Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 115 / 3

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

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The ambient matrix space has dimension $4n^2$, while the space of skew-symmetric $2n\times 2n$ matrices has dimensionBecause $\Omega$ is a regular value, the codimension of its level set is the dimension of the target. Hence

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Ambient matrix multiplicationis bilinear and therefore smooth. Its restriction to the embedded submanifold $G\times G$ is smooth, and part a shows that its image lies in $G$. By the defining smooth structure on an embedded submanifold, this restriction is a smooth map $G\times G\to G$.

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