Return to Question

I would like to understand what is the "outer-automorphism group" $Out$ of $SO(p,q)$ and $O(p,q)$, where $p+q >0$ and $pq \neq 0$. My working definition of $Out$ is as follows:

Let us denote by $Aut(G)$ the automorphism group of a Lie group $G$. I take the inner-automorphism group $Inn(G)$ of $G$ to be all elements $K\in Aut(G)$ for which there exists a $g\in G$ such that $K = Ad_{g}$, namely $K(h) = g h g^{-1}$ for all $h\in G$. $Inn(G)$ is a normal subgroup of $Aut(G)$ and then $Out(G) = Aut(G)/Inn(G)$ is a group which I define to be the outer-morphism group of $G$. I have not been able to find what $Out(G)$ is for $G = SO(p,q), O(p,q)$.

I have seen that there are many references dealing with the outer-automorphism group of complex Lie algebras, which can be read off from their Dynkin diagram. However, $\mathfrak{so}(p,q)\simeq\mathfrak{o}(p,q)$ is not a complex Lie algebra but it is the real form of the corresponding complex Lie algebra. I don't know how can the outer-automorphism group of a simple real Lie algebra can be obtained. Wikipedia says that in fact the characterization of the outer-automorphism group of a real simple Lie algebra was obtained as recently as in 2010! In any case I am not interested in the outer-automorphism group of a real Lie algebra but of the full Lie group $SO(p,q)$ and $O(p,q)$. If I am not mistaken, for $q=0$ and $p = even$ we have $O(p,0) = SO(p,0)\rtimes\mathbb{Z}_{2}$, where $\mathbb{Z}_{2}$ is the outer-automorphism group of $SO(p,0)$, so $Out(SO(p,0)) = \mathbb{Z}_{2}$.

Thanks.

I would like to understand what is the "outer-automorphism group" $Out$ of $SO(p,q)$ and $O(p,q)$, where $p+q >0$ and $pq \neq 0$. My working definition of $Out$ is as follows:

Let us denote by $Aut(G)$ the automorphism group of a Lie group $G$. I take the inner-automorphism group $Inn(G)$ of $G$ to be all elements $K\in Aut(G)$ for which there exists a $g\in G$ such that $K = Ad_{g}$, namely $K(h) = g h g^{-1}$ for all $h\in G$. $Inn(G)$ is a normal subgroup of $Aut(G)$ and then $Out(G) = Aut(G)/Inn(G)$ is a group which I define to be the outer-morphism group of $G$. I have not been able to find what $Out(G)$ is for $G = SO(p,q), O(p,q)$.

I have noticed that there are many references dealing with the outer-automorphism group of complex Lie algebras, which can be read off from their Dynkin diagram. However, $\mathfrak{so}(p,q)\simeq\mathfrak{o}(p,q)$ is not a complex Lie algebra but a real form. I don't know how the outer-automorphism group of a simple real Lie algebra can be computed in general. In fact, Wikipedia says that the characterization of the outer-automorphism group of a real simple Lie algebra in terms of a short exact sequence involving the full and inner autmorphisms groups (a result classical for complex Lie algebras) was only obtained as recently as in 2010! In any case, I expect the answer to my question to be even more involved since I am not interested in the outer-automorphism group of a real Lie algebra but of the full real Lie group, in my case $SO(p,q)$ and $O(p,q)$. If I am not mistaken, for $q=0$ and $p = even$ we have $O(p,0) = SO(p,0)\rtimes\mathbb{Z}_{2}$, where $\mathbb{Z}_{2}$ is the outer-automorphism group of $SO(p,0)$, so $Out(SO(p,0)) = \mathbb{Z}_{2}$.

Thanks.

I would like to know what is the "outer-automorphism group" $Out$ of $SO(p,q)$ and $O(p,q)$, where $p+q >0$. My working definition of $Out$ is as follows:

Let us denote by $Aut(G)$ the automorphism group of a Lie group $G$. I take the inner-automorphism group $Inn(G)$ of $G$ to be all elements $K\in Aut(G)$ for which there exists a $g\in G$ such that $K = Ad_{g}$, namely $K(h) = g h g^{-1}$ for all $h\in G$. $Inn(G)$ is a normal subgroup of $Aut(G)$ and then $Out(G) = Aut(G)/Inn(G)$ is a group which I define to be the outer-morphism group of $G$. I have not been able to find what $Out(G)$ is for $G = SO(p,q), O(p,q)$.

I have seen that there are many references dealing with the outer-automorphism group of complex Lie algebras, which can be read off from their Dynkin diagram. However, $\mathfrak{so}(p,q)\simeq\mathfrak{o}(p,q)$ is not a complex Lie algebra but it is the real form of the corresponding complex Lie algebra. I don't know how can the outer-automorphism group of a simple real Lie algebra can be obtained. Wikipedia says that in fact the characterization of the outer-automorphism group of a real simple Lie algebra was obtained as recently as in 2010! In any case I am not interested in the outer-automorphism group of a real Lie algebra but of the full Lie group $SO(p,q)$ and $O(p,q)$. If I am not mistaken, for $q=0$ and $p = even$ we have $O(p,0) = SO(p,0)\rtimes\mathbb{Z}_{2}$, where $\mathbb{Z}_{2}$ is the outer-automorphism group of $SO(p,0)$, so $Out(SO(p,0)) = \mathbb{Z}_{2}$.

Thanks.

I would like to understand what is the "outer-automorphism group" $Out$ of $SO(p,q)$ and $O(p,q)$, where $p+q >0$ and $pq \neq 0$. My working definition of $Out$ is as follows:

Let us denote by $Aut(G)$ the automorphism group of a Lie group $G$. I take the inner-automorphism group $Inn(G)$ of $G$ to be all elements $K\in Aut(G)$ for which there exists a $g\in G$ such that $K = Ad_{g}$, namely $K(h) = g h g^{-1}$ for all $h\in G$. $Inn(G)$ is a normal subgroup of $Aut(G)$ and then $Out(G) = Aut(G)/Inn(G)$ is a group which I define to be the outer-morphism group of $G$. I have not been able to find what $Out(G)$ is for $G = SO(p,q), O(p,q)$.

I have seen that there are many references dealing with the outer-automorphism group of complex Lie algebras, which can be read off from their Dynkin diagram. However, $\mathfrak{so}(p,q)\simeq\mathfrak{o}(p,q)$ is not a complex Lie algebra but it is the real form of the corresponding complex Lie algebra. I don't know how can the outer-automorphism group of a simple real Lie algebra can be obtained. Wikipedia says that in fact the characterization of the outer-automorphism group of a real simple Lie algebra was obtained as recently as in 2010! In any case I am not interested in the outer-automorphism group of a real Lie algebra but of the full Lie group $SO(p,q)$ and $O(p,q)$. If I am not mistaken, for $q=0$ and $p = even$ we have $O(p,0) = SO(p,0)\rtimes\mathbb{Z}_{2}$, where $\mathbb{Z}_{2}$ is the outer-automorphism group of $SO(p,0)$, so $Out(SO(p,0)) = \mathbb{Z}_{2}$.

Thanks.

I would like to know what is the "outer-automorphism group" $Out$ of $SO(p,q)$ and $O(p,q)$, where $p+q >0$. My working definition of $Out$ is as follows:

Let us denote by $Aut(G)$ the automorphism group of a Lie group $G$. I take the inner-automorphism group $Inn(G)$ of $G$ to be all elements $K\in Aut(G)$ for which there exists a $g\in G$ such that $K = Ad_{g}$, namely $K(h) = g h g^{-1}$ for all $h\in G$. $Inn(G)$ is a normal subgroup of $Aut(G)$ and then $Out(G) = Aut(G)/Inn(G)$ is a group which I define to be the outer-morphism group of $G$. I have not been able to find what $Out(G)$ is for $G = SO(p,q), O(p,q)$. I have seen that there are many references dealing with the outer-automorphism group of complex Lie algebras, which can be read off from their Dynkin diagram. However, $\mathfrak{so}(p,q)\simeq\mathfrak{o}(p,q)$ is not a complex Lie algebra but it is the real form of the corresponding complex Lie algebra. I don't know how can the outer-automorphism group of a simple real Lie algebra can be obtained. Wikipedia says that in fact the characterization of the outer-automorphism group of a real simple Lie algebra was obtained as recently as in 2010! In any case I am not interested in the outer-automorphism group of a real Lie algebra but of the full Lie group $SO(p,q)$ and $O(p,q)$. If I am not mistaken, for $q=0$ and $p = even$ we have $O(p,0) = SO(p,0)\rtimes\mathbb{Z}_{2}$, where $\mathbb{Z}_{2}$ is the outer-automorphism group of $SO(p,0)$, so $Out(SO(p,0)) = \mathbb{Z}_{2}$.

Thanks.

I would like to know what is the "outer-automorphism group" $Out$ of $SO(p,q)$ and $O(p,q)$, where $p+q >0$. My working definition of $Out$ is as follows:

Let us denote by $Aut(G)$ the automorphism group of a Lie group $G$. I take the inner-automorphism group $Inn(G)$ of $G$ to be all elements $K\in Aut(G)$ for which there exists a $g\in G$ such that $K = Ad_{g}$, namely $K(h) = g h g^{-1}$ for all $h\in G$. $Inn(G)$ is a normal subgroup of $Aut(G)$ and then $Out(G) = Aut(G)/Inn(G)$ is a group which I define to be the outer-morphism group of $G$. I have not been able to find what $Out(G)$ is for $G = SO(p,q), O(p,q)$. 

I have seen that there are many references dealing with the outer-automorphism group of complex Lie algebras, which can be read off from their Dynkin diagram. However, $\mathfrak{so}(p,q)\simeq\mathfrak{o}(p,q)$ is not a complex Lie algebra but it is the real form of the corresponding complex Lie algebra. I don't know how can the outer-automorphism group of a simple real Lie algebra can be obtained. Wikipedia says that in fact the characterization of the outer-automorphism group of a real simple Lie algebra was obtained as recently as in 2010! In any case I am not interested in the outer-automorphism group of a real Lie algebra but of the full Lie group $SO(p,q)$ and $O(p,q)$. If I am not mistaken, for $q=0$ and $p = even$ we have $O(p,0) = SO(p,0)\rtimes\mathbb{Z}_{2}$, where $\mathbb{Z}_{2}$ is the outer-automorphism group of $SO(p,0)$, so $Out(SO(p,0)) = \mathbb{Z}_{2}$.

Thanks.