Re: SCALAR Inner Products ?

From: LD (spam_at_go.home)
Date: 03/24/05 

Date: Thu, 24 Mar 2005 14:52:37 +0100

Terry Padden wrote:

> You are just repeating the definition - not answering Why ?

Sorry, I misread your question: I've told you why an inner product is
*called* scalar, whereas you indeed asked why it *is* a scalar.

> If the Outer Product can be a multi-component/dimension (pseudo)-VECTOR,
> why should we not define the corresponding "Inner" Product as a VECTOR of
> some kind ? Is there some mathematical logic which must constrain it to
> 0-dimensions ?

Here is a possible argument: if E is the space of vectors in 3 dimensions
and if you want a map from ExE to E (as opposed to a map from ExE to the
real numbers), surely you want it to be bilinear and invariant under
rotations. So technically you want a linear map E@E->E ("@"=tensor product)
that is compatible with rotations. In particular, it must be surjective,
and its kernel, which must therefore have dimension 6, must also be
invariant under rotations. But the representation theory of the rotation
group tells you that the 9-dimensional space E@E only has three invariant
subspaces of respective dimensions 1, 3, 5, plus three others obtained as a
direct sum of two of these. So there is a *unique* invariant subspace of
dimension 6, which happens to be the kernel of the outer product map.

Similarly, there is a unique invariant subspace of dimension 8, which
happens to be the kernel of the inner product map.

LD

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