Let $V=V_0 \oplus V_1$ be a $\mathbb Z_2$-graded vector space over $\mathbb C$. Suppose $V$ has an even non-degenerate bilinear form $(-, -)$ which is symmetric on $V_0$, skew symmetric on $V_1$, and satisfies $(V_0, V_1) = (V_1, V_0) = 0$.
The orthosymplectic Lie superalgebra $\mathfrak{osp}(m,2n)$ is the $\mathbb Z_2$-graded subalgebra of $\mathfrak{gl}(m,2n)$ defined by $\{X \in \mathfrak{gl}(m,2n): (Xv,w)+(-1)^{|X||v|}(v,Xw)=0\}$. Note that $\mathfrak{osp}(m,2n)_0=\mathfrak{so}(m) \oplus \mathfrak{sp}(2n)$.
The corresponding supergroup $OSp(V)$ is called the orthosymplectic supergroup in the literature. There are many confusing definitions for the orthosymplectic supergroup. One defines over the grassmann algebra and other definition says it is the set of $\mathbb C$-points of the orthosymplectic supergroup scheme. One more definition says it is the Harish-Chandra super pair $(OSp(V)_0,\mathfrak{osp}(m,2n))$ where $(OSp(V)_0= O(V_0) \times Sp(V_1)$.
My first question is can't we just define it as the sub-supergroup of the general linear supergroup $GL(m,2n)$ on those elements which preserve the above bilinear form ?
My second question is as the adjoint representation of $OSp(V)$, what is $\mathfrak{osp}(m,2n)$ isomorphic to ?
