Responses from Goodwillie and Korbas on Grassmanians.....DMD
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Subject: Re: 3 more on Grassmanians
Date: Mon, 2 Oct 2000 23:30:49 -0400 (EDT)
From: "Tom Goodwillie,304 Kassar,863-2590,617-926-3565" <tomg@math.brown.edu>

>
> Naively it seems that if one has a fibre bundle
> $$
>         F ----> E ----> B
> $$
> of Riemannian manifolds, a metric can be chosen on the total
> space $E$ which is locally the product of th metrics on the fibre
> $F$ and base $B$. Hence if all manifolds are compact, then the
> volume of $E$ is the product of the volumens of $B$ and $F$.

Depends how you interpret the question. If the hypothesis is merely that
E is fibered over B, all fibers are diffeomorphic to F, and
Riemannian metrics on B and on F are given, then you can't do very much.
I mean, you can't expect E to admit a metric whose restriction
to every fiber is isometric to F. For example, you could pick a
metric on F=S^2 such that there is a unique point of maximal curvature;
then a compatible metric on E would determine a section of the
map E -> B, ruling out some bundles.

Or if you don't like that example you can make one involving a
perfectly ordinary metric on F=S^n and a bundle of n-spheres that
does not arise from a vector bundle. (Such things exist for large n.)

Or just think of a bundle of 2-tori over S^1 involving
a diffeomorphism of the torus that is not isotopic
to an isometry for any metric.

But if E is fibered over B and a metric on B is given and
a metric on the vector bundle "tangent space of the fiber"
on E ("kernel" below) is given, then that determines a metric on E as soon a
splitting of the sequence

0 -> kernel -> TE -> p^*TB -> 0

of vector bundles is chosen (which can always be done).
And if the fibers of E over B are all isometric (or even same volume), then
vol(E)=vol(B)vol(F).

> Moreover if $F$ is a compact Lie group acting freely
> and isometrically on $E$, then a metric on $B$ can be chosen so
> that a similar volume formula holds. IS THIS TRUE?

I guess so. Not just "can be chosen" but "is determined".
(Assuming that F also had a given (invariant) metric, and that
the maps g\mapsto gx from F to E are isometric embeddings.)

>
> For example, applying this method to the Hopf fibration
> $$
>         S^1 ---> S^3 ---> CP^1
> $$
> with the standard metrics on $S^1$ and $S^3$ leads to the
> conclusion that the induced metric on $CP^1$ is isometric to a
> 2-sphere of radius 1/2.
>

That seems fitting, when you think that RP^1 is a circle of
radius 1/2 (in the sense that its double cover is a circle
of radius 1).

 -- Tom Goodwillie
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Date: Tue, 03 Oct 2000 09:32:46 +0200
From: Julius Korbas <korbas@fmph.uniba.sk>
Subject: Re: 3 more on Grassmanians

Two  minor observations on D. Ravenel's questions:

1) There seems to be a little discrepancy in interpretating the
Grassmannians. Participants of the discussion up to now seem
to have considered Grassmannians of ALL vector subspaces
of a fixed dimension in a given Euclidean space (also D. Ravenel
in his message of Sept 28 says: "Let G(m,n) denote the Grassmannian
of m-dimensional subspaces of R^n..."). But now (in his message of
Oct 2),  D. Ravenel identifies
$$ G(m,n)=SO(n)/SO(m)\times SO(n-m),$$
which in fact is the Grassmannian of ORIENTED m-subspaces
in R^n.
2) Taking in account that the question about volumes is for the
Grassmannians of  ORIENTED m-subspaces in R^n, it perhaps
can help (in checking particular cases) that the Grassmannian
of ORIENTED 2-subspaces in R^4 is diffeomorphic to
$S^2\times S^2$.
  Best regards,
                                      Julius Korbas
                                (korbas@fmph.uniba.sk)
                                    KATC MFF UK,
                                    Mlynska dolina,
                                     SK-842 48 Bratislava,
                                        Slovakia