we are after. But a square that commutes lets you compute a missing side by going around the other way: instead of crossing the bottom directly ($A\mapsto f(A)$, unknown), go up the left leg, across the known top, and down the right leg. Explicitly, to evaluate $f(A)$:

1. Up the left leg (the only hard step — the leg must be traversed backwards). Find a generator that reaches $A$: a matrix $X\in\mathfrak{gl}(n,\mathbb C)$ with $e^X=A$. Any such $X$ is called a logarithm of $A$ and written $X=\log A$. Mind the types: $A$ is a group element; $\log A$ is an algebra element — an honest $n\times n$ matrix (generally not unique; that is gap (i) below).

2. Across the top arrow. Apply the known map $\phi$ to it: the number $\phi(X)=a\operatorname{tr}X+b\,\overline{\operatorname{tr}X}\in\mathbb C$. (Recall from the type table of section 1: $\phi$ is not a Lie algebra — it is the linear map $\mathfrak{gl}(n,\mathbb C)\to\mathbb C$, the algebra-level shadow of $f$.)

3. Down the right leg. Exponentiate the number: $e^{\phi(X)}\in\mathbb C^*$.

4. The square commutes, so the detour lands on the right answer: by the flow formula of section 3, $f(A)=f(e^X)=e^{\phi(X)}$.

For the normalized charge $a=1,\ b=0$ (i.e. $\phi=\operatorname{tr}$) the detour reads, symbol by symbol,

i.e. the direct formula

— three known maps composed, replacing the one unknown map. It is worth checking on transformations you know:

• $A=\operatorname{diag}(d_1,d_2)$: take $X=\operatorname{diag}(\log d_1,\log d_2)$, so $\operatorname{tr}X=\log d_1+\log d_2$ and $e^{\operatorname{tr}X}=d_1d_2$. ✓

• a rotation $R(\theta)=e^{\theta(E_{21}-E_{12})}$: the generator is traceless, so $\det R=e^0=1$ — rotations preserve volume because their generator is traceless. Same for boosts/transvections.

• a rotation–scaling $A=r\begin{pmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{pmatrix}$: one checks $A=e^X$ with $X=\begin{pmatrix}\log r&-\theta\\ \theta&\log r\end{pmatrix}$, so $\operatorname{tr}X=2\log r$ and $\det A=e^{2\log r}=r^2$ — the area-expansion factor of the spiral map, with the rotation angle dropping out of the trace.

But two honest gaps separate the boxed formula from a theorem. Gap (i): $\log A$ is multivalued — the recipe must not depend on which $X$ we pick. Gap (ii): the formula is not closed-form — it requires producing a logarithm, and even the statement “every $A\in GL(n,\mathbb C)$ has one” is a theorem. This section closes gap (i) by integrating the two generator families of section 2 by hand; section 6 closes gap (ii).

5.1 Transvections: $f=1$, automatically¶

The off-diagonal generators are traceless, $\operatorname{tr}E_{ij}=0$, so the flow formula gives

No ambiguity arises (the answer is the same for every choice), and Step 2 of the elementary note falls out in one line: an abelian charge cannot register a traceless, boost-like generator. By multiplicativity, multiplying any matrix by transvections never changes $f$.

5.2 Diagonal matrices: the multivaluedness bites, and quantizes the charge¶

Every $d_k\in\mathbb C^*$ is an exponential: pick logs $z_k$ with $d_k=e^{z_k}$ — here the branch choice is genuinely free, $z_k\to z_k+2\pi i$ gives the same $d_k$. Then $\operatorname{diag}(d_1,\dots,d_n)=\exp\big(\operatorname{diag}(z_1,\dots,z_n)\big)$, and since $\operatorname{tr}\operatorname{diag}(z)=\sum_k z_k$, the flow formula gives

The left side is one fixed number; it cannot depend on which branches we chose. Shift a single $z_k\to z_k+2\pi i$: the right side picks up the factor $e^{2\pi i(a-b)}$. Single-valuedness therefore forces the quantization

— the same mechanism that quantizes angular momentum ($e^{ik\varphi}$ must be single-valued on the circle) and the $U(1)$ charge of section 1. The topology responsible: $\mathbb C^*\cong\mathbb R_{>0}\times U(1)$ contains a circle, $GL(n,\mathbb C)$ contains a non-contractible loop (wind the phase of one diagonal entry by $2\pi$), and a representation must map winding numbers to winding numbers integrally. This is global information that the algebra — which only sees an infinitesimal neighborhood of $I$ — cannot know; it is exactly what the bare generator analysis of section 4 was missing.

Given the quantization, the recipe

does not depend on the branch of $\log w$ (a branch shift changes the exponent by $2\pi i m(a-b)\in 2\pi i\,\mathbb Z$), and $g$ is a continuous homomorphism $\mathbb C^*\to\mathbb C^*$ — precisely the family $|w|^s(w/|w|)^k$ derived independently in Homomorphisms from C-star to C-star, now reconstructed from the two real-linear components of $\phi$. Since $\sum_k z_k$ is a log of the product $d_1\cdots d_n$, the display above says exactly

This is Step 3 of the elementary note: the trace doesn’t care which diagonal slot a number sits in, so $f$ sees only the product.

6. Closing gap (ii): the antisymmetric singlet derives the Leibniz formula¶

The boxed formula $e^{\operatorname{tr}\log A}$ requires a logarithm of $A$. We now construct the same function in closed form — an explicit polynomial in the entries of $A$ whose generator is $\operatorname{tr}$ — so that no logarithm is ever needed. The physicist’s candidate is the $\varepsilon$-tensor.

In $n$ dimensions, a totally antisymmetric object with $n$ indices has exactly one independent component: $\psi_{i_1\cdots i_n}=c\,\varepsilon_{i_1\cdots i_n}$. Equivalently, the wedge $e_1\wedge\cdots\wedge e_n$ spans a one-dimensional space (the “singlet”). Let $A$ act on every slot. The result is again totally antisymmetric, hence a multiple of the singlet, and that multiple is a number we define as $L(A)$:

Now just expand in components, $Ae_k=\sum_j A_{jk}\,e_j$, using multilinearity of the wedge:

A wedge with a repeated vector vanishes by antisymmetry, so only tuples $(j_1,\dots,j_n)$ that are permutations $\sigma$ of $(1,\dots,n)$ survive, and reordering costs the parity sign: $e_{\sigma(1)}\wedge\cdots\wedge e_{\sigma(n)}=\operatorname{sgn}(\sigma)\,e_1\wedge\cdots\wedge e_n$. Hence

This is the Leibniz formula — derived, not posited: it is nothing but the single matrix element of $A$ acting on the one-dimensional antisymmetric singlet. (The row form $\sum_\sigma\operatorname{sgn}(\sigma)\prod_k A_{k,\sigma(k)}$ is the same sum under $\sigma\mapsto\sigma^{-1}$, which preserves the sign.) Its properties now cost one line each, with no determinant theory anywhere:

• Multiplicative. Slot-wise action of $B$ maps the singlet line to itself with scalar $L(B)$; following with $A$ multiplies by $L(A)$; the composite is the slot-wise action of $AB$. Scalars on a line compose by multiplication: $L(AB)=L(A)L(B)$.

• Diagonals. $\operatorname{diag}(d)\,e_k=d_k e_k$, so each slot contributes its factor: $L\big(\operatorname{diag}(d)\big)=d_1\cdots d_n$.

• Transvections. $T_{ij}(c)$ fixes every $e_k$ except $e_j\mapsto e_j+c\,e_i$; the extra term inserts a second copy of $e_i$ into the wedge and is killed by antisymmetry: $L\big(T_{ij}(c)\big)=1$.

• Generator. Expanding each slot of $(I+\varepsilon X)e_1\wedge\cdots\wedge(I+\varepsilon X)e_n$ to first order, the slot-$k$ term $Xe_k=\sum_j X_{jk}e_j$ survives the wedge only for $j=k$ (all other basis vectors are already present), so $L(I+\varepsilon X)=1+\varepsilon\sum_k X_{kk}+O(\varepsilon^2)=1+\varepsilon\operatorname{tr}X$. So $L$ is an integral of the trace — the explicit polynomial with $dL_I=\operatorname{tr}$ — and by the flow formula of section 3, $L(e^X)=e^{\operatorname{tr}X}$: on the image of $\exp$, $L$ coincides with $e^{\operatorname{tr}\log A}$, with no branch to choose.

7. Patching the square together on the whole group¶

Sections 5.1–5.2 evaluated $f$ on the two generator families; $L$ of section 6 is defined everywhere outright. What remains is to connect them on an arbitrary $A\in GL(n,\mathbb C)$, and for that we borrow exactly one result from the elementary note — the Gauss factorization Lemma of Determinant From Homomorphism (plain row reduction, no determinants used; we do not repeat the proof):

Apply $f$ using multiplicativity, 5.1 and 5.2:

Apply $L$ using its one-line properties (peel the transvections, then the diagonal):

The second display does double duty: it shows the pivot product $d_1\cdots d_n$ — a priori dependent on the chosen row reduction — equals the unambiguous polynomial $L(A)$, hence is the same for every factorization; and substituting it into the first display gives the final result:

Every ingredient was constructed, none assumed: $a,b$ came from differentiating $f$ (sections 3–4); the quantization $a-b\in\mathbb Z$ from single-valuedness on the diagonal subgroup (section 5.2); $g$ was built from $(a,b)$; $L$ was built as the antisymmetric singlet (section 6); and the patching used only row reduction and multiplicativity. Finally, imposing homogeneity $f(\lambda I)=\lambda^n$: with $\lambda=e^z$, the flow formula gives $f(\lambda I)=e^{n(az+b\bar z)}\overset{!}{=}e^{nz}$ for all $z$, forcing $a=1$, $b=0$, i.e. $g=\mathrm{id}$ and

(Aside: the eigenvalue formula $f(A)=\prod_i\lambda_i(A)$ of Part III, Step 5 in determinant_over_C.md expresses the same result, but needs the surjectivity of $\exp$ over $\mathbb C$ — gap (ii) closed by analysis instead of algebra. The route above trades that analytic input for the Gauss factorization, the same trade the elementary note makes, and produces the Leibniz formula constructively along the way.)

8. The role of $g$: the charge of the representation¶

The derivation above did not — and could not — produce $f=\det$ outright: the boxed result of section 7 is $f=g\circ L$, and only the normalization $f(\lambda I)=\lambda^n$ collapsed $g$ to the identity. This is not a defect of the method; the factor $g$ has a precise Lie-theoretic identity, worth spelling out. $g$ is a one-dimensional representation of the target group $\mathbb C^*$ itself, and it carries a charge — a label, in the exact sense of the $U(1)$ charge of gauge theory, that section 8.3 makes precise. The determinant carries no freedom; it is the universal, canonical piece. All the freedom of $f$ lives in $g$.

8.1 The square factors into two stacked squares¶

What sections 4–7 actually established is that the commuting square of section 1 splits in the middle:

Top row: $\phi=\bar\phi\circ\operatorname{tr}$, where $\bar\phi(z)=az+b\bar z$ is a functional on the one-dimensional quotient $\mathfrak{gl}_n/\mathfrak{sl}_n\cong\mathbb C$. Bottom row: $f=g\circ\det$. The left square is rigid — no parameters, the same for every $f$; it is pure structure ($[\mathfrak{gl},\mathfrak{gl}]=\mathfrak{sl}$, and the trace is the quotient map). The right square is where the entire $(a,b)$ freedom sits, and its bottom arrow is $g$. So $g$ is literally “the part of $f$ that remains after the universal part has been factored out.”

8.2 Group level: $g$ is $f$ seen on the quotient¶

The commutator argument forces $f$ to be trivial on $SL(n,\mathbb C)=[GL,GL]$. A function trivial on $SL$ depends on $A$ only through its coset $A\cdot SL$ — i.e., it descends to the quotient group $GL(n,\mathbb C)/SL(n,\mathbb C)$. The determinant is exactly the identification of that quotient with $\mathbb C^*$. So the statement $f=g\circ\det$ decomposes as

This is the most Lie-theoretic answer to “where does $g$ come from”: group theory can prove that $f$ factors through the abelianization, and can identify the abelianization with $\mathbb C^*$ — but it cannot choose a representation of $\mathbb C^*$. The target group has many self-homomorphisms ($w\mapsto w^2$, $w\mapsto\bar w$, $w\mapsto|w|^s(w/|w|)^k$, …), all equally lawful, and composing any homomorphism with one of them produces another homomorphism. No structural principle prefers $w\mapsto w$ among them. That residual choice is $g$.

8.3 What “charge” means here, and how it works¶

A word of precision first, because it is easy to conflate two things: $g$ is a representation; the charge is the label that names it. Saying loosely “$g$ is the charge” is shorthand for “$g$ is the character carrying a particular charge.” The two are in bijection — each charge names exactly one $g$ — but they are different kinds of object (a number versus a map), and the whole point of the word “charge” is the structure that label carries. Here is that structure, built up from the example you already know.

The model: the $U(1)$ charge. In electromagnetism a field $\psi$ has a charge $q\in\mathbb Z$. The integer $q$ does not act on anything by itself; it labels how $\psi$ transforms under the gauge group: a gauge rotation $e^{i\theta}\in U(1)$ multiplies $\psi$ by $e^{iq\theta}$. The map $\rho_q:e^{i\theta}\mapsto e^{iq\theta}$ is the one-dimensional representation; $q$ is its charge. Three properties make $q$ deserve the name — and all three will recur verbatim for $g$:

• (quantization) $q$ must be an integer, because $e^{i\theta}$ and $e^{i(\theta+2\pi)}$ are the same group element, so $e^{iq\theta}$ must be single-valued on the circle. This is not imposed; it is forced by the topology of $U(1)$.

• (additivity) charges add under tensoring: $\psi_1\otimes\psi_2$ transforms by $e^{i(q_1+q_2)\theta}$, so it has charge $q_1+q_2$. The product of representations corresponds to the sum of labels.

• (coupling to the generator) writing $e^{i\theta}=e^{i\theta}$ as the exponential of the generator, $q$ is the eigenvalue with which $\psi$ couples to it: $\hat Q\psi=q\psi$. The charge is the coefficient in front of the generator.