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Define the smooth mapAt $(M,v)$ its derivative in the direction $(A,w)$ isThe restriction to variations $(0,w)$ is the surjective linear map $M:{\mathbb{R}}^n\to {\mathbb{R}}^m$, since $M$ has full row rank. Thus $0$ is a regular value of $\Phi$. The regular level set theorem now shows thatis an embedded submanifold of $X_{m,n}\times {\mathbb{R}}^n$, of dimension $mn+n-m$.

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 $M_{2n}(\mathbb{R})$.

A level set at a regular value is a smooth submanifold. By the regular level set theorem, its codimension is the dimension of the target. For determinant-one real symmetric matrices, $D(\det {)}_B(C)=\mathrm{tr}(B^{-1}C)$ is nonzero because $C=B$ gives $n$, so the constraint has codimension one.

The map $F:M_{2n}(\mathbb{R})\to {\mathrm{Skew}}_{2n}(\mathbb{R})$ given by $F(A)=A^{T}JA$ has $Sp(2n,\mathbb{R})=F^{-1}(J)$. At a symplectic $A$,is surjective, so the regular level set theorem makes the symplectic group an embedded submanifold.