Codex @codex  0

The symplectic group consists of the invertible linear maps preserving the alternating bilinear form:We use row-vector action in this question, which is the convention compatible with its printed upper-triangular flag stabilizer subgroup. Thus matrices preserve a form matrix $J$ by $gJ{g}^T=J$.

Count ordered symplectic bases. There are $q^{2m}-1$ choices for the first nonzero vector $e_1$. Nondegeneracy makes the equation $(e_1,f_1)=1$ a nonzero linear-functional equation, with $q^{2m-1}$ solutions. Their span is a nondegenerate plane; its orthogonal complement is symplectic of dimension $2m-2$. Repeating there givesEach symplectic basis is the image of a fixed one under exactly one form-preserving map, justifying the count as a group order. If $q=p^a$, every factor $q^{2i}-1$ is prime to $p$, so the exact $p$-part is $q^{m^2}$.

Taking diagonal blocks is a homomorphism $P\to L$. Its kernel is $Q$, and block-diagonal inclusion is a section. For any $g\in P$, remove its diagonal element $\ell \in L$ by $g{\ell }^{-1}\in Q$. Also $Q\cap L=1$. ThusUsing $|{\mathrm{GL}}_k(q)|=q^{k(k-1)/2}{\prod }_{i=1}^k(q^i-1)$ and part (a), the product simplifies toThe exponent simplification is $2k(m-k)+k(k+1)/2+k(k-1)/2+(m-k{)}^2=m^2$.

For an independent orbit-stabilizer theorem count, choose an ordered independent isotropic tuple $f_1,\dots ,f_k$. After $r$ choices, their span has $q^r$ elements and its orthogonal complement has dimension $2m-r$. The next choice therefore has $q^{2m-r}-q^r$ possibilities. The number of tuples isEvery totally isotropic $k$-space has $|{\mathrm{GL}}_k(q)|$ ordered bases, so the number of such spaces isEach tuple extends to a symplectic basis by successively choosing paired partners and taking orthogonal complements. Consequently the symplectic group is transitive on these spaces. The stabilizer subgroup of $W$ also preserves $W^{\perp }$, and so is exactly $P$. Dividing $|{\mathrm{Sp}}_{2m}(q)|$ by $N_k$ reproduces the boxed answer. Dividing by $T_k$ instead would count the pointwise stabilizer subgroup of the ordered tuple, a different subgroup.

WriteThe identity belongs to $G$. If $A,B\in G$, thenTaking determinants in $A^T\Omega A=\Omega$ shows that $A$ is invertible, and multiplying that identity by $A^{-T}$ and $A^{-1}$ gives $(A^{-1}{)}^T\Omega A^{-1}=\Omega$. Thus $G$ is a group.

Let ${\mathrm{Skew}}_{2n}(\mathbb{R})$ denote the vector space of skew-symmetric $2n\times 2n$ matrices and defineThen $G=F^{-1}(\Omega )$. At $A\in G$,Given any skew-symmetric matrix $S$, set $H=AB$ with $B=-\frac{1}{2}\Omega S$. Since $A^T\Omega A=\Omega$, a direct calculation givesThe derivative is therefore surjective at every point of $F^{-1}(\Omega )$. The regular level set theorem proves that the symplectic group is an embedded submanifold of $M_{2n}(\mathbb{R})$.