[proofplan] We prove the correspondence in both directions using the quotient construction of the universal enveloping algebra. A left $U(\mathfrak g)$-module restricts along $\iota: \mathfrak g \to U(\mathfrak g)$, and the defining relation in $U(\mathfrak g)$ guarantees that the restricted action preserves Lie brackets. Conversely, a Lie algebra representation extends multiplicatively to the tensor algebra $T(\mathfrak g)$; the representation identity forces the enveloping-algebra relations to act by zero, so the action descends to the quotient $U(\mathfrak g)$. The two constructions are inverse because both are determined by the action of the generators $\iota(\mathfrak g)$. [/proofplan] [step:Restrict a left $U(\mathfrak g)$-module to a Lie algebra representation] Let \begin{align*} \mu: U(\mathfrak g) \times V &\to V \\ (a,v) &\mapsto a \cdot v \end{align*} be a left $U(\mathfrak g)$-module structure on $V$. Define a $k$-[linear map](/page/Linear%20Map) \begin{align*} \rho_\mu: \mathfrak g &\to \mathfrak{gl}(V) \\ x &\mapsto \bigl(v \mapsto \iota(x)\cdot v\bigr). \end{align*} We verify that $\rho_\mu$ is a Lie algebra homomorphism. For $x,y \in \mathfrak g$ and $v \in V$, the defining relation in $U(\mathfrak g)$ is \begin{align*} \iota(x)\iota(y)-\iota(y)\iota(x)=\iota([x,y]). \end{align*} Using associativity of the left module action, we compute \begin{align*} [\rho_\mu(x),\rho_\mu(y)](v) &= \rho_\mu(x)(\rho_\mu(y)(v))-\rho_\mu(y)(\rho_\mu(x)(v)) \\ &= \iota(x)\cdot(\iota(y)\cdot v)-\iota(y)\cdot(\iota(x)\cdot v) \\ &= (\iota(x)\iota(y))\cdot v-(\iota(y)\iota(x))\cdot v \\ &= (\iota(x)\iota(y)-\iota(y)\iota(x))\cdot v \\ &= \iota([x,y])\cdot v \\ &= \rho_\mu([x,y])(v). \end{align*} Since this holds for every $v \in V$, we have \begin{align*} [\rho_\mu(x),\rho_\mu(y)] = \rho_\mu([x,y]). \end{align*} Thus $\rho_\mu: \mathfrak g \to \mathfrak{gl}(V)$ is a representation of $\mathfrak g$ on $V$. [guided] Start with a left $U(\mathfrak g)$-module structure \begin{align*} \mu: U(\mathfrak g) \times V &\to V \\ (a,v) &\mapsto a \cdot v. \end{align*} Because $\mathfrak g$ maps into $U(\mathfrak g)$ through the canonical Lie algebra map $\iota: \mathfrak g \to U(\mathfrak g)$, every element $x \in \mathfrak g$ acts on $V$ by first viewing it as $\iota(x) \in U(\mathfrak g)$ and then using the module action. This defines \begin{align*} \rho_\mu: \mathfrak g &\to \mathfrak{gl}(V) \\ x &\mapsto \bigl(v \mapsto \iota(x)\cdot v\bigr). \end{align*} The map $\rho_\mu$ is $k$-linear because $\iota$ is $k$-linear and the module action is $k$-linear in the algebra variable. The only point to check is compatibility with Lie brackets. The universal enveloping algebra is constructed so that multiplication in $U(\mathfrak g)$ remembers the Lie bracket through the relation \begin{align*} \iota(x)\iota(y)-\iota(y)\iota(x)=\iota([x,y]) \end{align*} for all $x,y \in \mathfrak g$. For $v \in V$, associativity of the module action gives \begin{align*} \iota(x)\cdot(\iota(y)\cdot v) = (\iota(x)\iota(y))\cdot v, \end{align*} and similarly with $x$ and $y$ interchanged. Therefore \begin{align*} [\rho_\mu(x),\rho_\mu(y)](v) &= \rho_\mu(x)(\rho_\mu(y)(v))-\rho_\mu(y)(\rho_\mu(x)(v)) \\ &= \iota(x)\cdot(\iota(y)\cdot v)-\iota(y)\cdot(\iota(x)\cdot v) \\ &= (\iota(x)\iota(y))\cdot v-(\iota(y)\iota(x))\cdot v \\ &= (\iota(x)\iota(y)-\iota(y)\iota(x))\cdot v \\ &= \iota([x,y])\cdot v \\ &= \rho_\mu([x,y])(v). \end{align*} Since two endomorphisms of $V$ are equal exactly when they agree on every $v \in V$, this proves \begin{align*} [\rho_\mu(x),\rho_\mu(y)] = \rho_\mu([x,y]). \end{align*} Thus the restricted action is a Lie algebra representation. [/guided] [/step] [step:Extend a Lie algebra representation to the tensor algebra] Now let \begin{align*} \rho: \mathfrak g &\to \mathfrak{gl}(V) \end{align*} be a Lie algebra representation. Let $T(\mathfrak g)$ denote the tensor algebra of the $k$-[vector space](/page/Vector%20Space) $\mathfrak g$, and let $\operatorname{End}_k(V)$ denote the unital associative $k$-algebra of $k$-linear maps $V \to V$, with multiplication given by composition. Define a $k$-linear map \begin{align*} \widetilde{\rho}: T(\mathfrak g) &\to \operatorname{End}_k(V) \end{align*} as follows: on the degree-zero summand $k \subset T(\mathfrak g)$, set \begin{align*} \widetilde{\rho}(\lambda)=\lambda \operatorname{id}_V \end{align*} for $\lambda \in k$, and on a pure tensor $x_1 \otimes \cdots \otimes x_m \in \mathfrak g^{\otimes m}$ with $m \geq 1$, set \begin{align*} \widetilde{\rho}(x_1 \otimes \cdots \otimes x_m) = \rho(x_1)\circ \cdots \circ \rho(x_m). \end{align*} Extending by $k$-linearity gives a $k$-algebra homomorphism because multiplication in $T(\mathfrak g)$ is tensor concatenation and multiplication in $\operatorname{End}_k(V)$ is composition of endomorphisms. Explicitly, for pure tensors \begin{align*} s &= x_1 \otimes \cdots \otimes x_m, \\ t &= y_1 \otimes \cdots \otimes y_n, \end{align*} we have \begin{align*} \widetilde{\rho}(st) &= \widetilde{\rho}(x_1 \otimes \cdots \otimes x_m \otimes y_1 \otimes \cdots \otimes y_n) \\ &= \rho(x_1)\circ \cdots \circ \rho(x_m)\circ \rho(y_1)\circ \cdots \circ \rho(y_n) \\ &= \widetilde{\rho}(s)\circ \widetilde{\rho}(t). \end{align*} Thus $\widetilde{\rho}$ is the associative algebra action of $T(\mathfrak g)$ generated by the operators $\rho(x)$. [guided] The tensor algebra $T(\mathfrak g)$ is the associative algebra freely generated by the vector space $\mathfrak g$. Let $\operatorname{End}_k(V)$ denote the unital associative $k$-algebra of $k$-linear maps $V \to V$, with multiplication given by composition. To turn the Lie algebra action $\rho$ into an associative algebra action, we let a tensor $x_1 \otimes \cdots \otimes x_m$ act by applying the operators $\rho(x_m)$, then $\rho(x_{m-1})$, and so on, which is written as the composition \begin{align*} \rho(x_1)\circ \cdots \circ \rho(x_m). \end{align*} Formally, define \begin{align*} \widetilde{\rho}: T(\mathfrak g) &\to \operatorname{End}_k(V) \end{align*} by \begin{align*} \widetilde{\rho}(\lambda)=\lambda \operatorname{id}_V \end{align*} for $\lambda \in k$ and \begin{align*} \widetilde{\rho}(x_1 \otimes \cdots \otimes x_m) = \rho(x_1)\circ \cdots \circ \rho(x_m) \end{align*} for $x_1,\dots,x_m \in \mathfrak g$ and $m \geq 1$. We then extend this rule by $k$-linearity to all of $T(\mathfrak g)$. We check multiplicativity because this is what will make the action a module action. Multiplication in $T(\mathfrak g)$ is concatenation of tensors. If \begin{align*} s &= x_1 \otimes \cdots \otimes x_m, \\ t &= y_1 \otimes \cdots \otimes y_n, \end{align*} then \begin{align*} st = x_1 \otimes \cdots \otimes x_m \otimes y_1 \otimes \cdots \otimes y_n. \end{align*} Therefore \begin{align*} \widetilde{\rho}(st) &= \widetilde{\rho}(x_1 \otimes \cdots \otimes x_m \otimes y_1 \otimes \cdots \otimes y_n) \\ &= \rho(x_1)\circ \cdots \circ \rho(x_m)\circ \rho(y_1)\circ \cdots \circ \rho(y_n) \\ &= \widetilde{\rho}(s)\circ \widetilde{\rho}(t). \end{align*} The degree-zero rule gives the identity and scalar action, so $\widetilde{\rho}$ is a unital associative algebra homomorphism from $T(\mathfrak g)$ to $\operatorname{End}_k(V)$. [/guided] [/step] [step:Show the tensor algebra action kills the enveloping algebra relations] Let $J$ be the two-sided ideal of $T(\mathfrak g)$ generated by all elements \begin{align*} x \otimes y-y \otimes x-[x,y], \end{align*} where $x,y \in \mathfrak g$ and where $[x,y]$ is regarded as an element of the degree-one summand $\mathfrak g \subset T(\mathfrak g)$. By definition, \begin{align*} U(\mathfrak g)=T(\mathfrak g)/J. \end{align*} For $x,y \in \mathfrak g$, the representation identity for $\rho$ gives \begin{align*} \rho([x,y])=[\rho(x),\rho(y)] = \rho(x)\circ \rho(y)-\rho(y)\circ \rho(x). \end{align*} Hence \begin{align*} \widetilde{\rho}(x \otimes y-y \otimes x-[x,y]) &= \rho(x)\circ \rho(y)-\rho(y)\circ \rho(x)-\rho([x,y]) \\ &= 0. \end{align*} Since $\widetilde{\rho}$ is an algebra homomorphism and $J$ is generated as a two-sided ideal by these elements, every element of $J$ lies in $\ker \widetilde{\rho}$. [guided] The universal enveloping algebra is not the whole tensor algebra. It is the quotient that imposes the rule that commutators in the associative algebra agree with Lie brackets in $\mathfrak g$. Let $J$ be the two-sided ideal of $T(\mathfrak g)$ generated by the elements \begin{align*} x \otimes y-y \otimes x-[x,y] \end{align*} for $x,y \in \mathfrak g$, with $[x,y]$ placed in the degree-one copy of $\mathfrak g$ inside $T(\mathfrak g)$. Then \begin{align*} U(\mathfrak g)=T(\mathfrak g)/J. \end{align*} To descend the tensor algebra action to this quotient, we must prove that every element of $J$ acts as zero. It is enough to check the listed generators, because $\widetilde{\rho}$ is an algebra homomorphism and the kernel of an algebra homomorphism is a two-sided ideal. For $x,y \in \mathfrak g$, the defining property of a Lie algebra representation is \begin{align*} \rho([x,y])=[\rho(x),\rho(y)]. \end{align*} The bracket on $\mathfrak{gl}(V)$ is the commutator bracket, so this means \begin{align*} \rho([x,y])=\rho(x)\circ \rho(y)-\rho(y)\circ \rho(x). \end{align*} Therefore \begin{align*} \widetilde{\rho}(x \otimes y-y \otimes x-[x,y]) &= \rho(x)\circ \rho(y)-\rho(y)\circ \rho(x)-\rho([x,y]) \\ &= 0. \end{align*} Thus every defining relation of $U(\mathfrak g)$ acts by zero on $V$, and hence $J \subset \ker \widetilde{\rho}$. [/guided] [/step] [step:Descend the tensor algebra action to a left $U(\mathfrak g)$-module action] Let \begin{align*} q: T(\mathfrak g) &\to U(\mathfrak g)=T(\mathfrak g)/J \end{align*} be the quotient map. Since $J \subset \ker \widetilde{\rho}$, there is a unique $k$-algebra homomorphism \begin{align*} \overline{\rho}: U(\mathfrak g) &\to \operatorname{End}_k(V) \end{align*} such that \begin{align*} \overline{\rho}(q(a))=\widetilde{\rho}(a) \end{align*} for every $a \in T(\mathfrak g)$. Define \begin{align*} \mu_\rho: U(\mathfrak g) \times V &\to V \\ (u,v) &\mapsto \overline{\rho}(u)(v). \end{align*} Because $\overline{\rho}$ is a unital algebra homomorphism, $\mu_\rho$ is a left $U(\mathfrak g)$-module structure on $V$. For $x \in \mathfrak g$ and $v \in V$, \begin{align*} \iota(x)\cdot v = \overline{\rho}(\iota(x))(v) = \rho(x)(v), \end{align*} so the restricted action is exactly the original representation. [/step] [step:Verify that the two constructions are inverse] Starting with a left $U(\mathfrak g)$-module structure $\mu$, restricting to $\rho_\mu$ and then extending back gives a $U(\mathfrak g)$-module structure whose associated algebra homomorphism \begin{align*} U(\mathfrak g) &\to \operatorname{End}_k(V) \end{align*} agrees with the original action homomorphism on every generator $\iota(x)$ with $x \in \mathfrak g$. Since $U(\mathfrak g)$ is generated as a unital associative algebra by $\iota(\mathfrak g)$, the two algebra homomorphisms are equal. Conversely, starting with a representation $\rho: \mathfrak g \to \mathfrak{gl}(V)$, extending it to $U(\mathfrak g)$ and then restricting along $\iota$ gives \begin{align*} x \mapsto \bigl(v \mapsto \overline{\rho}(\iota(x))(v)\bigr) = \bigl(v \mapsto \rho(x)(v)\bigr), \end{align*} so the original representation is recovered. Therefore the two constructions define inverse canonical bijections between representations of $\mathfrak g$ on $V$ and left $U(\mathfrak g)$-module structures on $V$. [/step]