Classical bilinear-form equations linearize to matrix spaces

By [F1], each tangent vector at the identity is a based $k$-morphism $\mathrm{Spec}D\to X$. The coordinate-ring correspondence [F3, F4], polynomial universal property [F5], and quotient property [F6] identify these with matrices over $D$ satisfying the defining equations and reducing to identity. By [F2], every such matrix has unique form $A=I+ϵB$ for a matrix $B$ over $k$.

For $O_n$, substitution from step 1.1 gives $(I+ϵB{)}^{\mathsf{T}}(I+ϵB)=I+ϵ(B^{\mathsf{T}}+B)$, so preservation of $I_n$ is equivalent to $B^{\mathsf{T}}+B=0$. In characteristic not two, the diagonal equations give $2b_{ii}=0$ and hence $b_{ii}=0$; the off-diagonal equations give $b_{ji}=-b_{ij}$. In the Leibniz expansion of $\det (I+ϵB)$, the identity permutation contributes ${\prod }_i(1+ϵb_{ii})=1+ϵ{\sum }_i{b}_{ii}$; any nonidentity permutation moves at least two indices, so every nonzero term has at least two $ϵ$ factors and vanishes. Thus [F8, F9] give $\det (I+ϵB)=1+ϵ\mathrm{tr}(B)$. Every skew matrix here has trace zero, so it satisfies the additional $S{O}_n$ determinant equation, in both directions, and the two tangent spaces coincide. The entries $b_{ij}$ for $i<j$ are free, giving the skew-matrix dimension $n(n-1)\mathrm{/}2$.

For the symplectic scheme, step 1.1 gives $(I+ϵB{)}^{\mathsf{T}}J(I+ϵB)=J+ϵ(B^{\mathsf{T}}J+JB)$, so preservation of $J$ is equivalent to $B^{\mathsf{T}}J+JB=0$. Write $B=\left(\begin{array}{cc}P& Q\\ R& S\end{array}\right)$ in $n\times n$ blocks. By [F7], $B^{\mathsf{T}}J+JB=\left(\begin{array}{cc}R-R^{\mathsf{T}}& P^{\mathsf{T}}+S\\ -S^{\mathsf{T}}-P& Q^{\mathsf{T}}-Q\end{array}\right)$. This vanishes exactly when $R^{\mathsf{T}}=R$, $S=-P^{\mathsf{T}}$, and $Q^{\mathsf{T}}=Q$; the lower-left equation follows from $S=-P^{\mathsf{T}}$. For every characteristic, $\mathrm{tr}B=\mathrm{tr}P-\mathrm{tr}(P^{\mathsf{T}})=0$, so by the determinant calculation in step 2.1 the equation $\det A=1$ adds no first-order condition. This proves both inclusions in the asserted tangent-space description.

The block $P$ ranges over $M_n(k)$ and contributes $n^2$ dimensions by [F11]. For a symmetric block, the map from $k^{n(n+1)\mathrm{/}2}$ that fills the diagonal and upper-triangular entries freely and copies each off-diagonal entry into its transposed position is a linear bijection, using the entrywise vector-space operations in [F7]: symmetry forces exactly those copied entries and leaves the chosen coordinates arbitrary. Thus each of $Q$ and $R$ contributes $n+n(n-1)\mathrm{/}2=n(n+1)\mathrm{/}2$ dimensions. Therefore the symplectic tangent dimension is $n^2+2\cdot n(n+1)\mathrm{/}2=2n^2+n=n(2n+1)$. For $n=1$, this is the three-parameter family $\left(\begin{array}{cc}p& q\\ r& -p\end{array}\right)$.

All three schemes contain the identity, so none is empty; the zero tangent vector is $B=0$. For $n=1$ and $\mathrm{char}k\mathrm{\ne }2$, the orthogonal tangent space is zero, while the symplectic tangent space has dimension $3$. The orthogonal characteristic restriction is necessary: in characteristic $2$, for $n=1$, $(1+ϵb{)}^2=1$ for every $b$, so $T_1{O}_1=k$, whereas $\det (1+ϵb)=1$ forces $b=0$ for $S{O}_1$. The symplectic block and determinant calculations in step 3.1 remain valid in characteristic $2$. The identity corresponds to $B=0$, both tangent descriptions are equation equivalences, and the proof uses only finite entrywise calculations, with no basis choices or AC/DC. The counts are tangent-space dimensions only; no global group dimension or smoothness follows.