# Let $h,h'$ be hermitian metric over vector space $V$, then grassmanian $G_n(V_h)\to G_n(V_{h'})$ is always continuous?

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.

• 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. – Tyrone Jul 16 '19 at 10:48
• @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? – user45765 Jul 16 '19 at 11:31
• @Tyrone Your underlying assumption is metrics here are norms. Why so? – user45765 Jul 16 '19 at 11:35
• 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. – Tyrone Jul 16 '19 at 13:51
• @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. – 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.