# Show that $\mathfrak{so}(p,q,\mathbb{R}) = I_{p,q} \mathfrak{so}(n,\mathbb{R})$.

I found this statement in another stackexchange post but couldn't get the $$\mathfrak{so}(p,q,\mathbb{R}) \subseteq I_{p,q} \mathfrak{so}(n,\mathbb{R})$$ direction. For the reverse, I could show that $$\left(I_{p,q} x\right)^t I_{p,q} = - I_{p,q} \left(I_{p,q} x\right)$$ (assuming $$x\in \mathfrak{so}(n,\mathbb{R})$$), but I'm stuck on getting from $$x^t I_{p,q} = -I_{p,q} x$$ to $$x^t = -x$$.

For reference, $$I_{p,q}$$ is defined as

$$\begin{pmatrix} I_p & 0_{p\times q} \\ 0_{q\times p} & -I_q\end{pmatrix}.$$

Any help would be appreciated.

• Your converse statement doesn't look like your original. You want to prove that if $x$ satisfies the $\mathfrak{so}(p,q)$ condition then $I_{p,q}^{-1}x=I_{p,q}x$ satisfies the $\mathfrak{so}(n)$ condition. Commented Aug 10 at 1:23
• @Callum I'm confused, why? I thought $I_{p,q} \mathfrak{so}(n,\mathbb{R})$ would be the set of $I_{p,q} x$, where $x^t = -x$. Why would its elements satisfy $(I_{p,q} x)^t = -(I_{p,q} x)$? Maybe I'm getting confused about what $I_{p,q} \mathfrak{so}(n,\mathbb{R})$ actually means. Commented Aug 10 at 1:55
• You misunderstand me. If $y$ is in $\mathfrak{so}(p,q)$ then $y^tI_{p,q} = - I_{p,q}y$ but now we write $y = I_{p,q}x$ for some $x$ (Indeed simply set $x=I_{p,q}y$ since $I_{p,q}$ is its own inverse) then our equation becomes $x^tI_{p,q}I_{p,q} = - I_{p,q}I_{p,q}x$ which simplifies to $x^t = -x$ so that $x \in \mathfrak{so}(n)$. Thus $y\in I_{p,q}\mathfrak{so}(n)$ and $\mathfrak{so}(p,q)\subseteq I_{p,q}\mathfrak{so}(n)$ as required. Commented Aug 10 at 13:00
• @Callum Ah, yep, that makes sense. I also got the same from Chris' answer. Thank you both. Commented Aug 10 at 13:27

You don’t need to show the other inclusion. $$I_{p,q}$$ is an isomorphism so you have that the dimensions of the two vector spaces are the same, and it is a fact that if $$U\subset V$$ and their dimensions are equal then $$U=V$$

Let $$X\in so(p,q)$$, and $$\eta$$ be the pseudo Euclidean norm of signature $$(p,q)$$, Then by definition: $$X^T\eta+\eta X=0$$ Now note that $$\eta X$$ is by definition in the Lie algebra $$so(n)$$. It follows that $$X=\eta^2X=\eta(\eta X)$$ hence lies in $$\eta so(n).$$

• In the post I linked, the goal is to show that the two vector spaces have the same dimension, so we can't use that here. Equivalence needs to be shown explicitly. Commented Aug 10 at 3:23
• @Borealis you know they are the same dimension as their Lie groups are the same dimension as smooth manifolds (which is easily seen via sards theorem) Commented Aug 10 at 3:27
• I appreciate the answer and the effort, but the purpose of my question was to prove the statement in the context of that post. Since it was stated in that context, I assume the equivalence can be seen without recourse to dimensional arguments, and it is such a solution that I am looking for. Commented Aug 10 at 3:32
• @Borealis please check the edit Commented Aug 10 at 3:45