As Alain Valette says, a centreless connected complex Lie group $G$ has an injective homomorphism into $GL_n({\mathbb C})$. However, it need not be algebraic. To see this, consider the semi-direct product $G={\mathbb C}^2 \rtimes {\mathbb C}$. Here $z\in {\mathbb C}$ acts on the standard basis $e_1,e_2$ by the characters $e^{2\pi i z}$ and $e^{2\pi i z/\sqrt{2}}$ respectively. If $G$ can be given the structure of an algebraic group, then these two characters on ${\mathbb C}$ would become algebraically dependent, which cannot be.

Incidentally, this $G$ is not closed in its adjoint "embedding", since the closure contains $S^1\times S^1$ in the diagonal part. That is $G\subset {\mathbb C}^2\times D_2$ where $D_2$ is the group of diagonals in $GL({\mathbb C}^2)$.

As Alain Valette says, a centreless connected complex Lie group $G$ has an injective homomorphism into $GL_n({\mathbb C})$. However, it need not be algebraic. To see this, consider the semi-direct product $G={\mathbb C}^2 \rtimes {\mathbb C}$. Here $z\in {\mathbb C}$ acts on the standard basis $e_1,e_2$ by the characters $e^{2\pi i z}$ and $e^{2\pi i z/\sqrt{2}}$ respectively. If $G$ can be given the structure of an algebraic group, then these two characters on ${\mathbb C}$ would become algebraically dependent, which cannot be.

Incidentally, this $G$ is not closed in its adjoint "embedding", since the closure contains $S^1\times S^1$ in the diagonal part. That is $G\subset {\mathbb C}^2\rtimes D_2$ where $D_2$ is the group of diagonals in $GL({\mathbb C}^2)$.

As Alain Valette says, a centreless connected complex Lie group $G$ has an injective homomorphism into $GL_n({\mathbb C})$. However, it need not be algebraic. To see this, consider the semi-direct product $G={\mathbb C}^2 \rtimes {\mathbb C}$. Here $z\in {\mathbb C}$ acts on the standard basis $e_1,e_2$ by the characters $e^{2\pi i z}$ and $e^{2\pi i z/\sqrt{2}}$ respectively. If $G$ can be given the structure of an algebraic group, then these two characters on ${\mathbb C}$ would become algebraically independent, which cannot be.

Incidentally, this $G$ is not closed in its adjoint "embedding", since the closure contains $S^1\times S^1$ in the diagonal part. That is $G\subset {\mathbb C}^2\times D_2$ where $D_2$ is the group of diagonals in $GL({\mathbb C}^2)$.

As Alain Valette says, a centreless connected complex Lie group $G$ has an injective homomorphism into $GL_n({\mathbb C})$. However, it need not be algebraic. To see this, consider the semi-direct product $G={\mathbb C}^2 \rtimes {\mathbb C}$. Here $z\in {\mathbb C}$ acts on the standard basis $e_1,e_2$ by the characters $e^{2\pi i z}$ and $e^{2\pi i z/\sqrt{2}}$ respectively. If $G$ can be given the structure of an algebraic group, then these two characters on ${\mathbb C}$ would become algebraically dependent, which cannot be.

Incidentally, this $G$ is not closed in its adjoint "embedding", since the closure contains $S^1\times S^1$ in the diagonal part. That is $G\subset {\mathbb C}^2\times D_2$ where $D_2$ is the group of diagonals in $GL({\mathbb C}^2)$.