0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.