NetNews Usenet Archive 1992 #16/NN_1992_16.iso/spool/sci/math/9359

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From: [e-mail] (Hartmut Frommert)
Subject: Re: Non-Euclidean geometry with complex metric tensor?
Message-ID: <[e-mail]>
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Organization: Dept. of Physics, University of Constance
References: <[e-mail]> <[e-mail]> <[e-mail]> <allenk.711662004@harry>
Date: Tue, 21 Jul 1992 14:05:52 GMT
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[e-mail] (Allen Knutson) writes:

>[e-mail] (Hartmut Frommert) makes a bold claim:

>>In complex vectorspaces (and manifolds, via tangent space) you can introduce
>>3 kinds of metrics, which are defined as semilinear (i.e. linear or
>>antilinear) isomorphisms g from the vectorspace, say V, into its dual ~V:
>>     g: V --> ~V, v |-> g(v)  [g(v): V --> C, v' |-> [g(v)](v') =: g(v,v')]

(**)    >>with       g(a*u+b*v) = s(a)*g(u)+s(b)*g(v), s \in \{ id, cc \}.

>>(cc: complex conjugation)
>>These are
>>1. (pseudo)orthogonal metrics g_o:   g_o(u,v) =: (u|v) = (v|u)
>>2. symplectic metrics g_s:           g_s(u,v) =: [u|v] = - [v|u]
>>3. (pseudo)unitary metrics g_u:      g_u(u,v) =: <u|v> = cc(<v|u>)
>>Proof that these are the only possibilities is a not too difficult but
>>lengthy exercise (in the real case, *3.* coincides with *1.*, so that there
>>are only (pseudo)orthogonal and symplectic metrics).

>There is an important assumption here: (a|b)=0 implies (b|a)=0, which is
>presumably desirable in geometrical applications, but not implied by
>anything above. With it, it is true that these are the only possibilities
>(Artin, Geometric Algebra).                                    Allen K.

The point you mention is not too difficult to proof (if you are able to get,
see
   Sexl/Urbandtke: Relativity, Groups, and Particles, Springer-Verlag 1982
esp. the outlines in section 7.5).

Since you request, I sketch the outline: Assuming the semi-linear mapping
bijective mapping g, you can write
   g(u,v)=f(g(v,u))=f(f(g(u,v)))         [i.e. $f\circ f=id$].
Vectorspace isomorphy implies g(0)=0, which implies f(f(0))=0. Expanding
   f: C --> C, x |-> f(x)=f_0+f_{1,id}*x+f_{1,cc}*cc(x)+...,
this implies (f_0)^2=0, thereby
   f_0==0.
In addition, since
   f(f(x))=x  for all x \in C,
you have
   f_{n,s_1,..,s_n}==0  iff  n <> 1
and
   (f_{1,id})^2 + |f_{1,cc}|^2 = 1 .
This has the obvious solution
   f_{1,id}=cos(alpha), f_{1,cc}=sin(alpha)*exp(i*phi) .
If you imply the condition (**) for only one of the s, you have either
   f_{1,cc}=0, f_{1,id} \in \{ 1, -1 \}, i.e.
        g(u,v) = g(v,u)   ([pseudo]orthogonal)     or
        g(u,v) = - g(v,u) (symplectic)
or
   f_{1,id}=0, f_{1,cc}=exp(i*phi), i.e.
        g(u,v) = exp(i*phi) * cc(g(v,u)),
     which is equivalent to
        exp(-i*phi/2)*g(u,v) = cc(exp(-i*phi/2)*g(v,u))
     so that g'=exp(-i*phi/2)*g is a [pseudo]unitary metric.

So in the real case there only remain the [pseudo]orthogonal and symplectic,
in the complex case also the [pseudo]unitary metrics.

Hope this clears.
--
 Hartmut Frommert, LS Dehnen, Physics, | E-Mail: <[e-mail]>
 Univ of Constance, P.O.Box 55 60,     |     or  <[e-mail]>
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