Topological Vector Bundles and Grassmannian Classification — Examples

Write $S^0=\{a,b\}$. A clutching map is a pair $(g(a),g(b))$ of nonzero real numbers. A gauge on either interval has boundary values in the same sign component, so the sign of $g(a)g(b{)}^{-1}$ is unchanged. Conversely, multiply by a constant gauge and use paths within ${\mathbb{R}}_{>0}$ or ${\mathbb{R}}_{<0}$ to normalize the pair to $(1,1)$ or $(1,-1)$. Thus [F1] gives at most and at least these two classes.

For $(1,1)$ the two trivial intervals glue their fiber coordinates without a sign change, producing $S^1\times \mathbb{R}$. For $(1,-1)$, cut the circle at one equatorial point; the remaining interval bundle closes by $(0,t)\sim (1,-t)$, which is the Möbius line. These two witnesses realize the normalized classes.

The invariant in step 1.1 has values $+1$ and $-1$ on the two witnesses in step 2.1, so they are not isomorphic; exhaustion in step 1.1 shows there are no others.

For finite $N$, by definition $\mathbb{F}{P}^N$ is the quotient of ${\mathbb{F}}^{N+1}\setminus \{0\}$ by nonzero scalar multiplication, so a point is exactly a line $\mathrm{\ell }\subseteq {\mathbb{F}}^{N+1}$. Hence $\mathbb{F}{P}^N={\mathrm{Gr}}_1({\mathbb{F}}^{N+1})$, and the set $\{(\mathrm{\ell },v):v\in \mathrm{\ell }\}$ is exactly the tautological bundle in [F1]. For $N=\mathrm{\infty }$, both projective space and its tautological line are the filtered unions of these finite stages inside ${\mathbb{F}}^{\mathrm{\infty }}$; no expression ${\mathbb{F}}^{\mathrm{\infty }+1}$ is used.

For finite $N$, the displayed bundle inclusion sends its fiber over $\mathrm{\ell }$ to the same line $\mathrm{\ell }$ in ${\mathbb{F}}^{N+1}\subseteq {\mathbb{F}}^{\mathrm{\infty }}$. Taking image planes therefore sends $\mathrm{\ell }$ to itself under the standard finite-stage inclusion, and pulling back ${\gamma }_1$ returns the original pairs $(\mathrm{\ell },v)$. At $N=\mathrm{\infty }$ the same assertion is the identity on the filtered union.

For $N=\mathrm{\infty }$, the finite-stage inclusions unite to the identity on ${\mathrm{Gr}}_1({\mathbb{F}}^{\mathrm{\infty }})$. The classifying-space notation in [F2] gives $B\mathrm{O}(1)$ and $B\mathrm{U}(1)$, while step 1.1 gives $\mathbb{R}{P}^{\mathrm{\infty }}$ and $\mathbb{C}{P}^{\mathrm{\infty }}$. All identifications are direct and use no choice principle.

On the plus disk use points $[z:1]$, $\mathrm{\mid }z\mathrm{\mid }\le 1$, and the tautological frame $s_{+}(z)=(z,1)$. On the minus disk use $[1:w]$, $\mathrm{\mid }w\mathrm{\mid }\le 1$, with $w=z^{-1}$ on the equator, and frame $s_{-}(w)=(1,w)$. These vectors span the represented lines by [F2].

For $\mathrm{\mid }z\mathrm{\mid }=1$, one has $s_{+}(z)=(z,1)=z(1,z^{-1})=z{s}_{-}(z^{-1})$. Thus a physical vector with plus coefficient $v$ has minus coefficient $zv$. By [F1], the clutching function is exactly $g(z)=z$, not its inverse.

Interchanging the plus and minus charts reverses the coordinate change, so [F1] gives $g^{-1}(z)=z^{-1}$. Dualizing a line bundle inverts its scalar transition functions, hence this second clutching is ${\gamma }^{\ast }$. This completes the sign calculation without a choice principle.

Suppose first that $n>0$. Every $A\in {\mathrm{GL}}_n(\mathbb{C})$ has polar form $A=UP$ with $U$ unitary and $P$ positive definite. The path $U((1-t)P+tI)$ joins $A$ to $U$ through invertible matrices. By the finite-dimensional spectral theorem, $U=W\mathrm{diag}(e^{i{\theta }_1},\dots ,e^{i{\theta }_n})W^{\ast }$ for real angles ${\theta }_j$, and $W\mathrm{diag}(e^{i(1-t){\theta }_1},\dots ,e^{i(1-t){\theta }_n})W^{\ast }$ joins $U$ to $I$. Hence ${\mathrm{GL}}_n(\mathbb{C})$ is path connected.

The two values of $g$ can therefore be joined independently to $I$, producing a homotopy $S^0\times I\to {\mathrm{GL}}_n(\mathbb{C})$ from $g$ to the constant identity map. By [F1], $E$ is isomorphic to the identity-clutched bundle, which is $S^1\times {\mathbb{C}}^n$.

If $n=0$, then ${\mathrm{GL}}_0(\mathbb{C})$ is a point and the same conclusion is forced. The real argument fails at step 1.1 because ${\mathrm{GL}}_1(\mathbb{R})={\mathbb{R}}^{\times }$ has two components; the clutching values in different components give the Möbius line. No choice principle is used.

Let $p:E\to X$ have rank zero. Every fiber is the one-element vector space ${\mathbb{F}}^0=\{0\}$ by [F1], so $p$ is bijective. Each bundle chart is a homeomorphism $p^{-1}(U)\cong U\times \{0\}\cong U$ over $U$, and these local inverses glue to the inverse of $p$. Thus $p$ is a bundle isomorphism to ${\mathrm{i}\mathrm{d}}_X:X\to X$, and any bundle map over $X$ between two such bundles is forced fiberwise. There is exactly one rank-zero isomorphism class.

By [F2], there is exactly one map $X\to {\mathrm{Gr}}_0({\mathbb{F}}^{\mathrm{\infty }})$ and exactly one homotopy class of such maps. Pulling back the zero tautological bundle gives the bundle in step 1.1, so the two singleton sets correspond.

Now let $X=\mathrm{\varnothing }$ and allow any $n$. A map $E\to \mathrm{\varnothing }$ exists only when $E=\mathrm{\varnothing }$, and local triviality is vacuous, so this is the unique rank-$n$ bundle. There is also exactly one function from $\mathrm{\varnothing }$ to ${\mathrm{Gr}}_n({\mathbb{F}}^{\mathrm{\infty }})$ and exactly one homotopy between any two such functions. Hence both classification sets are again singletons. No choice principle is used.

The map $R:\mathbb{R}\mathrm{/}\mathbb{Z}\to \mathrm{SO}(2)$ defined by $R([t])=\left(\begin{array}{cc}\cos (2\pi t)& -\sin (2\pi t)\\ \sin (2\pi t)& \cos (2\pi t)\end{array}\right)$ is a continuous group isomorphism with continuous inverse obtained from the oriented angle of the first column. Hence [F2] gives ${\pi }_1(\mathrm{SO}(2),I)\cong \mathbb{Z}$, with the class of $t\mapsto R([mt])$ corresponding to $m$.

Apply [F1] with $n=k=2$. Since $\mathrm{SO}(2)$ is path connected, the unbased set $[S^1,\mathrm{SO}(2)]$ is identified with its fundamental group; it is abelian, so changing the path used to the basepoint causes no conjugacy ambiguity. Step 1.1 therefore assigns exactly one integer $m$ to each oriented bundle, and every $m$ is realized by clutching with $t\mapsto R([mt])$. In particular $m=0$ gives the trivial oriented bundle.

Take the reflection $r=\mathrm{diag}(1,-1)$. Direct multiplication gives $rR([t])r^{-1}=R([-t])$. Thus the orientation-reversal action from [F1] sends the loop of winding $m$ to the loop of winding $-m$, as asserted. The calculation uses no choice principle.

Suppose for contradiction that a rank-$r$ bundle $\eta$ satisfies $\gamma \oplus \eta \cong \mathbb{R}{P}^{\mathrm{\infty }}\times {\mathbb{R}}^{r+1}$. The inclusion of the first summand followed by this isomorphism is a fiberwise-linear embedding $e:\gamma \hookrightarrow \mathbb{R}{P}^{\mathrm{\infty }}\times {\mathbb{R}}^{r+1}$.

The base is in the scope of [F2]. It is paracompact by [F2] and a Hausdorff CW complex by [F3] and [F6]. To see compact generation directly, if $A$ pulls back to a closed set under every compact-Hausdorff test map, then its pullback under every characteristic disk is closed; since a characteristic disk surjects onto its closed cell and the latter is Hausdorff, $A$ meets every closed cell in a closed set, so the weak topology makes $A$ closed. Hausdorffness makes compact images closed, hence the space is also weak Hausdorff. Thus it is CGWH.

The rank-one Schubert symbols in [F3] give exactly one cell in every nonnegative dimension, with $\mathbb{R}{P}^m$ as the $m$-skeleton. By [F4], the differential between any two such cells is zero over ${\mathbb{F}}_2$, since it already occurs in a sufficiently large finite skeleton. Hence [F5] gives $H_k(\mathbb{R}{P}^{\mathrm{\infty }};{\mathbb{F}}_2)\cong {\mathbb{F}}_2$ for every $k\ge 0$, while $H_k(\mathbb{R}{P}^r;{\mathbb{F}}_2)=0$ for $k>r$.

Send $x$ to the image line $e({\gamma }_x)\subseteq {\mathbb{R}}^{r+1}$. In a local nonzero frame $s$ for $\gamma$, this line is represented by the continuous nonzero vector $e(s(x))$, so the resulting map $f:\mathbb{R}{P}^{\mathrm{\infty }}\to {\mathrm{Gr}}_1({\mathbb{R}}^{r+1})=\mathbb{R}{P}^r$ is continuous and $f^{\ast }{\gamma }_r\cong \gamma$. If $j:\mathbb{R}{P}^r\hookrightarrow \mathbb{R}{P}^{\mathrm{\infty }}$ is the standard inclusion, then $(jf{)}^{\ast }\gamma \cong \gamma$.

The identity also pulls $\gamma$ back to itself by [F1]. The injective direction of the classification bijection [F2], applied using step 1.2, therefore gives $jf\simeq {\mathrm{id}}_{\mathbb{R}{P}^{\mathrm{\infty }}}$. This is the sole use of AC in the counterexample.

Put $k=r+1$. On $H_k(-;{\mathbb{F}}_2)$, the map $(jf{)}_{\ast }=j_{\ast }{f}_{\ast }$ is zero because it factors through the zero group $H_k(\mathbb{R}{P}^r;{\mathbb{F}}_2)$ from step 1.3. But [F5] and the homotopy in step 3.1 say that $(jf{)}_{\ast }$ is the identity on the nonzero group $H_k(\mathbb{R}{P}^{\mathrm{\infty }};{\mathbb{F}}_2)\cong {\mathbb{F}}_2$, a contradiction.

Therefore the assumed finite-rank complement $\eta$ cannot exist. The witness is paracompact Hausdorff and CGWH by step 1.2, so it specifically shows that those hypotheses do not replace compactness in the finite-rank complement theorem.

The tangent projection $TL\to L$ is a locally trivial rank-one real vector bundle: its linear charts are the derivatives of the smooth coordinate charts on the one-manifold $L$. The base is CGWH and fails only the library's separate second-countability convention, by [F1]. Thus $TL$ satisfies exactly the topological hypotheses in the false claim.

Suppose $TL$ were numerable. By [F2] it would have a continuous positive-definite fiber inner product $g$. We show directly that such a $g$ metrizes $L$. Any two points of the connected one-manifold $L$ can be joined by a piecewise smooth path: the points reachable from a fixed point by finite chains of coordinate intervals form a nonempty open-and-closed set. Define $d_g(p,q)$ as the infimum of the $g$-lengths of these paths. It is finite, symmetric, and satisfies the triangle inequality.

To prove positivity and identify the topology, fix $p\mathrm{\ne }q$ and choose a coordinate interval $U$ about $p$, not containing $q$, with a smaller closed coordinate interval $K\subset U$ whose interior contains $p$. In the coordinate $t$, write $g=a(t)d{t}^2$. On compact $K$, continuity and positivity give $0<m\le a\le M$. Every path from $p$ to $q$ must first leave $K$, so its coordinate variation before leaving is at least the positive coordinate distance from $p$ to $\mathrm{\partial }K$; its length is therefore bounded below by that distance times $\sqrt{m}$. Thus $d_g(p,q)>0$. Conversely, within a still smaller coordinate interval, straight coordinate segments have length at most $\sqrt{M}$ times their coordinate displacement, while the preceding lower bound forces sufficiently small $d_g$-balls to stay in any prescribed coordinate neighborhood. Hence $d_g$ induces exactly the topology of $L$, contradicting the nonmetrizability in [F1]. Therefore $TL$ is not numerable. The implication from a supplied numeration to $g$, and this metric-topology argument, make no choices beyond the supplied data.

By [F3], AC supplies a numeration of the tautological line ${\gamma }_1\to {\mathrm{Gr}}_1({\mathbb{R}}^{\mathrm{\infty }})$, and pulling this fixed numeration back along any map $f:L\to {\mathrm{Gr}}_1({\mathbb{R}}^{\mathrm{\infty }})$ gives a numeration of $f^{\ast }{\gamma }_1$. Therefore $TL\cong f^{\ast }{\gamma }_1$ would contradict step 2.1. No such classifying map exists. This is the sole nonlocal use of AC, recorded by [A1]; the contradiction after a hypothetical numeration is choice-free.

Consequently the locally trivial rank-one bundle $TL$ over the CGWH space $L$ is the required witness, and the false claim fails precisely because it omitted numerability.