Subject: Re: question aby Grassmanians
Date: Wed, 27 Sep 2000 22:25:58 -0400 (EDT)
From: "Tom Goodwillie,304 Kassar,863-2590,617-926-3565" <tomg@math.brown.edu>

How about identifying a vector subspace of R^n with the orthogonal
projection onto it? Thus the Grassmanian becomes a submanifold of
the vector space of n by n real matrices, namely the set of all idempotent
symmetric matrices of a given rank. That gives it a metric, which
seems like a pretty nhice one ...

Tom Goodwillie