Past exam of the short options course of the University of Oxford / 2014 / 4 / e

Use the six-dimensional vector space of real symmetric $3\times 3$ matrices as the target, rather than all nine matrix entries independently. DefineIts derivative isThe given $J$ is symmetric with $J^2=I$. Every $A\in G$ is invertible, since taking determinants of $A^{T}JA=J$ gives $(\det A{)}^2=1$. For any symmetric target $S$, chooseThen $A^{T}JH=\frac{1}{2}JJS=S/2$, and its transpose is also $S/2$. Thus $D{F}_A(H)=S$ and the derivative is surjective at every point of $G$.

The regular level set theorem now givesIt also proves smooth embeddedness in the entire ambient vector space of matrices. The target symmetry is what gives the correct codimension.

The tangent space at the identity is the kernel of a linear map from part (i):This says precisely that $JH$ is skew-symmetric. Since $J^{-1}=J$, multiplication by $J$ identifies this tangent space with the three-dimensional space of skew-symmetric $3\times 3$ matrices. In entries it isSubstitution verifies the equation directly and displays its three independent parameters.