Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 115 / 1

An $m$ by $n$ matrix has matrix rank $m$ exactly when at least one of its $m$ by $m$ minors has nonzero determinant. For each multi-index $I$, the setis open in the vector space of all real $m$ by $n$ matrices, because the determinant is continuous. Their union is $X_{m,n}$, so $X_{m,n}$ is itself open in ${\mathbb{R}}^{mn}$. It therefore inherits the standard smooth manifold structure and has dimension $mn$. This is the full-row-rank matrix manifold.

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 $mn+n-m$.

Fix $I$ and write $I^{\prime }$ for the complementary column indices. Over $U_I$, decompose $M=(M_I,M_{I^{\prime }})$ and $v=(v_I,v_{I^{\prime }})$. The equation $Mv=0$ is equivalent toMatrix inversion is smooth on the invertible matrices, sois a smooth, fiberwise-linear trivialization $U_I\times {\mathbb{R}}^{n-m}\to {\pi }^{-1}(U_I)$. The sets $U_I$ cover the base, proving that $\pi :E_{m,n}\to X_{m,n}$ is a vector bundle of rank $n-m$. It is the kernel bundle of a constant-rank family of linear maps, and its rank also follows from the rank-nullity theorem.

For $x\in S^2\subset {\mathbb{R}}^3$, let $F(x)=x^T$, viewed as a nonzero $1$ by $3$ matrix. This defines a smooth map $F:S^2\to X_{1,3}$. Its pulled-back fiber isso the evident fiberwise identity gives an isomorphism of vector bundles $F^{\ast }{E}_{1,3}\cong T{S}^2$.

If $E_{1,3}$ were a trivial vector bundle, its pullback would be trivial. A trivial rank-two bundle has a nowhere-zero section, whereas the assumed form of the Hairy ball theorem says that the tangent bundle $T{S}^2$ does not. Hence $E_{1,3}$ is nontrivial.