4 fixed rotation matrix; deleted 1 characters in body

It is not clear to me what kind of answer is expected. Generally speaking, subgroups of Lie groups can be classified by Lie correspondence combined with combinatorial analysis resulting from structure theory of semisimple Lie groups. Below I address two particular cases.

If $K$ is a compact connected semisimple subgroup of $SO(2)\times SO(n)$ then its projection onto the first factor is trivial and the question is reduced to the $SO(n)$ case. (Closed) Lie subgroups of $SO(n)$ are precisely (compact) Lie groups with a faithful $n$-dimensional real orthogonal representation, so there are quite a few of them (the maximal connected ones were classified long time ago by Dynkin). If you need a complete description for small values of $n$, the Atlas of Lie groups is very handy.

In the other extreme case where $K=SO(2)$ you are, in effect, asking about the maps

$$f: SO(2)\to SO(2)\times SO(n).$$

They can be classified by passing to the Lie algebras. More precisely, the differential of $f$ is a linear map $so(2)\to so(2)\oplus so(n).$ Identifying $so(2)$ with $\mathbb{R}$ and $so(n)$ with the skew-symmetric matrices, it may be viewed as a pair $(d,A),$ where $d$ is an integer and $A$ is a skew-symmetric matrix whose eigenvalues are integral multiples of $i.$ Explicitly,

$$f:R(\varphi)\mapsto (R(d\varphi), \exp(\varphi A)),$$

where

$$R(\varphi)=\begin{bmatrix} \cos(\varphi) phantom{-}\cos(\varphi) & \sin(\varphi) \cr -\sin(\varphi) & \cos(\varphi) \end{bmatrix}$$

is the counterclockwise rotation by $\varphi$ and $\exp$ is the matrix exponential function.

The maps $f$ and $f'$ associated with non-zero pairs $(d,A)$ and $(d',A')$ have the same image if and only if the pairs are proportional. The case $d=0$ corresponds to an $SO(2)$ subgroup of the second factor $SO(n).$ In the case $d=1$, the subgroup $f(K)$ is the graph of a map $SO(2)\to SO(n).$

Note that for the original question about subgroups of $SO(2,n)$ one must impose further equivalences in the case $n=2$, because different subgroups of $SO(2)\times SO(2)$ can be conjugate in $SO(2,2)$.

3 formatting (matrices), notation (exp)

It is not clear to me what kind of answer is expected. Generally speaking, subgroups of Lie groups can be classified by Lie correspondence combined with combinatorial analysis resulting from structure theory of semisimple Lie groups. Below I address two particular cases.

If $K$ is a compact connected semisimple subgroup of $SO(2)\times SO(n)$ then its projection onto the first factor is trivial and the question is reduced to the $SO(n)$ case. (Closed) Lie subgroups of $SO(n)$ are precisely (compact) Lie groups with a faithful $n$-dimensional real orthogonal representation, so there are quite a few of them (the maximal connected ones were classified long time ago by Dynkin). If you need a complete description for small values of $n$, the Atlas of Lie groups is very handy.

In the other extreme case where $K=SO(2)$ you are, in effect, asking about the maps

$$f: SO(2)\to SO(2)\times SO(n).$$

They can be classified by passing to the Lie algebras. More precisely, the differential of $f$ is a linear map $so(2)\to so(2)\oplus so(n).$ Identifying $so(2)$ with $\mathbb{R}$ and $so(n)$ with the skew-symmetric matrices, it may be viewed as a pair $(d,A),$ where $d$ is an integer and $A$ is a skew-symmetric matrix whose eigenvalues are integral multiples of $i.$ Explicitly,

$$f:\begin{bmatrixf:R(\varphi)\mapsto (R(d\varphi), \exp(\varphi A)),$$

where

$$R(\varphi)=\begin{bmatrix} \cos(\varphi) & \sin(\varphi) \ -\sin(\varphi) & \cos(\varphi) \end{bmatrix} \mapsto \left(\begin{bmatrix} \cos(n\varphi) & \sin(n\varphi) \ -\sin(n\varphi) & \cos(n\varphi) \end{bmatrix},\exp(\varphi A)\right).$$

is the counterclockwise rotation by $\varphi$ and $\exp$ is the matrix exponential function.

The maps $f$ and $f'$ associated with non-zero pairs $(d,A)$ and $(d',A')$ have the same image if and only if the pairs are proportional. The case $d=0$ corresponds to an $SO(2)$ subgroup of the second factor $SO(n).$ In the case $d=1$, the subgroup $f(K)$ is the graph of a map $SO(2)\to SO(n).$

Note that for the original question about subgroups of $SO(2,n)$ one must impose further equivalences in the case $n=2$, because different subgroups of $SO(2)\times SO(2)$ can be conjugate in $SO(2,2)$.

2 typos

It is not clear to me what kind of answer is expected. Generally speaking, subgroups of Lie groups can be classified by Lie correspondence combined with combinatorial analysis resulting from structure theory of semisimple Lie groups. Below I address two particular cases.

If $K$ is a compact connected semisimple subgroup of $SO(2)\times SO(n)$ then its projection onto the first factor is trivial and the question is reduced to the $SO(n)$ case. (Closed) Lie subgroups of $SO(n)$ are precisely (compact) Lie groups with a faithful $n$-dimensional real orthogonal representation, so there are quite a few of them (the maximal connected ones were classified long time ago by Dynkin). If you need a complete description for small values of $n$, atlas the Atlas of Lie groups is very handy.

In the other extreme case when where $K=SO(2)$ you are, in effect, asking about the maps

$$f: SO(2)\to SO(2)\times SO(n).$$

They can be classified by passing to the Lie algebras. More precisely, the differential of $f$ is a linear map $so(2)\to so(2)\oplus so(n).$ Identifying $so(2)$ with $\mathbb{R}$ and $so(n)$ with the skew-symmetric matrices, it may be viewed as a pair $(d,A),$ where $d$ is an integer and $A$ is a skew-symmetric matrix whose eigenvalues are integral multiples of $i.$ Explicitly,

$$f:\begin{bmatrix} \cos(\varphi) & \sin(\varphi) \ -\sin(\varphi) & \cos(\varphi) \end{bmatrix} \mapsto \left(\begin{bmatrix} \cos(n\varphi) & \sin(n\varphi) \ -\sin(n\varphi) & \cos(n\varphi) \end{bmatrix},\exp(A)\right). end{bmatrix},\exp(\varphi A)\right).$$ The maps $f$ and $f'$ associated with non-zero pairs $(d,A)$ and $(d',A')$ have the same image if and only if the pairs are proportional. The case $d=0$ corresponds to a an $SO(2)$ subgroup of the second factor $SO(n).$ In the case $d=1$, the subgroup $f(K)$ is the graph of a map $SO(2)\to SO(n).$

Note that for the original question about subgroups of $SO(2,n)$ one must impose further equivalences in the case $n=2$, because different subgroups of $SO(2)\times SO(2)$ can be conjugate in $SO(2,2)$.

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