Re: Vector Spaces as Manifolds. Quick follow-up.

On Dec 27, 6:41 pm, galathaea <galath...@xxxxxxxxx> wrote:
On Dec 27, 11:11 am, "J.K" <J...@xxxxxxxxx> wrote:



On 26-12-2007 21:58, J.K wrote:

Anyone know if this is the structure Lee is
refering to?

Jose Carlos Santos wrote:

Sort of, but is is more simple than that. Just fix
some basis
{v_1, ..., v_n} and consider the map from R^n onto V
defined by

(a_1,...,a_n) |-> a_1v_1 + ... a_nv_n.

This is a chart. And that is all that you need; the
set whose only
element is this chart happens to be an atlas. There
is no need to
think about what would happen if you replace that
basis by another
one.

Actually, it is quite useful to check that all the charts
obtained in this way are compatible. Otherwise, you do not
know that there is one preferred manifold structure...

A follow-up, please:
Don't we need to check 2nd countability,
Hausdorffness, etc. Does it follow immediately?

real spaces are hausdorff
if that is what you are asking

for one thing
they're metric so automatically hausdorff

you only need to prove second countability

Well, finite dimensional real vector spaces are second
countable, too, as all separable metric spaces are.

-- m

.

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