Re: skew symmetric map/comples geometry
- From: Jannick Asmus <jannick.news@xxxxxx>
- Date: Sun, 22 Jan 2006 10:43:09 +0100
On 21.01.2006 15:28, singau wrote:
> l.s.,
>
>
> in my study of some basic complex geometry, I got a little lost in
> linear algebra:
>
> Let V be a real linear space with inproduct S, let J be an R-linear
> isomorphism s.t. J^2= -I. It's eigenvalues are \pm i, in particular,
> they are purely imaginiary. Can we conclude that J is skew-symmetric,
> i.e. that S(Ja,b)= -S(a,Jb)?
No. :-( Here is a counterexample: Take V=R^2 euclidean and J=[1 2
(first row), -1 -1 (second row)].
I constructed this example by applying a non-orthogonal basis
transformation to J1=[0 -1 (first row), 1 0 (second row)].
So the properties of J to be (1) skew-symmetric w.r.t. a given scalar
product and (2) J^2=-I are independent of each other.
It might be helpful to know that V can be equipped with a C-vector space
structure by setting i v := J(v) for any v \in V. Hence, V is
necessarily even dimensional.
>
> I want to use this to show that a real vectorspace V with inproduct S
> and complex structure J can be viewed a a complex linear space with
> hermitean inproduct H=S+iA , where A(a,b)=S(Ja,b).
>
> thanks in advance,
>
> singau
>
J. :-)
.
- Follow-Ups:
- Re: skew symmetric map/comples geometry
- From: singau
- Re: skew symmetric map/comples geometry
- References:
- skew symmetric map/comples geometry
- From: singau
- skew symmetric map/comples geometry
- Prev by Date: Re: Real-time Calculation of Standard Deviation
- Next by Date: Re: An infinite number question
- Previous by thread: skew symmetric map/comples geometry
- Next by thread: Re: skew symmetric map/comples geometry
- Index(es):
Relevant Pages
|
