Let $G_{r}(V)$ be the Grassmannian of a complex vector space $V$ consists of subspaces of codimension $r$. It is well known that $$TG_{r}(V)=Hom(S,Q)$$ where $S$ is the tautological subbundle and $Q=G_{r}\times V/S$.
I have some trouble with understanding Demailly's proof in his book "complex analytic and differential geometry". Indeed, since $Gl(V)$ acts transitively on $G_{r}(V)$, given $x\in G_{r}(V)$, let $H_{x}$ be the isotropy subgroup of $x$, then $G_{r}(V)$ is isomorphic to $Gl(V)/H_{x}=M$. So far so good.
Then we have $$
T_{x}G_{r}(V)=T_{H_{x}}M=Hom(V,V)/\{u;u(x)\subset x\}=Hom(x,V/x)=Hom(S_{x},Q_{x}).
$$
His proof ends here. My question is that how to show that this pointwise isomorphism induces a bundle isomorphism?
$\begingroup$
$\endgroup$
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
5
I think this follows from a standard fact about bundles. Let $E, E'$ be two bundles over the same space $X$ with structure maps $\pi, \pi'$. If there is a bundle map $f: E \to E'$ which is an isomorphism on each fiber, then $f$ is an isomorphism of bundles.
This fact applies in your case, as the two bundles on on the left and right ends of your chain of equalities are not literally equal, they're just naturally identified. So you've got two different bundles on $G_r$ and a pointwise isomorphism between them.
To convince yourself of the fact, notice that the only thing we need to upgrade $f$ to a bundle isomorphism is that it has to be a homeomorphism on the total spaces, which is a local property since $f$ is already a bijection, so we can assume that the bundles are trivial. We're reduced to the case of $f: X \times E \to X \times E'$, where $E$ is now simply a vector space. This map can be written down as $(x,v) \mapsto (x, g(x)\cdot v)$, where $g$ is a continuous map $X \to GL(E)$. This map is surely continuous with continuous inverse, and we assumed $f$ was an isomorphism on fibers, so it's a homeomorphism.
-
$\begingroup$ What I need is for "f" to be a holomorphic map between "E" and "E'". The problem is why under local trivialization "f" looks like the way you discribed? $\endgroup$ Commented Jul 25, 2023 at 6:02
-
$\begingroup$ It seems not obvious to me that the function "g" is even continuous. $\endgroup$ Commented Jul 25, 2023 at 6:15
-
$\begingroup$ @Tom $g$ is continuous because $f$ is continuous. If you fix a basis $e_1, \dotsc, e_n$ of $E$, then as a matrix $g(x) = (a_{ij}(x))_{ij}$ where $f(x, e_i) = \sum a_ij(x) e_j$. $\endgroup$ Commented Jul 25, 2023 at 8:03
-
$\begingroup$ @red_trumpet I my question, f is given by pointwise vector space isomorphism, there is no garantee that f is a bundle morphism. How do you show that f is continuous or holomorphic? $\endgroup$ Commented Jul 25, 2023 at 12:03
-
$\begingroup$ @A. Thomas Yerger Could you explain in detail why locally the g is continuous? $\endgroup$ Commented Jul 26, 2023 at 1:21
$\begingroup$
$\endgroup$
Here is my own explaination to the proof, welcome for comments and corrections. Identify $G_{r}(V)$ with $M$, it suffices to focus on the tangent bundle of $M$. Let $F=M\times Hom(V,V)$ be the trivial bundle over $M$. Define a subbundle $P$ with $P_{x}=\{u\in Hom(V,V);u(x)\subset x$}. It is not difficult to check that $P$ is a subbundle of $F$ by trivialization.
Then we get following exact sequence of bundles:
$$
0\rightarrow P\rightarrow F \rightarrow \overline{F}\rightarrow 0
$$
where $\overline{F}$ is the quotient bundle of $F$ over $P$.
I guess $\overline{F}$ is isomorphic to $Hom(S,Q)$ by construction because the bundle structure of $Q$ is also obtained by taking quotients and we have done nothing to $S$. And $\overline{F}$ is also isomorphic to $TM$ due to the pointwise isomorphism $T_{x}M\cong Hom(V,V)/\{u;u(x)\subset x\}$? But how does the pointwise isomorphism turn into a bundle morphism? I guess it is related to the construction of the homogenous space $M$ which I am not familiar with.