Consider the real algebraic group $SO(p,q)$, this is the automorphism group of the vector space $\mathbb{R}^n$ of dimension $n=p+q$ over $\mathbb{R}$, endowed with the diagonal quadratic form with $p$ pluses and $q$ minuses on the diagonal and with a nonzero skew symmetric $n$-form.


Now consider the corresponding spinor group $G=Spin(p,q)$. Can one say that it is the group of automorphisms of a some object over $\mathbb{R}$, which is not very difficult to describe?



Spin group as an automorphism group
To expand on my comment to the question, we have the following algebraic construction (I think originally due to Robert Brown, 'A characterization of spin representations'): 


Let $V$ be a quadratic space over a field $k$ of characteristic $\neq 2$. Attached to this is the Clifford algebra $C=C(V)$: it is equipped with a $\mathbb{Z}/2\mathbb{Z}$-grading $C=C^+\oplus C^-$ and an embedding $V\hookrightarrow C^-$. The general Spin group $GSpin(V)$ consists of units in $C^+$ that preserve $V$ under conjugation. The group $Spin(V)$ is the sub-group of elements that have trivial spinor norm. So, to describe $Spin(V)$ as an automorphism group, it suffices to do so for $GSpin(V)$.


Let $H$ be the graded vector space $C$ viewed as a representation of $GSpin(V)$ via left multiplication: it is also a right $C$-module via right multiplication. Then $GSpin(V)$ clearly lies within the group $U(H)$ of $C$-equivariant, grading preserving automorphisms of $H$. 


Set $E=End(H)$: this is a representation of $GSpin(V)$ via conjugation. Define a bilinear form $\{,\}:E\times E\to k$ by $$\{f,g\}=\frac{1}{2^{dim(V)}}trace(fg).$$ 


Now choose a basis $\{v\_i\}$ for $V$, and let $A=(v\_i\cdot v\_j)\_{i,j}$ be the inner product matrix attached to this basis. Set $(b\_{i,j})=B=A^{-1}$. Define an endomorphism $\pi:E\to E$ by the formula:


$$\pi(f)(h)=\sum\_{i,j}b\_{i,j}\{v\_i,f\}v\_jh.$$


Clearly, the image of $\pi$ is $V\subset E$, where $V$ acts on $H$ via left multiplication. Let $G'\subset U(H)$ be the stabilizer of the endomorphism $\pi$. Then $G'$ preserves $V$ via conjugation, and is therefore contained in $GSpin(V)$. On the other hand, it is not hard to see that $GSpin(V)$ stabilizes $\pi$. 


So we see that $GSpin(V)$ can be described as the group of $C$-equivariant, grading preserving automorphisms of $H$ that also stabilize $\pi$.


