Closed subgroups of $\operatorname{GL}_n(\mathbb C)$ with Lie algebra $\mathfrak{so}_n(\mathbb C)$ What is the classification of (Zariski) closed subgroups in $\operatorname{GL}_n(\mathbb C)$ (viewed as a linear algebraic group) with Lie algebra $\mathfrak{so}_n(\mathbb C)$?
Is it true that every such group is a product of $\operatorname{SO}_n(\mathbb C)$ with a cyclic subgroup generated by $\mu I$, where
$\mu$ is some root of unity and $I$ is the identity matrix?
Correction in response to Mikhail Borovoi's comment: Is it true that for $n \neq 8$ every such group is a product of either $\operatorname{SO}_n(\mathbb C)$ or $\operatorname{O}_n(\mathbb C)$ with a cyclic subgroup generated by $\mu I$, where $\mu$ is some root of unity and $I$ is the identity matrix?
 A: Such a subgroup contains $\mathrm{SO}_n(\mathbf{C})$ and is contained in the normalizer $N_n$ of $\mathfrak{so}_n(\mathbf{C})$.
What is $N_n$? It contains scalars acts on the 1-dimensional space of bilinear forms preserved by $\mathfrak{so}_n(\mathbf{C})$. It follows that $N_n=D\mathrm{O}_n(\mathbf{C})$ where $D\simeq\mathbf{C}^*$ is the group of scalars. If $n$ is odd, it follows that $N_n=D\mathrm{SO}_n(\mathbf{C})$ thus the description is exactly the one described: every closed subgroup with Lie algebra $\mathfrak{so}_n(\mathbf{C})$ is generated by $\mu\mathrm{Id}_n$ for some root of unity $\mu$.
If $n$ is even, it still holds that the possible groups are parameterized by finite subgroups of $D\mathrm{O}_n(\mathbf{C})/\mathrm{SO}_n(\mathbf{C})$. This group is abelian and is direct product of $D\mathrm{SO}_n(\mathbf{C})/\mathrm{SO}_n(\mathbf{C})\simeq\mathbf{C}^*$ by a group $W_2$ of order 2 (generated by any matrix of determinant $-1$ in $\mathrm{O}_n(\mathbf{C})$. So the resulting subgroups $H$ are parameterized by finite subgroups of $\mathbf{C}^*\times W_2$ and only those corresponding to a subgroup of $\mathbf{C}^\times \{1\}$ have the given form (generated by $\mathrm{SO}_n(\mathbf{C})$ and scalar matrices). In particular, $H/\mathrm{SO}_n(\mathbf{C})$ can be non-cyclic.
If $H$ contains $\mathrm{O}_n(\mathbf{C})$, it therefore happens that $H$ is generated by $\mathrm{O}_n(\mathbf{C})$ and $\mu I_n$ for some root of unity.
However when $n$ is even, it can happen that $H$ doesn't contain $\mathrm{O}_n(\mathbf{C})$, and also is not generated by $\mathrm{SO}_n(\mathbf{C})$ and scalar matrices. The simplest example is the group generated by $\mathrm{SO}_2(\mathbf{C})$ and the matrix $\begin{pmatrix}0 & i\\ i & 0\end{pmatrix}$ (note that this one is contained in $\mathrm{SL}_2(\mathbf{C})$).
(Note: $n=8$ is not a special case. Indeed, in general the "unusual" behavior of $\mathfrak{so}_n$ for $n=8$ disappears when $\mathrm{so}_n$ it is endowed with its standard $n$-dimensional representation.)
A: There's nothing special about $\mathbb C$ here except that it is a characteristic-$0$ field, so I will work over an arbitrary such field.
The identity component of a group $G$ with Lie algebra $\mathfrak{so}_n$ is $\operatorname{SO}_n$, so you're asking about the finite extensions of $\operatorname{SO}_n$ in $\operatorname{GL}_n$.
If $n = 1$, these are the finite subgroups of $\operatorname{GL}_1$, which are as you want.
If $n = 2$, these are the finite extensions of the (algebraic) torus $\operatorname{SO}_2$.  Note that $\operatorname{SO}_2$ is a maximal torus in $\operatorname{GL}_2$, hence self-centralising.  There is thus an injection from $\pi_0(G)$ to the Weyl group $\operatorname N_{\operatorname{GL}_2}(\operatorname{SO}_2)/{\operatorname{SO}_2}$, which is cyclic of order $2$.  It is easy to check that the representatives of the non-identity coset of the Weyl group lie in $\operatorname O_2$, so your conjecture works in this case.
There is an injection from $\pi_0(G)/{\operatorname C_G(\operatorname{SO}_n)}$ to $\operatorname{Out}(\operatorname{SO}_n)$, which, since $n \ne 8$, is cyclic of order $1$ (for $n > 2$ even) or $2$ (for $n > 2$ odd).
We have that $\operatorname C_G(\operatorname{SO}_n)$ is contained in $\operatorname C_{\operatorname{GL}_n}(\operatorname{SO}_n)$.  For $n > 2$, the latter equals $\operatorname Z(\operatorname{GL}_n)$ (as can be tested by computing with diagonal matrices, for example).  Thus, for $n > 2$ odd, we are done.
For $n > 2$ even (with $n \ne 8$), the inner-automorphism map is an isomorphism $\operatorname O_n/{\operatorname{SO}_n} \to \operatorname{Out}(\operatorname{SO}_n)$.  We thus have that $G$ is contained in $\operatorname Z(\operatorname{GL}_n)\cdot\operatorname O_n$, as desired.
