Let $V$ be a finite dimensional vector spaces over complex number and choose hermitian metrics $h,h'$ over $V$. Let $G_n(V)$ be the set of codimension $n$ hyperplanes of $V$. Since $V$ has 2 metrics, one can induce 2 topologies $G_n(V_h)$ and $G_n(V_{h'})$ under the following procedure.

Given a codimension $n$ hyperplane of $V$, one can use either $h$ or $h'$ to find complementary vector space $U_h,U_{h'}$ which gives rise 2 different projection operators in $End(V_h)$ and $End(V_{h'})$ respectively.(Map $V\to U_h$ or $V\to U_{h'}$.) Thus one give $G_n(V_i)\to End(V_i)$ induced topology on $G_n(V_i)$ for $i=h,h'$ via the maps.

Now consider trivial vector bundle $G_n(V_h)\times V$ and classifying bundle $E=\frac{G_n(V_{h'})\times V}{F}$ where $F=\{(g,v)\vert g\in G_n(V_{h'}), v\in g\}$.

$\textbf{Q:}$ Why is $G_n(V_h)\times V\to E$ is a bundle homomoprhism?(What is the base? There are 2 different bases $G_n(V_i)$ for $i=h,h'$ but topology is assumed to be different before proving coincidence.) Why is this a continuous map to start with? I am aware of algebraic construction of grassmanian which will automatically give $G_n(V)$ topology independent metric on $V$. The whole point of the discussion is to yield a map $Id:G_n(V_h)\to G_n(V_{h'})$ as continuous map.

Ref. Atiyah K-Theory, Chpt 1, pg 28, 3rd paragraph.

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    $\begingroup$ Any two metrics induce the same topology on $\mathbb{R}^{2n}$ and hence on $\mathbb{C}^n$, and hence on $V$, so that the two topologies on $G_n(V)$ coincide. $\endgroup$ – Tyrone Jul 16 '19 at 10:48
  • $\begingroup$ @Tyrone Yep. I should have used that notion but how did Atiyah reach his conclusion there is a "bundle morphism $G_n(V_h)\times V\to E$" and what is the base here? $\endgroup$ – user45765 Jul 16 '19 at 11:31
  • $\begingroup$ @Tyrone Your underlying assumption is metrics here are norms. Why so? $\endgroup$ – user45765 Jul 16 '19 at 11:35
  • $\begingroup$ His reasoning here seems backwards to me, although I am not following the book. It seems easier to me to show that the topology on $G_n(V)$ is uniquely defined by the linear structure of $V$, and then get the epimorphism from this. $\endgroup$ – Tyrone Jul 16 '19 at 13:51
  • $\begingroup$ @Tyrone I could see there is such a map independent of metric on $V$ as it is algebraic map to projective space and this topology has to agree with projective embedding which is independent of norm. The whole point of introducing the metric on $V$ is to induce topology on $End(V)$ in which $G_n(V)$ is embeded as projective operators. Then later he has to show this structure is independent of the choice of metric. $\endgroup$ – user45765 Jul 16 '19 at 15:30

You have to recall the definition of the topology on $G_n(V)$. A metric $h$ on $V$ is a hermitian metric, i.e. a complex scalar product on $V$. Using this scalar product, each subspace $W \subset V$ determines the orthogonal projection $p_W^h : V \to V$ onto $W$. This is unique linear endomorphism such that $p_W^h(V) = W$, $p_W^h(w) = w$ for all $w \in W$ and such that $W, \ker(p_W^h)$ are orhogonal subspaces with respect to $h$.

The assignment $W \mapsto p_W^h$ yields an injective function $p^h : G_n(V) \to End(V)$ and one gives $G_n(V)$ the unique topology such that $p^h$ becomes a homeomorphism between $G_n(V)$ and $p^h(G_n(V)) \subset End(V)$. This topology could theoretically depend on $h$ because we do not know how $p^h(G_n(V))$ looks like. It is definitely not a linear subspace of $End(V)$ in which case the independence on $h$ would be clear.

The projection $\pi : G_n(V) \times V \to G_n(V)$ gives us a bundle over $G_n(V)$ with total space $G_n(V) \times V$ and fiber $V$. We thus obtain a quotient bundle $\pi' : (G_n(V) \times V) / F \to G_n(V)$ over $G_n(V)$ with total space $E = (G_n(V) \times V)/F$. This is the classifying bundle over $G_n(V)$. Atiyah denotes it simply by $E$.

Note that as a set we have $F = \bigcup_{W \in G_n(V)} \{ W \} \times W$. The fiber over the point $W \in G_n(V)$ is the linear subspace $W \subset V$. Moreover, the quotient map $q : G_n(V) \times V \to E$ forms a bundle map (in fact a bundle epimorphism). This is true for any choice of $h$.

Now the essential point is the construction on the bottom of p.27. Given a bundle epimorphism $\varphi : X \times V \to E'$, where $E'$ is any bundle over $X$, we obtain an induced map $f = \varphi' : X \to G_n(V) = G_n(V_h)$ which is continuous for any choice of $h$.

Consider two metrics $h, h'$. Then $q_{h'} : G_n(V_{h'}) \times V \to E_{h'}$ induces a continuous $q'_{h'} : G_n(V_{h'}) \to G_n(V_{h})$ which is the identity. Similarly we see that the identity $G_n(V_{h}) \to G_n(V_{h'})$ is continuous. Therefore $G_n(V_{h'})=G_n(V_{h})$ as topological spaces.

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