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$\DeclareMathOperator\Lie{Lie}\DeclareMathOperator\SL{SL}$I am just writing some details on the comment by BCnrd. We use various statements from chapter 7 page 105ff in Borel's 'Linear Algebraic Groups, second enlarged edition'.

We start with a unipotent element $g$ in a linear algebraic group $G$. After conjugation we may assume that $g$ is upper triangular (with $1$'s on diagonal). We can apply the matrix logarithm $$ X\mathrel{:=} \log (g) \mathrel{:=} \sum_{n=1}^{\infty} (-1)^{k+1} \frac{(g-I)^n}{n}, $$ which is well defined in characteristic 0 since there are only finitely many powers of $(g-I)$ (so it is a finite sum). The matrix $X$ is nilpotent and we can use it to construct an exponential function from the additive group to $G$ (as in 7.4(1) Borel) \begin{align} \operatorname{G}_{a} & \to G \\ t &\mapsto \exp (t\cdot X) = \sum_{n=0}^{\infty} \frac{(tX)^n}{n!} \end{align} which is an algebraic morphism in characteristic 0 since there are only finitely many powers of $X$. Since we are in characteristic 0, it is also injective. Since $g=\exp(1\cdot X)$, and we are in char. 0, the closed one-dimensional algebraic subgroup of $G$ generated by $g$ is $U:=\exp (\operatorname{G}_a)$. The Lie algebra $\Lie(U)$ consists of nilpotent elements and we can apply Jacobson–Morozow to get an $\mathfrak{sl}_2$-triple (let's call its span $M$) that contains $\Lie(U)$. Following Borel, we define the closed algebraic subgroup $$ \mathcal{A}(M) := \bigcap_{\text{$H$ closed subgroup of $G$} \\ M \subset \Lie(H) \subset \Lie(G)} H $$ and take its commutator group $$ \SL_g \mathrel{:=} [\mathcal{A}(M),\mathcal{A}(M)] \mathrel{:=} \left\{ ghg^{-1}h^{-1} : g,h \in \mathcal{A}(M) \right\}. $$ Using Prop 7.8 and Cor 7.9 in Borel, we get $$ \Lie(\SL_g) = \Lie([\mathcal{A}(M),\mathcal{A}(M)]) = [\Lie(\mathcal{A}(M)),\Lie(\mathcal{A}(M))] = [M, M] = M, $$ where the last equality follows since $M$ is the span of a Lie-triple system. The Lie-algebra of $\SL_g$ is thus isomorphic to $\mathfrak{sl}_2$ and since we are in char. 0, this means that $\SL_g$ has to be isomorphic to $\SL_2$ (I think for algebraic isomorphisms it cannot be $\operatorname{PSL}_2$).

In other words, an algebraic exponential map $ \operatorname{G}_a \to G$, $t \mapsto \exp(t\cdot \tilde{X})$ factors through $\SL_2$ via $$ t \mapsto \left( \begin{matrix} 1 & t \\ 0 & 1\end{matrix}\right), $$ so indeed, the original element $g$ comes from a unipotent upper triangular element in $\SL_2$.

$\DeclareMathOperator\Lie{Lie}\DeclareMathOperator\SL{SL}$I am just writing some details on the comment by BCnrd. We use various statements from chapter 7 page 105ff in Borel's 'Linear Algebraic Groups, second enlarged edition'.

We start with a unipotent element $g$ in a linear algebraic group $G$. After conjugation we may assume that $g$ is upper triangular (with $1$'s on diagonal). We can apply the matrix logarithm $$ X\mathrel{:=} \log (g) \mathrel{:=} \sum_{n=1}^{\infty} (-1)^{k+1} \frac{(g-I)^n}{n}, $$ which is well defined in characteristic 0 since there are only finitely many powers of $(g-I)$ (so it is a finite sum). The matrix $X$ is nilpotent and we can use it to construct an exponential function from the additive group to $G$ (as in 7.4(1) Borel) \begin{align} \operatorname{G}_{a} & \to G \\ t &\mapsto \exp (t\cdot X) = \sum_{n=0}^{\infty} \frac{(tX)^n}{n!} \end{align} which is an algebraic morphism in characteristic 0 since there are only finitely many powers of $X$. Since we are in characteristic 0, it is also injective. Since $g=\exp(1\cdot X)$, and we are in char. 0, the closed one-dimensional algebraic subgroup of $G$ generated by $g$ is $U:=\exp (\operatorname{G}_a)$. The Lie algebra $\Lie(U)$ consists of nilpotent elements and we can apply Jacobson–Morozow to get an $\mathfrak{sl}_2$-triple (let's call its span $M$) that contains $\Lie(U)$. Following Borel, we define the closed algebraic subgroup $$ \mathcal{A}(M) := \bigcap_{\text{$H$ closed subgroup of $G$} \\ M \subset \Lie(H) \subset \Lie(G)} H $$ and take its commutator group $$ \SL_g \mathrel{:=} [\mathcal{A}(M),\mathcal{A}(M)] \mathrel{:=} \left< ghg^{-1}h^{-1} : g,h \in \mathcal{A}(M) \right>. $$ Using Prop 7.8 and Cor 7.9 in Borel, we get $$ \Lie(\SL_g) = \Lie([\mathcal{A}(M),\mathcal{A}(M)]) = [\Lie(\mathcal{A}(M)),\Lie(\mathcal{A}(M))] = [M, M] = M, $$ where the last equality follows since $M$ is the span of a Lie-triple system. The Lie-algebra of $\SL_g$ is thus isomorphic to $\mathfrak{sl}_2$ and since we are in char. 0, this means that $\SL_g$ has to be isomorphic to $\SL_2$ (or contain a subgroup isomorphic to $\SL_2$).

In other words, an algebraic exponential map $ \operatorname{G}_a \to G$, $t \mapsto \exp(t\cdot \tilde{X})$ factors through $\SL_2$ via $$ t \mapsto \left( \begin{matrix} 1 & t \\ 0 & 1\end{matrix}\right), $$ so indeed, the original element $g$ comes from a unipotent upper triangular element in $\SL_2$.

$\DeclareMathOperator\Lie{Lie}\DeclareMathOperator\SL{SL}$I am just writing some details on the comment by BCnrd. We use various statements from chapter 7 page 105ff in Borel's 'Linear Algebraic Groups, second enlarged edition'.

We start with a unipotent element $g$ in a linear algebraic group $G$. After conjugation we may assume that $g$ is upper triangular (with $1$'s on diagonal). We can apply the matrix logarithm $$ X\mathrel{:=} \log (g) \mathrel{:=} \sum_{n=1}^{\infty} (-1)^{k+1} \frac{(g-I)^n}{n}, $$ which is well defined in any field since there are only finitely many powers of $(g-I)$ (so it is a finite sum). The matrix $X$ is nilpotent and we can use it to construct an exponential function from the additive group to $G$ (as in 7.4(1) Borel) \begin{align} \operatorname{G}_{a} & \to G \\ t &\mapsto \exp (t\cdot X) = \sum_{n=0}^{\infty} \frac{(tX)^n}{n!} \end{align} which is an algebraic function since there are only finitely many powers of $X$. Since we are in characteristic 0, it is also injective. Since $g=\exp(1\cdot X)$, and we are in char. 0, the closed one-dimensional algebraic subgroup of $G$ generated by $g$ is $U:=\exp (\operatorname{G}_a)$. The Lie algebra $\Lie(U)$ consists of nilpotent elements and we can apply Jacobson–Morozow to get an $\mathfrak{sl}_2$-triple (let's call its span $M$) that contains $\Lie(U)$. Following Borel, we define the closed algebraic subgroup $$ \mathcal{A}(M) := \bigcap_{\text{$H$ closed subgroup of $G$} \\ M \subset \Lie(H) \subset \Lie(G)} H $$ and take its commutator group $$ \SL_g \mathrel{:=} [\mathcal{A}(M),\mathcal{A}(M)] \mathrel{:=} \left\{ ghg^{-1}h^{-1} : g,h \in \mathcal{A}(M) \right\}. $$ Using Prop 7.8 and Cor 7.9 in Borel, we get $$ \Lie(\SL_g) = \Lie([\mathcal{A}(M),\mathcal{A}(M)]) = [\Lie(\mathcal{A}(M)),\Lie(\mathcal{A}(M))] = [M, M] = M, $$ where the last equality follows since $M$ is the span of a Lie-triple system. The Lie-algebra of $\SL_g$ is thus isomorphic to $\mathfrak{sl}_2$ and since we are in char. 0, this means that $\SL_g$ has to be isomorphic to $\SL_2$ (I think for algebraic isomorphisms it cannot be $\operatorname{PSL}_2$).

In other words, an algebraic exponential map $ \operatorname{G}_a \to G$, $t \mapsto \exp(t\cdot \tilde{X})$ factors through $\SL_2$ via $$ t \mapsto \left( \begin{matrix} 1 & t \\ 0 & 1\end{matrix}\right), $$ so indeed, the original element $g$ comes from a unipotent upper triangular element in $\SL_2$.

$\DeclareMathOperator\Lie{Lie}\DeclareMathOperator\SL{SL}$I am just writing some details on the comment by BCnrd. We use various statements from chapter 7 page 105ff in Borel's 'Linear Algebraic Groups, second enlarged edition'.

We start with a unipotent element $g$ in a linear algebraic group $G$. After conjugation we may assume that $g$ is upper triangular (with $1$'s on diagonal). We can apply the matrix logarithm $$ X\mathrel{:=} \log (g) \mathrel{:=} \sum_{n=1}^{\infty} (-1)^{k+1} \frac{(g-I)^n}{n}, $$ which is well defined in characteristic 0 since there are only finitely many powers of $(g-I)$ (so it is a finite sum). The matrix $X$ is nilpotent and we can use it to construct an exponential function from the additive group to $G$ (as in 7.4(1) Borel) \begin{align} \operatorname{G}_{a} & \to G \\ t &\mapsto \exp (t\cdot X) = \sum_{n=0}^{\infty} \frac{(tX)^n}{n!} \end{align} which is an algebraic morphism in characteristic 0 since there are only finitely many powers of $X$. Since we are in characteristic 0, it is also injective. Since $g=\exp(1\cdot X)$, and we are in char. 0, the closed one-dimensional algebraic subgroup of $G$ generated by $g$ is $U:=\exp (\operatorname{G}_a)$. The Lie algebra $\Lie(U)$ consists of nilpotent elements and we can apply Jacobson–Morozow to get an $\mathfrak{sl}_2$-triple (let's call its span $M$) that contains $\Lie(U)$. Following Borel, we define the closed algebraic subgroup $$ \mathcal{A}(M) := \bigcap_{\text{$H$ closed subgroup of $G$} \\ M \subset \Lie(H) \subset \Lie(G)} H $$ and take its commutator group $$ \SL_g \mathrel{:=} [\mathcal{A}(M),\mathcal{A}(M)] \mathrel{:=} \left\{ ghg^{-1}h^{-1} : g,h \in \mathcal{A}(M) \right\}. $$ Using Prop 7.8 and Cor 7.9 in Borel, we get $$ \Lie(\SL_g) = \Lie([\mathcal{A}(M),\mathcal{A}(M)]) = [\Lie(\mathcal{A}(M)),\Lie(\mathcal{A}(M))] = [M, M] = M, $$ where the last equality follows since $M$ is the span of a Lie-triple system. The Lie-algebra of $\SL_g$ is thus isomorphic to $\mathfrak{sl}_2$ and since we are in char. 0, this means that $\SL_g$ has to be isomorphic to $\SL_2$ (I think for algebraic isomorphisms it cannot be $\operatorname{PSL}_2$).

In other words, an algebraic exponential map $ \operatorname{G}_a \to G$, $t \mapsto \exp(t\cdot \tilde{X})$ factors through $\SL_2$ via $$ t \mapsto \left( \begin{matrix} 1 & t \\ 0 & 1\end{matrix}\right), $$ so indeed, the original element $g$ comes from a unipotent upper triangular element in $\SL_2$.

I am just writing some details on the comment by BCnrd. We use various statements from chapter 7 page 105ff in Borel's 'Linear Algebraic Groups, second enlarged edition'.

We start with a unipotent element $g$ in a linear algebraic group $G$. After conjugation we may assume that $g$ is upper triangular (with $1$'s on diagonal). We can apply the matrix logarithm $$ X:= \log (g) := \sum_{n=1}^{\infty} (-1)^{k+1} \frac{(g-I)^n}{n}, $$ which is well defined in any field since there are only finitely many powers of $(g-I)$ (so it is a finite sum). The matrix $X$ is nilpotent and we can use it to construct an exponential function from the additive group to $G$ (as in 7.4(1) Borel) \begin{align} \operatorname{G}_{a} & \to G \\ t &\mapsto \exp (t\cdot X) = \sum_{n=0}^{\infty} \frac{(tX)^n}{n!} \end{align} which is an algebraic function since there are only finitely many powers of $X$. Since we are in characteristic 0, it is also injective. Since $g=\exp(1\cdot X)$, and we are in char. 0, the closed one-dimensional algebraic subgroup of $G$ generated by $g$ is $U:=\exp (\operatorname{G}_a)$. The Lie algebra $Lie(U)$ consists of nilpotent elements and we can apply Jacobson-Morozow to get an $\mathfrak{sl}_2$-triple (let's call its span $M$) that contains $Lie(U)$. Following Borel, we define the closed algebraic subgroup $$ \mathcal{A}(M) := \bigcap_{H \text{ closed subgroup of }G\\ M \subset Lie(H) \subset Lie(G)} H $$ and take its commutator group $$ SL_g := [\mathcal{A}(M),\mathcal{A}(M)] := \left\{ ghg^{-1}h^{-1} : g,h \in \mathcal{A}(M) \right\}. $$ Using Prop 7.8 and Cor 7.9 in Borel, we get $$ Lie(SL_g) = Lie([\mathcal{A}(M),\mathcal{A}(M)]) = [Lie(\mathcal{A}(M)),Lie(\mathcal{A}(M))] = [M, M] = M, $$ where the last equality follows since $M$ is the span of a Lie-triple system. The Lie-Algebra of $SL_g$ is thus isomorphic to $\mathfrak{sl}_2$ and since we are in char. 0, this means that $SL_g$ has to be isomorphic to $SL_2$ (I think for algebraic isomorphisms it cannot be $PSL_2$).

In other words, an algebraic exponential map $ \operatorname{G}_a \to G, t \mapsto \exp(t\cdot \tilde{X})$ factors through $SL_2$ via $$ t \mapsto \left( \begin{matrix} 1 & t \\ 0 & 1\end{matrix}\right), $$ so indeed, the original element $g$ comes from a unipotent upper triangular element in $SL_2$.

$\DeclareMathOperator\Lie{Lie}\DeclareMathOperator\SL{SL}$I am just writing some details on the comment by BCnrd. We use various statements from chapter 7 page 105ff in Borel's 'Linear Algebraic Groups, second enlarged edition'.

We start with a unipotent element $g$ in a linear algebraic group $G$. After conjugation we may assume that $g$ is upper triangular (with $1$'s on diagonal). We can apply the matrix logarithm $$ X\mathrel{:=} \log (g) \mathrel{:=} \sum_{n=1}^{\infty} (-1)^{k+1} \frac{(g-I)^n}{n}, $$ which is well defined in any field since there are only finitely many powers of $(g-I)$ (so it is a finite sum). The matrix $X$ is nilpotent and we can use it to construct an exponential function from the additive group to $G$ (as in 7.4(1) Borel) \begin{align} \operatorname{G}_{a} & \to G \\ t &\mapsto \exp (t\cdot X) = \sum_{n=0}^{\infty} \frac{(tX)^n}{n!} \end{align} which is an algebraic function since there are only finitely many powers of $X$. Since we are in characteristic 0, it is also injective. Since $g=\exp(1\cdot X)$, and we are in char. 0, the closed one-dimensional algebraic subgroup of $G$ generated by $g$ is $U:=\exp (\operatorname{G}_a)$. The Lie algebra $\Lie(U)$ consists of nilpotent elements and we can apply Jacobson–Morozow to get an $\mathfrak{sl}_2$-triple (let's call its span $M$) that contains $\Lie(U)$. Following Borel, we define the closed algebraic subgroup $$ \mathcal{A}(M) := \bigcap_{\text{$H$ closed subgroup of $G$} \\ M \subset \Lie(H) \subset \Lie(G)} H $$ and take its commutator group $$ \SL_g \mathrel{:=} [\mathcal{A}(M),\mathcal{A}(M)] \mathrel{:=} \left\{ ghg^{-1}h^{-1} : g,h \in \mathcal{A}(M) \right\}. $$ Using Prop 7.8 and Cor 7.9 in Borel, we get $$ \Lie(\SL_g) = \Lie([\mathcal{A}(M),\mathcal{A}(M)]) = [\Lie(\mathcal{A}(M)),\Lie(\mathcal{A}(M))] = [M, M] = M, $$ where the last equality follows since $M$ is the span of a Lie-triple system. The Lie-algebra of $\SL_g$ is thus isomorphic to $\mathfrak{sl}_2$ and since we are in char. 0, this means that $\SL_g$ has to be isomorphic to $\SL_2$ (I think for algebraic isomorphisms it cannot be $\operatorname{PSL}_2$).

In other words, an algebraic exponential map $ \operatorname{G}_a \to G$, $t \mapsto \exp(t\cdot \tilde{X})$ factors through $\SL_2$ via $$ t \mapsto \left( \begin{matrix} 1 & t \\ 0 & 1\end{matrix}\right), $$ so indeed, the original element $g$ comes from a unipotent upper triangular element in $\SL_2$.